CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11

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Tails of a convergent improper integral tend to zero

Statement

If $\int_{a}^{\infty} f$ converges, then $\lim_{R \to \infty} \int_{R}^{\infty} f = 0.$ Equivalently, for every $ε > 0$ all sufficiently remote proper tails $\int_{u}^{v} f$ have absolute value below $ε$ . The corresponding tails tend to zero at $- \infty$ and at either finite singular endpoint.

Facts & Assumptions

Given: A convergent one-ended improper integral of $f$ .

[L1]

A convergent integral may be split at every finite truncation (Improper convergence is independent of finite truncations and split points).

[L2]

Its proper remote tails satisfy the Cauchy criterion (Cauchy criterion for improper integrals).

Proof

technique · direct

1.1

Write $I = \int_{a}^{\infty} f$ . By [L1], $\int_{R}^{\infty} f = I - \int_{a}^{R} f$ , which tends to zero by the definition of $I$ .

2.1

The epsilon formulation is exactly [L2]. Reversing orientation or replacing infinite truncations by one-sided finite truncations proves all other stated forms.

L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

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