Improper Integrals

An improper integral is evaluated using a limit because the interval of integration is unbounded or the integrand becomes unbounded. Some improper integrals have finite values; others diverge. An infinite endpoint does not, by itself, tell you which will happen.

The integrand is the function being integrated. In \(\int_a^b f(x)\,dx\), the integrand is \(f(x)\), and a and b are the limits of integration.

An infinite interval

For a continuous function on \([a,\infty)\), define

\[\int_a^\infty f(x)\,dx=\lim_{R\to\infty}\int_a^R f(x)\,dx,\]

provided that the limit exists as a finite real number. If it does, the improper integral converges. If no finite limit exists, the integral diverges.

The calculation uses ordinary definite integrals with finite endpoints, followed by a limit. Infinity is not a real number to substitute into an antiderivative.

Example: a convergent integral

Evaluate \(\int_1^\infty x^{-2}\,dx\). First replace infinity with a finite upper endpoint:

\[\int_1^R x^{-2}\,dx=\left[-\frac1x\right]_1^R=1-\frac1R.\]

Now take the limit:

\[\int_1^\infty\frac1{x^2}\,dx=\lim_{R\to\infty}\left(1-\frac1R\right)=1.\]

The region extends indefinitely to the right, but its height decreases quickly enough for its total area to be finite.

Example: a divergent integral

For \(\int_1^\infty x^{-1}\,dx\),

\[\int_1^R\frac1x\,dx=\ln R,\]

and \(\ln R\to\infty\) as \(R\to\infty\). The integral diverges. Although \(1/x\to0\), that fact alone does not guarantee a finite integral.

An unbounded integrand on a finite interval

An integral can be improper even when both endpoints are finite. For example, \(1/\sqrt{x}\) becomes unbounded as \(x\to0^+\). Define the integral using a one-sided limit:

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

Evaluating gives

\[\lim_{\varepsilon\to0^+}\left[2\sqrt{x}\right]_\varepsilon^1=\lim_{\varepsilon\to0^+}\left(2-2\sqrt{\varepsilon}\right)=2.\]

This improper integral converges despite the unbounded integrand.

When an integral must be split

If the integrand becomes unbounded at an interior point c, split the interval there:

\[\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx.\]

Both one-sided improper integrals must converge separately. The same principle applies to an integral over the entire real line: split it at a finite point and check each infinite tail.

For example, \(\int_{-1}^1 1/x\,dx\) does not converge as an ordinary improper integral. The left-hand integral tends to \(-\infty\), and the right-hand integral tends to \(+\infty\). These cannot be added to obtain zero. A symmetric limiting procedure gives a Cauchy principal value of zero, which is a different concept.

The p-test

Two useful benchmarks are

\[\int_1^\infty\frac1{x^p}\,dx\quad\text{converges exactly when }p>1,\]

and

\[\int_0^1\frac1{x^p}\,dx\quad\text{converges exactly when }p<1.\]

Notice that the conditions differ: the first tests behavior at infinity, while the second tests behavior near zero. At \(p=1\), both integrals diverge.

A reliable method

Find every infinite endpoint and every point where the integrand is unbounded.
Split the integral when necessary.
Replace each problematic endpoint with a finite variable and write the appropriate one-sided limit.
Evaluate the finite integral, then take the limit.
Declare convergence only if every required limit is finite.

Some problems require comparison tests rather than an elementary antiderivative. The meaning of convergence remains the same.

For the finite-interval foundation, see definite integrals.