Topic 4 of 7

Improper integrals

Step 16 · Calculus 2 topics

Make sense of integrals over infinite intervals or across vertical asymptotes as limits, and decide whether they converge by comparison.

Infinite limits of integration

An integral out to infinity is defined as a limit: integrate up to b, then let b grow. If the limit is a finite number the integral converges; otherwise it diverges.

\int_{a}^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_{a}^{b} f(x)\,dx

Example: Converges

Find $\int_{1}^{\infty} \frac{1}{x^{2}}\,dx$ .

$\int_{1}^{b} x^{-2}\,dx = 1 - \frac{1}{b}$ .
As b grows, 1 ÷ b goes to 0.

Answer1

Example: Diverges

Find $\int_{1}^{\infty} \frac{1}{x}\,dx$ .

The integral up to b is ln b.
ln b grows without bound.

AnswerIt diverges

y = 1 ÷ x (solid) and y = 1 ÷ x² (dashed) from x = 1. Both tails shrink toward 0, but only the area under 1 ÷ x² is finite.

Example: Exponential decay

Find $\int_{0}^{\infty} e^{-2x}\,dx$ .

$\int_{0}^{b} e^{-2x}\,dx = \frac{1}{2}\left(1 - e^{-2b}\right)$ .
$e^{-2b}$ goes to 0.

Answer $\frac{1}{2}$

Integrands that blow up

If f has a vertical asymptote at an endpoint, approach that endpoint with a limit. If the asymptote is inside the interval, split the integral there: both parts must converge.

Example: At an endpoint

Find $\int_{0}^{4} \frac{1}{\sqrt{x}}\,dx$ .

$\int_{t}^{4} x^{-1/2}\,dx = 4 - 2\sqrt{t}$ .
As t approaches 0 from the right, this approaches 4.

Answer4

Example: Inside the interval

Is $\int_{-1}^{1} \frac{1}{x^{2}}\,dx$ equal to −2?

Blindly using the antiderivative −1 ÷ x gives −1 − 1, which is −2.
But 1 ÷ x² is positive, so the integral cannot be negative: there is an asymptote at 0.
Split at 0. The integral from 0 to 1 already diverges.

AnswerNo: it diverges

Comparing improper integrals

Often you only need to know whether an integral converges. If 0 ≤ f ≤ g and the integral of g converges, so does the integral of f. If f ≥ g ≥ 0 and the integral of g diverges, so does the integral of f.

Example: Below a convergent one

Does $\int_{1}^{\infty} \frac{dx}{x^{2} + 5}$ converge?

$\frac{1}{x^{2} + 5} < \frac{1}{x^{2}}$ for every x.
The integral of 1 ÷ x² from 1 converges.

AnswerYes, it converges

Example: Above a divergent one

Does $\int_{2}^{\infty} \frac{dx}{\ln x}$ converge?

ln x < x, so $\frac{1}{\ln x} > \frac{1}{x}$ for x ≥ 2.
The integral of 1 ÷ x diverges.

AnswerNo, it diverges

Try it yourself

Work each one on paper first, then open the answer to check.

Find $\int_{1}^{\infty} \frac{1}{x^{3}}\,dx$ .
Show answer
$\frac{1}{2}$
Does $\int_{1}^{\infty} \frac{1}{\sqrt{x}}\,dx$ converge?
Show answer
No: p is one half, which is not more than 1
Find $\int_{0}^{\infty} e^{-x}\,dx$ .
Show answer
1
Find $\int_{0}^{1} \frac{1}{\sqrt[3]{x}}\,dx$ .
Show answer
$\frac{3}{2}$
Does $\int_{0}^{2} \frac{1}{(x - 1)^{2}}\,dx$ converge?
Show answer
No: it blows up at x = 1, inside the interval
Does $\int_{1}^{\infty} \frac{\sin^{2} x}{x^{2}}\,dx$ converge?
Show answer
Yes: the integrand is at most $\frac{1}{x^{2}}$