False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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FALSE: existence of a Cauchy principal value forces improper convergence

Statement

False claim: if $PV ⁣ \int_{- \infty}^{\infty} f (x) d x$ exists, then $\int_{- \infty}^{\infty} f (x) d x$ exists as an ordinary improper integral.

Facts & Assumptions

Given: The odd rational function $f (x) = x / (1 + x^{2})$ .

[L1]

Principal value is defined by symmetric truncation and does not by itself assert separate one-sided convergence (Cauchy principal values at a finite singularity and on the real line, This page keeps Cauchy principal values distinct from genuine improper convergence).

[L2]

Ordinary improper convergence implies principal-value convergence, but the implication is not stated as an equivalence (Separate improper convergence implies convergence of the principal value).

Refutation

technique · direct

1.1

Because $f$ is odd, [given, L1] $\int_{- R}^{R} \frac{x}{1 + x^{2}} d x = 0$ for every $R > 0$ , so gives $PV ⁣ \int_{- \infty}^{\infty} \frac{x}{1 + x^{2}} d x = 0.$

2.1

On $(0, \infty)$ one has $x / (1 + x^{2}) \sim 1 / x$ , so $\int_{1}^{\infty} \frac{x}{1 + x^{2}} d x = \frac{1}{2} \log (1 + x^{2}) ∣_{1}^{\infty}$ diverges to $+ \infty$ , and by oddness the left tail diverges to $- \infty$ . Therefore the ordinary improper integral does not exist.

step 1.1L2algebra

3.1

So the existence of a principal value does not force ordinary improper convergence.

step 1.1step 2.1∎

Depends on

Used by

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Sources

L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 4 §5.3 (standard reference, not scraped)
W. F. Trench, Introduction to Real Analysis, Section 3.4 (standard reference, not scraped)