# 瑕積分基礎 · 脈衝 R04

本批已通過 Schema、去重、來源快照與獨立答案重算；仍未標記為人工審定。

本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留；AI 覆核、有限計算與 Lean 形式證明分開標示。

batch-20261014-advance-20261112-r04-improper-integral-foundations · 2026-10-14 · 100 題

欄序為來源綁定的呈現補充；原始 JSONL 與數學答案不變。

[欄序與來源綁定資料](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":1,"exponent":2,"lower":1}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=1/x^2, [1,infinity), 1, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=1/x^2, [1,infinity), 1, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0001/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 2. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":2,"exponent":3,"lower":2}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=2/x^3, [2,infinity), 1/4, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=2/x^3, [2,infinity), 1/4, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0002/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 3. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":3,"exponent":4,"lower":3}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=3/x^4, [3,infinity), 1/27, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=3/x^4, [3,infinity), 1/27, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0003/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 4. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":4,"exponent":5,"lower":4}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=4/x^5, [4,infinity), 1/256, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=4/x^5, [4,infinity), 1/256, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0004/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 5. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":5,"exponent":2,"lower":5}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=5/x^2, [5,infinity), 1, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=5/x^2, [5,infinity), 1, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0005/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 6. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":6,"exponent":3,"lower":1}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=6/x^3, [1,infinity), 3, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=6/x^3, [1,infinity), 3, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0006/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 7. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":7,"exponent":4,"lower":2}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=7/x^4, [2,infinity), 7/24, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=7/x^4, [2,infinity), 7/24, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0007/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 8. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":8,"exponent":5,"lower":3}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=8/x^5, [3,infinity), 2/81, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=8/x^5, [3,infinity), 2/81, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0008/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 9. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":9,"exponent":2,"lower":4}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=9/x^2, [4,infinity), 9/4, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=9/x^2, [4,infinity), 9/4, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0009/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 10. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":10,"exponent":3,"lower":5}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=10/x^3, [5,infinity), 1/5, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=10/x^3, [5,infinity), 1/5, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0010/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 11. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":11,"exponent":4,"lower":1}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=11/x^4, [1,infinity), 11/3, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=11/x^4, [1,infinity), 11/3, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0011/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 12. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":12,"exponent":5,"lower":2}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=12/x^5, [2,infinity), 3/16, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=12/x^5, [2,infinity), 3/16, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0012/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 13. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":13,"exponent":2,"lower":3}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=13/x^2, [3,infinity), 13/3, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=13/x^2, [3,infinity), 13/3, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0013/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 14. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":14,"exponent":3,"lower":4}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=14/x^3, [4,infinity), 7/16, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=14/x^3, [4,infinity), 7/16, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0014/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 15. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":15,"exponent":4,"lower":5}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=15/x^4, [5,infinity), 1/25, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=15/x^4, [5,infinity), 1/25, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0015/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 16. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":16,"exponent":5,"lower":1}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=16/x^5, [1,infinity), 4, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=16/x^5, [1,infinity), 4, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0016/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 17. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":17,"exponent":2,"lower":2}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=17/x^2, [2,infinity), 17/2, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=17/x^2, [2,infinity), 17/2, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0017/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 18. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":18,"exponent":3,"lower":3}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=18/x^3, [3,infinity), 1, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=18/x^3, [3,infinity), 1, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0018/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 19. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":19,"exponent":4,"lower":4}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=19/x^4, [4,infinity), 19/192, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=19/x^4, [4,infinity), 19/192, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0019/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 20. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":20,"exponent":5,"lower":5}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=20/x^5, [5,infinity), 1/125, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=20/x^5, [5,infinity), 1/125, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0020/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 21. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":21,"exponent":2,"lower":1}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=21/x^2, [1,infinity), 21, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=21/x^2, [1,infinity), 21, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0021/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 22. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":22,"exponent":3,"lower":2}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=22/x^3, [2,infinity), 11/4, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=22/x^3, [2,infinity), 11/4, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0022/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 23. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":23,"exponent":4,"lower":3}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=23/x^4, [3,infinity), 23/81, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=23/x^4, [3,infinity), 23/81, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0023/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 24. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":24,"exponent":5,"lower":4}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=24/x^5, [4,infinity), 3/128, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=24/x^5, [4,infinity), 3/128, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0024/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 25. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-one-sided-tail；資料={"scale":25,"exponent":2,"lower":5}。請依 [integrand, interval, integral, converges, certificate] 欄序回答。

查看答案與解釋(f(x)=25/x^2, [5,infinity), 5, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)

Wave6第六層核對=(f(x)=25/x^2, [5,infinity), 5, 是, limit_of_integrals_from_lower_to_b_as_b_tends_to_infinity)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0025/)

瑕積分 · qf-mod43-improper-one-sided-tail · candidate





## 26. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":1,"right_scale":2,"left_power_numerator":0,"left_power_denominator":1,"right_power_numerator":1,"right_power_denominator":2,"singularity":-12}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-13,-11], -12, left:1*(c-x)^(-0/1), right:2*(x-c)^(-1/2), 1, 4, 是, 是, 5, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-13,-11], -12, left:1*(c-x)^(-0/1), right:2*(x-c)^(-1/2), 1, 4, 是, 是, 5, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0026/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 27. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":2,"right_scale":3,"left_power_numerator":1,"left_power_denominator":2,"right_power_numerator":2,"right_power_denominator":3,"singularity":-11}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-12,-10], -11, left:2*(c-x)^(-1/2), right:3*(x-c)^(-2/3), 4, 9, 是, 是, 13, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-12,-10], -11, left:2*(c-x)^(-1/2), right:3*(x-c)^(-2/3), 4, 9, 是, 是, 13, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0027/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 28. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":3,"right_scale":4,"left_power_numerator":2,"left_power_denominator":3,"right_power_numerator":3,"right_power_denominator":4,"singularity":-10}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-11,-9], -10, left:3*(c-x)^(-2/3), right:4*(x-c)^(-3/4), 9, 16, 是, 是, 25, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-11,-9], -10, left:3*(c-x)^(-2/3), right:4*(x-c)^(-3/4), 9, 16, 是, 是, 25, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0028/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 29. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":4,"right_scale":5,"left_power_numerator":3,"left_power_denominator":4,"right_power_numerator":0,"right_power_denominator":1,"singularity":-9}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-10,-8], -9, left:4*(c-x)^(-3/4), right:5*(x-c)^(-0/1), 16, 5, 是, 是, 21, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-10,-8], -9, left:4*(c-x)^(-3/4), right:5*(x-c)^(-0/1), 16, 5, 是, 是, 21, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0029/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 30. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":5,"right_scale":6,"left_power_numerator":0,"left_power_denominator":1,"right_power_numerator":1,"right_power_denominator":2,"singularity":-8}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-9,-7], -8, left:5*(c-x)^(-0/1), right:6*(x-c)^(-1/2), 5, 12, 是, 是, 17, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-9,-7], -8, left:5*(c-x)^(-0/1), right:6*(x-c)^(-1/2), 5, 12, 是, 是, 17, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0030/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 31. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":6,"right_scale":7,"left_power_numerator":1,"left_power_denominator":2,"right_power_numerator":2,"right_power_denominator":3,"singularity":-7}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-8,-6], -7, left:6*(c-x)^(-1/2), right:7*(x-c)^(-2/3), 12, 21, 是, 是, 33, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-8,-6], -7, left:6*(c-x)^(-1/2), right:7*(x-c)^(-2/3), 12, 21, 是, 是, 33, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0031/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 32. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":7,"right_scale":8,"left_power_numerator":2,"left_power_denominator":3,"right_power_numerator":3,"right_power_denominator":4,"singularity":-6}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-7,-5], -6, left:7*(c-x)^(-2/3), right:8*(x-c)^(-3/4), 21, 32, 是, 是, 53, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-7,-5], -6, left:7*(c-x)^(-2/3), right:8*(x-c)^(-3/4), 21, 32, 是, 是, 53, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0032/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 33. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":8,"right_scale":9,"left_power_numerator":3,"left_power_denominator":4,"right_power_numerator":0,"right_power_denominator":1,"singularity":-5}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-6,-4], -5, left:8*(c-x)^(-3/4), right:9*(x-c)^(-0/1), 32, 9, 是, 是, 41, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-6,-4], -5, left:8*(c-x)^(-3/4), right:9*(x-c)^(-0/1), 32, 9, 是, 是, 41, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0033/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 34. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":9,"right_scale":10,"left_power_numerator":0,"left_power_denominator":1,"right_power_numerator":1,"right_power_denominator":2,"singularity":-4}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-5,-3], -4, left:9*(c-x)^(-0/1), right:10*(x-c)^(-1/2), 9, 20, 是, 是, 29, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-5,-3], -4, left:9*(c-x)^(-0/1), right:10*(x-c)^(-1/2), 9, 20, 是, 是, 29, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0034/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 35. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":10,"right_scale":11,"left_power_numerator":1,"left_power_denominator":2,"right_power_numerator":2,"right_power_denominator":3,"singularity":-3}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-4,-2], -3, left:10*(c-x)^(-1/2), right:11*(x-c)^(-2/3), 20, 33, 是, 是, 53, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-4,-2], -3, left:10*(c-x)^(-1/2), right:11*(x-c)^(-2/3), 20, 33, 是, 是, 53, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0035/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 36. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":11,"right_scale":12,"left_power_numerator":2,"left_power_denominator":3,"right_power_numerator":3,"right_power_denominator":4,"singularity":-2}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-3,-1], -2, left:11*(c-x)^(-2/3), right:12*(x-c)^(-3/4), 33, 48, 是, 是, 81, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-3,-1], -2, left:11*(c-x)^(-2/3), right:12*(x-c)^(-3/4), 33, 48, 是, 是, 81, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0036/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 37. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":12,"right_scale":13,"left_power_numerator":3,"left_power_denominator":4,"right_power_numerator":0,"right_power_denominator":1,"singularity":-1}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-2,0], -1, left:12*(c-x)^(-3/4), right:13*(x-c)^(-0/1), 48, 13, 是, 是, 61, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-2,0], -1, left:12*(c-x)^(-3/4), right:13*(x-c)^(-0/1), 48, 13, 是, 是, 61, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0037/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 38. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":13,"right_scale":14,"left_power_numerator":0,"left_power_denominator":1,"right_power_numerator":1,"right_power_denominator":2,"singularity":0}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([-1,1], 0, left:13*(c-x)^(-0/1), right:14*(x-c)^(-1/2), 13, 28, 是, 是, 41, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([-1,1], 0, left:13*(c-x)^(-0/1), right:14*(x-c)^(-1/2), 13, 28, 是, 是, 41, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0038/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 39. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":14,"right_scale":15,"left_power_numerator":1,"left_power_denominator":2,"right_power_numerator":2,"right_power_denominator":3,"singularity":1}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([0,2], 1, left:14*(c-x)^(-1/2), right:15*(x-c)^(-2/3), 28, 45, 是, 是, 73, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([0,2], 1, left:14*(c-x)^(-1/2), right:15*(x-c)^(-2/3), 28, 45, 是, 是, 73, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0039/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 40. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":15,"right_scale":16,"left_power_numerator":2,"left_power_denominator":3,"right_power_numerator":3,"right_power_denominator":4,"singularity":2}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([1,3], 2, left:15*(c-x)^(-2/3), right:16*(x-c)^(-3/4), 45, 64, 是, 是, 109, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([1,3], 2, left:15*(c-x)^(-2/3), right:16*(x-c)^(-3/4), 45, 64, 是, 是, 109, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0040/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 41. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":16,"right_scale":17,"left_power_numerator":3,"left_power_denominator":4,"right_power_numerator":0,"right_power_denominator":1,"singularity":3}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([2,4], 3, left:16*(c-x)^(-3/4), right:17*(x-c)^(-0/1), 64, 17, 是, 是, 81, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([2,4], 3, left:16*(c-x)^(-3/4), right:17*(x-c)^(-0/1), 64, 17, 是, 是, 81, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0041/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 42. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":17,"right_scale":18,"left_power_numerator":0,"left_power_denominator":1,"right_power_numerator":1,"right_power_denominator":2,"singularity":4}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([3,5], 4, left:17*(c-x)^(-0/1), right:18*(x-c)^(-1/2), 17, 36, 是, 是, 53, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([3,5], 4, left:17*(c-x)^(-0/1), right:18*(x-c)^(-1/2), 17, 36, 是, 是, 53, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0042/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 43. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":18,"right_scale":19,"left_power_numerator":1,"left_power_denominator":2,"right_power_numerator":2,"right_power_denominator":3,"singularity":5}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([4,6], 5, left:18*(c-x)^(-1/2), right:19*(x-c)^(-2/3), 36, 57, 是, 是, 93, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([4,6], 5, left:18*(c-x)^(-1/2), right:19*(x-c)^(-2/3), 36, 57, 是, 是, 93, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0043/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 44. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":19,"right_scale":20,"left_power_numerator":2,"left_power_denominator":3,"right_power_numerator":3,"right_power_denominator":4,"singularity":6}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([5,7], 6, left:19*(c-x)^(-2/3), right:20*(x-c)^(-3/4), 57, 80, 是, 是, 137, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([5,7], 6, left:19*(c-x)^(-2/3), right:20*(x-c)^(-3/4), 57, 80, 是, 是, 137, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0044/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 45. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":20,"right_scale":21,"left_power_numerator":3,"left_power_denominator":4,"right_power_numerator":0,"right_power_denominator":1,"singularity":7}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([6,8], 7, left:20*(c-x)^(-3/4), right:21*(x-c)^(-0/1), 80, 21, 是, 是, 101, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([6,8], 7, left:20*(c-x)^(-3/4), right:21*(x-c)^(-0/1), 80, 21, 是, 是, 101, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0045/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 46. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":21,"right_scale":22,"left_power_numerator":0,"left_power_denominator":1,"right_power_numerator":1,"right_power_denominator":2,"singularity":8}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([7,9], 8, left:21*(c-x)^(-0/1), right:22*(x-c)^(-1/2), 21, 44, 是, 是, 65, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([7,9], 8, left:21*(c-x)^(-0/1), right:22*(x-c)^(-1/2), 21, 44, 是, 是, 65, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0046/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 47. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":22,"right_scale":23,"left_power_numerator":1,"left_power_denominator":2,"right_power_numerator":2,"right_power_denominator":3,"singularity":9}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([8,10], 9, left:22*(c-x)^(-1/2), right:23*(x-c)^(-2/3), 44, 69, 是, 是, 113, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([8,10], 9, left:22*(c-x)^(-1/2), right:23*(x-c)^(-2/3), 44, 69, 是, 是, 113, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0047/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 48. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":23,"right_scale":24,"left_power_numerator":2,"left_power_denominator":3,"right_power_numerator":3,"right_power_denominator":4,"singularity":10}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([9,11], 10, left:23*(c-x)^(-2/3), right:24*(x-c)^(-3/4), 69, 96, 是, 是, 165, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([9,11], 10, left:23*(c-x)^(-2/3), right:24*(x-c)^(-3/4), 69, 96, 是, 是, 165, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0048/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 49. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":24,"right_scale":25,"left_power_numerator":3,"left_power_denominator":4,"right_power_numerator":0,"right_power_denominator":1,"singularity":11}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([10,12], 11, left:24*(c-x)^(-3/4), right:25*(x-c)^(-0/1), 96, 25, 是, 是, 121, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([10,12], 11, left:24*(c-x)^(-3/4), right:25*(x-c)^(-0/1), 96, 25, 是, 是, 121, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0049/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 50. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-interior-singular-split；資料={"left_scale":25,"right_scale":26,"left_power_numerator":0,"left_power_denominator":1,"right_power_numerator":1,"right_power_denominator":2,"singularity":12}。請依 [interval, singularity, left_integrand, right_integrand, left_integral, right_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋([11,13], 12, left:25*(c-x)^(-0/1), right:26*(x-c)^(-1/2), 25, 52, 是, 是, 77, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)

Wave6第六層核對=([11,13], 12, left:25*(c-x)^(-0/1), right:26*(x-c)^(-1/2), 25, 52, 是, 是, 77, interior_singularity_is_split_and_both_one_sided_limits_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0050/)

瑕積分 · qf-mod43-improper-interior-singular-split · candidate





## 51. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":1,"right_scale":2,"exponent":2}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:1/abs(x)^2;-1<x<1:0;x>=1:2/x^2, (-infinity,infinity), 1, 2, 是, 是, 3, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:1/abs(x)^2;-1<x<1:0;x>=1:2/x^2, (-infinity,infinity), 1, 2, 是, 是, 3, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0051/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 52. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":2,"right_scale":3,"exponent":3}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:2/abs(x)^3;-1<x<1:0;x>=1:3/x^3, (-infinity,infinity), 1, 3/2, 是, 是, 5/2, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:2/abs(x)^3;-1<x<1:0;x>=1:3/x^3, (-infinity,infinity), 1, 3/2, 是, 是, 5/2, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0052/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 53. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":3,"right_scale":4,"exponent":4}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:3/abs(x)^4;-1<x<1:0;x>=1:4/x^4, (-infinity,infinity), 1, 4/3, 是, 是, 7/3, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:3/abs(x)^4;-1<x<1:0;x>=1:4/x^4, (-infinity,infinity), 1, 4/3, 是, 是, 7/3, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0053/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 54. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":4,"right_scale":5,"exponent":5}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:4/abs(x)^5;-1<x<1:0;x>=1:5/x^5, (-infinity,infinity), 1, 5/4, 是, 是, 9/4, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:4/abs(x)^5;-1<x<1:0;x>=1:5/x^5, (-infinity,infinity), 1, 5/4, 是, 是, 9/4, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0054/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 55. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":5,"right_scale":6,"exponent":2}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:5/abs(x)^2;-1<x<1:0;x>=1:6/x^2, (-infinity,infinity), 5, 6, 是, 是, 11, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:5/abs(x)^2;-1<x<1:0;x>=1:6/x^2, (-infinity,infinity), 5, 6, 是, 是, 11, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0055/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 56. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":6,"right_scale":7,"exponent":3}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:6/abs(x)^3;-1<x<1:0;x>=1:7/x^3, (-infinity,infinity), 3, 7/2, 是, 是, 13/2, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:6/abs(x)^3;-1<x<1:0;x>=1:7/x^3, (-infinity,infinity), 3, 7/2, 是, 是, 13/2, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0056/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 57. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":7,"right_scale":8,"exponent":4}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:7/abs(x)^4;-1<x<1:0;x>=1:8/x^4, (-infinity,infinity), 7/3, 8/3, 是, 是, 5, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:7/abs(x)^4;-1<x<1:0;x>=1:8/x^4, (-infinity,infinity), 7/3, 8/3, 是, 是, 5, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0057/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 58. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":8,"right_scale":9,"exponent":5}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:8/abs(x)^5;-1<x<1:0;x>=1:9/x^5, (-infinity,infinity), 2, 9/4, 是, 是, 17/4, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:8/abs(x)^5;-1<x<1:0;x>=1:9/x^5, (-infinity,infinity), 2, 9/4, 是, 是, 17/4, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0058/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 59. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":9,"right_scale":10,"exponent":2}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:9/abs(x)^2;-1<x<1:0;x>=1:10/x^2, (-infinity,infinity), 9, 10, 是, 是, 19, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:9/abs(x)^2;-1<x<1:0;x>=1:10/x^2, (-infinity,infinity), 9, 10, 是, 是, 19, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0059/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 60. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":10,"right_scale":11,"exponent":3}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:10/abs(x)^3;-1<x<1:0;x>=1:11/x^3, (-infinity,infinity), 5, 11/2, 是, 是, 21/2, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:10/abs(x)^3;-1<x<1:0;x>=1:11/x^3, (-infinity,infinity), 5, 11/2, 是, 是, 21/2, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0060/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 61. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":11,"right_scale":12,"exponent":4}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:11/abs(x)^4;-1<x<1:0;x>=1:12/x^4, (-infinity,infinity), 11/3, 4, 是, 是, 23/3, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:11/abs(x)^4;-1<x<1:0;x>=1:12/x^4, (-infinity,infinity), 11/3, 4, 是, 是, 23/3, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0061/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 62. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":12,"right_scale":13,"exponent":5}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:12/abs(x)^5;-1<x<1:0;x>=1:13/x^5, (-infinity,infinity), 3, 13/4, 是, 是, 25/4, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:12/abs(x)^5;-1<x<1:0;x>=1:13/x^5, (-infinity,infinity), 3, 13/4, 是, 是, 25/4, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0062/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 63. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":13,"right_scale":14,"exponent":2}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:13/abs(x)^2;-1<x<1:0;x>=1:14/x^2, (-infinity,infinity), 13, 14, 是, 是, 27, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:13/abs(x)^2;-1<x<1:0;x>=1:14/x^2, (-infinity,infinity), 13, 14, 是, 是, 27, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0063/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 64. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":14,"right_scale":15,"exponent":3}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:14/abs(x)^3;-1<x<1:0;x>=1:15/x^3, (-infinity,infinity), 7, 15/2, 是, 是, 29/2, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:14/abs(x)^3;-1<x<1:0;x>=1:15/x^3, (-infinity,infinity), 7, 15/2, 是, 是, 29/2, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0064/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 65. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":15,"right_scale":16,"exponent":4}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:15/abs(x)^4;-1<x<1:0;x>=1:16/x^4, (-infinity,infinity), 5, 16/3, 是, 是, 31/3, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:15/abs(x)^4;-1<x<1:0;x>=1:16/x^4, (-infinity,infinity), 5, 16/3, 是, 是, 31/3, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0065/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 66. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":16,"right_scale":17,"exponent":5}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:16/abs(x)^5;-1<x<1:0;x>=1:17/x^5, (-infinity,infinity), 4, 17/4, 是, 是, 33/4, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:16/abs(x)^5;-1<x<1:0;x>=1:17/x^5, (-infinity,infinity), 4, 17/4, 是, 是, 33/4, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0066/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 67. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":17,"right_scale":18,"exponent":2}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:17/abs(x)^2;-1<x<1:0;x>=1:18/x^2, (-infinity,infinity), 17, 18, 是, 是, 35, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:17/abs(x)^2;-1<x<1:0;x>=1:18/x^2, (-infinity,infinity), 17, 18, 是, 是, 35, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0067/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 68. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":18,"right_scale":19,"exponent":3}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:18/abs(x)^3;-1<x<1:0;x>=1:19/x^3, (-infinity,infinity), 9, 19/2, 是, 是, 37/2, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:18/abs(x)^3;-1<x<1:0;x>=1:19/x^3, (-infinity,infinity), 9, 19/2, 是, 是, 37/2, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0068/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 69. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":19,"right_scale":20,"exponent":4}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:19/abs(x)^4;-1<x<1:0;x>=1:20/x^4, (-infinity,infinity), 19/3, 20/3, 是, 是, 13, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:19/abs(x)^4;-1<x<1:0;x>=1:20/x^4, (-infinity,infinity), 19/3, 20/3, 是, 是, 13, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0069/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 70. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":20,"right_scale":21,"exponent":5}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:20/abs(x)^5;-1<x<1:0;x>=1:21/x^5, (-infinity,infinity), 5, 21/4, 是, 是, 41/4, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:20/abs(x)^5;-1<x<1:0;x>=1:21/x^5, (-infinity,infinity), 5, 21/4, 是, 是, 41/4, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0070/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 71. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":21,"right_scale":22,"exponent":2}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:21/abs(x)^2;-1<x<1:0;x>=1:22/x^2, (-infinity,infinity), 21, 22, 是, 是, 43, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:21/abs(x)^2;-1<x<1:0;x>=1:22/x^2, (-infinity,infinity), 21, 22, 是, 是, 43, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0071/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 72. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":22,"right_scale":23,"exponent":3}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:22/abs(x)^3;-1<x<1:0;x>=1:23/x^3, (-infinity,infinity), 11, 23/2, 是, 是, 45/2, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:22/abs(x)^3;-1<x<1:0;x>=1:23/x^3, (-infinity,infinity), 11, 23/2, 是, 是, 45/2, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0072/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 73. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":23,"right_scale":24,"exponent":4}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:23/abs(x)^4;-1<x<1:0;x>=1:24/x^4, (-infinity,infinity), 23/3, 8, 是, 是, 47/3, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:23/abs(x)^4;-1<x<1:0;x>=1:24/x^4, (-infinity,infinity), 23/3, 8, 是, 是, 47/3, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0073/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 74. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":24,"right_scale":25,"exponent":5}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:24/abs(x)^5;-1<x<1:0;x>=1:25/x^5, (-infinity,infinity), 6, 25/4, 是, 是, 49/4, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:24/abs(x)^5;-1<x<1:0;x>=1:25/x^5, (-infinity,infinity), 6, 25/4, 是, 是, 49/4, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0074/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 75. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-two-sided-independent；資料={"left_scale":25,"right_scale":26,"exponent":2}。請依 [piecewise_integrand, interval, left_tail_integral, right_tail_integral, left_converges, right_converges, sum, certificate] 欄序回答。

查看答案與解釋(x<=-1:25/abs(x)^2;-1<x<1:0;x>=1:26/x^2, (-infinity,infinity), 25, 26, 是, 是, 51, both_one_sided_tail_integrals_converge_independently)

Wave6第六層核對=(x<=-1:25/abs(x)^2;-1<x<1:0;x>=1:26/x^2, (-infinity,infinity), 25, 26, 是, 是, 51, both_one_sided_tail_integrals_converge_independently)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0075/)

瑕積分 · qf-mod43-improper-two-sided-independent · candidate





## 76. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":2,"epsilon_denominator":2}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 2, 1/2, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 2, 1/2, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0076/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 77. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":3,"epsilon_denominator":3}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 3, 1/3, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 3, 1/3, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0077/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 78. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":4,"epsilon_denominator":4}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 4, 1/4, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 4, 1/4, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0078/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 79. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":5,"epsilon_denominator":5}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 5, 1/5, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 5, 1/5, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0079/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 80. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":6,"epsilon_denominator":6}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 6, 1/6, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 6, 1/6, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0080/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 81. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":7,"epsilon_denominator":7}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 7, 1/7, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 7, 1/7, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0081/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 82. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":8,"epsilon_denominator":8}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 8, 1/8, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 8, 1/8, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0082/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 83. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":9,"epsilon_denominator":9}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 9, 1/9, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 9, 1/9, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0083/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 84. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":10,"epsilon_denominator":10}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 10, 1/10, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 10, 1/10, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0084/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 85. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":11,"epsilon_denominator":11}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 11, 1/11, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 11, 1/11, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0085/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 86. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":12,"epsilon_denominator":12}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 12, 1/12, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 12, 1/12, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0086/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 87. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":13,"epsilon_denominator":13}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 13, 1/13, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 13, 1/13, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0087/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 88. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":14,"epsilon_denominator":14}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 14, 1/14, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 14, 1/14, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0088/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 89. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":15,"epsilon_denominator":15}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 15, 1/15, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 15, 1/15, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0089/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 90. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":16,"epsilon_denominator":16}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 16, 1/16, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 16, 1/16, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0090/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 91. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":17,"epsilon_denominator":17}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 17, 1/17, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 17, 1/17, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0091/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 92. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":18,"epsilon_denominator":18}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 18, 1/18, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 18, 1/18, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0092/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 93. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":19,"epsilon_denominator":19}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 19, 1/19, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 19, 1/19, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0093/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 94. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":20,"epsilon_denominator":20}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 20, 1/20, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 20, 1/20, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0094/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 95. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":21,"epsilon_denominator":21}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 21, 1/21, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 21, 1/21, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0095/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 96. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":22,"epsilon_denominator":22}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 22, 1/22, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 22, 1/22, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0096/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 97. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":23,"epsilon_denominator":23}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 23, 1/23, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 23, 1/23, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0097/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 98. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":24,"epsilon_denominator":24}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 24, 1/24, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 24, 1/24, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0098/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 99. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":25,"epsilon_denominator":25}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 25, 1/25, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 25, 1/25, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0099/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate





## 100. 瑕積分

題幹明示integrand、區間、cutoff與每個無界端或奇點的one-sided limit責任。操作=improper-principal-value-boundary；資料={"cutoff":26,"epsilon_denominator":26}。請依 [integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary] 欄序回答。

對稱截斷值0與對稱主值為0，不等於通常瑕積分收斂；每個無界端與奇點仍須逐側核對。

查看答案與解釋(f(x)=1/x, (-infinity,infinity) with singularity at 0, 26, 1/26, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)

Wave6第六層核對=(f(x)=1/x, (-infinity,infinity) with singularity at 0, 26, 1/26, 0, 是, 否, 否, symmetric_principal_value_is_zero_but_the_required_one_sided_limits_diverge_and_finite_cutoffs_do_not_prove_convergence)；依完整明示資料獨立重算；one-sided family固定scale/x^p於[lower,infinity)；two-sided family固定zero-middle piecewise tails；interior family在[c-1,c+1]逐側明示powers；two-sided improper integral與interior singularity都要求各分段獨立收斂；symmetric principal value或有限cutoffs不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261112-r04-improper-integral-0100/)

瑕積分 · qf-mod43-improper-principal-value-boundary · candidate



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