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

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Abel's test for improper integrals

Statement

Suppose $\int_{a}^{\infty} f$ converges and $g : [a, \infty) \to R$ is bounded, monotone, and locally Riemann integrable. Then $\int_{a}^{\infty} f g$ converges. The analogous assertion holds at every other one-sided singular end.

Facts & Assumptions

Given: A convergent improper integral of $f$ and a bounded monotone multiplier $g$ .

[L1]

A bounded monotone function has a finite limit at the relevant end (Complete ordered field (least-upper-bound property), Lower bound, bounded below, bounded set, Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences).

[L2]

Convergence of $\int f$ bounds its truncation primitive near the singular end, while on the remaining compact interval the integral function is Lipschitz and hence bounded (Improper integrals over unbounded intervals, The integral function of a bounded integrable $f$ is Lipschitz, hence uniformly continuous).

[L3]

Dirichlet's test applies to a nonnegative monotone multiplier tending to zero (Dirichlet's test for improper integrals).

[L4]

Convergent improper integrals are linear (Linearity of convergent improper integrals).

Proof

technique · direct

1.1

Suppose first that $g$ is nondecreasing and let $L$ be the supremum of its bounded range. Given $ε > 0$ , the definition of supremum gives $x_{0}$ with $L - ε < g (x_{0}) \leq L$ ; monotonicity then gives $L - ε < g (x) \leq L$ for every $x \geq x_{0}$ , so $g (x) \to L$ . The infimum argument handles a nonincreasing $g$ . If $g$ is nonincreasing put $h = g - L$ ; if it is nondecreasing put $h = L - g$ . In either case $h \geq 0$ , $h$ is nonincreasing toward the singular end, and $h \to 0$ .

2.1

The primitive of $f$ is bounded by [L2], so [L3] makes $\int f h$ converge. Since $f g = L f \pm f h$ , linearity [L4] and convergence of $\int f$ prove convergence of $\int f g$ . The oriented endpoint variants are identical.

L2L3L4∎

Depends on

Used by

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Dependency tree · two levels

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Sources

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