CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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The sine integral converges conditionally but not absolutely

Statement refuted

Refuted claim: if $\int_{0}^{\infty} \frac{\sin x}{x} d x$ converges, then $\int_{0}^{\infty} ∣ \frac{\sin x}{x} ∣ d x$ must also converge.

Facts & Assumptions

Given: The function $f (x) = \sin x / x$ on $(0, \infty)$ .

[L1]

Dirichlet's test makes $\int_{1}^{\infty} \sin x / x d x$ converge (Dirichlet's test for improper integrals).

[L2]

A uniform positive amount of absolute mass on infinitely many disjoint tails forces divergence of the absolute integral (Uniform oscillatory tail mass forces failure of absolute convergence).

Counterexample

technique · direct

1.1

By, the oscillatory integral of $\sin x / x$ converges on $[1, \infty)$ , [L1] so the half-line sine integral is conditionally convergent.

1.2

On each interval [L2, algebra] $I_{k} = [k π + π / 6, k π + 5 π / 6]$ one has $∣ \sin x ∣ \geq 1 / 2$ , while $x \leq (k + 1) π$ . Therefore $\int_{I_{k}} ∣ \frac{\sin x}{x} ∣ d x \geq \frac{1}{2} \cdot \frac{2 π / 3}{(k + 1) π} = \frac{1}{3 (k + 1)} .$ The lower bounds have divergent harmonic sum, so implies $\int_{1}^{\infty} ∣ \sin x ∣ / x d x = \infty$ .

2.1

Thus $\sin x / x$ gives a convergent improper integral whose absolute-value integral diverges.

step 1.1step 1.2∎

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Sources

W. F. Trench, Introduction to Real Analysis, Section 3.4 (standard reference, not scraped)