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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.

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