Improper Integrals

An infinite interval can enclose a finite amount of area. An integrand can also become unbounded at an endpoint while its accumulated value stays finite. Improper integrals make these possibilities precise by taking limits of integrals over ordinary finite intervals.

The calculation has two jobs: evaluate the finite-cutoff integral, then determine whether its limit is a finite number. A formula that skips the second job has not established convergence.

On this page

What makes an integral improper?

This page considers integrals with an infinite bound or an integrand that becomes unbounded at an endpoint or interior point. Away from the problematic points, assume the integrand is integrable on each closed finite subinterval used in the calculation.

For one infinite upper endpoint, the definition is

\int_a^\infty f(x)\,\mathrm{d}x=\lim_{b\to\infty}\int_a^b f(x)\,\mathrm{d}x.

The improper integral converges when this limit exists as a finite real number. A limit of $+\infty$ or $-\infty$ describes divergence, not a finite integral value. Failure to approach any limit also means divergence.

An infinite endpoint is not a number to substitute into an antiderivative. Integrate with a finite cutoff first. The OpenStax treatment of improper integrals develops these definitions for infinite intervals and finite singularities.

A bounded jump discontinuity is a different issue: it need not make a Riemann integral improper. Likewise, an isolated removable hole can often be filled without changing an integral. The essential concern here is the infinite interval or unbounded behaviour.

Infinite intervals

A convergent example

For $b>1$ , the integral of $x^{-2}$ from $1$ to $b$ is

\int_1^b\frac{\mathrm{d}x}{x^2}=[-1/x]_1^b=1-\frac{1}{b}.

As $b\to\infty$ , the missing tail $1/b$ tends to zero. Therefore

\int_1^\infty\frac{\mathrm{d}x}{x^2}=1.

The interval keeps growing, but each additional tail contributes less. For example, the area from $10$ onward is $1/10$ , and from $100$ onward is $1/100$ .

A similar-looking divergent example

For $1/x$ on the same interval,

\int_1^b\frac{\mathrm{d}x}{x}=\ln b.

Since $\ln b\to\infty$ , the improper integral diverges. Both $1/x$ and $1/x^2$ approach zero, so a shrinking integrand alone does not settle the convergence question.

Two infinite endpoints require two limits

An integral over the entire real line is defined by choosing a finite splitting point $c$ and requiring both tails to converge:

\int_{-\infty}^{\infty}f(x)\,\mathrm{d}x=\int_{-\infty}^{c}f(x)\,\mathrm{d}x+\int_c^{\infty}f(x)\,\mathrm{d}x.

For $f(x)=e^{-\lvert x\rvert}$ , split at $0$ . The left integrand is $e^x$ , the right integrand is $e^{-x}$ , and each integral equals $1$ . The total is $2$ . If either tail fails to converge, an ordinary two-sided improper integral does not acquire a value through cancellation with the other tail.

Unbounded integrands at finite points

At a problematic left endpoint $a$ , approach from within the integration interval:

\int_a^b f(x)\,\mathrm{d}x=\lim_{\varepsilon\to0^+}\int_{a+\varepsilon}^{b}f(x)\,\mathrm{d}x.

For the example $1/\sqrt{x}$ on $(0,1]$ ,

\int_\varepsilon^1\frac{\mathrm{d}x}{\sqrt{x}}=[2\sqrt{x}]_\varepsilon^1=2-2\sqrt{\varepsilon}.

The limit is $2$ . The function is unbounded near zero, yet the area near that endpoint has a finite total. By comparison, $\int_\varepsilon^1 1/x\,\mathrm{d}x=-\ln\varepsilon\to\infty$ .

For an interior singularity $c$ , split at $c$ . Take the left limit toward $c$ from below and the right limit toward $c$ from above. Both pieces must be finite before adding them.

For example, inserting $-1$ and $1$ directly into $-1/x$ would suggest $\int_{-1}^1x^{-2}\,\mathrm{d}x=-2$ . That cannot be the integral of a positive function. The error is crossing the singularity at zero. Each correctly defined one-sided integral diverges to $+\infty$ .

The p-integral benchmarks

For real $p\ne1$ , an antiderivative of $x^{-p}$ on $x>0$ is $x^{1-p}/(1-p)$ . The same power has different limiting behaviour near zero and near infinity.

Two power-integral benchmarks
Integral	Finite convergence condition	Value under that condition
$\int_1^\infty x^{-p}\,\mathrm{d}x$	$p>1$	$1/(p-1)$
$\int_0^1 x^{-p}\,\mathrm{d}x$	$p<1$	$1/(1-p)$

At infinity, $b^{1-p}$ tends to zero only when $p>1$ . At zero, $\varepsilon^{1-p}$ tends to zero only when $p<1$ . The logarithmic case $p=1$ diverges in both settings.

For example, $p=2$ gives convergence at infinity but divergence near zero. The exponent $p=1/2$ gives the reverse behaviour. No single power $x^{-p}$ has a finite integral over the whole interval $(0,\infty)$ , because that would require both incompatible conditions at once.

Comparison can decide convergence without an antiderivative

For non-negative integrands, a convergent upper bound or a divergent lower bound can settle the question. The relevant functions must be integrable over the finite intervals involved. The comparison inequality only needs to hold sufficiently far out in the tail, or sufficiently near the singularity, provided the rest contributes a finite amount.

Convergence by an upper bound

For $x\ge1$ ,

0\le\frac{1}{x^2+1}\le\frac{1}{x^2}.

The right-hand integral from $1$ to infinity converges, so the middle integral converges too. This proves finiteness without calculating its exact value. Being smaller than a divergent integral would not prove convergence.

Divergence by a lower bound

For $x\ge1$ , the inequality $x^2+1\le2x^2$ gives

\frac{1}{\sqrt{x^2+1}}\ge\frac{1}{\sqrt{2}\,x}.

The integral of the right side diverges, so the left side also has infinite area on that tail. Being larger than a convergent integral would not prove divergence.

For a signed integrand, convergence of the integral of its absolute value is a sufficient condition for convergence. A comparison theorem for non-negative functions should not be applied to a sign-changing expression without first addressing its signs.

Improper integral versus principal value

For $1/x$ on $[-1,1]$ , the independent one-sided cutoffs give

\int_{-1}^{-\varepsilon}\frac{\mathrm{d}x}{x}=\ln\varepsilon\to-\infty.

\int_{\delta}^{1}\frac{\mathrm{d}x}{x}=-\ln\delta\to+\infty.

Neither side converges as a finite integral. Their sum is therefore not an ordinary improper integral with value zero.

If one forces the cutoffs to move symmetrically, taking $\delta=\varepsilon$ , the finite-cutoff sum is zero for every positive cutoff. Its limit is the Cauchy principal value, which is a different definition. The dependence on the way the cutoffs are linked is visible if $\delta=2\varepsilon$ : the sum becomes $-\ln2$ instead.

The ordinary definition avoids that ambiguity by requiring the two sides to converge independently. Do not combine $+\infty$ and $-\infty$ as though they were finite quantities.

An integral with two kinds of improper behaviour

The illustration shows $f(x)=1/[(x+1)\sqrt{x}]$ for positive $x$ . It has an unbounded endpoint at zero and an infinite interval to the right. Unlike a generic decreasing curve, its formula tells us exactly which convergence questions to check.

A positive curve rises without bound near zero and has a shaded tail extending to the right. — The source illustrates $f(x)=1/[(x+1)\sqrt{x}]$ , whose improper integral over $(0,\infty)$ is $\pi$ .
Figure explained
The vertical scale is clipped near zero and the horizontal scale displays only a finite portion of the tail. The blue region represents the positive accumulation. For exact reference, $f(1/4)=8/5$ , $f(1)=1/2$ and $f(4)=1/10$ . The curve in this source is schematic; the limiting calculation in the text determines the value.
KSmrq, Improper integral.svg. CC BY-SA 3.0. English number labels selected; white-background raster export. Source curve and axes retained.

With $u=\sqrt{x}$ , we have $x=u^2$ and $\mathrm{d}x=2u\,\mathrm{d}u$ . On finite positive bounds, the transformed integrand is $2/(1+u^2)$ . Taking the independently justified endpoint limits gives

\int_0^\infty\frac{\mathrm{d}x}{(x+1)\sqrt{x}}=2\lim_{U\to\infty}\arctan U-2\lim_{v\to0^+}\arctan v=\pi.

Near zero it behaves like $x^{-1/2}$ , and far out it behaves like $x^{-3/2}$ . Both comparisons fit the appropriate convergent power benchmark. The illustration is a schematic view of this function; use the formula, not measurements of the pixels, for calculation.

Practice questions

Does $\int_1^\infty1/x^2\,\mathrm{d}x$ converge?
Does $\int_0^1 1/\sqrt{x}\,\mathrm{d}x$ converge?
Why is $\int_{-1}^1 1/x\,\mathrm{d}x$ not zero as an ordinary improper integral?

Check your answers

Yes. The finite-cutoff value is $1-1/b$ . Its limit as $b\to\infty$ is $1$ .
Yes. The cutoff integral is $2-2\sqrt{\varepsilon}$ . Taking $\varepsilon\to0^+$ gives $2$ .
The left piece tends to $-\infty$ and the right piece to $+\infty$ . They do not converge independently. Symmetric cancellation gives a principal value of zero, not ordinary improper convergence.

Sources and further reading

OpenStax Calculus Volume 2: Improper Integrals. Cutoff definitions, convergence, comparison and power benchmarks.
Wolfram Language: Calculus. Independent symbolic calculations and limit verification.