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Comparison Test for Improper Integrals

Proposition

Suppose ff and gg are Riemann integrable on every finite closed subinterval away from the specified improper endpoint, with no other singularities, and 0≤f(x)≤g(x)0\leq f(x)\leq g(x) near that endpoint of an improper integral. Then:

  1. If ∫g(x) dx\displaystyle\int g(x)\,\mathrm{d}x converges, then ∫f(x) dx\displaystyle\int f(x)\,\mathrm{d}x converges.
  2. If ∫f(x) dx\displaystyle\int f(x)\,\mathrm{d}x diverges, then ∫g(x) dx\displaystyle\int g(x)\,\mathrm{d}x diverges.

Limit comparison test. If f,g>0f,g>0 and

lim⁡f(x)g(x)=c∈(0,+∞)\lim \frac{f(x)}{g(x)}=c\in(0,+\infty)

at the improper endpoint, then the two improper integrals have the same convergence behaviour.

Absolute convergence. If hh is Riemann integrable on those finite closed subintervals and ∫∣h(x)∣ dx\displaystyle\int |h(x)|\,\mathrm{d}x converges, then ∫h(x) dx\displaystyle\int h(x)\,\mathrm{d}x converges.

Interpretation

The comparison test decides convergence of an improper integral by measuring the integrand against one whose behaviour is already known, usually a power function settled by the pp-test.

The limit form is more flexible: it only asks that the two integrands have the same order of magnitude at the problem endpoint, not that one dominates the other pointwise.

Push the upper limit bb out and compare the two areas: a convergent larger integral caps the smaller one, a divergent smaller one drags the larger along, and in the limit form only the ratio f/gf/g matters.

Only the two stated directions give a conclusion: a smaller integrand under a convergent one converges, and a larger one over a divergent one diverges. The bound 1x2+x≤1x\frac{1}{x^2+x}\le\frac{1}{x} on [1,∞)[1,\infty) says nothing, since ∫1∞dxx\int_1^\infty\frac{\mathrm{d}x}{x} diverges. Here the limit form helps: 1/(x2+x)1/x2=x2x2+x→1\frac{1/(x^2+x)}{1/x^2}=\frac{x^2}{x^2+x}\to1, so ∫1∞dxx2+x\int_1^\infty\frac{\mathrm{d}x}{x^2+x} behaves like ∫1∞dxx2\int_1^\infty\frac{\mathrm{d}x}{x^2} and converges. Nonnegativity is essential: −x≤1x2-x\le\frac{1}{x^2} on [1,∞)[1,\infty) and the larger integral converges, yet ∫1∞(−x) dx\int_1^\infty(-x)\,\mathrm{d}x diverges. Only the behaviour near the problem endpoint matters, because on a bounded piece away from it the integrals are ordinary.

The same reasoning is the comparison test for series. It also shows that the gamma function is well defined: near 00, tx−1e−t≤tx−1t^{x-1}\mathrm{e}^{-t}\le t^{x-1}, and near ∞\infty the factor e−t\mathrm{e}^{-t} makes the integrand smaller than t−2t^{-2}.

Key Results

  • Comparison requires the integrands to be nonnegative near the improper endpoint; the absolute-convergence statement is what handles sign changes.
  • A finite nonzero limit ratio cc transfers convergence and divergence in both directions.
  • Comparison is normally paired with the pp-test, which supplies the reference integrals ∫a+∞t−p dt\int_a^{+\infty}t^{-p}\,\mathrm{d}t and ∫0at−p dt.\int_0^{a}t^{-p}\,\mathrm{d}t\text{.}
  • The corresponding statement for series is the comparison test for series; the integral test is what connects the two settings.

Examples

  • For x≥1x\geq1 we have 0≤1x2+x≤1x2.\displaystyle 0\leq\frac{1}{x^2+x}\leq\frac{1}{x^2}\text{.} Since ∫1+∞x−2 dx\displaystyle\int_1^{+\infty}x^{-2}\,\mathrm{d}x converges, so does ∫1+∞dxx2+x.\displaystyle\int_1^{+\infty}\frac{\mathrm{d}x}{x^2+x}\text{.}
  • For 0<x≤10<x\leq1 we have 0≤1x(1+x)≤1x.\displaystyle 0\leq\frac{1}{\sqrt{x(1+x)}}\leq\frac{1}{\sqrt{x}}\text{.} The pp-test at 00 then shows that ∫01dxx(1+x)\displaystyle\int_0^1\frac{\mathrm{d}x}{\sqrt{x(1+x)}} converges.

Prerequisites

Sources

Aliases: direct comparison test for integrals Keywords: limit comparison test, absolute convergence of improper integrals Symbols: element of, infinity, integral, square root

Last updated: 20 October 2026 03:02