Math Core

Can a region that stretches out forever have a finite area? Surprisingly, yes. Improper integrals give a precise meaning to integrals over infinite intervals, and to integrals of functions that blow up inside the interval. You will use them again in the series unit, where the integral test compares an infinite sum to an improper integral.

What makes an integral improper

The Fundamental Theorem of Calculus requires a finite interval $[a, b]$ and a function that is continuous on it. An integral is improper when either condition fails:

Type 1: a limit of integration is infinite, as in $\displaystyle\int_1^{\infty} \frac{dx}{x^2}$ .
Type 2: the integrand has an infinite discontinuity (a vertical asymptote) at an endpoint or inside the interval, as in $\displaystyle\int_0^1 \frac{dx}{\sqrt{x}}$ .

In both cases you can't just plug in. Instead, integrate over a safe interval and take a limit.

Definition

Improper integral

For an infinite upper limit:

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

If $f$ has a vertical asymptote at $x = a$ :

\int_a^c f(x)\,dx = \lim_{t \to a^+} \int_t^c f(x)\,dx.

If the limit exists and is finite, the integral converges to that value. Otherwise it diverges.

Lower infinite limits and asymptotes at the right endpoint work the same way. If there is trouble at both ends, or at a point inside the interval, split the integral into pieces so that each has one trouble spot. The original integral converges only if every piece converges.

Infinite intervals

Worked example: A convergent Type 1 integral

Evaluate $\displaystyle\int_1^{\infty} \frac{dx}{x^2}$ .

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

The region under $y = \dfrac{1}{x^2}$ to the right of $x = 1$ is infinitely long, yet its area is exactly 1.

The shaded region under y = 1/x² for x ≥ 1 has area 1. The dashed curve y = 1/x looks similar, but the area under it for x ≥ 1 is infinite.Open in grapher →

Worked example: A divergent Type 1 integral

Evaluate $\displaystyle\int_1^{\infty} \frac{dx}{x}$ , or show that it diverges.

\int_1^{\infty} \frac{dx}{x} = \lim_{b \to \infty}\Big[\ln x\Big]_1^b = \lim_{b \to \infty} \ln b = \infty.

The integral diverges. Even though $\dfrac{1}{x} \to 0$ , it doesn't shrink fast enough for the area to stay finite.

Those two examples are part of a pattern that is worth memorizing.

p-integrals

\int_1^{\infty} \frac{dx}{x^p} \ \text{ converges (to } \tfrac{1}{p - 1}\text{) if } p > 1, \text{ and diverges if } p \le 1.

\int_0^{1} \frac{dx}{x^p} \ \text{ converges (to } \tfrac{1}{1 - p}\text{) if } p < 1, \text{ and diverges if } p \ge 1.

Notice the direction flips. Near infinity you need a function that dies off quickly (big $p$ ). Near zero you need a function that doesn't blow up too fast (small $p$ ). The borderline $p = 1$ diverges in both cases.

Infinite discontinuities

Worked example: A convergent Type 2 integral

Evaluate $\displaystyle\int_0^1 \frac{dx}{\sqrt{x}}$ .

The integrand has a vertical asymptote at $x = 0$ .

\int_0^1 x^{-1/2}\,dx = \lim_{t \to 0^+}\Big[2\sqrt{x}\Big]_t^1 = \lim_{t \to 0^+}\left(2 - 2\sqrt{t}\right) = 2.

Common mistake

Always scan the interval for vertical asymptotes before you apply the Fundamental Theorem. Blindly computing $\displaystyle\int_{-1}^{1} \frac{dx}{x^2} = \left[-\frac{1}{x}\right]_{-1}^{1} = -1 - 1 = -2$ is wrong. A positive function can't have a negative integral! The integrand blows up at $x = 0$ , and $\displaystyle\int_0^1 \frac{dx}{x^2}$ diverges by the $p$ -integral rule, so the whole integral diverges.

Limits you'll need

Evaluating an improper integral often comes down to a limit at infinity. These facts come up constantly:

$\displaystyle\lim_{b \to \infty} e^{-b} = 0$ and $\displaystyle\lim_{b \to \infty} \arctan b = \frac{\pi}{2}$ .
$\displaystyle\lim_{b \to \infty} \ln b = \infty$ .
Exponentials beat powers, and powers beat logs. For example, $\displaystyle\lim_{b \to \infty} \frac{b}{e^b} = 0$ and $\displaystyle\lim_{b \to \infty} \frac{\ln b}{b} = 0$ . You can confirm these with L'Hospital's Rule.

Worked example: Parts and a limit together

Evaluate $\displaystyle\int_0^{\infty} x e^{-x}\,dx$ .

By parts with $u = x$ and $dv = e^{-x}\,dx$ : $\displaystyle\int x e^{-x}\,dx = -x e^{-x} - e^{-x} + C$ .

\int_0^{\infty} x e^{-x}\,dx = \lim_{b \to \infty}\Big[-x e^{-x} - e^{-x}\Big]_0^b = \lim_{b \to \infty}\left(-\frac{b}{e^b} - \frac{1}{e^b}\right) - (0 - 1) = 0 - 0 + 1 = 1.

The limit $\dfrac{b}{e^b} \to 0$ follows from L'Hospital's Rule: $\displaystyle\lim_{b \to \infty}\frac{b}{e^b} = \lim_{b \to \infty}\frac{1}{e^b} = 0$ .

Tip

Write the limit every time. On the AP exam, an answer like $\left[-\dfrac{1}{x}\right]_1^{\infty}$ with $\infty$ plugged in directly does not earn full credit. The notation $\displaystyle\lim_{b \to \infty}$ shows you know why the integral has a value.