第 23668 題：瑕積分

原題與已知條件

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 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 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 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。

• mko-improper-integral

逐步推導

source operation=improper-principal-value-boundary;original snapshot and mathematics retained

explicit ordered source output fields=[integrand, domain, cutoff, epsilon, symmetric_truncated_value, symmetric_principal_value_zero, ordinary_improper_converges, finite_cutoffs_certificate, boundary]

ORIGINAL f(x)=1/x;B=19>=2;epsilon=1/19<=1/2

for EVERY B>epsilon>0,positive truncated integral=log(B)-log(epsilon),negative truncated integral=its negative;sum0,so symmetric principal value0

ORIGINAL whole integrand1/x on both infinite ends and both sides of singularity0

integral_epsilon^1=-log(epsilon) tends+infinity and integral_-1^-epsilon tends-infinity;integral_1^B=log(B) tends+infinity and integral_-B^-1 tends-infinity;ordinary improper convergence=否,finite_cutoffs_certificate=否

逐步反向核對

• S003 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S002 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。

結論

(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)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。