Improper Integrals

The definite integral is defined for a function that is continuous on a closed, bounded interval. Two very common situations fall outside that definition: the interval itself runs off to infinity, or the function becomes unbounded somewhere inside the interval. Neither case can be handled by the ordinary Newton–Leibniz formula, because in both cases the quantity that formula evaluates is not yet known to exist.

Improper integrals are the extension that covers these cases. The idea is always the same: replace the missing endpoint by a finite cutoff, integrate as usual, and then ask whether the limit of the results exists. That is the whole definition — and the whole difficulty. Most of the work on this page is in deciding, carefully and case by case, whether a given limit exists.

Definition

Improper Integrals of the First Kind

Let $f$ be continuous on $[a, +\infty)$. Then

$$\int_a^{+\infty} f(x)\,dx = \lim_{b \to +\infty} \int_a^b f(x)\,dx.$$

The improper integral converges if this limit exists as a finite number, and diverges otherwise. The other unbounded-interval cases follow the same pattern: $\int_{-\infty}^{b} f(x)\,dx = \lim_{a \to -\infty} \int_a^b f(x)\,dx$ on a left-unbounded interval, and $\int_{-\infty}^{+\infty} f(x)\,dx = \int_{-\infty}^{c} f(x)\,dx + \int_c^{+\infty} f(x)\,dx$ for any interior point $c$, which converges only if both pieces converge.

Improper Integrals of the Second Kind

Let $f$ be continuous on $[a,b)$ and unbounded at $x = b$. Then

$$\int_a^b f(x)\,dx = \lim_{\varepsilon \to 0^+} \int_a^{b-\varepsilon} f(x)\,dx.$$

Again the integral converges if the limit exists and is finite. If the integrand is unbounded at the left endpoint the truncation is $\lim_{\varepsilon \to 0^+} \int_{a+\varepsilon}^b f(x)\,dx$, and if it is unbounded at an interior point $c$ the integral is $\int_a^c f(x)\,dx + \int_c^b f(x)\,dx$, converging only if both pieces converge. The truncations are always made from inside the interval, so an improper integral is always a limit of proper integrals.

Convergence Tests

The Comparison Test

Let $f$ and $g$ be continuous and nonnegative on $[a, +\infty)$ with $f(x) \leq g(x)$ there.

1. If $\int_a^{+\infty} g(x)\,dx$ converges, then $\int_a^{+\infty} f(x)\,dx$ converges.
2. If $\int_a^{+\infty} f(x)\,dx$ diverges, then $\int_a^{+\infty} g(x)\,dx$ diverges.

The test works because the partial integrals are monotone in the integrand: if $0 \leq f \leq g$ then $\int_a^b f \leq \int_a^b g$ for every $b$, so a finite limit of the larger family bounds the smaller one, and a divergent smaller family rules out the larger one.

The Limit Comparison Test

Let $f$ and $g$ be continuous and positive on $[a, +\infty)$ with $\lim_{x \to +\infty} \frac{f(x)}{g(x)} = L$.

1. If $0 < L < +\infty$, then $\int_a^{+\infty} f(x)\,dx$ and $\int_a^{+\infty} g(x)\,dx$ converge or diverge together.
2. If $L = 0$ and $\int_a^{+\infty} g(x)\,dx$ converges, then $\int_a^{+\infty} f(x)\,dx$ converges.
3. If $L = +\infty$ and $\int_a^{+\infty} g(x)\,dx$ diverges, then $\int_a^{+\infty} f(x)\,dx$ diverges.

This is the workhorse test, because most integrands of interest are asymptotic to a simple benchmark function.

The p-Integrals as Benchmarks

Two families supply the benchmarks: $\int_1^{+\infty} \frac{1}{x^p}\,dx$ converges if and only if $p > 1$, in which case it equals $\frac{1}{p-1}$; and $\int_0^1 \frac{1}{x^p}\,dx$ converges if and only if $p < 1$, in which case it equals $\frac{1}{1-p}$. Everything else is compared against these. As an illustration, $\int_1^{+\infty} \frac{1}{x \ln x}\,dx$ diverges, because the substitution $u = \ln x$ turns it into $\int_0^{+\infty} \frac{1}{u}\,du$, which is itself improper at $u = 0$.

Worked Examples

Example 1. Compute $\int_1^{+\infty} \frac{1}{x^2}\,dx$.

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

Example 2. Decide whether $\int_1^{+\infty} \frac{1}{x}\,dx$ converges. The truncated value is $\left[\ln x\right]_1^b = \ln b$, and $\lim_{b \to +\infty} \ln b = +\infty$, so the integral diverges. Contrast this with Example 1: the two integrands differ only in the exponent, and that single digit decides convergence.

Example 3. Compute $\int_0^1 \frac{1}{\sqrt{x}}\,dx$, which is improper because the integrand is unbounded at $x = 0$.

$$\int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{\varepsilon \to 0^+} \int_{\varepsilon}^1 x^{-1/2}\,dx = \lim_{\varepsilon \to 0^+} \left[2\sqrt{x}\right]_{\varepsilon}^1 = \lim_{\varepsilon \to 0^+} \left(2 - 2\sqrt{\varepsilon}\right) = 2.$$

Example 4. Compute $\int_0^1 \frac{1}{\sqrt{1-x^2}}\,dx$, which is improper at $x = 1$. The truncated integral is $\left[\arcsin x\right]_0^{1-\varepsilon} = \arcsin(1-\varepsilon)$, and since $\arcsin(1) = \frac{\pi}{2}$ the limit is $\frac{\pi}{2}$.

Properties of Improper Integrals

Absolute and Conditional Convergence

An improper integral $\int_a^{+\infty} f(x)\,dx$ is absolutely convergent if $\int_a^{+\infty} \lvert f(x)\rvert\,dx$ converges. An absolutely convergent improper integral always converges, because $\left|\int_a^b f\right| \leq \int_a^b \lvert f\rvert$ bounds the truncated values by a convergent sequence.

The converse fails. The standard counterexample is $\int_1^{+\infty} \frac{\sin x}{x}\,dx$, which converges although $\int_1^{+\infty} \frac{\lvert \sin x\rvert}{x}\,dx$ diverges; proving both halves of that statement requires tools beyond this page, such as Dirichlet’s test. The practical consequence is already clear: a divergent absolute integral says nothing about the integral itself, so it can never be used to declare divergence.

Applications

Probability. The Gaussian normalization $\int_{-\infty}^{+\infty} e^{-x^2/2}\,dx = \sqrt{2\pi}$ and the exponential density normalization $\int_0^{+\infty} \lambda e^{-\lambda x}\,dx = 1$ (for $\lambda > 0$) are improper integrals, and both are finite only because the integrand decays fast enough.

Physics. The gravitational potential energy of a body of mass $m$ at distance $r$ from a body of mass $M$ is $W = \int_r^{+\infty} \frac{GMm}{x^2}\,dx = \frac{GMm}{r}$. Similarly, the electric field at a point a distance $r$ from a uniformly charged ray that begins at distance $r_0$ (with $r$ strictly less than $r_0$) is $E(r) = \int_{r_0}^{+\infty} \frac{\lambda\,dx}{4\pi\varepsilon_0 (x-r)^2} = \frac{\lambda}{4\pi\varepsilon_0 (r_0 - r)}$, where $\lambda$ is the linear charge density.

Analysis. The gamma function $\Gamma(s) = \int_0^{+\infty} x^{s-1} e^{-x}\,dx$ and the beta function $B(p,q) = \int_0^1 x^{p-1}(1-x)^{q-1}\,dx$ are improper integrals for most of their natural parameter ranges, and their convergence conditions are part of their definitions.

Common Errors

• Assuming every improper integral converges. Each one must be tested; $\int_1^{+\infty} \frac{1}{x}\,dx$ is the standard counterexample.
• Ignoring the unbounded points. Every singularity of the integrand inside the interval must be located and treated separately.
• Applying Newton–Leibniz directly. The formula evaluates a proper integral; for an improper one the limit must come first.
• Integrating across a singularity without splitting. At an interior unbounded point the integral converges only if both sides converge.

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Exercises

Exercise 1

Compute the improper integral $\int_1^{+\infty} \frac{1}{x^3}\,dx$.

Exercise 2

Decide whether the improper integral $\int_0^1 \frac{1}{x^2}\,dx$ converges.

Exercise 3

Compute the improper integral $\int_0^{+\infty} e^{-2x}\,dx$.

Exercise 4

For which values of $p$ does the improper integral $\int_1^{+\infty} \frac{1}{x^p}\,dx$ converge, and what is its value in those cases?

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Summary

An improper integral is not a new kind of integral; it is the same definite integral with a limit attached, and the entire content of the topic lies in deciding whether that limit exists. The workflow is always the same three steps: locate the unbounded endpoint or the unbounded point of the integrand, truncate the interval from inside, and evaluate the limit of the resulting proper integrals. When the integrand has several singularities, split at each one and require every piece to converge. Comparison against the benchmark p-integrals settles the great majority of questions, and the limit comparison test turns most remaining integrands into something comparable. Finally, keep two ideas apart that students most often blur: absolute convergence is stronger than convergence, and convergence is about the existence of a limit, not about the limit being large.

Symbols Used in This Article