Improper Integrals: Convergence Tests and Worked Examples

An integral can have a familiar antiderivative and still fail to have a finite value. The missing step is a limit. If the interval extends forever, or the function becomes unbounded at an endpoint or inside the interval, you cannot simply substitute the printed bounds into an antiderivative. First replace the problematic boundary with an ordinary variable. Then ask what happens as that variable approaches the boundary.

That is what makes an integral improper. The label describes the setup, not the answer. Some improper integrals converge to a finite number. Others diverge. This guide puts paired examples side by side so you can see which step makes the difference. If the usual definite-integral notation is still new, read our introduction to integrals first. The limits guide explains the operation that decides every example here.

First identify what makes the integral improper

There are two common cases. An infinite interval has a bound such as $+\infty$ or $-\infty$. An unbounded integrand has a point where the function shoots without bound, even if the written bounds are finite. That point may be an endpoint or lie inside the interval.

OpenStax Calculus, Volume 2, §3.7 defines both cases by limits. Its distinction matters because writing $\infty$ as if it were a number, or evaluating across an interior singularity in one stroke, can produce a plausible but false answer.

A finite limit means convergence. A limit that grows without bound, or otherwise fails to settle to a finite value, means divergence. The same antiderivative can appear in both outcomes, so finding it is only part of the work.

A four-step method you can reuse

1. Find every suspect boundary. Include an infinite end, a zero denominator, a logarithm leaving its domain, or another place the function is unbounded. A singularity inside the interval requires a split before any integration.

2. Replace one suspect boundary with a finite variable. Write the resulting proper integral. For example, replace $\int_1^{\infty} f(x)\,dx$ with $\int_1^R f(x)\,dx$ for a finite $R>1$.

3. Evaluate the proper integral. An antiderivative is useful here. Check its domain and signs before substituting the finite bounds. If you cannot integrate directly, a comparison may still decide convergence.

4. Take the limit and state the verdict. Do not report just the expression at finite $R$ or $\varepsilon$. Write the limiting result and say convergent or divergent. If an interior point was split, do this independently on both sides.

The order makes the argument auditable. A reader can see whether you checked the right boundary and whether the final number is actually a limit.

At infinity: one finite area, one infinite area

Start with a convergent example:

$$\int_1^{\infty}\frac{1}{x^2}\,dx =\lim_{R\to\infty}\int_1^R x^{-2}\,dx =\lim_{R\to\infty}\left[-\frac1x\right]_1^R =\lim_{R\to\infty}\left(1-\frac1R\right)=1.$$

The interval never ends, but the added area becomes small enough that the total approaches 1. Now change only the power in the denominator:

$$\int_1^{\infty}\frac1x\,dx =\lim_{R\to\infty}\int_1^R\frac1x\,dx =\lim_{R\to\infty}[\ln x]_1^R =\lim_{R\to\infty}\ln R=+\infty.$$

This integral diverges. The fact that $1/x$ approaches zero does not make its accumulated area finite. It decreases too slowly. The comparison is the useful picture: both integrands approach zero, both have elementary antiderivatives, and only one yields a finite integral limit.

For $\int_1^{\infty}x^{-p}\,dx$, the boundary lies at $p=1$. It converges when $p>1$ and diverges when $p\le1$. That shortcut follows from the limit calculation, including the logarithmic case at $p=1$. Use the underlying limit if you are unsure whether the exponent or interval fits the shortcut.

At an endpoint: the same boundary from the other side

An integral can be improper even when both printed bounds are finite. At $x=0$, $x^{-1/2}$ is unbounded. Replace the lower bound with $\varepsilon>0$:

$$\int_0^1\frac1{\sqrt{x}}\,dx =\lim_{\varepsilon\to0^+}\int_{\varepsilon}^1 x^{-1/2}\,dx =\lim_{\varepsilon\to0^+}[2\sqrt{x}]_{\varepsilon}^1 =\lim_{\varepsilon\to0^+}2(1-\sqrt{\varepsilon})=2.$$

The function becomes unbounded at the endpoint, yet the integral converges. Compare $1/x$ on the same interval:

$$\int_0^1\frac1x\,dx =\lim_{\varepsilon\to0^+}\int_{\varepsilon}^1\frac1x\,dx =\lim_{\varepsilon\to0^+}(0-\ln\varepsilon)=+\infty.$$

This one diverges. Do not substitute zero into $\ln x$, because $\ln0$ is not defined. The limit is the actual mathematical question. The endpoint pair mirrors the pair at infinity: a singularity itself does not settle convergence, and a working antiderivative does not settle it either.

Inside the interval: never cancel across the hole

The most dangerous shortcut appears when a singularity lies between two ordinary-looking endpoints. Consider

$$\int_{-1}^{1}\frac1x\,dx.$$

The antiderivative $\ln|x|$ has the same value at $-1$ and $1$. It is tempting to write zero. That step crosses $x=0$, where $1/x$ is undefined. The fundamental theorem does not license one subtraction across that break. Test the two sides separately before assigning a value to the whole expression:

$$\int_{-1}^{0}\frac1x\,dx\qquad\text{and}\qquad\int_0^1\frac1x\,dx.$$

Each side needs its own limit. On the left, with $\varepsilon>0$, $\int_{-1}^{-\varepsilon}1/x\,dx=\ln\varepsilon\to-\infty$. On the right, $\int_{\varepsilon}^{1}1/x\,dx=-\ln\varepsilon\to+\infty$. Neither side converges to a finite value. Therefore the original improper integral diverges.

A symmetric cutoff can be made to produce zero for every positive $\varepsilon$, but that is a different construction called a Cauchy principal value. It pairs two diverging sides and lets them cancel. It is not the value of the ordinary improper integral. The University of South Carolina worksheet states the separate-side requirement for an interior singularity.

If the antiderivative is hard, compare

Not every improper integral has an antiderivative you want to compute. For a nonnegative integrand, you may compare its size with one whose behavior you already know. For example, for $x\ge1$,

$$0<\frac1{x^2+1}\le\frac1{x^2}.$$

Because $\int_1^{\infty}1/x^2\,dx$ converges, the smaller nonnegative integral $\int_1^{\infty}1/(x^2+1)\,dx$ converges too. This proves convergence without finding its exact value. OpenStax's comparison theorem in §3.7 spells out the conditions. The direction of the inequality matters: a small function below a divergent one tells you nothing by itself, while a nonnegative function below a convergent one inherits convergence.

Check your answer before handing it in

Before calling an answer finished, name the suspect boundary and inspect your final line. A lone antiderivative, a substituted $\infty$, or an undefined $\ln0$ is an unfinished solution. At an interior singularity, two finite one-sided answers are required. A symmetric cancellation is not a substitute.

Try two quick checks. First, does $\int_2^{\infty}1/x^2\,dx$ converge? The same calculation gives $1/2$. Second, does $\int_0^1x^{-2}\,dx$ converge? Its endpoint limit grows without bound, so it diverges. These are different boundaries and different verdicts even though the same power appears. If either result surprises you, write the finite-bound version and let the limit show why.

Math Zen's limits guide supports the final step. The L'Hôpital guide covers a tool for eligible limit forms, but it is not a replacement for identifying and splitting the improper boundary. Once you can point to every problematic point and write the correct one-sided limits, the calculation becomes ordinary calculus again.