Improper Integrals and Convergence

Replace a dangerous endpoint with a limit

An infinite interval or an unbounded integrand makes an integral improper. For example,

\int_a^\infty f(x)dx=\lim_{R\to\infty}\int_a^Rf(x)dx.

The integral converges only when this limit is finite. A singular endpoint similarly requires a one-sided limit. An interior singularity requires two integrals, each converging separately. On the whole real line, the two infinite tails must also converge separately.

Visual guide

VISUAL GUIDESimilar tails can have different total areas

1/x: divergent area
1/x²: convergent area

Both 1/x and 1/x² approach zero, but their integrals from 1 to infinity differ: the first diverges while the second equals 1. Shrinking height alone does not decide whether the infinite tail has finite area.

Worked example: the two power tests

For $p\ne1$ , integrate $x^{-p}$ using $x^{1-p}/(1-p)$ . On $[1,\infty)$ the limit is finite exactly when $p>1$ , with value $1/(p-1)$ . When $p=1$ , the logarithm grows without bound.

Near zero the condition reverses: $\int_0^1x^{-p}dx$ converges exactly when $p<1$ , with value $1/(1-p)$ . Thus $1/\sqrt x$ is integrable near zero but not over an infinite tail. The location of the problematic endpoint matters as much as the exponent.

Worked example: cancellation is insufficient

For $\int_{-1}^1dx/x$ , splitting at zero gives a negatively divergent left integral and a positively divergent right integral. The improper integral does not exist. Symmetric truncation gives zero, but this is a Cauchy principal value, a different notion. It cannot replace the definition without explicitly changing the problem.

Comparison without an antiderivative

For nonnegative functions with $0\le f\le g$ on the relevant tail, convergence of $\int g$ implies convergence of $\int f$ . Divergence of $\int f$ implies divergence of $\int g$ . For example, $0\le e^{-x^2}\le e^{-x}$ for $x\ge1$ , so the Gaussian tail converges even though its antiderivative is not elementary.

Limit comparison applies when $f/g\to L$ with $0<L<\infty$ : the two nonnegative integrals have the same convergence behavior. A limit of zero or infinity does not automatically give this two-way conclusion. Absolute convergence, meaning convergence of $\int|f|$ , is sufficient for convergence of $\int f$ .

Practice

Evaluate $\int_1^\infty x^{-3}dx$ .
Does $\int_0^1x^{-3/2}dx$ converge?
Decide whether $\int_1^\infty dx/(x^2+4)$ converges by comparison.

Show worked solutions

$\lim_{R\to\infty}[-1/(2x^2)]_1^R=1/2$ .
No. The exponent $p=3/2$ is too large at zero.
Yes: $0<1/(x^2+4)\le1/x^2$ , whose tail integral converges.