# Declared Power Integrands, Independent One-Sided Limits and Principal-Value Boundaries This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed. This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately. batch-20261014-advance-20261112-r04-improper-integral-foundations · 2026-10-14 · 100 questions This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt. Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged. [Field-order and source-binding data](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json) ## 1. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=1/x^2 on [1,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=1/x^2, [1,infinity), 1, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0001/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 2. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=2/x^3 on [2,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=2/x^3, [2,infinity), 1/4, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0002/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 3. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=3/x^4 on [3,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=3/x^4, [3,infinity), 1/27, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0003/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 4. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=4/x^5 on [4,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=4/x^5, [4,infinity), 1/256, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0004/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 5. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=5/x^2 on [5,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=5/x^2, [5,infinity), 1, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0005/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 6. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=6/x^3 on [1,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=6/x^3, [1,infinity), 3, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0006/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 7. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=7/x^4 on [2,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=7/x^4, [2,infinity), 7/24, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0007/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 8. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=8/x^5 on [3,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=8/x^5, [3,infinity), 2/81, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0008/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 9. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=9/x^2 on [4,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=9/x^2, [4,infinity), 9/4, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0009/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 10. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=10/x^3 on [5,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=10/x^3, [5,infinity), 1/5, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0010/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 11. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=11/x^4 on [1,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=11/x^4, [1,infinity), 11/3, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0011/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 12. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=12/x^5 on [2,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=12/x^5, [2,infinity), 3/16, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0012/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 13. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=13/x^2 on [3,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=13/x^2, [3,infinity), 13/3, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0013/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 14. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=14/x^3 on [4,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=14/x^3, [4,infinity), 7/16, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0014/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 15. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=15/x^4 on [5,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=15/x^4, [5,infinity), 1/25, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0015/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 16. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=16/x^5 on [1,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=16/x^5, [1,infinity), 4, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0016/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 17. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=17/x^2 on [2,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=17/x^2, [2,infinity), 17/2, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0017/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 18. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=18/x^3 on [3,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=18/x^3, [3,infinity), 1, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0018/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 19. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=19/x^4 on [4,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=19/x^4, [4,infinity), 19/192, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0019/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 20. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=20/x^5 on [5,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=20/x^5, [5,infinity), 1/125, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0020/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 21. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=21/x^2 on [1,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=21/x^2, [1,infinity), 21, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0021/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 22. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=22/x^3 on [2,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=22/x^3, [2,infinity), 11/4, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0022/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 23. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=23/x^4 on [3,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=23/x^4, [3,infinity), 23/81, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0023/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 24. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=24/x^5 on [4,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=24/x^5, [4,infinity), 3/128, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0024/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 25. The One-Sided Tail Is a Limit of Finite-Cutoff Integrals Use f(x)=25/x^2 on [5,infinity), with integer exponent≥2 and lower≥1. Give (integrand, interval, integral, converges, certificate). View answer and explanation(f(x)=25/x^2, [5,infinity), 5, Yes, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0025/) Improper integral · qf-mod43-improper-one-sided-tail · candidate ## 26. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0026/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 27. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0027/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 28. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0028/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 29. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0029/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 30. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0030/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 31. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0031/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 32. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0032/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 33. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0033/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 34. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0034/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 35. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0035/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 36. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0036/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 37. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0037/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 38. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([-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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0038/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 39. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0039/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 40. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0040/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 41. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0041/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 42. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0042/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 43. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0043/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 44. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0044/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 45. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0045/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 46. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0046/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 47. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0047/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 48. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0048/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 49. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0049/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 50. Both Sides of the Interior Singularity Need Independent Limits 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). View answer and explanation([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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0050/) Improper integral · qf-mod43-improper-interior-singular-split · candidate ## 51. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=1/abs(x)^2 for x≤−1, f(x)=0 for −1=1:2/x^2, (-infinity,infinity), 1, 2, Yes, Yes, 3, both_one_sided_tail_integrals_converge_independently) 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=1:2/x^2, (-infinity,infinity), 1, 2, Yes, Yes, 3, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0051/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 52. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=2/abs(x)^3 for x≤−1, f(x)=0 for −1=1:3/x^3, (-infinity,infinity), 1, 3/2, Yes, Yes, 5/2, both_one_sided_tail_integrals_converge_independently) 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=1:3/x^3, (-infinity,infinity), 1, 3/2, Yes, Yes, 5/2, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0052/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 53. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=3/abs(x)^4 for x≤−1, f(x)=0 for −1=1:4/x^4, (-infinity,infinity), 1, 4/3, Yes, Yes, 7/3, both_one_sided_tail_integrals_converge_independently) 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=1:4/x^4, (-infinity,infinity), 1, 4/3, Yes, Yes, 7/3, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0053/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 54. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=4/abs(x)^5 for x≤−1, f(x)=0 for −1=1:5/x^5, (-infinity,infinity), 1, 5/4, Yes, Yes, 9/4, both_one_sided_tail_integrals_converge_independently) 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=1:5/x^5, (-infinity,infinity), 1, 5/4, Yes, Yes, 9/4, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0054/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 55. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=5/abs(x)^2 for x≤−1, f(x)=0 for −1=1:6/x^2, (-infinity,infinity), 5, 6, Yes, Yes, 11, both_one_sided_tail_integrals_converge_independently) 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=1:6/x^2, (-infinity,infinity), 5, 6, Yes, Yes, 11, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0055/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 56. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=6/abs(x)^3 for x≤−1, f(x)=0 for −1=1:7/x^3, (-infinity,infinity), 3, 7/2, Yes, Yes, 13/2, both_one_sided_tail_integrals_converge_independently) 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=1:7/x^3, (-infinity,infinity), 3, 7/2, Yes, Yes, 13/2, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0056/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 57. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=7/abs(x)^4 for x≤−1, f(x)=0 for −1=1:8/x^4, (-infinity,infinity), 7/3, 8/3, Yes, Yes, 5, both_one_sided_tail_integrals_converge_independently) 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=1:8/x^4, (-infinity,infinity), 7/3, 8/3, Yes, Yes, 5, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0057/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 58. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=8/abs(x)^5 for x≤−1, f(x)=0 for −1=1:9/x^5, (-infinity,infinity), 2, 9/4, Yes, Yes, 17/4, both_one_sided_tail_integrals_converge_independently) 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=1:9/x^5, (-infinity,infinity), 2, 9/4, Yes, Yes, 17/4, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0058/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 59. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=9/abs(x)^2 for x≤−1, f(x)=0 for −1=1:10/x^2, (-infinity,infinity), 9, 10, Yes, Yes, 19, both_one_sided_tail_integrals_converge_independently) 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=1:10/x^2, (-infinity,infinity), 9, 10, Yes, Yes, 19, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0059/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 60. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=10/abs(x)^3 for x≤−1, f(x)=0 for −1=1:11/x^3, (-infinity,infinity), 5, 11/2, Yes, Yes, 21/2, both_one_sided_tail_integrals_converge_independently) 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=1:11/x^3, (-infinity,infinity), 5, 11/2, Yes, Yes, 21/2, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0060/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 61. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=11/abs(x)^4 for x≤−1, f(x)=0 for −1=1:12/x^4, (-infinity,infinity), 11/3, 4, Yes, Yes, 23/3, both_one_sided_tail_integrals_converge_independently) 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=1:12/x^4, (-infinity,infinity), 11/3, 4, Yes, Yes, 23/3, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0061/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 62. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=12/abs(x)^5 for x≤−1, f(x)=0 for −1=1:13/x^5, (-infinity,infinity), 3, 13/4, Yes, Yes, 25/4, both_one_sided_tail_integrals_converge_independently) 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=1:13/x^5, (-infinity,infinity), 3, 13/4, Yes, Yes, 25/4, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0062/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 63. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=13/abs(x)^2 for x≤−1, f(x)=0 for −1=1:14/x^2, (-infinity,infinity), 13, 14, Yes, Yes, 27, both_one_sided_tail_integrals_converge_independently) 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=1:14/x^2, (-infinity,infinity), 13, 14, Yes, Yes, 27, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0063/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 64. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=14/abs(x)^3 for x≤−1, f(x)=0 for −1=1:15/x^3, (-infinity,infinity), 7, 15/2, Yes, Yes, 29/2, both_one_sided_tail_integrals_converge_independently) 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=1:15/x^3, (-infinity,infinity), 7, 15/2, Yes, Yes, 29/2, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0064/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 65. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=15/abs(x)^4 for x≤−1, f(x)=0 for −1=1:16/x^4, (-infinity,infinity), 5, 16/3, Yes, Yes, 31/3, both_one_sided_tail_integrals_converge_independently) 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=1:16/x^4, (-infinity,infinity), 5, 16/3, Yes, Yes, 31/3, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0065/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 66. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=16/abs(x)^5 for x≤−1, f(x)=0 for −1=1:17/x^5, (-infinity,infinity), 4, 17/4, Yes, Yes, 33/4, both_one_sided_tail_integrals_converge_independently) 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=1:17/x^5, (-infinity,infinity), 4, 17/4, Yes, Yes, 33/4, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0066/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 67. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=17/abs(x)^2 for x≤−1, f(x)=0 for −1=1:18/x^2, (-infinity,infinity), 17, 18, Yes, Yes, 35, both_one_sided_tail_integrals_converge_independently) 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=1:18/x^2, (-infinity,infinity), 17, 18, Yes, Yes, 35, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0067/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 68. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=18/abs(x)^3 for x≤−1, f(x)=0 for −1=1:19/x^3, (-infinity,infinity), 9, 19/2, Yes, Yes, 37/2, both_one_sided_tail_integrals_converge_independently) 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=1:19/x^3, (-infinity,infinity), 9, 19/2, Yes, Yes, 37/2, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0068/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 69. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=19/abs(x)^4 for x≤−1, f(x)=0 for −1=1:20/x^4, (-infinity,infinity), 19/3, 20/3, Yes, Yes, 13, both_one_sided_tail_integrals_converge_independently) 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=1:20/x^4, (-infinity,infinity), 19/3, 20/3, Yes, Yes, 13, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0069/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 70. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=20/abs(x)^5 for x≤−1, f(x)=0 for −1=1:21/x^5, (-infinity,infinity), 5, 21/4, Yes, Yes, 41/4, both_one_sided_tail_integrals_converge_independently) 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=1:21/x^5, (-infinity,infinity), 5, 21/4, Yes, Yes, 41/4, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0070/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 71. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=21/abs(x)^2 for x≤−1, f(x)=0 for −1=1:22/x^2, (-infinity,infinity), 21, 22, Yes, Yes, 43, both_one_sided_tail_integrals_converge_independently) 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=1:22/x^2, (-infinity,infinity), 21, 22, Yes, Yes, 43, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0071/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 72. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=22/abs(x)^3 for x≤−1, f(x)=0 for −1=1:23/x^3, (-infinity,infinity), 11, 23/2, Yes, Yes, 45/2, both_one_sided_tail_integrals_converge_independently) 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=1:23/x^3, (-infinity,infinity), 11, 23/2, Yes, Yes, 45/2, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0072/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 73. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=23/abs(x)^4 for x≤−1, f(x)=0 for −1=1:24/x^4, (-infinity,infinity), 23/3, 8, Yes, Yes, 47/3, both_one_sided_tail_integrals_converge_independently) 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=1:24/x^4, (-infinity,infinity), 23/3, 8, Yes, Yes, 47/3, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0073/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 74. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=24/abs(x)^5 for x≤−1, f(x)=0 for −1=1:25/x^5, (-infinity,infinity), 6, 25/4, Yes, Yes, 49/4, both_one_sided_tail_integrals_converge_independently) 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=1:25/x^5, (-infinity,infinity), 6, 25/4, Yes, Yes, 49/4, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0074/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 75. Two Infinite Tails Must Converge Separately before Their Values Are Added Use f(x)=25/abs(x)^2 for x≤−1, f(x)=0 for −1=1:26/x^2, (-infinity,infinity), 25, 26, Yes, Yes, 51, both_one_sided_tail_integrals_converge_independently) 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=1:26/x^2, (-infinity,infinity), 25, 26, Yes, Yes, 51, both_one_sided_tail_integrals_converge_independently). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0075/) Improper integral · qf-mod43-improper-two-sided-independent · candidate ## 76. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0076/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 77. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0077/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 78. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0078/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 79. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0079/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 80. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0080/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 81. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0081/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 82. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0082/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 83. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0083/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 84. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0084/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 85. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0085/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 86. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0086/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 87. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0087/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 88. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0088/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 89. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0089/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 90. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0090/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 91. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0091/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 92. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0092/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 93. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0093/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 94. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0094/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 95. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0095/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 96. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0096/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 97. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0097/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 98. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0098/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 99. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0099/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate ## 100. A Zero Symmetric Principal Value Does Not Give Ordinary Improper Convergence 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). Zero symmetric truncations and a zero symmetric principal value do not imply ordinary improper convergence; each infinite endpoint and singular side has its own obligation. View answer and explanation(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) 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). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261112-r04-improper-integral-0100/) Improper integral · qf-mod43-improper-principal-value-boundary · candidate [Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261112-r04-improper-integral-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261014-advance-20261112-r04-improper-integral-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261112-r04-improper-integral-foundations/validation-report.json)