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Limit comparison for positive improper integrals

Statement

Let $f,g\ge0$ 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 $L/2<f/g<3L/2$ sufficiently near the singular end. Thus $(L/2)g\le f\le(3L/2)g$ . Applying [L2] in both directions proves the equivalence.

L1L2

2.1

If $L=0$ , eventually $f/g\le1$ , so $f\le g$ . If $f/g\to+\infty$ , eventually $g\le f$ . The one-way conclusions again follow from [L2].

L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 65 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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