Improper Integrals

Can a region that goes on forever have a finite area? Surprisingly, yes. The region under $y = \dfrac{1}{x^2}$ from $x = 1$ out to infinity has area exactly $1$. An improper integral is an integral where the interval is infinite or the function blows up to infinity somewhere in it. You handle it by computing an ordinary integral and then taking a limit. Improper integrals are BC-only, and they lead straight into the integral test and p-series for infinite series.

Key ideas

Infinite limits of integration

Replace $\infty$ with a letter, integrate, then let the letter go to infinity:

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

If the limit is a finite number, the integral converges to that number. If the limit is infinite or doesn’t exist, the integral diverges. Likewise, $\displaystyle\int_{-\infty}^{b} f(x)\,dx = \lim_{a \to -\infty} \int_a^b f(x)\,dx$.

For $\displaystyle\int_{-\infty}^{\infty}$, split at any convenient point, for example $0$. The whole integral converges only if both halves converge.

Infinite discontinuities

If $f$ has a vertical asymptote at an endpoint, approach that endpoint with a one-sided limit. For an asymptote at $x = a$:

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

If the asymptote is inside the interval, at $x = c$ with $a \lt c \lt b$, split there:

$$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$$

and both pieces must converge. These are easy to miss, because the integral looks ordinary. Always check the integrand for zeros in a denominator (or logs of $0$) before using the Fundamental Theorem.

Writing it correctly

On the AP exam, you must show the limit. Writing $\Big[-\tfrac{1}{x}\Big]_1^\infty$ and “plugging in $\infty$” loses the point. Write $\displaystyle\lim_{b \to \infty}\Big[-\tfrac{1}{x}\Big]_1^b$ instead. Limits you’ll use often: $\displaystyle\lim_{b \to \infty} \frac{1}{b^n} = 0$ for $n \gt 0$, $\displaystyle\lim_{b \to \infty} e^{-b} = 0$, $\displaystyle\lim_{b \to \infty} \ln b = \infty$, and $\displaystyle\lim_{b \to \infty} \arctan b = \frac{\pi}{2}$. Sometimes you’ll need L’Hôpital’s rule.

The p-integral

For $p \ne 1$, $\displaystyle\int_1^b x^{-p}\,dx = \frac{b^{1 - p} - 1}{1 - p}$. As $b \to \infty$, $b^{1 - p}$ goes to $0$ if $p \gt 1$ and to $\infty$ if $p \lt 1$. For $p = 1$ you get $\ln b \to \infty$. So:

$$\int_1^\infty \frac{1}{x^p}\,dx \;\begin{cases} \text{converges to } \dfrac{1}{p - 1} & \text{if } p \gt 1 \\[2mm] \text{diverges} & \text{if } p \le 1 \end{cases}$$

This is the integral behind the p-series test.

The curves look alike, but $\int_1^\infty \frac{1}{x^2}\,dx = 1$ while $\int_1^\infty \frac{1}{x}\,dx$ diverges.

Worked examples

Example 1: Converge or diverge?

Determine whether each integral converges. If it does, find its value. (a) $\displaystyle\int_1^\infty \frac{1}{x^2}\,dx$ (b) $\displaystyle\int_1^\infty \frac{1}{x}\,dx$

Solution.

(a)

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

The integral converges to $1$.

(b)

$$\int_1^\infty \frac{1}{x}\,dx = \lim_{b \to \infty} \Big[\ln x\Big]_1^b = \lim_{b \to \infty} \ln b = \infty$$

The integral diverges. Both curves approach the $x$-axis, but $\dfrac{1}{x}$ does so too slowly.

Example 2: An infinite interval with substitution

Evaluate $\displaystyle\int_0^\infty x e^{-x^2}\,dx$.

Solution. First find the antiderivative with $u = -x^2$, $du = -2x\,dx$: $\displaystyle\int x e^{-x^2}\,dx = -\tfrac{1}{2}e^{-x^2} + C$.

$$\int_0^\infty x e^{-x^2}\,dx = \lim_{b \to \infty}\Big[-\tfrac{1}{2}e^{-x^2}\Big]_0^b = \lim_{b \to \infty}\left(-\tfrac{1}{2}e^{-b^2} + \tfrac{1}{2}\right) = \frac{1}{2}$$

The integral converges to $\tfrac{1}{2}$.

Example 3: A vertical asymptote at an endpoint

Evaluate $\displaystyle\int_0^4 \frac{1}{\sqrt{x}}\,dx$.

Solution. The integrand is undefined at $x = 0$ and goes to $\infty$ as $x \to 0^+$, so this is improper.

$$\int_0^4 x^{-1/2}\,dx = \lim_{t \to 0^+} \Big[2\sqrt{x}\Big]_t^4 = \lim_{t \to 0^+} \left(4 - 2\sqrt{t}\right) = 4$$

It converges to $4$, even though the region is infinitely tall.

The region under $y = \frac{1}{\sqrt{x}}$ from $0$ to $4$ is infinitely tall but has finite area $4$.

Example 4: A hidden asymptote in the middle

Evaluate $\displaystyle\int_{-1}^{2} \frac{1}{x^2}\,dx$, or show it diverges.

Solution. The integrand has a vertical asymptote at $x = 0$, inside the interval. Split there:

$$\int_{-1}^{2} \frac{1}{x^2}\,dx = \int_{-1}^{0} \frac{1}{x^2}\,dx + \int_{0}^{2} \frac{1}{x^2}\,dx$$

Check the right piece:

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

One piece diverges, so the whole integral diverges. (You don’t need to check the other piece.)

Notice what goes wrong if you ignore the asymptote: $\Big[-\tfrac{1}{x}\Big]_{-1}^{2} = -\tfrac{1}{2} - 1 = -\tfrac{3}{2}$. A negative answer for the integral of a positive function is a sure sign something broke.

Common mistakes

Plugging in infinity. Write the limit. "$-\tfrac{1}{\infty} = 0$" is not acceptable work on the AP exam; $\displaystyle\lim_{b \to \infty}\left(-\tfrac{1}{b}\right) = 0$ is.

Missing an interior asymptote. Before evaluating any definite integral, check whether the integrand is undefined somewhere between the limits. If it is, split the integral there and use limits.

Thinking “the function goes to 0, so the integral converges”. $\dfrac{1}{x} \to 0$, but $\displaystyle\int_1^\infty \frac{1}{x}\,dx$ diverges. The function has to shrink fast enough.

Getting the p-integral rule backwards. $\displaystyle\int_1^\infty \frac{1}{x^p}\,dx$ converges for $p \gt 1$ (and diverges for $p \le 1$, including $p = 1$). Remember the two key cases: $\dfrac{1}{x^2}$ converges, $\dfrac{1}{x}$ diverges.

Letting two infinities cancel. For $\displaystyle\int_{-\infty}^{\infty}$ or an interior asymptote, each piece must converge on its own. If one piece is $\infty$, the integral diverges; don’t argue that $\infty - \infty = 0$.

Practice

1. (Warm-up) Evaluate $\displaystyle\int_1^\infty \frac{1}{x^3}\,dx$.

Solution

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

Converges to $\tfrac{1}{2}$, matching $\dfrac{1}{p - 1}$ with $p = 3$.

2. (Warm-up) A medicine enters the bloodstream at a rate of $r(t) = 40e^{-0.5t}$ milligrams per hour, where $t$ is in hours. Find the total amount that ever enters the bloodstream, $\displaystyle\int_0^\infty r(t)\,dt$.

Solution

$$\int_0^\infty 40e^{-0.5t}\,dt = \lim_{b \to \infty}\Big[-80e^{-0.5t}\Big]_0^b = \lim_{b \to \infty}\left(-80e^{-0.5b} + 80\right) = 80$$

A total of $80$ mg.

3. (Warm-up) Evaluate $\displaystyle\int_0^1 \frac{1}{\sqrt[3]{x}}\,dx$.

Solution

There’s an asymptote at $x = 0$.

$$\int_0^1 x^{-1/3}\,dx = \lim_{t \to 0^+}\Big[\tfrac{3}{2}x^{2/3}\Big]_t^1 = \lim_{t \to 0^+}\left(\tfrac{3}{2} - \tfrac{3}{2}t^{2/3}\right) = \frac{3}{2}$$

4. (Core) Show that $\displaystyle\int_1^\infty \frac{1}{\sqrt{x}}\,dx$ diverges.

Solution

$$\int_1^\infty x^{-1/2}\,dx = \lim_{b \to \infty}\Big[2\sqrt{x}\Big]_1^b = \lim_{b \to \infty}\left(2\sqrt{b} - 2\right) = \infty$$

The limit is infinite, so the integral diverges. (This is a p-integral with $p = \tfrac{1}{2} \le 1$.)

5. (Core) Evaluate $\displaystyle\int_0^\infty \frac{1}{1 + x^2}\,dx$.

Solution

$$\lim_{b \to \infty}\Big[\arctan x\Big]_0^b = \lim_{b \to \infty}\arctan b - 0 = \frac{\pi}{2}$$

6. (Core) Evaluate $\displaystyle\int_e^\infty \frac{1}{x(\ln x)^2}\,dx$.

Solution

Substitute $u = \ln x$, $du = \tfrac{1}{x}\,dx$: the antiderivative is $-\dfrac{1}{\ln x}$.

$$\lim_{b \to \infty}\Big[-\frac{1}{\ln x}\Big]_e^b = \lim_{b \to \infty}\left(-\frac{1}{\ln b} + \frac{1}{\ln e}\right) = 0 + 1 = 1$$

7. (Core) Evaluate $\displaystyle\int_2^\infty \frac{1}{x^2 - 1}\,dx$.

Solution

Use partial fractions: $\dfrac{1}{(x - 1)(x + 1)} = \dfrac{1/2}{x - 1} - \dfrac{1/2}{x + 1}$, so an antiderivative is $\tfrac{1}{2}\ln\left|\dfrac{x - 1}{x + 1}\right|$.

$$\lim_{b \to \infty}\frac{1}{2}\Big[\ln\frac{x - 1}{x + 1}\Big]_2^b = \frac{1}{2}\left(\lim_{b \to \infty}\ln\frac{b - 1}{b + 1} - \ln\frac{1}{3}\right) = \frac{1}{2}\left(\ln 1 + \ln 3\right) = \frac{1}{2}\ln 3$$

since $\dfrac{b - 1}{b + 1} \to 1$. The value is $\tfrac{1}{2}\ln 3 \approx 0.549$.

8. (Challenge) Evaluate $\displaystyle\int_0^3 \frac{1}{(x - 1)^{2/3}}\,dx$, or show it diverges.

Solution

There’s an asymptote at $x = 1$, inside the interval, so split. An antiderivative is $3(x - 1)^{1/3}$.

$$\int_0^1 (x - 1)^{-2/3}\,dx = \lim_{t \to 1^-}\Big[3(x - 1)^{1/3}\Big]_0^t = 0 - 3(-1) = 3$$$$\int_1^3 (x - 1)^{-2/3}\,dx = \lim_{s \to 1^+}\Big[3(x - 1)^{1/3}\Big]_s^3 = 3\sqrt[3]{2} - 0 = 3\sqrt[3]{2}$$

Both pieces converge, so the integral converges to $3 + 3\sqrt[3]{2} \approx 6.780$.

9. (Challenge) Evaluate $\displaystyle\int_0^\infty x e^{-x}\,dx$.

Solution

By integration by parts ($u = x$, $dv = e^{-x}\,dx$), an antiderivative is $-x e^{-x} - e^{-x}$.

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

The limit $\displaystyle\lim_{b \to \infty} \frac{b}{e^b}$ has the form $\dfrac{\infty}{\infty}$; by L’Hôpital’s rule it equals $\displaystyle\lim_{b \to \infty} \frac{1}{e^b} = 0$. So the integral converges to $0 - 0 + 1 = 1$.