How to evaluate improper integrals with limits

An improper integral asks what happens as we move a finite bound toward infinity or toward a point where the integrand is unbounded. We’ll set up that limit and see why one integral gives a finite value while another diverges. The video introduces how to recognize the two types.

Replace the improper bound with a finite cutoff

If the upper bound is $\infty$ , integrate to a finite $b$ , then let $b \to \infty$ . If the lower bound is $- \infty$ , integrate from a finite $a$ and let $a \to - \infty$ . The improper integral converges when its limit exists as a finite real number.

If both bounds are infinite, split at a finite point $c$ where the integrand is defined. The integral to the left of $c$ and the integral to the right each need their own finite limit before we add them.

Example: An infinite upper bound can converge

First integrate over [0, b]. Then let the finite cutoff b increase without bound.

\begin{array}{l} \int_{0}^{\infty} \frac{1}{1 + x^{2}} d x \\ = \lim_{b \to \infty} \int_{0}^{b} \frac{1}{1 + x^{2}} d x \\ = \lim_{b \to \infty} [\arctan (b) - \arctan (0)] \\ = \frac{π}{2} \end{array}

The limit is finite, so this integral converges to $\frac{π}{2}$ . The integrand is continuous everywhere; the infinite interval is what makes this integral improper.

Approach an unbounded endpoint from inside the interval

If $f$ is unbounded near the lower endpoint $a$ , use a cutoff $c > a$ and let $c \to a^{+}$ . If $f$ is unbounded near the upper endpoint $b$ , use a cutoff $c < b$ and let $c \to b^{-}$ .

Example: A lower endpoint singularity can diverge

For $\frac{1}{x^{3}}$ from $0$ to $2$ , the integrand grows without bound as $x \to 0^{+}$ . We can still integrate from $c > 0$ to $2$ .

\begin{array}{l} \int_{0}^{2} \frac{1}{x^{3}} d x \\ = \lim_{c \to 0^{+}} \int_{c}^{2} \frac{1}{x^{3}} d x \\ = \lim_{c \to 0^{+}} (- \frac{1}{8} + \frac{1}{2 c^{2}}) \end{array}

As $c \to 0^{+}$ , the term $\frac{1}{2 c^{2}}$ grows without bound. The integral diverges to positive infinity.

Example: An unbounded upper endpoint can still give a finite area

For $\frac{1}{\sqrt{9 - x}}$ from $0$ to $9$ , the upper cutoff $b$ approaches $9$ from the left. The antiderivative is $- 2 \sqrt{9 - x}$ .

\begin{array}{l} \int_{0}^{9} \frac{1}{\sqrt{9 - x}} d x \\ = \lim_{b \to 9^{-}} (6 - 2 \sqrt{9 - b}) \\ = 6 \end{array}

Split at a singularity inside the interval

If the integrand is unbounded at an interior point, split the integral there and approach that point from each side. Both pieces have to converge separately. For example, split $\frac{1}{x^{2}}$ on $[- 1, 1]$ at $x = 0$ . The right piece already diverges, since

\begin{array}{l} \int_{0}^{1} \frac{1}{x^{2}} d x \\ = \lim_{c \to 0^{+}} (\frac{1}{c} - 1) \end{array}

That limit grows without bound, so the whole integral diverges. This same separate-limit rule applies when both endpoints are infinite.

Learn mathKrista KingJanuary 20, 2021math, learn online, online course, online math, calculus 2, calculus ii, calc 2, calc ii, single variable calculus, single variable calc, improper integrals, improper, integrals with infinity, infinite integrals, integrals with discontinuities, integrals, integration