{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0001","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e5fa34ee721eb5427fec4c3dd0d50cfda215d3829550b33a111b79c5becfe60a","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124001,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":2,\"lower\":1,\"scale\":1}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=1/x^2 on [1,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=1/x^2, [1,infinity), 1, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=1/x^2, [1,infinity), 1, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0002","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b0d98aea7ce1796b6ee112fe2a04f328e6eeb4de3687e39d8871807ee6f0f2d4","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124002,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":3,\"lower\":2,\"scale\":2}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=2/x^3 on [2,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=2/x^3, [2,infinity), 1/4, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=2/x^3, [2,infinity), 1/4, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0003","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"916d8d109663d20d147f462ec03c8a7c1d47097cee0c0b3b4e0f8c61c88a574d","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124003,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":4,\"lower\":3,\"scale\":3}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=3/x^4 on [3,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=3/x^4, [3,infinity), 1/27, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=3/x^4, [3,infinity), 1/27, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0004","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d612fcf18475d2989e4886e972f9201d731395c7f1c40420ad03ab1c1f9747ae","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124004,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":5,\"lower\":4,\"scale\":4}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=4/x^5 on [4,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=4/x^5, [4,infinity), 1/256, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=4/x^5, [4,infinity), 1/256, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0005","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c960c76f71f501b2045bc9249f32afba26890a47811671a3f67648965d834b52","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124005,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":2,\"lower\":5,\"scale\":5}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=5/x^2 on [5,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=5/x^2, [5,infinity), 1, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=5/x^2, [5,infinity), 1, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0006","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"acd4cd6d1c1f196d5f4d75a466fb6c42c55d90bd443b4af43dd907270029486b","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124006,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":3,\"lower\":1,\"scale\":6}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=6/x^3 on [1,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=6/x^3, [1,infinity), 3, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=6/x^3, [1,infinity), 3, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0007","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7c579c6fa0d90ec790434fae17d04be90b4f213530a78bf3c9f547989fa37dce","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124007,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":4,\"lower\":2,\"scale\":7}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=7/x^4 on [2,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=7/x^4, [2,infinity), 7/24, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=7/x^4, [2,infinity), 7/24, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0008","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ea321de65eb8e4b22d56021804166f99491de5f550b640470ca74a98f7ca489","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124008,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":5,\"lower\":3,\"scale\":8}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=8/x^5 on [3,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=8/x^5, [3,infinity), 2/81, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=8/x^5, [3,infinity), 2/81, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0009","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25aa496e2dd4edb7c67e52692afe1fc53d52fc3d63cf3ec8213593eeb7d312f3","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124009,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":2,\"lower\":4,\"scale\":9}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=9/x^2 on [4,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=9/x^2, [4,infinity), 9/4, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=9/x^2, [4,infinity), 9/4, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0010","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e34427b8289132b7a4ae32e2d8f5bb2fa652f430605f731a4084e5d7d035daeb","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124010,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":3,\"lower\":5,\"scale\":10}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=10/x^3 on [5,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=10/x^3, [5,infinity), 1/5, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=10/x^3, [5,infinity), 1/5, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0011","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5c7b58f03cb57c2373d8b1753cfb443f8f9ac52595ed5fffe8e899b967560dc","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124011,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":4,\"lower\":1,\"scale\":11}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=11/x^4 on [1,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=11/x^4, [1,infinity), 11/3, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=11/x^4, [1,infinity), 11/3, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0012","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b02b2fdfa92eead1b4ec62717eb8feb653fa018a7bac0815a74623f4713cea6","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124012,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":5,\"lower\":2,\"scale\":12}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=12/x^5 on [2,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=12/x^5, [2,infinity), 3/16, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=12/x^5, [2,infinity), 3/16, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0013","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e4a61c65961477c12f34dbd833090b920261e3da1f498c0a4f25f1d4f5c821e7","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124013,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":2,\"lower\":3,\"scale\":13}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=13/x^2 on [3,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=13/x^2, [3,infinity), 13/3, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=13/x^2, [3,infinity), 13/3, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0014","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"556d9d9fb840e711a325e2e740bf4e3c77b8507568d1cdeb427dc4a185ac12a1","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124014,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":3,\"lower\":4,\"scale\":14}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=14/x^3 on [4,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=14/x^3, [4,infinity), 7/16, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=14/x^3, [4,infinity), 7/16, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0015","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ce8d8749210afbf69211912881851a684618fb56ad9376a419316b3a96539f2","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124015,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":4,\"lower\":5,\"scale\":15}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=15/x^4 on [5,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=15/x^4, [5,infinity), 1/25, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=15/x^4, [5,infinity), 1/25, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0016","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6c03b92d078bf26be36521627f36415416cac88e2f98c62c695cf38e0376306f","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124016,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":5,\"lower\":1,\"scale\":16}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=16/x^5 on [1,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=16/x^5, [1,infinity), 4, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=16/x^5, [1,infinity), 4, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0017","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"899eb9112923fd1ec0c5d39702090b2ceb9e37cafecb24b429f8311ef0abb16c","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124017,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":2,\"lower\":2,\"scale\":17}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=17/x^2 on [2,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=17/x^2, [2,infinity), 17/2, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=17/x^2, [2,infinity), 17/2, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0018","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"284ff14fce02ae7ccf10363da9ee18d6d895da01d2bd1fc14f16c1d6d476b13c","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124018,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":3,\"lower\":3,\"scale\":18}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=18/x^3 on [3,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=18/x^3, [3,infinity), 1, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=18/x^3, [3,infinity), 1, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0019","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c39d0b89a9e419ea79800bd0d0aeb698dc6ccc5ed77eb44b5e13da53af78c40","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124019,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":4,\"lower\":4,\"scale\":19}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=19/x^4 on [4,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=19/x^4, [4,infinity), 19/192, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=19/x^4, [4,infinity), 19/192, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0020","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b625ffc0c0237b16405de144d3bb0b59d6f54a99252795cb789fede2cd1816e8","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124020,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":5,\"lower\":5,\"scale\":20}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=20/x^5 on [5,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=20/x^5, [5,infinity), 1/125, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=20/x^5, [5,infinity), 1/125, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0021","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e0727089b136fe0f84209cb8628bf675de1c962bfc0fadc0e611ba70fe2d4a2","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124021,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":2,\"lower\":1,\"scale\":21}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=21/x^2 on [1,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=21/x^2, [1,infinity), 21, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=21/x^2, [1,infinity), 21, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0022","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f8cbfddbe22eec8fe87de9fec8997813c1c49f23d233ba732e4dadfe9c2c810a","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124022,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":3,\"lower\":2,\"scale\":22}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=22/x^3 on [2,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=22/x^3, [2,infinity), 11/4, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=22/x^3, [2,infinity), 11/4, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0023","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"310a1b45eebecf95f58a370d4e7925eb07cb03e80d555faa0a4500772e3ec343","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124023,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":4,\"lower\":3,\"scale\":23}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=23/x^4 on [3,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=23/x^4, [3,infinity), 23/81, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=23/x^4, [3,infinity), 23/81, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0024","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1cfa572dfac7291034fa887308aa57f99514d7f61c63d166044a6aea7d8708f2","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124024,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":5,\"lower\":4,\"scale\":24}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=24/x^5 on [4,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=24/x^5, [4,infinity), 3/128, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=24/x^5, [4,infinity), 3/128, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0025","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c83ba2348ab710899843862c7207b6d4f24fbd609ace951c285589a0b7d7cee","family_id":"qf-mod43-improper-one-sided-tail","seed":202611124025,"semantic_template":"qf-mod43-improper-one-sided-tail/{\"exponent\":2,\"lower\":5,\"scale\":25}"},"fields":{"title":"The One-Sided Tail Is a Limit of Finite-Cutoff Integrals","stem":"Use f(x)=25/x^2 on [5,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate).","answer":{"kind":"expression","value":"(f(x)=25/x^2, [5,infinity), 5, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)","explanation":"Take the upper cutoff b→infinity in the exact antiderivative; the value is scale/((exponent−1)*lower^(exponent−1)). Signed or zero scale is allowed. This symbolic all-cutoff argument, not a finite simulation, supplies convergence for the declared family. The exact result, retaining the requested order, is (f(x)=25/x^2, [5,infinity), 5, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)."},"skills":["The One-Sided Tail Is a Limit of Finite-Cutoff Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0026","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c851729e212f1c3c8b07bf89e2732aa5ebb0782b7170765030dec6016fe14424","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124026,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":1,\"left_power_numerator\":0,\"left_scale\":1,\"right_power_denominator\":2,\"right_power_numerator\":1,\"right_scale\":2,\"singularity\":-12}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=-12, interval [c−1,c+1], left integrand=1*(c−x)^(−0/1) and right integrand=2*(x−c)^(−1/2). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-13,-11], -12, left:1*(c-x)^(-0/1), right:2*(x-c)^(-1/2), 1, 4, Yes, Yes, 5, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-13,-11], -12, left:1*(c-x)^(-0/1), right:2*(x-c)^(-1/2), 1, 4, Yes, Yes, 5, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0027","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0412c59ab37cfc6f959158cd53dfef8c82dca83fceb8331285e7f9720c90bc5c","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124027,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":2,\"left_power_numerator\":1,\"left_scale\":2,\"right_power_denominator\":3,\"right_power_numerator\":2,\"right_scale\":3,\"singularity\":-11}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=-11, interval [c−1,c+1], left integrand=2*(c−x)^(−1/2) and right integrand=3*(x−c)^(−2/3). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-12,-10], -11, left:2*(c-x)^(-1/2), right:3*(x-c)^(-2/3), 4, 9, Yes, Yes, 13, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-12,-10], -11, left:2*(c-x)^(-1/2), right:3*(x-c)^(-2/3), 4, 9, Yes, Yes, 13, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0028","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a89d7b3be86f762e1972a59a08910ec203a16245d03032ef54043dab3a05b50a","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124028,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":3,\"left_power_numerator\":2,\"left_scale\":3,\"right_power_denominator\":4,\"right_power_numerator\":3,\"right_scale\":4,\"singularity\":-10}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=-10, interval [c−1,c+1], left integrand=3*(c−x)^(−2/3) and right integrand=4*(x−c)^(−3/4). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-11,-9], -10, left:3*(c-x)^(-2/3), right:4*(x-c)^(-3/4), 9, 16, Yes, Yes, 25, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-11,-9], -10, left:3*(c-x)^(-2/3), right:4*(x-c)^(-3/4), 9, 16, Yes, Yes, 25, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0029","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e51001ae4a1e5824b7330c1e1f0c927b034be7e08c3641c3986005c337b8635","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124029,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":4,\"left_power_numerator\":3,\"left_scale\":4,\"right_power_denominator\":1,\"right_power_numerator\":0,\"right_scale\":5,\"singularity\":-9}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=-9, interval [c−1,c+1], left integrand=4*(c−x)^(−3/4) and right integrand=5*(x−c)^(−0/1). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-10,-8], -9, left:4*(c-x)^(-3/4), right:5*(x-c)^(-0/1), 16, 5, Yes, Yes, 21, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-10,-8], -9, left:4*(c-x)^(-3/4), right:5*(x-c)^(-0/1), 16, 5, Yes, Yes, 21, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0030","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a96545d63d19ef016c963876f71c41700dc944c02335e59ffd3a98a062f5b68e","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124030,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":1,\"left_power_numerator\":0,\"left_scale\":5,\"right_power_denominator\":2,\"right_power_numerator\":1,\"right_scale\":6,\"singularity\":-8}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=-8, interval [c−1,c+1], left integrand=5*(c−x)^(−0/1) and right integrand=6*(x−c)^(−1/2). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-9,-7], -8, left:5*(c-x)^(-0/1), right:6*(x-c)^(-1/2), 5, 12, Yes, Yes, 17, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-9,-7], -8, left:5*(c-x)^(-0/1), right:6*(x-c)^(-1/2), 5, 12, Yes, Yes, 17, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0031","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87a3bebbcf26461b66d44e6eb8abf816d61f128a037a3e1d90f8452ee53eb5fd","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124031,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":2,\"left_power_numerator\":1,\"left_scale\":6,\"right_power_denominator\":3,\"right_power_numerator\":2,\"right_scale\":7,\"singularity\":-7}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=-7, interval [c−1,c+1], left integrand=6*(c−x)^(−1/2) and right integrand=7*(x−c)^(−2/3). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-8,-6], -7, left:6*(c-x)^(-1/2), right:7*(x-c)^(-2/3), 12, 21, Yes, Yes, 33, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-8,-6], -7, left:6*(c-x)^(-1/2), right:7*(x-c)^(-2/3), 12, 21, Yes, Yes, 33, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0032","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8f573571d13a4490c86b4cde28a4f4625d3179a5ecafbb7876c810140ff2eb7a","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124032,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":3,\"left_power_numerator\":2,\"left_scale\":7,\"right_power_denominator\":4,\"right_power_numerator\":3,\"right_scale\":8,\"singularity\":-6}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=-6, interval [c−1,c+1], left integrand=7*(c−x)^(−2/3) and right integrand=8*(x−c)^(−3/4). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-7,-5], -6, left:7*(c-x)^(-2/3), right:8*(x-c)^(-3/4), 21, 32, Yes, Yes, 53, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-7,-5], -6, left:7*(c-x)^(-2/3), right:8*(x-c)^(-3/4), 21, 32, Yes, Yes, 53, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0033","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27e4b748cceb72e517a8835c756788bbf2f06b34c37a20cd4d2fe03f3ec1d50e","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124033,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":4,\"left_power_numerator\":3,\"left_scale\":8,\"right_power_denominator\":1,\"right_power_numerator\":0,\"right_scale\":9,\"singularity\":-5}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=-5, interval [c−1,c+1], left integrand=8*(c−x)^(−3/4) and right integrand=9*(x−c)^(−0/1). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-6,-4], -5, left:8*(c-x)^(-3/4), right:9*(x-c)^(-0/1), 32, 9, Yes, Yes, 41, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-6,-4], -5, left:8*(c-x)^(-3/4), right:9*(x-c)^(-0/1), 32, 9, Yes, Yes, 41, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0034","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1ff192c5eefdf7cccd77ef4559c50f1d39cd65e95a3570076b012a2aeb54a72","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124034,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":1,\"left_power_numerator\":0,\"left_scale\":9,\"right_power_denominator\":2,\"right_power_numerator\":1,\"right_scale\":10,\"singularity\":-4}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=-4, interval [c−1,c+1], left integrand=9*(c−x)^(−0/1) and right integrand=10*(x−c)^(−1/2). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-5,-3], -4, left:9*(c-x)^(-0/1), right:10*(x-c)^(-1/2), 9, 20, Yes, Yes, 29, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-5,-3], -4, left:9*(c-x)^(-0/1), right:10*(x-c)^(-1/2), 9, 20, Yes, Yes, 29, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0035","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ded1a20687219063d8c5b8e263e4109c5f100d5e902e1c56d5842bdb5f5f111","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124035,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":2,\"left_power_numerator\":1,\"left_scale\":10,\"right_power_denominator\":3,\"right_power_numerator\":2,\"right_scale\":11,\"singularity\":-3}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=-3, interval [c−1,c+1], left integrand=10*(c−x)^(−1/2) and right integrand=11*(x−c)^(−2/3). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-4,-2], -3, left:10*(c-x)^(-1/2), right:11*(x-c)^(-2/3), 20, 33, Yes, Yes, 53, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-4,-2], -3, left:10*(c-x)^(-1/2), right:11*(x-c)^(-2/3), 20, 33, Yes, Yes, 53, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0036","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3be94cfd09bc937df5f4aefb8e11e71f3003def1ba94c23cfd838c892685a0fc","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124036,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":3,\"left_power_numerator\":2,\"left_scale\":11,\"right_power_denominator\":4,\"right_power_numerator\":3,\"right_scale\":12,\"singularity\":-2}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=-2, interval [c−1,c+1], left integrand=11*(c−x)^(−2/3) and right integrand=12*(x−c)^(−3/4). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-3,-1], -2, left:11*(c-x)^(-2/3), right:12*(x-c)^(-3/4), 33, 48, Yes, Yes, 81, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-3,-1], -2, left:11*(c-x)^(-2/3), right:12*(x-c)^(-3/4), 33, 48, Yes, Yes, 81, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0037","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed61d7d2ed2e53dac70fba99d0a2e78e47dd49d47edb78f60f7548cd474cd18b","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124037,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":4,\"left_power_numerator\":3,\"left_scale\":12,\"right_power_denominator\":1,\"right_power_numerator\":0,\"right_scale\":13,\"singularity\":-1}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=-1, interval [c−1,c+1], left integrand=12*(c−x)^(−3/4) and right integrand=13*(x−c)^(−0/1). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-2,0], -1, left:12*(c-x)^(-3/4), right:13*(x-c)^(-0/1), 48, 13, Yes, Yes, 61, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-2,0], -1, left:12*(c-x)^(-3/4), right:13*(x-c)^(-0/1), 48, 13, Yes, Yes, 61, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0038","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b374aaeb6be5bfbf254d12aeedb92efe1b8aa75d1428a9cca8f6c240c948065","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124038,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":1,\"left_power_numerator\":0,\"left_scale\":13,\"right_power_denominator\":2,\"right_power_numerator\":1,\"right_scale\":14,\"singularity\":0}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=0, interval [c−1,c+1], left integrand=13*(c−x)^(−0/1) and right integrand=14*(x−c)^(−1/2). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([-1,1], 0, left:13*(c-x)^(-0/1), right:14*(x-c)^(-1/2), 13, 28, Yes, Yes, 41, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([-1,1], 0, left:13*(c-x)^(-0/1), right:14*(x-c)^(-1/2), 13, 28, Yes, Yes, 41, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0039","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7850139735f002bd980970a3ca4e52239671a7e785b88ae19abc21a552ec43e1","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124039,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":2,\"left_power_numerator\":1,\"left_scale\":14,\"right_power_denominator\":3,\"right_power_numerator\":2,\"right_scale\":15,\"singularity\":1}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=1, interval [c−1,c+1], left integrand=14*(c−x)^(−1/2) and right integrand=15*(x−c)^(−2/3). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([0,2], 1, left:14*(c-x)^(-1/2), right:15*(x-c)^(-2/3), 28, 45, Yes, Yes, 73, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([0,2], 1, left:14*(c-x)^(-1/2), right:15*(x-c)^(-2/3), 28, 45, Yes, Yes, 73, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0040","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9ae4954ceafe53e6bd19d593354f5ca4ff238fc5ec80e32099aee61697e7e32","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124040,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":3,\"left_power_numerator\":2,\"left_scale\":15,\"right_power_denominator\":4,\"right_power_numerator\":3,\"right_scale\":16,\"singularity\":2}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=2, interval [c−1,c+1], left integrand=15*(c−x)^(−2/3) and right integrand=16*(x−c)^(−3/4). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([1,3], 2, left:15*(c-x)^(-2/3), right:16*(x-c)^(-3/4), 45, 64, Yes, Yes, 109, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([1,3], 2, left:15*(c-x)^(-2/3), right:16*(x-c)^(-3/4), 45, 64, Yes, Yes, 109, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0041","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f57f96fc91163680ea551ec279d775bf3f380d05006598a4c73df23991d1157","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124041,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":4,\"left_power_numerator\":3,\"left_scale\":16,\"right_power_denominator\":1,\"right_power_numerator\":0,\"right_scale\":17,\"singularity\":3}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=3, interval [c−1,c+1], left integrand=16*(c−x)^(−3/4) and right integrand=17*(x−c)^(−0/1). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([2,4], 3, left:16*(c-x)^(-3/4), right:17*(x-c)^(-0/1), 64, 17, Yes, Yes, 81, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([2,4], 3, left:16*(c-x)^(-3/4), right:17*(x-c)^(-0/1), 64, 17, Yes, Yes, 81, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0042","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b3c41aa795a52369de2e0ff4829612ac259f15a2c3cb8b01881feaf4b6ca1a3","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124042,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":1,\"left_power_numerator\":0,\"left_scale\":17,\"right_power_denominator\":2,\"right_power_numerator\":1,\"right_scale\":18,\"singularity\":4}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=4, interval [c−1,c+1], left integrand=17*(c−x)^(−0/1) and right integrand=18*(x−c)^(−1/2). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([3,5], 4, left:17*(c-x)^(-0/1), right:18*(x-c)^(-1/2), 17, 36, Yes, Yes, 53, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([3,5], 4, left:17*(c-x)^(-0/1), right:18*(x-c)^(-1/2), 17, 36, Yes, Yes, 53, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0043","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb53e4a448593e679082d5018894d15caf215b79b390a6fe08953c17e49525e2","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124043,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":2,\"left_power_numerator\":1,\"left_scale\":18,\"right_power_denominator\":3,\"right_power_numerator\":2,\"right_scale\":19,\"singularity\":5}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=5, interval [c−1,c+1], left integrand=18*(c−x)^(−1/2) and right integrand=19*(x−c)^(−2/3). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([4,6], 5, left:18*(c-x)^(-1/2), right:19*(x-c)^(-2/3), 36, 57, Yes, Yes, 93, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([4,6], 5, left:18*(c-x)^(-1/2), right:19*(x-c)^(-2/3), 36, 57, Yes, Yes, 93, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0044","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e01b9fcdfa2cf5f82b70faf727a6df1611fdc4196bf88b358535731037374dec","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124044,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":3,\"left_power_numerator\":2,\"left_scale\":19,\"right_power_denominator\":4,\"right_power_numerator\":3,\"right_scale\":20,\"singularity\":6}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=6, interval [c−1,c+1], left integrand=19*(c−x)^(−2/3) and right integrand=20*(x−c)^(−3/4). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([5,7], 6, left:19*(c-x)^(-2/3), right:20*(x-c)^(-3/4), 57, 80, Yes, Yes, 137, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([5,7], 6, left:19*(c-x)^(-2/3), right:20*(x-c)^(-3/4), 57, 80, Yes, Yes, 137, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0045","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7533abf7e7c2fbc6c0c95c02011c96f5f6777c95d66510019d5da60e40c26703","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124045,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":4,\"left_power_numerator\":3,\"left_scale\":20,\"right_power_denominator\":1,\"right_power_numerator\":0,\"right_scale\":21,\"singularity\":7}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=7, interval [c−1,c+1], left integrand=20*(c−x)^(−3/4) and right integrand=21*(x−c)^(−0/1). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([6,8], 7, left:20*(c-x)^(-3/4), right:21*(x-c)^(-0/1), 80, 21, Yes, Yes, 101, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([6,8], 7, left:20*(c-x)^(-3/4), right:21*(x-c)^(-0/1), 80, 21, Yes, Yes, 101, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0046","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7111f4c30f1f04ecea499ff3a655ab022ffd865383035a3126d72c45fd85afa","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124046,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":1,\"left_power_numerator\":0,\"left_scale\":21,\"right_power_denominator\":2,\"right_power_numerator\":1,\"right_scale\":22,\"singularity\":8}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=8, interval [c−1,c+1], left integrand=21*(c−x)^(−0/1) and right integrand=22*(x−c)^(−1/2). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([7,9], 8, left:21*(c-x)^(-0/1), right:22*(x-c)^(-1/2), 21, 44, Yes, Yes, 65, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([7,9], 8, left:21*(c-x)^(-0/1), right:22*(x-c)^(-1/2), 21, 44, Yes, Yes, 65, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0047","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d97b6264fe9bd7d8e3d05742ef16b1ca07f2e03064dd4694948def65959bdbf","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124047,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":2,\"left_power_numerator\":1,\"left_scale\":22,\"right_power_denominator\":3,\"right_power_numerator\":2,\"right_scale\":23,\"singularity\":9}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=9, interval [c−1,c+1], left integrand=22*(c−x)^(−1/2) and right integrand=23*(x−c)^(−2/3). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([8,10], 9, left:22*(c-x)^(-1/2), right:23*(x-c)^(-2/3), 44, 69, Yes, Yes, 113, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([8,10], 9, left:22*(c-x)^(-1/2), right:23*(x-c)^(-2/3), 44, 69, Yes, Yes, 113, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0048","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11f1a03cbd21d8ccd64dc29c9c38f37034b6711426e25c77d4ca143425d104d1","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124048,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":3,\"left_power_numerator\":2,\"left_scale\":23,\"right_power_denominator\":4,\"right_power_numerator\":3,\"right_scale\":24,\"singularity\":10}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=10, interval [c−1,c+1], left integrand=23*(c−x)^(−2/3) and right integrand=24*(x−c)^(−3/4). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([9,11], 10, left:23*(c-x)^(-2/3), right:24*(x-c)^(-3/4), 69, 96, Yes, Yes, 165, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([9,11], 10, left:23*(c-x)^(-2/3), right:24*(x-c)^(-3/4), 69, 96, Yes, Yes, 165, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0049","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2945d65933c51ceb60e292715cf69e8cc4879bb18dd433f8d0ff4fd670e44463","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124049,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":4,\"left_power_numerator\":3,\"left_scale\":24,\"right_power_denominator\":1,\"right_power_numerator\":0,\"right_scale\":25,\"singularity\":11}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=11, interval [c−1,c+1], left integrand=24*(c−x)^(−3/4) and right integrand=25*(x−c)^(−0/1). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([10,12], 11, left:24*(c-x)^(-3/4), right:25*(x-c)^(-0/1), 96, 25, Yes, Yes, 121, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([10,12], 11, left:24*(c-x)^(-3/4), right:25*(x-c)^(-0/1), 96, 25, Yes, Yes, 121, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0050","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3ada1ad8f331b85d592215ef261e8515ec93d7181b947db6f8379addfa27bfc","family_id":"qf-mod43-improper-interior-singular-split","seed":202611124050,"semantic_template":"qf-mod43-improper-interior-singular-split/{\"left_power_denominator\":1,\"left_power_numerator\":0,\"left_scale\":25,\"right_power_denominator\":2,\"right_power_numerator\":1,\"right_scale\":26,\"singularity\":12}"},"fields":{"title":"Both Sides of the Interior Singularity Need Independent Limits","stem":"Use c=12, interval [c−1,c+1], left integrand=25*(c−x)^(−0/1) and right integrand=26*(x−c)^(−1/2). Both exponents are reduced rationals in [0,1). Give (interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"([11,13], 12, left:25*(c-x)^(-0/1), right:26*(x-c)^(-1/2), 25, 52, Yes, Yes, 77, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)","explanation":"Split at c. Compute each one-sided limit as its scale/(1−exponent) before summing. Keep both convergence flags; signed cancellation cannot substitute for either required limit. The exact result, retaining the requested order, is ([11,13], 12, left:25*(c-x)^(-0/1), right:26*(x-c)^(-1/2), 25, 52, Yes, Yes, 77, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)."},"skills":["Both Sides of the Interior Singularity Need Independent Limits"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0051","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dfe198932cb3368cb02ac8bd391c1bfc2f4f6e003dc10454217205a5c3225b75","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124051,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":2,\"left_scale\":1,\"right_scale\":2}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=1/abs(x)^2 for x≤−1, f(x)=0 for −1<x<1 and f(x)=2/x^2 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:1/abs(x)^2;-1<x<1:0;x>=1:2/x^2, (-infinity,infinity), 1, 2, Yes, Yes, 3, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:1/abs(x)^2;-1<x<1:0;x>=1:2/x^2, (-infinity,infinity), 1, 2, Yes, Yes, 3, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0052","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d9689fa5e8331ae9ec675a19acefc846a1dc2a337fe765377298540ba43a23a","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124052,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":3,\"left_scale\":2,\"right_scale\":3}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=2/abs(x)^3 for x≤−1, f(x)=0 for −1<x<1 and f(x)=3/x^3 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:2/abs(x)^3;-1<x<1:0;x>=1:3/x^3, (-infinity,infinity), 1, 3/2, Yes, Yes, 5/2, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:2/abs(x)^3;-1<x<1:0;x>=1:3/x^3, (-infinity,infinity), 1, 3/2, Yes, Yes, 5/2, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0053","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"448efa98ed76a8b90611b72c75eaf6584c126d51931f357f972ad098a02bae19","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124053,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":4,\"left_scale\":3,\"right_scale\":4}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=3/abs(x)^4 for x≤−1, f(x)=0 for −1<x<1 and f(x)=4/x^4 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:3/abs(x)^4;-1<x<1:0;x>=1:4/x^4, (-infinity,infinity), 1, 4/3, Yes, Yes, 7/3, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:3/abs(x)^4;-1<x<1:0;x>=1:4/x^4, (-infinity,infinity), 1, 4/3, Yes, Yes, 7/3, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0054","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c257fec6a41711633e50443be9e3340266e7bf9afe038724f3bba80f77caf60e","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124054,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":5,\"left_scale\":4,\"right_scale\":5}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=4/abs(x)^5 for x≤−1, f(x)=0 for −1<x<1 and f(x)=5/x^5 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:4/abs(x)^5;-1<x<1:0;x>=1:5/x^5, (-infinity,infinity), 1, 5/4, Yes, Yes, 9/4, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:4/abs(x)^5;-1<x<1:0;x>=1:5/x^5, (-infinity,infinity), 1, 5/4, Yes, Yes, 9/4, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0055","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca25bfa980ea481e63503dc8a3d0d7796e15a1bdf68d7c045be969b7e6a9f4f0","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124055,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":2,\"left_scale\":5,\"right_scale\":6}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=5/abs(x)^2 for x≤−1, f(x)=0 for −1<x<1 and f(x)=6/x^2 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:5/abs(x)^2;-1<x<1:0;x>=1:6/x^2, (-infinity,infinity), 5, 6, Yes, Yes, 11, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:5/abs(x)^2;-1<x<1:0;x>=1:6/x^2, (-infinity,infinity), 5, 6, Yes, Yes, 11, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0056","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3bdc2dd8d082bd93891538f55d9b752543597987563f68ad542d49d50b4af8f7","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124056,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":3,\"left_scale\":6,\"right_scale\":7}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=6/abs(x)^3 for x≤−1, f(x)=0 for −1<x<1 and f(x)=7/x^3 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:6/abs(x)^3;-1<x<1:0;x>=1:7/x^3, (-infinity,infinity), 3, 7/2, Yes, Yes, 13/2, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:6/abs(x)^3;-1<x<1:0;x>=1:7/x^3, (-infinity,infinity), 3, 7/2, Yes, Yes, 13/2, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0057","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"314e96136d18c8123237d17eae17ba3aaeaa914a54c4671fe27fc35c1a23a1cc","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124057,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":4,\"left_scale\":7,\"right_scale\":8}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=7/abs(x)^4 for x≤−1, f(x)=0 for −1<x<1 and f(x)=8/x^4 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:7/abs(x)^4;-1<x<1:0;x>=1:8/x^4, (-infinity,infinity), 7/3, 8/3, Yes, Yes, 5, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:7/abs(x)^4;-1<x<1:0;x>=1:8/x^4, (-infinity,infinity), 7/3, 8/3, Yes, Yes, 5, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0058","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"69c507880a41918064b27699fa7c475fee534590f35401791bc2e69c473e6e2f","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124058,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":5,\"left_scale\":8,\"right_scale\":9}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=8/abs(x)^5 for x≤−1, f(x)=0 for −1<x<1 and f(x)=9/x^5 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:8/abs(x)^5;-1<x<1:0;x>=1:9/x^5, (-infinity,infinity), 2, 9/4, Yes, Yes, 17/4, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:8/abs(x)^5;-1<x<1:0;x>=1:9/x^5, (-infinity,infinity), 2, 9/4, Yes, Yes, 17/4, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0059","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a739a794a4d2095e318ccef4528f0cd1b9a377f5bb4a3b67f929c5a5f2bd7dd6","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124059,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":2,\"left_scale\":9,\"right_scale\":10}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=9/abs(x)^2 for x≤−1, f(x)=0 for −1<x<1 and f(x)=10/x^2 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:9/abs(x)^2;-1<x<1:0;x>=1:10/x^2, (-infinity,infinity), 9, 10, Yes, Yes, 19, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:9/abs(x)^2;-1<x<1:0;x>=1:10/x^2, (-infinity,infinity), 9, 10, Yes, Yes, 19, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0060","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"684f8088b6066de182c9f8f3e95aeabe3d4fc4adfc521adfe50ddd57b3ba2820","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124060,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":3,\"left_scale\":10,\"right_scale\":11}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=10/abs(x)^3 for x≤−1, f(x)=0 for −1<x<1 and f(x)=11/x^3 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:10/abs(x)^3;-1<x<1:0;x>=1:11/x^3, (-infinity,infinity), 5, 11/2, Yes, Yes, 21/2, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:10/abs(x)^3;-1<x<1:0;x>=1:11/x^3, (-infinity,infinity), 5, 11/2, Yes, Yes, 21/2, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0061","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"360c150d9b79ed1e25df3071b34d8844d6619bfeef84fde3f8e23b315408cfb1","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124061,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":4,\"left_scale\":11,\"right_scale\":12}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=11/abs(x)^4 for x≤−1, f(x)=0 for −1<x<1 and f(x)=12/x^4 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:11/abs(x)^4;-1<x<1:0;x>=1:12/x^4, (-infinity,infinity), 11/3, 4, Yes, Yes, 23/3, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:11/abs(x)^4;-1<x<1:0;x>=1:12/x^4, (-infinity,infinity), 11/3, 4, Yes, Yes, 23/3, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0062","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26941ba456e100779663bd4527fd2c3e6a916d951bca59c81d0b9b06e53e19a1","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124062,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":5,\"left_scale\":12,\"right_scale\":13}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=12/abs(x)^5 for x≤−1, f(x)=0 for −1<x<1 and f(x)=13/x^5 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:12/abs(x)^5;-1<x<1:0;x>=1:13/x^5, (-infinity,infinity), 3, 13/4, Yes, Yes, 25/4, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:12/abs(x)^5;-1<x<1:0;x>=1:13/x^5, (-infinity,infinity), 3, 13/4, Yes, Yes, 25/4, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0063","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98abe93d9d09f02963b7008d2a2139bd124e58dc8431c0d55961fa1b012c3d5a","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124063,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":2,\"left_scale\":13,\"right_scale\":14}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=13/abs(x)^2 for x≤−1, f(x)=0 for −1<x<1 and f(x)=14/x^2 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:13/abs(x)^2;-1<x<1:0;x>=1:14/x^2, (-infinity,infinity), 13, 14, Yes, Yes, 27, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:13/abs(x)^2;-1<x<1:0;x>=1:14/x^2, (-infinity,infinity), 13, 14, Yes, Yes, 27, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0064","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c14fadc836401836bca60bcb77c4aab7cc52c2ee9a1138dd69dbf28105d9b9ae","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124064,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":3,\"left_scale\":14,\"right_scale\":15}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=14/abs(x)^3 for x≤−1, f(x)=0 for −1<x<1 and f(x)=15/x^3 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:14/abs(x)^3;-1<x<1:0;x>=1:15/x^3, (-infinity,infinity), 7, 15/2, Yes, Yes, 29/2, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:14/abs(x)^3;-1<x<1:0;x>=1:15/x^3, (-infinity,infinity), 7, 15/2, Yes, Yes, 29/2, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0065","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"338ff1331806edfa46fb14dcab865f3f8aeddb44a45ac097b3e1d878e072f5ec","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124065,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":4,\"left_scale\":15,\"right_scale\":16}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=15/abs(x)^4 for x≤−1, f(x)=0 for −1<x<1 and f(x)=16/x^4 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:15/abs(x)^4;-1<x<1:0;x>=1:16/x^4, (-infinity,infinity), 5, 16/3, Yes, Yes, 31/3, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:15/abs(x)^4;-1<x<1:0;x>=1:16/x^4, (-infinity,infinity), 5, 16/3, Yes, Yes, 31/3, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0066","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4bffb75e9a4cdb492cc1a55aa2cb0ae27d5cc7cdea7f0f1760efeb1862f9bf3f","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124066,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":5,\"left_scale\":16,\"right_scale\":17}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=16/abs(x)^5 for x≤−1, f(x)=0 for −1<x<1 and f(x)=17/x^5 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:16/abs(x)^5;-1<x<1:0;x>=1:17/x^5, (-infinity,infinity), 4, 17/4, Yes, Yes, 33/4, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:16/abs(x)^5;-1<x<1:0;x>=1:17/x^5, (-infinity,infinity), 4, 17/4, Yes, Yes, 33/4, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0067","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b375fbc9d90da768e164ebf8b376bdb677f6cd3350a4176eedca6eeba57bbcc","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124067,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":2,\"left_scale\":17,\"right_scale\":18}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=17/abs(x)^2 for x≤−1, f(x)=0 for −1<x<1 and f(x)=18/x^2 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:17/abs(x)^2;-1<x<1:0;x>=1:18/x^2, (-infinity,infinity), 17, 18, Yes, Yes, 35, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:17/abs(x)^2;-1<x<1:0;x>=1:18/x^2, (-infinity,infinity), 17, 18, Yes, Yes, 35, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0068","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d59e33a22d2ead253947bb07f235381e1fa49284cbb37c388fc1720206f7a9c","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124068,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":3,\"left_scale\":18,\"right_scale\":19}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=18/abs(x)^3 for x≤−1, f(x)=0 for −1<x<1 and f(x)=19/x^3 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:18/abs(x)^3;-1<x<1:0;x>=1:19/x^3, (-infinity,infinity), 9, 19/2, Yes, Yes, 37/2, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:18/abs(x)^3;-1<x<1:0;x>=1:19/x^3, (-infinity,infinity), 9, 19/2, Yes, Yes, 37/2, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0069","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2569d1fdd0f24616d03271345b5199897cac7ce3bf7b628fcfcbe426e436b0e9","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124069,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":4,\"left_scale\":19,\"right_scale\":20}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=19/abs(x)^4 for x≤−1, f(x)=0 for −1<x<1 and f(x)=20/x^4 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:19/abs(x)^4;-1<x<1:0;x>=1:20/x^4, (-infinity,infinity), 19/3, 20/3, Yes, Yes, 13, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:19/abs(x)^4;-1<x<1:0;x>=1:20/x^4, (-infinity,infinity), 19/3, 20/3, Yes, Yes, 13, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0070","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f624f42f3ce287235ef2c3365567ff8ef7ba3f007b0158376c812ad44a828413","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124070,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":5,\"left_scale\":20,\"right_scale\":21}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=20/abs(x)^5 for x≤−1, f(x)=0 for −1<x<1 and f(x)=21/x^5 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:20/abs(x)^5;-1<x<1:0;x>=1:21/x^5, (-infinity,infinity), 5, 21/4, Yes, Yes, 41/4, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:20/abs(x)^5;-1<x<1:0;x>=1:21/x^5, (-infinity,infinity), 5, 21/4, Yes, Yes, 41/4, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0071","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ef4776c2bd4098ce0951d7be28dc36f0776da93236891d74f0d05b266e0fba0","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124071,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":2,\"left_scale\":21,\"right_scale\":22}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=21/abs(x)^2 for x≤−1, f(x)=0 for −1<x<1 and f(x)=22/x^2 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:21/abs(x)^2;-1<x<1:0;x>=1:22/x^2, (-infinity,infinity), 21, 22, Yes, Yes, 43, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:21/abs(x)^2;-1<x<1:0;x>=1:22/x^2, (-infinity,infinity), 21, 22, Yes, Yes, 43, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0072","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d7b695dc89e654eacb3e5155e2e0bac8e82a64fe860904b223a9442fe33d221","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124072,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":3,\"left_scale\":22,\"right_scale\":23}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=22/abs(x)^3 for x≤−1, f(x)=0 for −1<x<1 and f(x)=23/x^3 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:22/abs(x)^3;-1<x<1:0;x>=1:23/x^3, (-infinity,infinity), 11, 23/2, Yes, Yes, 45/2, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:22/abs(x)^3;-1<x<1:0;x>=1:23/x^3, (-infinity,infinity), 11, 23/2, Yes, Yes, 45/2, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0073","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3cc90f85f9c6d2473248e20943cc27dd5b660396754050c3df49f300b8eb74a1","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124073,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":4,\"left_scale\":23,\"right_scale\":24}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=23/abs(x)^4 for x≤−1, f(x)=0 for −1<x<1 and f(x)=24/x^4 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:23/abs(x)^4;-1<x<1:0;x>=1:24/x^4, (-infinity,infinity), 23/3, 8, Yes, Yes, 47/3, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:23/abs(x)^4;-1<x<1:0;x>=1:24/x^4, (-infinity,infinity), 23/3, 8, Yes, Yes, 47/3, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0074","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7e6861b15e94ccd2110ed19b7aef8aadc97358eaf3a729c254d9082ff5b30391","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124074,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":5,\"left_scale\":24,\"right_scale\":25}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=24/abs(x)^5 for x≤−1, f(x)=0 for −1<x<1 and f(x)=25/x^5 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:24/abs(x)^5;-1<x<1:0;x>=1:25/x^5, (-infinity,infinity), 6, 25/4, Yes, Yes, 49/4, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:24/abs(x)^5;-1<x<1:0;x>=1:25/x^5, (-infinity,infinity), 6, 25/4, Yes, Yes, 49/4, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0075","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2c8bc57c1384bf57a43cd3104fb54bac152f8ed878253d26cd845086c49fad14","family_id":"qf-mod43-improper-two-sided-independent","seed":202611124075,"semantic_template":"qf-mod43-improper-two-sided-independent/{\"exponent\":2,\"left_scale\":25,\"right_scale\":26}"},"fields":{"title":"Two Infinite Tails Must Converge Separately before Their Values Are Added","stem":"Use f(x)=25/abs(x)^2 for x≤−1, f(x)=0 for −1<x<1 and f(x)=26/x^2 for x≥1, with integer exponent≥2. Give (piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate).","answer":{"kind":"expression","value":"(x<=-1:25/abs(x)^2;-1<x<1:0;x>=1:26/x^2, (-infinity,infinity), 25, 26, Yes, Yes, 51, both_one_sided_tail_integrals_converge_independently)","explanation":"Take the negative and positive infinite-tail limits independently. The zero middle contributes0; each tail equals its own scale/(exponent−1). Even if their finite values cancel, ordinary convergence comes from both separate obligations, not symmetry. The exact result, retaining the requested order, is (x<=-1:25/abs(x)^2;-1<x<1:0;x>=1:26/x^2, (-infinity,infinity), 25, 26, Yes, Yes, 51, both_one_sided_tail_integrals_converge_independently)."},"skills":["Two Infinite Tails Must Converge Separately before Their Values Are Added"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0076","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"075c83f515a10122bfc33b9c8972c8c468663f6fbe1aa83981eaa7abffe593aa","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124076,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":2,\"epsilon_denominator\":2}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=2 and exclusion epsilon=1/2. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 2, 1/2, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 2, 1/2, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0077","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"683eaf607987b0cd46eba1832a7bf01574c3d57c224b8198cc829b1730e5d123","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124077,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":3,\"epsilon_denominator\":3}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=3 and exclusion epsilon=1/3. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 3, 1/3, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 3, 1/3, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0078","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82b925b50c01027ffea36423f196c0e591edb7ac769d6c534a4a626a819f45c1","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124078,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":4,\"epsilon_denominator\":4}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=4 and exclusion epsilon=1/4. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 4, 1/4, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 4, 1/4, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0079","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f5d93e2721051e80fef444ec1a6aa634d70b7f07d48e2bd31d7e3ebdcb63ec0","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124079,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":5,\"epsilon_denominator\":5}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=5 and exclusion epsilon=1/5. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 5, 1/5, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 5, 1/5, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0080","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1aba382145a5f1a8bbd6de1ab70bdc4726b522764bb927a8af369a37c17d8504","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124080,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":6,\"epsilon_denominator\":6}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=6 and exclusion epsilon=1/6. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 6, 1/6, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 6, 1/6, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0081","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82d9d6c5b5c25a59767b7e0aacea49c0cc0ceb2d4c66468951f989aebd2f6e6e","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124081,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":7,\"epsilon_denominator\":7}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=7 and exclusion epsilon=1/7. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 7, 1/7, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 7, 1/7, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0082","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb044b17c6383e61b02b26e363fb07a4239d0b114fdba214519326a71d07f3f2","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124082,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":8,\"epsilon_denominator\":8}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=8 and exclusion epsilon=1/8. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 8, 1/8, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 8, 1/8, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0083","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"74aefef55b229d94fc47ad40b2bb193ce07bd422f411ac212297efe62014c427","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124083,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":9,\"epsilon_denominator\":9}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=9 and exclusion epsilon=1/9. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 9, 1/9, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 9, 1/9, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0084","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b8f78268edd43eb5f69cdb65246f0d81ad4de8aab84a6cd56ae6ac1d3a1d5f2","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124084,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":10,\"epsilon_denominator\":10}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=10 and exclusion epsilon=1/10. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 10, 1/10, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 10, 1/10, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0085","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c438094efd8337bfac88f55f934b17e0c77c6897b862b8a5c6aa719a51cbaf5e","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124085,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":11,\"epsilon_denominator\":11}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=11 and exclusion epsilon=1/11. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 11, 1/11, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 11, 1/11, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0086","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf761c7a5d5d4917734c4ea65e0e29c75f9cf889eca7c836bf8cfdd0bd6b193d","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124086,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":12,\"epsilon_denominator\":12}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=12 and exclusion epsilon=1/12. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 12, 1/12, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 12, 1/12, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0087","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"32371eb382bebb1b83f31c578b157f726f316c14d632da7ab24338e34f882885","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124087,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":13,\"epsilon_denominator\":13}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=13 and exclusion epsilon=1/13. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 13, 1/13, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 13, 1/13, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0088","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d73eb65796d4ed9cfc6daa87a46d451f13e2a2dccfcbce1dc9fd6eb3563aac5","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124088,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":14,\"epsilon_denominator\":14}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=14 and exclusion epsilon=1/14. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 14, 1/14, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 14, 1/14, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0089","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae7243255b3909339a3eb3812e8d08dfb4df71424ea2e397e032330e87edac05","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124089,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":15,\"epsilon_denominator\":15}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=15 and exclusion epsilon=1/15. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 15, 1/15, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 15, 1/15, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0090","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87c797aee9329ea1b8f3450b6077141253c0c3d460b119810d39f2b624d04ce9","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124090,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":16,\"epsilon_denominator\":16}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=16 and exclusion epsilon=1/16. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 16, 1/16, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 16, 1/16, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0091","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bdcce0d421978641cd71f92aa3db16770756292dfb958b8a7832419c8dab9be3","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124091,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":17,\"epsilon_denominator\":17}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=17 and exclusion epsilon=1/17. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 17, 1/17, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 17, 1/17, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0092","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cfdd89bdf71b15e96164c590cdca68667d3d1cf0e21f9e42d17c6c2f0f1c2f27","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124092,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":18,\"epsilon_denominator\":18}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=18 and exclusion epsilon=1/18. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 18, 1/18, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 18, 1/18, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0093","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b0d79ade956d900d3164c71e8ad60544810487eca17f73717b419a4923db8bdc","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124093,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":19,\"epsilon_denominator\":19}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=19 and exclusion epsilon=1/19. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 19, 1/19, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 19, 1/19, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0094","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5319b7e37b6fa5ba3858ed134d2cf9ea7ab66777b3dda77cbd1ee07ef5428283","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124094,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":20,\"epsilon_denominator\":20}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=20 and exclusion epsilon=1/20. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 20, 1/20, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 20, 1/20, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0095","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1572b52b8b49af6a6f1774d7dcd2678b28b66c4f4da8bed389408842917fe066","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124095,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":21,\"epsilon_denominator\":21}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=21 and exclusion epsilon=1/21. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 21, 1/21, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 21, 1/21, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0096","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ff0bdbf195782bc6383d610a63fd738db92244276c7b4f87469983e1dca94b9","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124096,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":22,\"epsilon_denominator\":22}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=22 and exclusion epsilon=1/22. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 22, 1/22, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 22, 1/22, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0097","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f1d037b2698941a92e060444ed2df5a23e41a2f58c84116c6b7e59382bb4800","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124097,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":23,\"epsilon_denominator\":23}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=23 and exclusion epsilon=1/23. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 23, 1/23, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 23, 1/23, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0098","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"64dbd745adfbda6f0ce68a113215a0e975e6dc7841fff94619efa56fd1508b15","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124098,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":24,\"epsilon_denominator\":24}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=24 and exclusion epsilon=1/24. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 24, 1/24, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 24, 1/24, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0099","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93bf76bbd5f7837ee908495a9554d6a415078c2cc3af88184d3826ec1823f5e1","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124099,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":25,\"epsilon_denominator\":25}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=25 and exclusion epsilon=1/25. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 25, 1/25, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 25, 1/25, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r04-improper-integral-0100","batch_id":"batch-20261014-advance-20261112-r04-improper-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b15a84a9f3ae08085af101df030ca0b01d0752821a52809b0dc97984661b199","family_id":"qf-mod43-improper-principal-value-boundary","seed":202611124100,"semantic_template":"qf-mod43-improper-principal-value-boundary/{\"cutoff\":26,\"epsilon_denominator\":26}"},"fields":{"title":"A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence","stem":"Use f(x)=1/x on the real line with singularity0, symmetric cutoff B=26 and exclusion epsilon=1/26. Give (integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary).","answer":{"kind":"expression","value":"(f(x)=1/x, (-infinity,infinity) with singularity at 0, 26, 1/26, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)","explanation":"The symmetric truncated integrals over [−B,−epsilon] and [epsilon,B] sum to0, and the symmetric principal value is0. The required one-sided limits at the singularity and infinite ends diverge, so ordinary improper convergence is No. Finitely many cutoffs cannot certify a universal convergence obligation. The exact result, retaining the requested order, is (f(x)=1/x, (-infinity,infinity) with singularity at 0, 26, 1/26, 0, Yes, No, No, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)."},"skills":["A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}