Improper integrals

AS · P1 · 10 min

Can a region that stretches off to infinity have a finite area? Surprisingly, often yes. An integral is called improper when one of its limits is infinite, or when the function being integrated becomes infinite at one of the limits. The syllabus includes "simple cases of 'improper' integrals", with $\displaystyle\int_0^1 x^{-\frac{1}{2}}\, dx$ and $\displaystyle\int_1^\infty x^{-2}\, dx$ as its examples. The method is short: replace the awkward limit with a letter, integrate as usual, then see what happens as the letter approaches the awkward value. These usually appear as one part of an area or volume question.

Infinite regions with finite area

Look at the curve $y = \dfrac{1}{x^2}$ for $x \ge 1$ . The region under it never ends, but the curve falls towards the $x$ -axis very quickly.

y = 1/x^2 fill 1 8 y = 1/x^2 x = 1

Find the area from $x = 1$ up to some large value $x = N$ :

\int_1^N x^{-2}\, dx = \Big[-x^{-1}\Big]_1^N = -\frac{1}{N} + 1 = 1 - \frac{1}{N}

Try a few values: $N = 10$ gives $0.9$ , $N = 100$ gives $0.99$ , $N = 1000$ gives $0.999$ . As $N \to \infty$ , $\dfrac{1}{N} \to 0$ , so the area approaches $1$ . However far right you go, the total area never exceeds $1$ . This is the same idea as the sum to infinity of a geometric progression: infinitely many pieces with a finite total.

Definition

For an infinite upper limit,

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

If the limit is a finite number, the integral converges (it "exists") and has that value. If the expression grows without bound, the integral diverges and has no value.

When the integrand is infinite at a limit

Now look at $y = \dfrac{1}{\sqrt{x}}$ between $x = 0$ and $x = 1$ . At $x = 0$ the function is undefined, and the curve shoots up the $y$ -axis. You cannot substitute $x = 0$ into anything that involves $x^{-\frac{1}{2}}$ .

y = 1/sqrt(x) fill 0.01 1 y = 1/sqrt(x) x = 1

Replace the bad limit $0$ by a small positive number $\varepsilon$ :

\int_\varepsilon^1 x^{-\frac{1}{2}}\, dx = \Big[2x^{\frac{1}{2}}\Big]_\varepsilon^1 = 2 - 2\sqrt{\varepsilon}

As $\varepsilon \to 0$ , $\sqrt{\varepsilon} \to 0$ , so the integral approaches $2$ . The region is infinitely tall but has area $2$ .

Definition

If $f(x)$ is undefined at $x = a$ but defined for $a < x \le b$ ,

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

and similarly for a function undefined at the upper limit $b$ .

In practice, once you have integrated, you will often find that the integral of the problem function is perfectly well-behaved at the bad point: $2x^{\frac{1}{2}}$ is $0$ at $x = 0$ , even though $x^{-\frac{1}{2}}$ is undefined there. That is a sign the integral converges.

Divergent integrals

Not every improper integral has a value.

\int_1^N x^{-\frac{1}{2}}\, dx = \Big[2x^{\frac{1}{2}}\Big]_1^N = 2\sqrt{N} - 2

As $N \to \infty$ , $2\sqrt{N}$ grows without limit, so $\displaystyle\int_1^\infty x^{-\frac{1}{2}}\, dx$ diverges. The curve $y = x^{-\frac{1}{2}}$ falls too slowly for the area to settle down.

Similarly, $\displaystyle\int_\varepsilon^1 x^{-2}\, dx = \dfrac{1}{\varepsilon} - 1$ , which grows without bound as $\varepsilon \to 0$ , so $\displaystyle\int_0^1 x^{-2}\, dx$ diverges: $\dfrac{1}{x^2}$ shoots up too fast near $0$ .

The pattern for powers

For a positive constant $p$ :

Integral	Converges when	Value
$\displaystyle\int_1^\infty x^{-p}\, dx$	$p > 1$	$\dfrac{1}{p - 1}$
$\displaystyle\int_0^1 x^{-p}\, dx$	$p < 1$	$\dfrac{1}{1 - p}$

Use this to check answers and to predict whether an integral exists. In the exam, always show the working with a limit, not just the table.

The case $p = 1$ (the function $\dfrac{1}{x}$ ) diverges both ways, but proving that needs logarithms, which are in Paper 3.

Evaluating an improper integral

Identify the problem: an infinite limit, or a limit where the integrand is undefined.
Replace that limit by a letter ( $N$ for $\infty$ , or $\varepsilon$ for the bad point).
Integrate and substitute the limits as usual, keeping the letter.
State what each term involving the letter tends to, for example "as $N \to \infty$ , $\dfrac{1}{N} \to 0$ ".
Give the value, or state that the integral diverges because a term grows without bound.

Worked examples

An infinite limit

Evaluate $\displaystyle\int_2^\infty \frac{3}{x^2}\, dx$ .

Solution

\int_2^N 3x^{-2}\, dx = \Big[-3x^{-1}\Big]_2^N = -\frac{3}{N} + \frac{3}{2}

As $N \to \infty$ , $\dfrac{3}{N} \to 0$ , so

\int_2^\infty \frac{3}{x^2}\, dx = \frac{3}{2}

An infinite integrand

Evaluate $\displaystyle\int_0^8 x^{-\frac{1}{3}}\, dx$ .

Solution

$x^{-\frac{1}{3}}$ is undefined at $x = 0$ . Using a lower limit $\varepsilon$ :

\int_\varepsilon^8 x^{-\frac{1}{3}}\, dx = \left[\tfrac{3}{2}x^{\frac{2}{3}}\right]_\varepsilon^8 = \tfrac{3}{2}(4) - \tfrac{3}{2}\varepsilon^{\frac{2}{3}} = 6 - \tfrac{3}{2}\varepsilon^{\frac{2}{3}}

As $\varepsilon \to 0$ , $\varepsilon^{\frac{2}{3}} \to 0$ , so the integral is $6$ .

A divergent integral

Show that $\displaystyle\int_1^\infty \frac{1}{\sqrt{x}}\, dx$ does not have a finite value.

Solution

\int_1^N x^{-\frac{1}{2}}\, dx = \Big[2\sqrt{x}\Big]_1^N = 2\sqrt{N} - 2

As $N \to \infty$ , $2\sqrt{N} \to \infty$ , so the integral increases without bound. It diverges and has no finite value.

A linear bracket

Evaluate (a) $\displaystyle\int_0^\infty \frac{1}{(2x + 1)^2}\, dx$ and (b) $\displaystyle\int_1^5 \frac{1}{\sqrt{x - 1}}\, dx$ .

Solution

(a) Using the $(ax + b)^n$ rule, $\displaystyle\int (2x + 1)^{-2}\, dx = \dfrac{(2x + 1)^{-1}}{2 \times (-1)} = -\dfrac{1}{2(2x + 1)}$ .

\int_0^N \frac{1}{(2x + 1)^2}\, dx = -\frac{1}{2(2N + 1)} + \frac{1}{2}

As $N \to \infty$ , the first term tends to $0$ , so the integral is $\dfrac{1}{2}$ .

(b) $\dfrac{1}{\sqrt{x - 1}}$ is undefined at $x = 1$ . $\displaystyle\int (x - 1)^{-\frac{1}{2}}\, dx = 2(x - 1)^{\frac{1}{2}}$ , so

\int_\varepsilon^5 \frac{1}{\sqrt{x - 1}}\, dx = 2\sqrt{4} - 2\sqrt{\varepsilon - 1}

with $\varepsilon$ slightly bigger than $1$ . As $\varepsilon \to 1$ , $\sqrt{\varepsilon - 1} \to 0$ , so the integral is $4$ .

Finding a limit from the area

The region bounded by the curve $y = \dfrac{8}{x^3}$ , the $x$ -axis and the line $x = k$ (where $k > 0$ ), extending infinitely to the right, has area $1$ . Find $k$ .

Solution

\int_k^N 8x^{-3}\, dx = \Big[-4x^{-2}\Big]_k^N = -\frac{4}{N^2} + \frac{4}{k^2}

As $N \to \infty$ , $\dfrac{4}{N^2} \to 0$ , so the area is $\dfrac{4}{k^2}$ . Setting $\dfrac{4}{k^2} = 1$ gives $k^2 = 4$ , and $k = 2$ .

An infinite solid

The region under the curve $y = \dfrac{1}{x^2}$ for $x \ge 1$ is rotated through $360^\circ$ about the $x$ -axis. Find the volume of the solid formed.

Solution

Using the formula from volumes of revolution, $V = \pi\displaystyle\int y^2\, dx$ , with $y^2 = x^{-4}$ :

\pi\int_1^N x^{-4}\, dx = \pi\left[-\frac{1}{3}x^{-3}\right]_1^N = \pi\left(\frac{1}{3} - \frac{1}{3N^3}\right)

As $N \to \infty$ , $\dfrac{1}{3N^3} \to 0$ , so the volume is $\dfrac{\pi}{3}$ .

Watch out

Substituting $\infty$ as if it were a number. Writing " $-\dfrac{3}{\infty} = 0$ " is not acceptable working. Use a letter and state what happens as it tends to infinity.

Not noticing the integral is improper. $\displaystyle\int_0^4 \frac{1}{\sqrt{x}}\, dx$ looks ordinary, but the integrand is undefined at $0$ . Check the lower limit before integrating.

Claiming a value for a divergent integral. If a term like $2\sqrt{N}$ appears, the integral has no value. Say it diverges.

Getting a negative area. If $\displaystyle\int_1^\infty x^{-2}\, dx$ comes out as $-1$ , you have subtracted the wrong way round: (upper) $-$ (lower), with the upper limit being $N$ .

Assuming an infinite region always has infinite area. It often does not, as $\dfrac{1}{x^2}$ shows.

Exam tip

Show the limiting process. Mark schemes typically award a method mark for integrating correctly, then an accuracy mark for the value with a statement like "as $N \to \infty$ , $\dfrac{1}{N} \to 0$ ". Some accept substituting directly if you write that the term " $\to 0$ ", but the letter method is always safe.
Improper integrals are often a final part of an area or volume question ("the region extends to infinity", "find the area of the region, which is unbounded"). Spot the words "infinitely" or "unbounded".
Check whether the curve has an asymptote at a limit: a vertical asymptote at a limit makes the integral improper.
Exact values: improper integrals in Paper 1 usually come out as simple fractions or multiples of $\pi$ .

Summary

An integral is improper if a limit is infinite, or the integrand is undefined at a limit.
Replace the problem limit by a letter, integrate, then let the letter tend to the problem value.
If the result tends to a finite value, the integral converges to that value; otherwise it diverges.
$\displaystyle\int_1^\infty x^{-2}\, dx = 1$ and $\displaystyle\int_0^1 x^{-\frac{1}{2}}\, dx = 2$ are the syllabus examples.
$\displaystyle\int_1^\infty x^{-p}\, dx$ converges only for $p > 1$ ; $\displaystyle\int_0^1 x^{-p}\, dx$ only for $p < 1$ .
State limits in words or symbols ("as $N \to \infty$ , $\dfrac{1}{N} \to 0$ "); do not treat $\infty$ as a number.

Practice questions

Question

Evaluate $\displaystyle\int_1^\infty x^{-\frac{3}{2}}\, dx$ .
Evaluate $\displaystyle\int_0^4 \frac{1}{\sqrt{x}}\, dx$ .
Evaluate $\displaystyle\int_1^\infty \left(\frac{1}{x^2} + \frac{1}{x^3}\right) dx$ .
Show that $\displaystyle\int_0^1 \frac{1}{x^2}\, dx$ does not exist.
Evaluate $\displaystyle\int_0^\infty \frac{4}{(2x + 1)^3}\, dx$ .
Find the area of the region bounded by the curve $y = \dfrac{1}{\sqrt{4 - x}}$ , the axes and the line $x = 4$ .
The region bounded by $y = \dfrac{k}{x^2}$ , the $x$ -axis and the line $x = 3$ , extending infinitely to the right, has area $2$ . Find $k$ .
(a) Show that the region under the curve $y = x^{-\frac{3}{4}}$ for $x \ge 1$ does not have a finite area. (b) Show that, when this region is rotated through $360^\circ$ about the $x$ -axis, the volume of the solid formed is finite, and find it.
The region under the curve $y = x^{-\frac{2}{3}}$ between $x = 0$ and $x = 1$ is unbounded. (a) Find its area. (b) Show that the solid formed by rotating this region through $360^\circ$ about the $x$ -axis does not have a finite volume.

Answers

$\displaystyle\int_1^N x^{-\frac{3}{2}}\, dx = \Big[-2x^{-\frac{1}{2}}\Big]_1^N = -\dfrac{2}{\sqrt{N}} + 2$ . As $N \to \infty$ , $\dfrac{2}{\sqrt{N}} \to 0$ , so the value is $2$ .
$\displaystyle\int_\varepsilon^4 x^{-\frac{1}{2}}\, dx = \Big[2\sqrt{x}\Big]_\varepsilon^4 = 4 - 2\sqrt{\varepsilon} \to 4$ as $\varepsilon \to 0$ . The value is $4$ .
$\displaystyle\int_1^N \left(x^{-2} + x^{-3}\right) dx = \left[-x^{-1} - \tfrac{1}{2}x^{-2}\right]_1^N = -\dfrac{1}{N} - \dfrac{1}{2N^2} + 1 + \dfrac{1}{2}$ . As $N \to \infty$ the terms in $N$ tend to $0$ , giving $\tfrac{3}{2}$ .
$\displaystyle\int_\varepsilon^1 x^{-2}\, dx = \Big[-x^{-1}\Big]_\varepsilon^1 = -1 + \dfrac{1}{\varepsilon}$ . As $\varepsilon \to 0$ , $\dfrac{1}{\varepsilon} \to \infty$ , so the integral diverges and does not exist.
$\displaystyle\int 4(2x + 1)^{-3}\, dx = \dfrac{4(2x + 1)^{-2}}{2 \times (-2)} = -\dfrac{1}{(2x + 1)^2}$ . So $\displaystyle\int_0^N = -\dfrac{1}{(2N + 1)^2} + 1 \to 1$ as $N \to \infty$ . The value is $1$ .
The integrand is undefined at $x = 4$ . $\displaystyle\int (4 - x)^{-\frac{1}{2}}\, dx = \dfrac{(4 - x)^{\frac{1}{2}}}{(-1) \times \tfrac{1}{2}} = -2\sqrt{4 - x}$ . With upper limit $\varepsilon$ slightly less than $4$ : $\Big[-2\sqrt{4 - x}\Big]_0^\varepsilon = -2\sqrt{4 - \varepsilon} + 4 \to 4$ as $\varepsilon \to 4$ . The area is $4$ .
$\displaystyle\int_3^N kx^{-2}\, dx = \Big[-kx^{-1}\Big]_3^N = -\dfrac{k}{N} + \dfrac{k}{3} \to \dfrac{k}{3}$ . So $\dfrac{k}{3} = 2$ and $k = 6$ .
(a) $\displaystyle\int_1^N x^{-\frac{3}{4}}\, dx = \Big[4x^{\frac{1}{4}}\Big]_1^N = 4N^{\frac{1}{4}} - 4$ , which tends to infinity as $N \to \infty$ . The area is not finite. (b) $V = \pi\displaystyle\int_1^N x^{-\frac{3}{2}}\, dx = \pi\left(2 - \dfrac{2}{\sqrt{N}}\right) \to 2\pi$ as $N \to \infty$ . The volume is $2\pi$ : an infinite area that sweeps out a finite volume.
(a) $\displaystyle\int_\varepsilon^1 x^{-\frac{2}{3}}\, dx = \Big[3x^{\frac{1}{3}}\Big]_\varepsilon^1 = 3 - 3\varepsilon^{\frac{1}{3}} \to 3$ as $\varepsilon \to 0$ . The area is $3$ . (b) $V = \pi\displaystyle\int_\varepsilon^1 x^{-\frac{4}{3}}\, dx = \pi\Big[-3x^{-\frac{1}{3}}\Big]_\varepsilon^1 = \pi\left(-3 + \dfrac{3}{\varepsilon^{1/3}}\right)$ . As $\varepsilon \to 0$ , $\dfrac{3}{\varepsilon^{1/3}} \to \infty$ , so the volume is not finite: here a finite area sweeps out an infinite volume.

Infinite regions with finite area

When the integrand is infinite at a limit

Divergent integrals

Worked examples

Practice questions

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