Type I Infinite Limits Of Integration

An integral is Type I when one or both limits of integration are infinite. We give such an integral meaning by replacing the infinite endpoint with a finite parameter, integrating in the ordinary way, and then letting that parameter run to infinity. If the resulting limit exists and is finite the integral converges; otherwise it diverges.

Item	Statement
Type I improper integral	A definite integral in which at least one limit of integration is $+ \infty$ or $- \infty$
Case (a), upper limit infinite	$\int_{a}^{\infty} f (x) d x = lim_{b \to \infty} \int_{a}^{b} f (x) d x$
Case (b), lower limit infinite	$\int_{- \infty}^{b} f (x) d x = lim_{a \to - \infty} \int_{a}^{b} f (x) d x$
Case (c), both limits infinite	$\int_{- \infty}^{\infty} f = \int_{- \infty}^{c} f + \int_{c}^{\infty} f$ , and both pieces must converge on their own
Converges or diverges	Converges if the limit exists and is finite; diverges otherwise, including when the limit is $\pm \infty$
Choice of the split point $c$	Any finite $c$ works, and every choice gives the same value
Classic trap	$lim_{b \to \infty} \int_{- b}^{b} f (x) d x$ is not the definition; that symmetric limit is the Cauchy principal value
Benchmark that converges	$\int_{1}^{\infty} \frac{d x}{x^{2}} = 1$
Benchmark that diverges	$\int_{1}^{\infty} \frac{d x}{x} = \infty$
Over the whole real line	$\int_{- \infty}^{\infty} \frac{d x}{1 + x^{2}} = π$
Exponential decay	$\int_{a}^{\infty} e^{- k x} d x = \frac{e^{- k a}}{k}$ for $k > 0$
Factorial integral	$\int_{0}^{\infty} x^{n} e^{- x} d x = n!$ for every positive integer $n$
Preview of the $p$ -test	$\int_{1}^{\infty} \frac{d x}{x^{p}}$ converges exactly when $p > 1$
Limits you will keep needing	$lim_{b \to \infty} b^{n} e^{- b} = 0$ and $lim_{b \to \infty} \frac{\ln b}{b} = 0$

The Definition

We extend one or both limits of integration to $\pm \infty$ by replacing the infinite endpoint with a finite parameter and taking a limit. If the limit exists and is finite the integral converges; otherwise it diverges.

The letters are worth fixing before we start. Throughout, $f$ is the integrand, $a$ and $b$ are the two limits of integration, and whichever of them stands in for an infinite endpoint is the parameter we push to infinity. In case (c) the extra letter $c$ is a finite number of our own choosing, used only to cut the real line into two halves.

If $f$ is continuous on $[a, \infty)$ , we define
\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int_a^\infty f(x)\,dx = \lim_{b\to\infty}\int_a^b f(x)\,dx,}
provided this limit exists and is finite.
If $f$ is continuous on $(- \infty, b]$ , we define
\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int_{-\infty}^b f(x)\,dx = \lim_{a\to-\infty}\int_a^b f(x)\,dx,}
provided this limit exists and is finite.
If $f$ is continuous on $(- \infty, \infty)$ , we pick any convenient number $c$ and define
\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int_{-\infty}^\infty f(x)\,dx = \int_{-\infty}^c f(x)\,dx + \int_c^\infty f(x)\,dx,}
provided both integrals on the right converge.

If the limit is finite, the integral converges; otherwise it diverges.

Read each of these as an instruction rather than as a formula: never integrate "to infinity" directly. Integrate to a finite endpoint first, get an honest number that depends on that endpoint, and only then take the limit. Case (b) is the mirror image of case (a), reflected through the $y$ -axis, so anything you learn about one transfers immediately to the other.

How to Evaluate a Type I Integral

Check continuity. Confirm that $f$ is continuous on the entire interval. If $f$ also blows up at some point of the interval, the integral is improper of Type II as well and must be split at that point first, before anything below applies.
Name the infinite endpoint. Replace $+ \infty$ by a letter $b$ and $- \infty$ by a letter $a$ . If both endpoints are infinite, split at a convenient finite $c$ first and handle the two halves separately.
Do an ordinary definite integral. Evaluate $\int_{a}^{b} f (x) d x$ by substitution, by parts, or by any other method. The answer is a formula in $a$ or $b$ .
Take the limit. Let $b \to \infty$ or $a \to - \infty$ . A finite limit means the integral converges to that number. An infinite limit, or no limit at all, means it diverges.

Step 3 is ordinary calculus. All the new content of this section lives in steps 2 and 4.

Reading the Definition Geometrically

Each of the three cases has a picture attached to it, and the pictures are the fastest way to remember which endpoint is moving.

Case (a). The left endpoint $a$ is nailed down and the right endpoint $b$ slides off to $+ \infty$ . The shaded strip is the ordinary integral $\int_{a}^{b} f (x) d x$ ; the question is whether its area settles down to a number as the strip grows without bound.

Case (a): $\int_{a}^{\infty} f (x) d x = lim_{b \to + \infty} \int_{a}^{b} f (x) d x$ . The shaded area equals $\int_{a}^{b} f (x) d x$ , and here it converges as $b \to + \infty$ .

Case (b). Now the right endpoint $b$ is fixed and the left endpoint $a$ slides off to $- \infty$ . Note that $a$ starts out to the left of the $y$ -axis and keeps moving left.

Case (b): $\int_{- \infty}^{b} f (x) d x = lim_{a \to - \infty} \int_{a}^{b} f (x) d x$ . Here $a < 0$ , to the left of the $y$ -axis, and it moves further toward $- \infty$ .

Case (c). Both endpoints run away at once, so we cut the line at a finite point $c$ and ask each half to converge separately. The picture below deliberately puts $c = 1$ rather than $c = 0$ , to make the point that $c$ is ours to choose, and it uses an $f$ that changes sign, to make the point that the definition never assumed $f \geq 0$ .

Case (c): $\int_{- \infty}^{\infty} f (x) d x$ , split at $c = 1$ . The function shown is negative just to the left of the origin and positive just to the right of it, so the two halves contribute areas of opposite sign. The definition applies to any continuous $f$ , not only to $f \geq 0$ .

Three remarks follow from these pictures.

The value of $c$ in part (c) does not matter, as long as both halves converge. Any finite value of $c$ gives the same integral, because moving $c$ transfers a single ordinary, finite integral from one half to the other.
If $f (x) \geq 0$ on $[a, \infty)$ , the improper integral $\int_{a}^{\infty} f (x) d x$ is the area of the unbounded region under $y = f (x)$ to the right of $x = a$ . It may seem surprising that such a region can have finite area, but the first example below shows that it can.
Nothing in the definition requires $f \geq 0$ . When $f$ changes sign, the integral is still a limit of ordinary signed areas.

A Warning About the Symmetric Limit

A common mistake is to evaluate $\int_{- \infty}^{\infty} f (x) d x$ by the single symmetric limit

lim_{b \to \infty} \int_{- b}^{b} f (x) d x .

This is not the definition, and it can produce a finite value even when the integral diverges. For example,

lim_{b \to \infty} \int_{- b}^{b} x d x = 0,

because $\int_{- b}^{b} x d x = \frac{b^{2}}{2} - \frac{b^{2}}{2} = 0$ for every single $b$ . Yet $\int_{0}^{\infty} x d x$ diverges, since $\int_{0}^{b} x d x = b^{2} / 2 \to \infty$ . The definition above requires both halves to converge independently, so $\int_{- \infty}^{\infty} x d x$ diverges.

The symmetric limit hides the divergence by cancelling a growing positive area against a growing negative area at exactly the same rate. If you pair the endpoints differently the cancellation fails: $\int_{- b}^{2 b} x d x = \frac{3 b^{2}}{2} \to + \infty$ . A quantity that changes when you change an arbitrary bookkeeping choice cannot be the value of the integral. The symmetric limit does have a name and a legitimate use: it is the Cauchy principal value, treated in the section on the Cauchy principal value.

Worked Examples

The first two examples are the pair every later test is calibrated against: two curves that both shrink to zero, only one of which encloses finite area.

Evaluate $\int_{1}^{\infty} \frac{d x}{x^{2}}$ .

Solution

\begin{aligned} \int_1^\infty \frac{dx}{x^2} &= \lim_{b\to\infty}\int_1^b x^{-2}\,dx \\ &= \lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^b \\ &= \lim_{b\to\infty}\left(1 - \frac{1}{b}\right) \\ &= 1. \end{aligned}

The integral converges to $1$ . Geometrically, the region under $y = 1 / x^{2}$ stretching infinitely far to the right has total area exactly $1$ : the piece beyond $x = b$ contributes only $1 / b$ , which we can make as small as we please.

Evaluate $\int_{1}^{\infty} \frac{d x}{x}$ .

Solution

\begin{aligned} \int_1^\infty \frac{dx}{x} &= \lim_{b\to\infty}\int_1^b \frac{dx}{x} \\ &= \lim_{b\to\infty}\bigl[\ln x\bigr]_1^b \\ &= \lim_{b\to\infty}\ln b = \infty. \end{aligned}

The integral diverges. The area accumulates without bound, although it does so very slowly: reaching an area of $10$ requires going out to $b = e^{10} \approx 22026$ .

The two integrals just computed have integrands that look nearly identical near $x = 1$ , yet one converges and the other does not. Explain the difference, and illustrate it.

Solution

The curves $y = 1 / x^{2}$ and $y = 1 / x$ both pass through $(1, 1)$ and both fall to zero as $x \to \infty$ , yet the first gives a finite area and the second does not. The difference is not whether the integrand tends to zero: both do. The difference is how fast each decays as $x \to \infty$ .

The tail estimates make this quantitative. Beyond $x = b$ the two remaining areas are

\int_{b}^{\infty} \frac{d x}{x^{2}} = \frac{1}{b}, \int_{b}^{B} \frac{d x}{x} = \ln \frac{B}{b} .

The first tail shrinks to zero as $b$ grows; the second never does, because $\ln (B / b)$ can be made arbitrarily large by taking $B$ large enough, no matter how far out $b$ already is. We will quantify the general pattern precisely in the section on the $p$ -integrals, where the $p$ -test states that $\int_{1}^{\infty} d x / x^{p}$ converges exactly when $p > 1$ .

Both curves pass through $(1, 1)$ and look nearly identical near $x = 1$ , yet only $1 / x^{2}$ has finite area over $[1, \infty)$ .

The next example is the standard doubly infinite integral, and it shows case (c) in action: split first, then take two separate limits.

Evaluate $\int_{- \infty}^{\infty} \frac{d x}{1 + x^{2}}$ .

Solution

Split at $c = 0$ :

\int_{- \infty}^{\infty} \frac{d x}{1 + x^{2}} = \int_{- \infty}^{0} \frac{d x}{1 + x^{2}} + \int_{0}^{\infty} \frac{d x}{1 + x^{2}} .

For the right half:

\begin{aligned} \int_0^\infty \frac{dx}{1+x^2} &= \lim_{b\to\infty}\bigl[\arctan x\bigr]_0^b \\ &= \lim_{b\to\infty}\arctan b \\ &= \frac{\pi}{2}. \end{aligned}

For the left half:

\begin{aligned} \int_{-\infty}^0 \frac{dx}{1+x^2} &= \lim_{a\to-\infty}\bigl[\arctan x\bigr]_a^0 \\ &= 0 - \lim_{a\to-\infty}\arctan a \\ &= 0 - \left(-\frac{\pi}{2}\right) \\ &= \frac{\pi}{2}. \end{aligned}

Both halves converge, so

\int_{- \infty}^{\infty} \frac{d x}{1 + x^{2}} = \frac{π}{2} + \frac{π}{2} = π .

The limits $lim_{x \to + \infty} \arctan x = \frac{π}{2}$ and $lim_{x \to - \infty} \arctan x = - \frac{π}{2}$ are visible as the horizontal asymptotes of $y = \arctan x$ . They are why the two halves of the integral each equal $π / 2$ .

Integrals That Need Integration by Parts

Step 3 of the procedure can be any technique at all. The two examples below are the ones worth practising, because integration by parts leaves behind boundary terms such as $- b e^{- b}$ that must themselves be sent to a limit. The rule of thumb is: carry the finite endpoint $b$ through every line, and take the limit only at the very end.

Evaluate $\int_{0}^{\infty} x e^{- x} d x$ .

Solution

Integrate by parts with $u = x$ and $d v = e^{- x} d x$ , so $d u = d x$ and $v = - e^{- x}$ . We first compute the integral over $[0, b]$ and then take the limit:

\begin{aligned} \int_0^b x\,e^{-x}\,dx &= \bigl[-xe^{-x}\bigr]_0^b + \int_0^b e^{-x}\,dx \\ &= -be^{-b} + 0 + \bigl[-e^{-x}\bigr]_0^b \\ &= -be^{-b} - e^{-b} + 1. \end{aligned}

The boundary term needs its own limit. By L'Hôpital's rule applied to $b / e^{b}$ , which is of the form $\infty / \infty$ ,

lim_{b \to \infty} b e^{- b} = lim_{b \to \infty} \frac{b}{e^{b}} = lim_{b \to \infty} \frac{1}{e^{b}} = 0,

and of course $e^{- b} \to 0$ as well. Therefore

\begin{aligned} \int_0^\infty x\,e^{-x}\,dx &= \lim_{b\to\infty}\left(1 - be^{-b} - e^{-b}\right) \\ &= 1. \end{aligned}

More generally, repeated integration by parts gives

\int_{0}^{\infty} x^{n} e^{- x} d x = n!

for every positive integer $n$ . Each application of integration by parts peels one factor of $x$ off the front and multiplies by the exponent, which is exactly how a factorial is built up.

This integral defines the Gamma function: $Γ (n + 1) = \int_{0}^{\infty} x^{n} e^{- x} d x$ . For positive integers $Γ (n + 1) = n!$ , but $Γ$ is also defined for all positive real numbers, extending the factorial to a continuous function. It is indispensable in probability, combinatorics, and mathematical physics.

The next example uses a trick worth naming: when integrating by parts twice returns the original integral, treat the integral as an unknown and solve for it algebraically.

Evaluate $\int_{0}^{\infty} e^{- x} \cos x d x$ .

Solution

Call the integral $I$ . We integrate by parts twice on the finite integral $\int_{0}^{b}$ , then take the limit as $b \to \infty$ .

First integration by parts. Let $u = \cos x$ and $d v = e^{- x} d x$ , so $d u = - \sin x d x$ and $v = - e^{- x}$ :

\int_{0}^{b} e^{- x} \cos x d x = [- e^{- x} \cos x]_{0}^{b} - \int_{0}^{b} e^{- x} \sin x d x .

Second integration by parts. Apply integration by parts to the remaining integral with $u = \sin x$ and $d v = e^{- x} d x$ , so $d u = \cos x d x$ and $v = - e^{- x}$ :

\int_{0}^{b} e^{- x} \sin x d x = [- e^{- x} \sin x]_{0}^{b} + \int_{0}^{b} e^{- x} \cos x d x .

Solving for the integral. Substituting the second result into the first:

\begin{aligned} \int_0^b e^{-x}\cos x\,dx &= \bigl[-e^{-x}\cos x\bigr]_0^b - \bigl[-e^{-x}\sin x\bigr]_0^b \\ &\qquad - \int_0^b e^{-x}\cos x\,dx. \end{aligned}

The same integral now appears on both sides, so move it to the left:

\begin{aligned} 2\int_0^b e^{-x}\cos x\,dx &= \Bigl[-e^{-x}(\cos x-\sin x)\Bigr]_0^b \\ &= -e^{-b}(\cos b-\sin b) + 1. \end{aligned}

Since $| \cos b - \sin b | \leq 2$ , we have $| e^{- b} (\cos b - \sin b) | \leq 2 e^{- b} \to 0$ . Taking $b \to \infty$ gives $2 I = 1$ , so

\int_{0}^{\infty} e^{- x} \cos x d x = \frac{1}{2} .

The oscillation of $\cos x$ never causes trouble here because the factor $e^{- x}$ crushes it: the integrand is trapped between $- e^{- x}$ and $e^{- x}$ , both of which have convergent integrals.

Exercises

Evaluate each of the following integrals, or show that it diverges.

$\int_{0}^{\infty} e^{- x} d x$
$\int_{- \infty}^{0} e^{x} d x$
$\int_{2}^{\infty} \frac{d x}{x^{3}}$
$\int_{1}^{\infty} \frac{d x}{\sqrt{x}}$

Answer

(a) $1$
(b) $1$
(c) $\frac{1}{8}$
(d) Diverges

Solution

(a) Integrate to $b$ first, then let $b \to \infty$ :

\begin{aligned} \int_0^\infty e^{-x}\,dx &= \lim_{b\to\infty}\bigl[-e^{-x}\bigr]_0^b \\ &= \lim_{b\to\infty}\left(1 - e^{-b}\right) \\ &= 1. \end{aligned}

(b) By definition:

\begin{aligned} \int_{-\infty}^0 e^{x}\,dx &= \lim_{a\to-\infty}\int_a^0 e^{x}\,dx \\ &= \lim_{a\to-\infty}\bigl[e^{x}\bigr]_a^0 \\ &= \lim_{a\to-\infty}(1 - e^{a}) \\ &= 1. \end{aligned}

This is the mirror image of part (a), as case (b) of the definition always is.

(c) An antiderivative of $x^{- 3}$ is $- \frac{1}{2 x^{2}}$ , which tends to $0$ as $x \to \infty$ , so a finite limit is in prospect:

\begin{aligned} \int_2^\infty \frac{dx}{x^3} &= \lim_{b\to\infty}\left[-\frac{1}{2x^2}\right]_2^b \\ &= \lim_{b\to\infty}\left(\frac{1}{8} - \frac{1}{2b^2}\right) \\ &= \frac{1}{8}. \end{aligned}

(d) Here $1 / \sqrt{x} = x^{- 1 / 2}$ , whose antiderivative is $2 \sqrt{x}$ . Unlike the antiderivative in part (c), this one grows without bound, so we should expect trouble:

\begin{aligned} \int_1^b \frac{dx}{\sqrt{x}} &= \bigl[2\sqrt{x}\bigr]_1^b \\ &= 2\sqrt{b} - 2. \end{aligned}

Since $2 \sqrt{b} - 2 \to \infty$ as $b \to \infty$ , the integral diverges. The integrand does tend to zero, but far too slowly.

Evaluate $\int_{0}^{\infty} x e^{- x^{2}} d x$ .

Hint: Let $u = x^{2}$ .

Answer

$\frac{1}{2}$ .

Solution

By definition, $\int_{0}^{\infty} x e^{- x^{2}} d x = lim_{b \to + \infty} \int_{0}^{b} x e^{- x^{2}} d x .$

We first evaluate the integral over the finite interval $[0, b]$ . Let $u = x^{2}$ ; then $d u = 2 x d x$ , so $x d x = \frac{1}{2} d u$ . When $x = 0$ , $u = 0$ ; when $x = b$ , $u = b^{2}$ . Substituting:

\begin{aligned} \int_0^b x\,e^{-x^2}\,dx &= \int_0^{b^2} e^{-u}\cdot\frac{du}{2} \\ &= \frac{1}{2}\int_0^{b^2} e^{-u}\,du \\ &= \frac{1}{2}\Bigl[-e^{-u}\Bigr]_0^{b^2} \\ &= \frac{1}{2}\bigl(1-e^{-b^2}\bigr). \end{aligned}

Since $e^{- b^{2}} \to 0$ as $b \to + \infty$ , we obtain

\int_{0}^{\infty} x e^{- x^{2}} d x = lim_{b \to + \infty} \frac{1}{2} (1 - e^{- b^{2}}) = \frac{1}{2} .

Show that $\int_{1}^{\infty} \sin x d x$ diverges.

Answer

Diverges.

Solution

Computing the antiderivative over the finite interval $[1, b]$ :

\begin{aligned} \int_1^b \sin x\,dx &= \bigl[-\cos x\bigr]_1^b \\ &= -\cos b + \cos 1 \\ &= \cos 1 - \cos b. \end{aligned}

As $b \to \infty$ , $\cos b$ does not approach any limit: it oscillates between $- 1$ and $1$ indefinitely. Therefore $lim_{b \to \infty} (\cos 1 - \cos b)$ does not exist, and the integral diverges.

Note that divergence here is not caused by the area growing to infinity. The partial integrals stay bounded between $\cos 1 - 1$ and $\cos 1 + 1$ ; they simply never settle on a value. A limit that fails to exist counts as divergence just as much as an infinite one.

Evaluate $\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x$ .

Hint: Integrate by parts.

Answer

$1$ .

Solution

Integrate by parts. Let $u = \ln x$ and $d v = x^{- 2} d x$ , so $d u = d x / x$ and $v = - 1 / x$ . Then:

\begin{aligned} \int_1^b \frac{\ln x}{x^2}\,dx &= \left[-\frac{\ln x}{x}\right]_1^b + \int_1^b \frac{1}{x}\cdot\frac{dx}{x} \\ &= \left(-\frac{\ln b}{b} + \frac{\ln 1}{1}\right) + \int_1^b x^{-2}\,dx \\ &= -\frac{\ln b}{b} + 0 + \left[-\frac{1}{x}\right]_1^b \\ &= -\frac{\ln b}{b} - \frac{1}{b} + 1. \end{aligned}

As $b \to \infty$ :

$\frac{\ln b}{b} \to 0$ , by L'Hôpital's rule: $\frac{\ln b}{b}$ has the form $\infty / \infty$ , and differentiating top and bottom gives $\frac{1 / b}{1} = \frac{1}{b} \to 0$ .
$\frac{1}{b} \to 0$ .

Therefore:

\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x = lim_{b \to \infty} (- \frac{\ln b}{b} - \frac{1}{b} + 1) = 1.

Evaluate $\int_{0}^{\infty} x^{2} e^{- x} d x$ .

Answer

$2$ .

Solution

Integrate by parts. Let $u = x^{2}$ and $d v = e^{- x} d x$ , so $d u = 2 x d x$ and $v = - e^{- x}$ :

\begin{aligned} \int_0^b x^2 e^{-x}\,dx &= \bigl[-x^2 e^{-x}\bigr]_0^b + 2\int_0^b x\,e^{-x}\,dx \\ &= -b^2 e^{-b} + 0 + 2\int_0^b x\,e^{-x}\,dx. \end{aligned}

We showed in the worked example on $\int_{0}^{\infty} x e^{- x} d x$ that this last integral tends to $1$ . It remains to check that $b^{2} e^{- b} \to 0$ as $b \to \infty$ . Applying L'Hôpital's rule twice to $b^{2} / e^{b}$ :

lim_{b \to \infty} \frac{b^{2}}{e^{b}} = lim_{b \to \infty} \frac{2 b}{e^{b}} = lim_{b \to \infty} \frac{2}{e^{b}} = 0.

Taking $b \to \infty$ :

\int_{0}^{\infty} x^{2} e^{- x} d x = 0 + 2 (1) = 2 = 2! .

This is the case $n = 2$ of the formula $\int_{0}^{\infty} x^{n} e^{- x} d x = n!$ .

Determine whether $\int_{0}^{\infty} e^{- x} \sin x d x$ converges or diverges. If it converges, find its value.

Hint: Use the same technique as in the worked example on $\int_{0}^{\infty} e^{- x} \cos x d x$ : integrate by parts twice.

Answer

Converges, with value $\frac{1}{2}$ .

Solution

We integrate by parts twice, carrying a finite upper limit $b$ throughout and taking the limit at the end. Let $I = \int_{0}^{\infty} e^{- x} \sin x d x$ .

First integration by parts. Let $u = \sin x$ and $d v = e^{- x} d x$ , so $d u = \cos x d x$ and $v = - e^{- x}$ :

\begin{aligned} \int_0^b e^{-x}\sin x\,dx &= \bigl[-e^{-x}\sin x\bigr]_0^b + \int_0^b e^{-x}\cos x\,dx \\ &= -e^{-b}\sin b + 0 + \int_0^b e^{-x}\cos x\,dx. \end{aligned}

Second integration by parts. Apply integration by parts to the remaining integral. Let $u = \cos x$ and $d v = e^{- x} d x$ , so $d u = - \sin x d x$ and $v = - e^{- x}$ :

\begin{aligned} \int_0^b e^{-x}\cos x\,dx &= \bigl[-e^{-x}\cos x\bigr]_0^b - \int_0^b (-e^{-x})(-\sin x)\,dx \\ &= (-e^{-b}\cos b + 1) - \int_0^b e^{-x}\sin x\,dx. \end{aligned}

Solving for the integral. Substituting this back into the first result:

\int_{0}^{b} e^{- x} \sin x d x = - e^{- b} \sin b + (- e^{- b} \cos b + 1) - \int_{0}^{b} e^{- x} \sin x d x .

Moving the integral from the right side to the left:

\begin{aligned} 2\int_0^b e^{-x}\sin x\,dx &= 1 - e^{-b}\sin b - e^{-b}\cos b \\ &= 1 - e^{-b}(\sin b + \cos b). \end{aligned}

Since $| \sin b + \cos b | \leq 2$ , we have $| e^{- b} (\sin b + \cos b) | \leq 2 e^{- b} \to 0$ as $b \to \infty$ . Taking the limit:

2 I = 1, so I = \frac{1}{2} .

Let $a$ be any real number and let $k$ be a constant. Show that $\int_{a}^{\infty} e^{- k x} d x$ converges when $k > 0$ and find its value. What happens when $k \leq 0$ ?

Answer

For $k > 0$ the integral converges to $\frac{e^{- k a}}{k}$ . For $k \leq 0$ it diverges.

Solution

Assume first that $k \neq 0$ , so that $e^{- k x}$ has the antiderivative $- \frac{1}{k} e^{- k x}$ . Integrating over the finite interval $[a, b]$ :

\begin{aligned} \int_a^b e^{-kx}\,dx &= \left[-\frac{1}{k}e^{-kx}\right]_a^b \\ &= \frac{e^{-ka} - e^{-kb}}{k}. \end{aligned}

Case $k > 0$ . Then $- k b \to - \infty$ as $b \to \infty$ , so $e^{- k b} \to 0$ and

\int_{a}^{\infty} e^{- k x} d x = lim_{b \to \infty} \frac{e^{- k a} - e^{- k b}}{k} = \frac{e^{- k a}}{k} .

Case $k < 0$ . Write $k = - m$ with $m > 0$ . Then $e^{- k b} = e^{m b} \to \infty$ , so the expression above has no finite limit and the integral diverges.

Case $k = 0$ . The integrand is the constant $1$ , and $\int_{a}^{b} 1 d x = b - a \to \infty$ , so the integral diverges.

As a check, setting $a = 0$ and $k = 1$ gives $e^{0} / 1 = 1$ , which agrees with part (a) of the first exercise. The formula also shows the useful qualitative fact that faster decay, meaning larger $k$ , gives a smaller total area.

Evaluate $\int_{- \infty}^{\infty} \frac{d x}{x^{2} + 4}$ .

Answer

$\frac{π}{2}$ .

Solution

The antiderivative comes from the standard form $\int \frac{d x}{x^{2} + α^{2}} = \frac{1}{α} \arctan \frac{x}{α} + C$ with $α = 2$ :

\int \frac{d x}{x^{2} + 4} = \frac{1}{2} \arctan \frac{x}{2} + C .

Because both limits are infinite, we must use case (c) and split. Take $c = 0$ .

Right half:

\begin{aligned} \int_0^\infty \frac{dx}{x^2+4} &= \lim_{b\to\infty}\frac{1}{2}\left[\arctan\frac{x}{2}\right]_0^b \\ &= \frac{1}{2}\left(\frac{\pi}{2} - 0\right) \\ &= \frac{\pi}{4}. \end{aligned}

Left half:

\begin{aligned} \int_{-\infty}^0 \frac{dx}{x^2+4} &= \lim_{a\to-\infty}\frac{1}{2}\left[\arctan\frac{x}{2}\right]_a^0 \\ &= \frac{1}{2}\left(0 - \left(-\frac{\pi}{2}\right)\right) \\ &= \frac{\pi}{4}. \end{aligned}

Both halves converge, so the integral converges and

\int_{- \infty}^{\infty} \frac{d x}{x^{2} + 4} = \frac{π}{4} + \frac{π}{4} = \frac{π}{2} .

Compare this with $\int_{- \infty}^{\infty} \frac{d x}{1 + x^{2}} = π$ : widening the bell by a factor of $2$ while lowering its peak by a factor of $4$ has halved the total area.

Evaluate $\int_{2}^{\infty} \frac{d x}{x \ln^{2} x}$ .

Hint: Let $u = \ln x$ .

Answer

$\frac{1}{\ln 2} \approx 1.4427$ .

Solution

The integrand is continuous on $[2, \infty)$ , since $\ln x \geq \ln 2 > 0$ there, so the only impropriety is the infinite upper limit. Work on $[2, b]$ with $b > 2$ .

Let $u = \ln x$ , so $d u = d x / x$ . When $x = 2$ , $u = \ln 2$ ; when $x = b$ , $u = \ln b$ . The substitution turns the integral into one of the simplest possible shapes, an inverse square in $u$ :

\begin{aligned} \int_2^b \frac{dx}{x\ln^2 x} &= \int_{\ln 2}^{\ln b} \frac{du}{u^2} \\ &= \left[-\frac{1}{u}\right]_{\ln 2}^{\ln b} \\ &= \frac{1}{\ln 2} - \frac{1}{\ln b}. \end{aligned}

As $b \to \infty$ we have $\ln b \to \infty$ , so $1 / \ln b \to 0$ and

\int_{2}^{\infty} \frac{d x}{x \ln^{2} x} = \frac{1}{\ln 2} \approx 1.4427 .

The exponent on the logarithm is decisive, and the same substitution shows why. Applied to $\int_{2}^{\infty} \frac{d x}{x \ln x}$ it produces $\int_{\ln 2}^{\ln b} \frac{d u}{u} = \ln (\ln b) - \ln (\ln 2)$ , which tends to $\infty$ , so that integral diverges. Dropping the exponent from $2$ to $1$ turns a convergent integral into a divergent one, even though the change to the integrand looks slight.

Evaluate $\int_{0}^{\infty} \frac{d x}{(1 + x)^{3}}$ .

Answer

$\frac{1}{2}$ .

Solution

Let $u = 1 + x$ , so $d u = d x$ . When $x = 0$ , $u = 1$ ; when $x = b$ , $u = 1 + b$ :

\begin{aligned} \int_0^b \frac{dx}{(1+x)^3} &= \int_1^{1+b} u^{-3}\,du \\ &= \left[-\frac{1}{2u^2}\right]_1^{1+b} \\ &= \frac{1}{2} - \frac{1}{2(1+b)^2}. \end{aligned}

As $b \to \infty$ the subtracted term tends to $0$ , so

\int_{0}^{\infty} \frac{d x}{(1 + x)^{3}} = lim_{b \to \infty} (\frac{1}{2} - \frac{1}{2 (1 + b)^{2}}) = \frac{1}{2} .

The shift by $1$ is what makes this integral behave: it moves the singularity of $x^{- 3}$ from $x = 0$ off the interval, leaving only the Type I impropriety at the top. Far out, $(1 + x)^{- 3}$ is nearly $x^{- 3}$ , and its antiderivative $- \frac{1}{2} (1 + x)^{- 2}$ dies away to $0$ , which is exactly the behaviour that produces a finite answer.

Determine whether $\int_{0}^{\infty} \frac{2 x d x}{1 + x^{2}}$ converges or diverges.

Answer

Diverges.

Solution

Let $u = 1 + x^{2}$ , so $d u = 2 x d x$ . When $x = 0$ , $u = 1$ ; when $x = b$ , $u = 1 + b^{2}$ :

\begin{aligned} \int_0^b \frac{2x\,dx}{1+x^2} &= \int_1^{1+b^2}\frac{du}{u} \\ &= \bigl[\ln u\bigr]_1^{1+b^2} \\ &= \ln\left(1+b^2\right). \end{aligned}

As $b \to \infty$ , $\ln (1 + b^{2}) \to \infty$ , so the integral diverges.

The reason is visible in the integrand: for large $x$ ,

\frac{2 x}{1 + x^{2}} \approx \frac{2 x}{x^{2}} = \frac{2}{x},

so the tail behaves like twice the tail of $\int^{\infty} d x / x$ , which we already know diverges. Note how little separates this from the convergent case: dropping the factor $2 x$ gives $\int_{0}^{\infty} \frac{d x}{1 + x^{2}} = π / 2$ , a perfectly finite number.

Evaluate $\int_{1}^{\infty} \frac{\arctan x}{1 + x^{2}} d x$ .

Answer

$\frac{3 π^{2}}{32} \approx 0.9253$ .

Solution

The factor $\frac{1}{1 + x^{2}}$ is exactly the derivative of $\arctan x$ , which invites the substitution $u = \arctan x$ , $d u = \frac{d x}{1 + x^{2}}$ . When $x = 1$ , $u = \arctan 1 = \frac{π}{4}$ ; when $x = b$ , $u = \arctan b$ . Then

\begin{aligned} \int_1^b \frac{\arctan x}{1+x^2}\,dx &= \int_{\pi/4}^{\arctan b} u\,du \\ &= \left[\frac{u^2}{2}\right]_{\pi/4}^{\arctan b} \\ &= \frac{(\arctan b)^2}{2} - \frac{1}{2}\left(\frac{\pi}{4}\right)^2 \\ &= \frac{(\arctan b)^2}{2} - \frac{\pi^2}{32}. \end{aligned}

As $b \to \infty$ , $\arctan b \to \frac{π}{2}$ , so

\begin{aligned} \int_1^\infty \frac{\arctan x}{1+x^2}\,dx &= \frac{1}{2}\left(\frac{\pi}{2}\right)^2 - \frac{\pi^2}{32} \\ &= \frac{\pi^2}{8} - \frac{\pi^2}{32} \\ &= \frac{4\pi^2 - \pi^2}{32} \\ &= \frac{3\pi^2}{32} \approx 0.9253. \end{aligned}

The limit was easy because $\arctan$ is bounded: the substitution turned an infinite interval in $x$ into the finite interval $[π / 4, π / 2)$ in $u$ , and a bounded integrand on a finite interval cannot misbehave.

The region under $y = 1 / x$ for $x \geq 1$ is revolved about the $x$ -axis. The resulting infinitely long solid is called Gabriel's Horn.

The horn has mouth radius $1$ at $x = 1$ , and its cross-sectional radius $1 / x$ shrinks to zero as $x$ grows.

Using the disc formula $V = π \int_{1}^{\infty} y^{2} d x$ , show that the volume is finite and find it.
Using the surface formula S = 2\pi\displaystyle\int_1^\infty y\sqrt{1+(y')^2}\,dx, show that the surface area is infinite.

Answer

(a) $V = π$ . (b) $S$ diverges, so the surface area is infinite.

Solution

(a) The volume. With $y = 1 / x$ we have $y^{2} = 1 / x^{2}$ , so

\begin{aligned} V &= \pi\int_1^\infty \frac{dx}{x^2} \\ &= \pi\lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^b \\ &= \pi\lim_{b\to\infty}\left(1-\frac{1}{b}\right) \\ &= \pi. \end{aligned}

This is just the first worked example multiplied by $π$ . The horn holds exactly $π$ cubic units.

(b) The surface area. Here y' = -1/x^2, so (y')^2 = 1/x^4 and

S = 2 π \int_{1}^{\infty} \frac{1}{x} \sqrt{1 + \frac{1}{x^{4}}} d x .

We do not need to evaluate this. Since $1 / x^{4} > 0$ for all $x \geq 1$ , the square root satisfies $\sqrt{1 + 1 / x^{4}} > 1$ , so the integrand is larger than $1 / x$ at every point. Therefore, for every $b > 1$ ,

\int_{1}^{b} \frac{1}{x} \sqrt{1 + \frac{1}{x^{4}}} d x > \int_{1}^{b} \frac{d x}{x} = \ln b .

Because $\ln b \to \infty$ , the partial integrals are unbounded, so $S$ diverges: the surface area is infinite.

This is the painter's paradox. The horn can be filled with $π$ units of paint, yet no finite amount of paint suffices to coat its inner surface. The paradox dissolves once you notice that filling uses volume, which weights the radius quadratically as $1 / x^{2}$ (convergent), while coating uses surface, which weights it only linearly as $1 / x$ (divergent). Physical paint has thickness, and past a certain point the horn is narrower than the paint layer.

Consider $\int_{- \infty}^{\infty} x d x$ .

Show that $lim_{b \to \infty} \int_{- b}^{b} x d x = 0$ .
Explain why part (a) does not show that $\int_{- \infty}^{\infty} x d x$ converges.
Compute $lim_{b \to \infty} \int_{- b}^{2 b} x d x$ and say what it tells you.

Answer

(a) $0$ .
(b) Because the definition requires each half, $\int_{- \infty}^{c}$ and $\int_{c}^{\infty}$ , to converge on its own, and $\int_{0}^{\infty} x d x$ diverges.
(c) $+ \infty$ . Different pairings of the two endpoints give different answers, so no single value can be assigned; the integral diverges.

Solution

(a) For each fixed $b > 0$ ,

\begin{aligned} \int_{-b}^{b} x\,dx &= \left[\frac{x^2}{2}\right]_{-b}^{b} \\ &= \frac{b^2}{2} - \frac{b^2}{2} \\ &= 0. \end{aligned}

Every term of the family is $0$ , so the limit as $b \to \infty$ is $0$ .

(b) The definition of a doubly infinite integral is case (c): choose a finite $c$ and require that both $\int_{- \infty}^{c} x d x$ and $\int_{c}^{\infty} x d x$ converge. Take $c = 0$ . The right half is

\int_{0}^{\infty} x d x = lim_{b \to \infty} {[\frac{x^{2}}{2}]}_{0}^{b} = lim_{b \to \infty} \frac{b^{2}}{2} = \infty,

which diverges. One divergent half is enough: $\int_{- \infty}^{\infty} x d x$ diverges, and the symmetric limit in part (a) has not contradicted this. What that limit did was cancel a positive area $b^{2} / 2$ against a negative area $- b^{2} / 2$ , both of which are racing to infinity, by insisting the two endpoints move at exactly the same speed.

(c) Letting the right endpoint run out twice as fast:

\begin{aligned} \int_{-b}^{2b} x\,dx &= \left[\frac{x^2}{2}\right]_{-b}^{2b} \\ &= \frac{4b^2}{2} - \frac{b^2}{2} \\ &= \frac{3b^2}{2}, \end{aligned}

which tends to $+ \infty$ . More generally, $\int_{- b}^{λ b} x d x = \frac{(λ^{2} - 1) b^{2}}{2}$ , whose limit is $- \infty$ for $0 < λ < 1$ , is $0$ for $λ = 1$ , and is $+ \infty$ for $λ > 1$ . The "value" therefore depends entirely on an arbitrary choice about how the two endpoints are tied together, which is precisely why the definition refuses to tie them together at all.

The symmetric answer $0$ is not meaningless, and later in this chapter it is given a name and a definition of its own. What parts (b) and (c) establish is only that it is not the same thing as convergence, and that the definition of $\int_{- \infty}^{\infty}$ deliberately declines to use it.

The Definition

How to Evaluate a Type I Integral

Reading the Definition Geometrically

A Warning About the Symmetric Limit

Worked Examples

Integrals That Need Integration by Parts

Exercises

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