Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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Abel's test for improper integrals

Statement

Suppose af\int_a^\infty f converges and g:[a,)Rg:[a,\infty)\to\mathbb R is bounded, monotone, and locally Riemann integrable. Then afg\int_a^\infty fg converges. The analogous assertion holds at every other one-sided singular end.

Facts & Assumptions

Given: A convergent improper integral of ff and a bounded monotone multiplier gg.

[L2]

Convergence of f\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 ff 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 gg is nondecreasing and let LL be the supremum of its bounded range. Given ε>0\varepsilon>0, the definition of supremum gives x0x_0 with Lε<g(x0)LL-\varepsilon<g(x_0)\le L; monotonicity then gives Lε<g(x)LL-\varepsilon<g(x)\le L for every xx0x\ge x_0, so g(x)Lg(x)\to L. The infimum argument handles a nonincreasing gg. If gg is nonincreasing put h=gLh=g-L; if it is nondecreasing put h=Lgh=L-g. In either case h0h\ge0, hh is nonincreasing toward the singular end, and h0h\to0.

L1
2.1

The primitive of ff is bounded by [L2], so [L3] makes fh\int fh converge. Since fg=Lf±fhfg=Lf\pm fh, linearity [L4] and convergence of f\int f prove convergence of fg\int fg. The oriented endpoint variants are identical.

L2L3L4

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