OBJECTIVES

When you finish this section, you should be able to:

1. Find integrals with an infinite limit of integration (p. 524)
2. Interpret an improper integral geometrically (p. 525)
3. Integrate functions over [\(a\), \(b\)] that are not defined at an endpoint (p. 527)
4. Use the Comparison Test for Improper Integrals (p. 529)

spanDEFINITIONspan Improper Integral

If a function \(f\) is continuous on the interval \([a,\,\infty)\), then \(\int_{a}^{\infty }f(x)\,dx,\) called an improper integral, is defined as \[\bbox[5px, border:1px solid black, #F9F7ED]{ \int_{a}^{\infty }f(x)~dx=\lim\limits_{b \rightarrow \infty} \int_{a}^{b}f(x)~dx} \]

provided the limit exists as a real number. If \(\lim\limits_{b\rightarrow \infty }\int_{a}^{b} f(x)\,dx\) exists as a real number, the improper integral \( \int_{a}^{\infty }f(x)\,dx\) is said to converge. If the limit does not exist or if the limit is infinite, the improper integral \( \int_{a}^{\infty } f(x)\,dx\) is said to diverge.

If a function \(f\) is continuous on the interval \((-\,\infty ,\,b ]\), then \(\int_{-\infty }^{b} f(x)\,dx,\) called an improper integral, is defined as \[\bbox[5px, border:1px solid black, #F9F7ED]{ \int_{-\infty }^{b}f(x)\,dx=\lim\limits_{\,a\,\rightarrow \,-\infty }\int_{a}^{b} f(x)\,dx } \]

provided the limit exists as a real number. If \(\lim\limits_{a\rightarrow -\infty }\int_{a}^{b} f(x)\,dx\) exists as a real number, the improper integral \( \int_{-\infty }^{b}f(x)\,dx\) converges. If the limit does not exist or if the limit is infinite, the improper integral \(\int_{-\infty }^{b} f(x)\,dx\) diverges.

Integrating Functions over Infinite Intervals

Determine whether the following improper integrals converge or diverge:

1. \(\int_{1}^{\infty }\dfrac{1}{x}\,dx\)
2. \(\int_{-\infty }^{0}e^{x}\,dx\)
3. \(\int_{\pi/2}^{\infty }\sin x \,dx\)

Solution 
(a) \(\int_{1}^{\infty }\dfrac{1}{x}\,dx{:} \lim\limits_{b\,\rightarrow \,\infty }\int_{1}^{b}\dfrac{1}{x} \,dx=\lim\limits_{b\,\rightarrow \,\infty }\big[\ln \,\vert x \vert \big] _{1}^{b}=\lim\limits_{b\,\rightarrow \,\infty }[ \ln b-\ln 1] = \infty\).

The limit is infinite, so \(\int_{1}^{\infty }\dfrac{1}{x}\,dx\) diverges.

(b) \(\int_{-\infty }^{0} e^{x}\,dx: \lim\limits_{a\, \rightarrow \,-\infty }\int_{a}^{0}e^{x}\,dx=\lim\limits_{a\,\rightarrow \,-\infty }\big[e^{x}\big]_{a}^{0}=\lim\limits_{a\,\rightarrow \,-\infty }(1-e^{a}) = 1-0=1.\) Since the limit exists, \(\int_{-\infty }^{0}e^{x}\,dx\) converges and equals \(1\).

(c) \[ \begin{eqnarray*} \int_{\pi/2}^{\infty }\sin x \,dx: \lim\limits_{b\,\rightarrow \,\infty }\int_{\pi/2}^{b}\sin x\, dx &=& \lim\limits_{b\,\rightarrow \,\infty }\,\big[ -\cos x \big]^b_{\pi/2}\\ &=& \lim\limits_{b\,\rightarrow \,\infty } [ - \cos b +0 ] \\ &=& - \lim\limits_{b\,\rightarrow \,\infty }\cos b \end{eqnarray*} \]

This limit does not exist, since as \(b\rightarrow \infty\), the value of \(\cos b\) oscillates between \(-1\) and \(1\). So, \( \int_{\pi/2}^{\infty }\sin x \,dx\) diverges.

NEED TO REVIEW?

Limits at infinity are discussed in Section 1.5, pp. 117-122.

NOW WORK

Problem 17.

If a function \(f\) is continuous for all \(x\) and if, for any number \( c\), both improper integrals \(\int_{-\infty }^{c}f(x)~dx\) and \(\int_{c}^{\infty }f(x)~dx\) converge, then the improper integral \(\int_{-\infty }^{\infty }f(x)~dx\) converges, and \[\bbox[5px, border:1px solid black, #F9F7ED]{ \int_{-\infty }^{\infty }f(x)~dx= \int_{-\infty}^{c}f(x)~dx+ \int_{c}^{\infty } f(x)~dx} \]

If either or both of the integrals on the right diverge, then the improper integral \(\int_{-\infty }^{\infty } f(x)\,dx\) diverges.

Integrating Functions over Infinite Intervals

Determine whether \(\int_{-\infty }^{\,\infty }4x^{3}\,dx\) converges or diverges.

Solution We begin by writing \(\int_{-\infty }^{\,\infty }4x^{3}\,dx=\int_{-\infty }^{\,0}4x^{3}\,dx+\int_{0}^{\,\infty }4x^{3}\,dx\) and evaluate each improper integral on the right. \[ \int_{-\infty }^{0}4x^{3}\,dx\hbox{:} \quad \lim_{a\,\rightarrow \,-\infty }\left( 4\int_{a}^{0}x^{3}\,dx\right) = \lim_{a\,\rightarrow \,-\infty }\left[ x^{4}\right] _{a}^{0}=\lim_{a\,\rightarrow -\infty }(0-a^{4})=-\infty \]

There is no need to continue. \(\int_{-\infty }^{\,\infty }4x^{3}\,dx\) diverges.

NOTE

The decision to use \(0\) to break up the integral is arbitrary. Usually, the choice made simplifies finding the integral.

NOW WORK

Problem 23.

Determining If the Area Under a Graph Is Defined

Determine if the area under the graph of \(y=\frac{1}{x^{2}}\) to the right of \(x=1\) is defined.

Solution See Figure 22. To determine if the area is defined, we examine \(\int_{1}^{\infty }\,\frac{1}{x^{2}}dx\). \[ \int_{1}^{\infty }\frac{1}{x^{2}}dx\hbox{:}\quad \lim_{b\,\rightarrow \,\infty }\int_{1}^{b}\frac{1}{x^{2}}dx= \lim_{b\,\rightarrow \,\infty } \left[ -\frac{1}{x}\right] _{1}^{b}= \lim_{b\,\rightarrow \,\infty }\left( - \frac{1}{b}+1\right) =1 \]

The area under the graph \(f(x)= \frac{1}{x^{2}}\) to the right of \(1\) is defined and equals \(1\) square unit.

NOW WORK

Problem 71.

THEOREM

\(\int_{1}^{\,\infty }\frac{dx}{x^{p}}\) converges if \(p \gt1\) and diverges if \(p \le 1\).

Proof

We consider two cases: \(p=1\) and \(p \ne 1\). \[ \begin{eqnarray*} p=1\hbox{;}\quad \int_{1}^{\infty }\frac{dx}{x^{p}}=\int_{1}^{\infty }\frac{dx}{x}\hbox{:}\quad \lim\limits_{b\rightarrow \infty }\int_{1}^{b}\frac{dx}{x}=\lim\limits_{b\rightarrow \infty }\ln b=\infty \end{eqnarray*} \] From Example 1(a) \[ \begin{eqnarray*} p \ne 1\hbox{;}\quad \int_{1}^{\infty }\frac{dx}{x^{p}}\hbox{:}\quad \lim\limits_{b\rightarrow \infty }\int_{1}^{b}\frac{dx}{x^{p}} &=&\lim\limits_{b\rightarrow \infty }\left[ \frac{x^{-p+1}}{-p+1}\right] _{1}^{b}\\ &=&\lim\limits_{b\rightarrow \infty }\left[\frac{1}{1-p}( b^{-p+1}-1) \right] \\ &=&\dfrac{1}{1-p}\lim\limits_{b\rightarrow \infty }\left( \frac{1}{b^{p-1}}-1\right) \end{eqnarray*} \]

If \(p \gt1\), then \(p-1 \gt0,\) and \(\lim\limits_{b\,\rightarrow \,\infty }\dfrac{1}{b^{p-1}}=0.\) So, \[ \int_{1}^{\infty }\frac{dx}{x^{p}}=\frac{1}{1-p}( 0-1) =\frac{1 }{p-1} \]

and the improper integral \(\int_{1}^{\infty }\frac{dx}{x^{p}}\), \(p>1\), converges.

If \(p \lt1,\) then \(p-1 \lt0,\) and \(\lim\limits_{b\,\rightarrow \,\infty }\dfrac{1}{ b^{p-1}}=\lim\limits_{b\,\rightarrow \,\infty }b^{1-p}=\infty\). So, the improper integral \(\int_{1}^{\infty }\dfrac{dx}{x^{p}}\), \(p \lt1,\) diverges.

NOTE Interesting Fact

The area under the graph of \(y=\dfrac{1}{x}\) to the right of 1 is not defined [see Example 1(a)], but the volume obtained by revolving the region with this same area about the \(x\)-axis is defined.

Finding the Volume of Gabriel’s Horn

Find the volume of the solid of revolution, called Gabriel’s Horn, that is generated by revolving the region bounded by the graph of \(y=\frac{1}{x}\) and the \(x\)-axis to the right of \(1\) about the \(x\)-axis. Use the disk method.

Figure 24 Gabriel’s Horn

Solution Figure 24 illustrates the region being revolved and the solid of revolution that it generates. Using the disk method, the volume \(V\) is \[ \begin{eqnarray*} V=\pi \int_{1}^{\infty }\!\!\left( \frac{1}{x}\right) ^{2} dx\hbox{:}\quad \pi \lim\limits_{b\,\rightarrow \,\infty }\int_{1}^{b}\!\!\frac{1}{x^{2}} \,dx&=&\pi \lim\limits_{b\rightarrow \infty }\left[ -\frac{1}{x}\right] _{1}^{b}\\[6pt] &=&\pi \lim\limits_{b\rightarrow \,\infty }\left( -\frac{1}{b} + 1\right) = \pi \end{eqnarray*} \]

The volume of the solid of revolution is \(\pi\) cubic units.

NOW WORK

Problem 73.

IN WORDS

These improper integrals, sometimes called improper integrals of the second kind, have finite limits of integration, but the integrand is not defined at one (but not both) of them.

spanDEFINITIONspan

If a function \(f\) is continuous on \((a, b]\), but is not defined at \(a,\) then \(\int_{a}^{b}\!f(x)\,dx\) is an improper integral, and \[\bbox[5px, border:1px solid black, #F9F7ED]{ \int_{a}^{b} f(x)\,dx=\lim\limits_{t\,\rightarrow \,a^{+}} \int_{t}^{b} f(x)\,dx } \]

provided the limit exists as a real number.

If a function \(f\) is continuous on \([a,b),\) but is not defined at \(b\), then \(\int_{a}^{b} f(x)\,dx\) is an improper integral, and \[\bbox[5px, border:1px solid black, #F9F7ED]{ \int_{a}^{b} f(x)\,dx=\lim\limits_{t\,\rightarrow \,b^{-}} \int_{a}^{t} f(x)\,dx } \]

provided the limit exists as a real number.

Determining If the Area Under a Graph Is Defined

Determine if the area under the graph of \(y=\dfrac{1}{\sqrt{x}}\) from \(0\) to \(4\) is defined.

Solution Figure 27 shows the area under the graph of \(y=\dfrac{1}{ \sqrt{x}}\) from \(0\) to \(4.\) The area is given by \(\int_{0}^{4}\dfrac{1}{ \sqrt{x}}dx.\) Since the integrand \(f(x)=\dfrac{1}{\sqrt{x}}\) is continuous on (0, 4] but is not defined at 0, \(\int_{0}^{4}\dfrac{1}{ \sqrt{x}}~dx\) is an improper integral. \[ \begin{eqnarray*} \int_{0}^{4}\dfrac{1}{\sqrt{x}}\,dx:\quad \lim_{t\,\rightarrow \,0^{+}}\int_{t}^{4}\dfrac{1}{\sqrt{x}}\,dx &=& \lim_{t\,\rightarrow \,0^{+}}\int_{t}^{4}x^{-1/2}dx=\lim_{t\,\rightarrow \,0^{{ +} }}\left[ \dfrac{x^{1/2}}{\dfrac{1}{2}}\right] _{t}^{4}\\[5pt] &=& \lim_{t\,\rightarrow \,0^{+}}\big( 2\cdot 2-2\sqrt{t}\,\big) =4-2\lim_{t\,\rightarrow \,0^{+}}\sqrt{t}=4 \end{eqnarray*} \] So, \(\int_{0}^{4}\dfrac{1}{\sqrt{ x}}\,dx\) converges, and the area under the graph of \(y=\dfrac{1}{\sqrt{x}}\) from \(0\) to \(4\) is defined and equals \(4\).

Figure 27 \(y=\dfrac{1}{\sqrt{x}}\), \(0 \lt x \le 4\).

NOW WORK

Problem 27.

Determining Whether an Improper Integral Converges or Diverges

Determine whether \(\int_{0}^{\pi /2}\tan x\,dx\) converges or diverges.

Solution The function \(f (x)\) = \(\tan x\) is continuous on \(\left[ 0,\,\dfrac{\pi }{2}\right) \) but is not defined at \(\dfrac{\pi }{2}\), so \(\int_{0}^{\pi /2}\tan x\,dx\) is an improper integral. \[ \begin{eqnarray*} \int_{0}^{\pi /2}\tan x~dx\hbox{:} \lim\limits_{t\,\rightarrow \left( \pi /2\right) ^{-}}\int_{0}^{t}\tan x\,dx&=&\lim\limits_{t\,\rightarrow \ (\pi /2) ^{-}}\big[ \ln \vert \sec x \vert\big] _{0}^{t}\\[4pt] &=&\lim\limits_{t\,\rightarrow (\pi /2) ^{-}}[\ln \vert \sec t \vert -\ln \vert \sec 0 \vert ] \\[4pt] &=& \lim\limits_{t\,\rightarrow (\pi /2) ^{-}}\ln (\sec t)=\infty \end{eqnarray*} \]

So, \(\int_{0}^{\pi /2}\tan x\,dx\) diverges.

NOW WORK

Problem 43.

spanDEFINITIONspan

If a function \(f\) is continuous on a closed interval \([a,b]\), except at the number \(c,\) \(a<c<b\), where \(f\) is not defined, the integral \(\int_{a}^{b} f(x)\,dx\) is an improper integral and \[\bbox[5px, border:1px solid black, #F9F7ED]{ \int_{a}^{b}f(x)\,dx = \int_{a}^{c}f(x)\,dx+ \int_{c}^{b}f (x)\,dx } \]

provided that both improper integrals on the right converge.

If either improper integral \(\int_{a}^{c}f(x)\,dx\) or \(\int_{c}^{b}f(x) \,dx\) diverges, then the improper integral \(\int_{a}^{b}f(x)\,dx\) also diverges.

Determining Whether an Improper Integral Converges or Diverges

Determine whether \(\int_{0}^{2}\dfrac{1}{(x-1)^{2}}\,dx\) converges or diverges.

Solution Since \(f(x)=\dfrac{1}{(x-1) ^{2}}\) is not defined at 1, the integral \(\int_{0}^{2}\dfrac{1}{(x-1)^{2}}\,dx\) is an improper integral on the interval [ 0, 2]. We write the integral as follows: \[ \int_{0}^{2}\dfrac{1}{(x-1)^{2}}\,dx=\int_{0}^{1}\dfrac{1}{(x-1)^{2}} \,dx+\int_{1}^{2}\dfrac{1}{(x-1)^{2}}\,dx \]

and investigate each of the two improper integrals on the right. \[ \int_{0}^{1}\!\!\dfrac{1}{(x-1)^{2}}\,dx:\quad \lim_{t\,\rightarrow \,1^{-}}\int_{0}^{t}\dfrac{1}{(x-1)^{2}}\,dx=\lim_{t\rightarrow 1^{-}}\left[ \dfrac{-1}{x-1}\right] _{0}^{t}\!=\lim_{t\rightarrow 1^{-}}\left( \dfrac{-1}{t-1} -1\!\right)\! =\infty \] \(\int_{0}^{1}\dfrac{1}{(x-1) ^{2}}\,dx\) diverges, so there is no need to investigate the second integral.

The improper integral \(\int_{0}^{2}\dfrac{1}{(x-1) ^{2}}\,dx\) diverges.

CAUTION

It is a common mistake to look at an improper integral like \(\int_{0}^{2}\dfrac{dx}{(x-1) ^{2}}\) and not notice that the integrand is undefined at \(1\). If that happens, and you use the Fundamental Theorem of Calculus, you obtain \[ \int_{0}^{2}\dfrac{dx}{(x-1)^{2}}=\left[ \dfrac{-1}{x-1}\right] _{0}^{2}=-1-1=-2 \] which is an incorrect answer. Always check the domain of the integrand before attempting to integrate.

NOW WORK

Problem 31.

THEOREM Comparison Test for Improper Integrals

Let \(f\) and \(g\) be two functions that are nonnegative and continuous on the interval \(\left[ a,\infty \right) \), and suppose \[ f( x) \ge g( x) \]

for all numbers \(x>c,\) where \(c \ge a.\)

• If \(\int_{a}^{\infty}f(x)\,dx\) converges, then \(\int_{a}^{\infty }g(x)\,dx\) also converges
• If \(\int_{a}^{\infty }g(x)\,dx\) diverges, then \(\int_{a}^{\infty}f(x)\,dx\) also diverges.

Using the Comparison Test for Improper Integrals

Determine whether \(\int_{1}^{\infty }e^{-x^{2}}dx\) converges or diverges.

Solution By definition, \(\int_{1}^{\infty }e^{-x^{2}}dx=\lim\limits_{b\rightarrow \infty }\int_{1}^{b}e^{-x^{2}}dx\) converges if the limit exists and equals a real number. Since \(e^{-x^{2}}\) has no antiderivative, we use the Comparison Test for Improper Integrals. We proceed as follows: For \(x \ge 1,\) \[ \begin{eqnarray*} \begin{array}{rcl@{\qquad}l} x^{2} &\geq & x \\[3pt] -x^{2} &\leq & -x \\[3pt] 0 &\lt&e^{-x^{2}}\leq e^{-x} & {\color{#0066A7}{\hbox{ Since }{e>1}\hbox{, if }{a\leq b}\hbox{, then }{e}^{a}{\leq e}^{b}.}} \end{array} \end{eqnarray*} \]

Figure 28 illustrates this.

Figure 28

Based on the Comparison Test, if \(\int_{1}^{\infty }e^{-x}dx\) converges, so does \(\int_{1}^{\infty }e^{-x^{2}}dx.\) We investigate \(\int_{1}^{\infty }e^{-x}dx.\) \[ \begin{eqnarray*} \int_{1}^{\infty }e^{-x}dx :\quad\!\!\!\! \lim\limits_{b\rightarrow \infty }\int_{1}^{b}e^{-x}dx &=& \lim\limits_{b\rightarrow \infty }\!\big[-e^{-x}\big] _{1}^{b}=\lim\limits_{b\rightarrow \infty }[-e^{-b}+e^{-1}] =\lim\limits_{b\rightarrow \infty }\left[ \dfrac{1}{e}-\dfrac{1}{e^{b}} \!\right] \\[5pt] &=& \lim\limits_{b\rightarrow \infty }\dfrac{1}{e}-\lim\limits_{b \rightarrow \infty }\dfrac{1}{e^{b}}=\dfrac{1}{e} \end{eqnarray*} \]

Since \(\int_{1}^{\infty }e^{-x}dx\) converges, we conclude that \( \int_{1}^{\infty}e^{-x^{2}}dx\) converges.

NOW WORK

Problem 67.