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

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Limit comparison for positive improper integrals

Statement

Let $f, g \geq 0$ eventually toward a singular end, with $g > 0$ there, and suppose $\lim \frac{f (x)}{g (x)} = L$ at that end for a finite $L > 0$ . Then the improper integrals of $f$ and $g$ either both converge or both diverge. The statement applies to infinite and finite one-sided endpoints.

If $L = 0$ , convergence of $\int g$ still implies convergence of $\int f$ ; if the ratio tends to $+ \infty$ , convergence of $\int f$ implies convergence of $\int g$ .

Facts & Assumptions

Given: Eventually positive functions with the stated quotient limit.

[L1]

The definitions of a limit at infinity and of a finite one-sided limit give the same eventual epsilon bound at their respective singular ends (Limits at $+ \infty$ and $- \infty$ , and infinite limits at a point, The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ ).

[L2]

Eventual pointwise comparison transfers convergence, and finite positive scalar multiples preserve it (Comparison tests for improper integrals, Linearity of convergent improper integrals).

Proof

technique · direct

1.1

By [L1], tolerance $L / 2$ gives $L / 2 < f / g < 3 L / 2$ sufficiently near the singular end. Thus $(L / 2) g \leq f \leq (3 L / 2) g$ . Applying [L2] in both directions proves the equivalence.

L1L2

2.1

If $L = 0$ , eventually $f / g \leq 1$ , so $f \leq g$ . If $f / g \to + \infty$ , eventually $g \leq f$ . The one-way conclusions again follow from [L2].

L2∎

Depends on

Used by

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Sources

William F. Trench, Introduction to Real Analysis, Theorem 3.4.7 (standard reference, not scraped)