When an infinite region has finite area, integrals from easy to advanced lesson

The region under $y = 1/x^2$ to the right of $x = 1$ never ends, yet its area turns out to be $1$ . The region under $y = 1/x$ also never ends and its height also shrinks toward $0$ , but its area is infinite. Since an antiderivative cannot be evaluated at infinity, the first task is to say what such an area means. This page defines it as a limit of ordinary integrals and then settles, for every real number $p$ , whether $\int_1^\infty x^{-p}\,dx$ is finite.

Key terms

Improper integral

Definition Let $a$ be a real number and let $f$ be integrable on $[a, R]$ for every $R > a$ , for example continuous on $[a, \infty)$ . The improper integral of $f$ over $[a, \infty)$ is defined as the limit

\begin{aligned} &\int_a^\infty f(x)\,dx \\[0.4em] &\quad = \lim_{R \to \infty} \int_a^R f(x)\,dx. \end{aligned}

An integral over $(-\infty, b]$ is defined in the same way, with the lower limit tending to $-\infty$ . An integral over $(-\infty, \infty)$ is split at a point $c$ into integrals over $(-\infty, c]$ and $[c, \infty)$ , and each of the two must converge on its own. An integral whose integrand is unbounded near an endpoint, such as $\int_0^1 x^{-1/2}\,dx$ , is also improper and is defined by a limit at that endpoint.

In plain words An infinite interval has no right endpoint to substitute into an antiderivative. So stop at a finite point $R$ , compute an ordinary integral, and watch what those finite totals do as $R$ moves off to infinity. The order matters. Compute the finite integral first and take the limit last.

Example For every $R > 0$ ,

\int_0^R e^{-x}\,dx = 1 - e^{-R},

and $e^{-R} \to 0$ as $R \to \infty$ . So the improper integral $\int_0^\infty e^{-x}\,dx$ equals $1$ .

Convergent integral

Definition An improper integral converges, or is convergent, when its defining limit exists and is a finite real number. That number is the value of the integral. For an integral over $[a, \infty)$ this means that $\int_a^R f(x)\,dx$ approaches one real number $L$ as $R \to \infty$ , and then $\int_a^\infty f(x)\,dx = L$ .

In plain words The finite totals settle on one number. An unbounded region can then have a finite area, because its heights far out shrink fast enough that the region beyond a distant point adds only a little.

Example For $R > 1$ ,

\int_1^R \frac{dx}{x^2} = \Bigl[-\frac{1}{x}\Bigr]_1^R = 1 - \frac{1}{R}.

At $R = 10$ the total is $0.9$ , at $R = 100$ it is $0.99$ , and as $R \to \infty$ it approaches $1$ . So $\int_1^\infty x^{-2}\,dx$ converges, and its value is $1$ .

Divergent integral

Definition An improper integral diverges, or is divergent, when its defining limit does not exist as a finite number. The finite totals may grow without bound, called divergence to $+\infty$ , fall without bound, called divergence to $-\infty$ , or keep changing without settling on any value. A divergent integral is not assigned a real value.

In plain words Heights that shrink to $0$ are not enough for a finite area. Small heights spread over an unlimited stretch can still pile up an unlimited total.

Example The height $1/\sqrt{x}$ tends to $0$ , yet for $R > 1$

\int_1^R \frac{dx}{\sqrt{x}} = 2\sqrt{R} - 2,

which grows without bound. So $\int_1^\infty x^{-1/2}\,dx$ diverges to $+\infty$ . The integral $\int_0^\infty \cos x\,dx$ diverges in a different way. Its finite totals $\int_0^R \cos x\,dx = \sin R$ keep swinging between $-1$ and $1$ and approach no limit.

The p-test for integrals

Definition Let $p$ be a real number. If $p > 1$ , then

\int_1^\infty x^{-p}\,dx = \frac{1}{p - 1},

and if $p \le 1$ , the integral diverges to $+\infty$ . The companion statement near $0$ says that $\int_0^1 x^{-p}\,dx$ converges exactly when $p < 1$ , with value $\frac{1}{1 - p}$ , and diverges to $+\infty$ when $p \ge 1$ . For $p \le 0$ that integrand is continuous on $[0, 1]$ , so the integral is an ordinary one.

In plain words Beyond $x = 1$ , the curve $x^{-p}$ encloses a finite area exactly when $p > 1$ , that is, when it falls faster than the boundary curve $1/x$ . The value $\frac{1}{p-1}$ belongs to the convergent side only. Near $0$ the comparison reverses, and the area is finite exactly when the curve grows more slowly than $1/x$ . No power has finite area over all of $(0, \infty)$ , because that would need $p > 1$ and $p < 1$ at once.

Example Here is the integral from the opening, worked from the definition. Let $p$ be real with $p \ne 1$ , and let $R > 1$ . Raising the exponent $-p$ by one gives

\begin{aligned} \int_1^R x^{-p}\,dx &= \left[\frac{x^{1-p}}{1-p}\right]_1^R \\[0.4em] &= \frac{R^{1-p} - 1}{1 - p}, \end{aligned}

since every real power of $1$ equals $1$ . If $p > 1$ , the exponent $1 - p$ is negative, so $R^{1-p} \to 0$ and the totals approach $\frac{-1}{1-p} = \frac{1}{p-1}$ . If $p < 1$ , the exponent and the denominator are both positive, so $R^{1-p} \to \infty$ and the totals grow without bound. That case includes $0 < p < 1$ , where the graph still falls toward $0$ , and $p \le 0$ , where the height stays constant or grows. The remaining value $p = 1$ is the next card. For instance, $p = 2$ gives $1$ , $p = \tfrac32$ gives $2$ and $p = 1.1$ gives $10$ .

Boundary case p equals 1

Definition At $p = 1$ the power rule antiderivative $x^{1-p}/(1-p)$ is undefined, because it divides by $0$ . The integrand is then $1/x$ , whose antiderivative on $(0, \infty)$ is $\ln x$ , so for $R > 1$

\int_1^R \frac{dx}{x} = \ln R,

which grows without bound. Hence $\int_1^\infty dx/x$ diverges to $+\infty$ . In the same way $\int_t^1 dx/x = -\ln t$ grows without bound as $t \to 0^+$ , so $\int_0^1 dx/x$ diverges too. The value $p = 1$ therefore lies on the divergent side of both statements of the p-test, which is why their conditions are the strict inequalities $p > 1$ and $p < 1$ .

In plain words The curve $1/x$ is the dividing line. It falls toward $0$ , but just too slowly. Each stretch from $x$ to $2x$ is twice as wide as the stretch before it, and the curve over it is half as high, so every such stretch holds the same area, and infinitely many equal areas have no finite total.

Example For every whole number $k \ge 0$ ,

\begin{aligned} \int_{2^k}^{2^{k+1}} \frac{dx}{x} &= \ln 2^{k+1} - \ln 2^k \\[0.4em] &= \ln 2. \end{aligned}

So the area from $1$ to $2^m$ is $m \ln 2$ , which passes every bound as $m$ grows. The growth is slow, since the area from $1$ to one million is only $\ln 10^6 \approx 13.8$ , but it never stops. Close to the boundary the outcomes split. The power $p = 1.1$ gives a finite area of $10$ , while $p = 0.9$ gives an infinite one.

Remaining tail

Definition Suppose $\int_a^\infty f(x)\,dx$ converges. For $R \ge a$ , its remaining tail beyond $R$ is

\begin{aligned} \int_R^\infty f(x)\,dx &= \int_a^\infty f(x)\,dx \\[0.4em] &\quad - \int_a^R f(x)\,dx. \end{aligned}

The tail tends to $0$ as $R \to \infty$ , which restates convergence. For the p-test with $p > 1$ and every $R > 0$ ,

\int_R^\infty x^{-p}\,dx = \frac{R^{1-p}}{p - 1}.

In plain words The tail is the part of the area that a computation stopped at $R$ leaves out. It says how far out to go for a required accuracy, which a graph cut off at some point cannot show.

Example For $p = 2$ the total is $1$ and the tail is $1/R$ , so stopping at $R = 10$ leaves out a tenth of the total and stopping at $R = 100$ leaves out a hundredth. Near the boundary the tail shrinks very slowly. For $p = 1.1$ the total is $10$ and the tail is $10R^{-0.1}$ . At $R = 1000$ the tail is still about $5.01$ , more than half of the total, and it falls to $1$ only at $R = 10^{10}$ .

Check yourself

Decide whether $\int_1^\infty \frac{4}{x^3}\,dx$ converges, and find its value if it does.

Show the answer

For $R > 1$ ,

\begin{aligned} \int_1^R \frac{4}{x^3}\,dx &= \Bigl[-\frac{2}{x^2}\Bigr]_1^R \\[0.4em] &= 2 - \frac{2}{R^2}, \end{aligned}

which approaches $2$ as $R \to \infty$ . The integral converges and equals $2$ . This agrees with the p-test for $p = 3$ , which gives $4 \cdot \frac{1}{3 - 1} = 2$ .

Find $\int_1^\infty x^{-3/2}\,dx$ . For which $R$ is the remaining tail beyond $R$ less than $0.01$ ?

Show the answer

Here $p = \tfrac32 > 1$ , so the integral converges, with value $\frac{1}{3/2 - 1} = 2$ . The tail beyond $R$ is

\int_R^\infty x^{-3/2}\,dx = \frac{R^{-1/2}}{1/2} = \frac{2}{\sqrt{R}}.

It is less than $0.01$ exactly when $\sqrt{R} > 200$ , that is, when $R > 40000$ . So the integral up to $R$ comes within $0.01$ of the total $2$ only once $R$ passes $40000$ .

A classmate uses the formula $\frac{1}{p-1}$ for $\int_1^\infty x^{-2/3}\,dx$ and gets $-3$ . What is wrong, and what is the correct conclusion?

Show the answer

The formula holds only for $p > 1$ , and here $p = \tfrac23$ . The negative result is already a warning sign, because the integrand is positive. Working from the definition, for $R > 1$

\begin{aligned} \int_1^R x^{-2/3}\,dx &= \Bigl[3x^{1/3}\Bigr]_1^R \\[0.4em] &= 3R^{1/3} - 3, \end{aligned}

which grows without bound. At $R = 1000$ it already equals $27$ . The integral diverges to $+\infty$ and has no value. The condition $p > 1$ is part of the result, not a detail that can be dropped.

When an infinite region has finite area

Key terms

Improper integral

Convergent integral

Divergent integral

The p-test for integrals

Boundary case p equals 1

Remaining tail

Check yourself

How this lesson connects

Used later in