Improper Integrals — AP® Calculus BC Notes

Every definite integral so far had finite limits and a continuous integrand on a closed interval. An improper integral breaks one of those rules: a limit of integration is infinite, as in $\int_1^\infty \frac{1}{x^2}\,dx$ , or the integrand is unbounded on the interval, as in $\int_0^1 \frac{1}{\sqrt{x}}\,dx$ . Calculus handles both the same way: integrate on a safe interval, then take a limit.

Infinite limits of integration

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

If the limit is a finite number, the improper integral converges to that number. If the limit is infinite or does not exist, the integral diverges. The same idea works at the other end: $\int_{-\infty}^{b} f(x)\,dx$ $= \lim_{a \to -\infty} \int_a^b f(x)\,dx$ .

Worked example 1. Evaluate $\int_1^\infty \frac{1}{x^2}\,dx$ or show that it diverges.

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

The limit is a finite number, so the integral converges to 1.

Worked example 2. Evaluate $\int_1^\infty \frac{1}{x}\,dx$ or show that it diverges.

$\lim_{b \to \infty} \int_1^b \frac{1}{x}\,dx$ $= \lim_{b \to \infty} \ln b$ $= \infty$ , so the integral diverges.

The figure puts the two side by side. Both graphs approach 0, yet one area levels off and the other grows without bound. A height that goes to 0 is not enough to make the area finite: only the limit of the integrals decides.

Two panels. Left: y = 1/x squared with the area under it from x = 1 to x = b shaded violet; that area is 1 − 1/b, which is 0.500 at b = 2, 0.900 at b = 10, 0.990 at b = 100 and 0.999 at b = 1000, so it approaches 1 and the improper integral converges. Right: y = 1/x with the area from 1 to b shaded vermillion; that area is ln b, which is 0.693, 2.303, 4.605 and 6.908 at the same values of b and grows without bound, so the improper integral diverges. — **Figure 1.** The area under 1/x² from 1 to b approaches 1; the area under 1/x is ln b, which grows without bound. LevlPrep original diagram.

Powers of x: the p-integrals

For $p \ne 1$ , $\int_1^b x^{-p}\,dx$ $= \frac{b^{1 - p} - 1}{1 - p}$ . When $p > 1$ , $b^{1 - p} \to 0$ and the integral converges to $\frac{1}{p - 1}$ . When $p < 1$ , $b^{1 - p} \to \infty$ and it diverges; $p = 1$ gives $\ln b$ , which also diverges.

Integral	$\int_1^b$ equals	As $b \to \infty$
$\int_1^\infty x^{-2}\,dx$	$1 - \frac{1}{b}$	converges to 1
$\int_1^\infty x^{-3/2}\,dx$	$2 - \frac{2}{\sqrt{b}}$	converges to 2
$\int_1^\infty x^{-1}\,dx$	$\ln b$	diverges
$\int_1^\infty x^{-1/2}\,dx$	$2\sqrt{b} - 2$	diverges
$\int_0^\infty e^{-x}\,dx$	$1 - e^{-b}$ (from 0)	converges to 1

Integrands that blow up

If $f$ is unbounded at an endpoint, put the limit at that endpoint. For an upper endpoint $c$ : $\int_a^c f(x)\,dx$ $= \lim_{b \to c^-} \int_a^b f(x)\,dx$ .

Worked example 3. Evaluate $\int_0^3 \frac{1}{\sqrt{3 - x}}\,dx$ .

The integrand is unbounded as $x \to 3^-$ . With $\int (3 - x)^{-1/2}\,dx$ $= -2\sqrt{3 - x}$ :

$\lim_{b \to 3^-}\left(-2\sqrt{3 - x}\right)\Big|_0^b$ $= \lim_{b \to 3^-}\left(-2\sqrt{3 - b} + 2\sqrt{3}\right)$ $= 2\sqrt{3}.$ The integral converges.

The dangerous case is a blow-up inside the interval. $\int_{-1}^1 \frac{1}{x^2}\,dx$ looks like $-\frac{1}{x}\Big|_{-1}^1 = -2$ , a negative answer for a positive integrand. The Fundamental Theorem does not apply, because $\frac{1}{x^2}$ is not continuous at 0. Look at the piece from 0 to 1: $\int_0^1 \frac{1}{x^2}\,dx$ $= \lim_{a \to 0^+}\left(\frac{1}{a} - 1\right)$ $= \infty$ , so the integral diverges. Before integrating, always check the integrand for vertical asymptotes on the interval.

Using parts and L'Hospital's rule

Worked example 4. Evaluate $\int_0^\infty xe^{-x}\,dx$ .

Parts with $u = x$ , $dv = e^{-x}\,dx$ : $\int xe^{-x}\,dx$ $= -xe^{-x} - e^{-x}$ $+ C$ .

$\lim_{b \to \infty}\left(-xe^{-x} - e^{-x}\right)\Big|_0^b$ $= \lim_{b \to \infty}\left(-\frac{b}{e^b} - \frac{1}{e^b}\right) - (0 - 1).$

$\frac{b}{e^b}$ has the form $\frac{\infty}{\infty}$ , so by L'Hospital's rule $\lim_{b \to \infty}\frac{b}{e^b}$ $= \lim_{b \to \infty}\frac{1}{e^b}$ $= 0$ . The integral converges to $0 - 0 + 1 = 1$ .

Writing it so it earns the point

Write $\lim_{b \to \infty}$ in front of every expression until the step where you take the limit. Dropping it in the middle makes a false statement such as $\int_1^b \frac{dx}{x^2}$ $= 1$ .
Never substitute $\infty$ : $-\frac{1}{\infty}$ and $e^{-\infty}$ are not numbers. Recent scoring guidelines give the point only when the notation is right all the way.
State the conclusion in words: "the integral converges to 1" or "the integral diverges".