5.5 Improper Integrals

Definite integrals so far have been defined only for continuous functions over finite closed intervals. There are times when you will need to perform integration despite those conditions not being met. For example, in quantum mechanics the Dirac delta function^[1] $\delta$ is defined on $\Reals$ by four properties:

tikzpicture[every node/.style=font=] [black!60,line width=1pt,-latex] (-2.5,0) -- (2.6,0) node [right] x; [black!60,line width=1pt,-latex] (0,0) -- (0,2) node [right] y; [below] at (0,-0.1) 0; [<->,linecolor,line width=1.5pt] (-2,0) -- (2,0); [fill=white] (0,0) circle (2.5pt); (0,2.5) circle (2.2pt); [above] at (0,2.6) ; tikzpicture — Figure 5.5.1 $y=\delta(x)$

$\delta(x) ~=~ 0\;$ for all $x \ne 0$
$\delta(0) ~=~ \infty$
$\displaystyle\int_{-\infty}^{\infty} \delta(x)~\dx ~=~ 1$
For any continuous function $f$ on $\Reals$,

$\displaystyle\int_{-\infty}^{\infty} f(x)\;\delta(x)~\dx ~=~ f(0)$.

Properties (3) and (4) provide examples of one type of improper integral: an integral over an infinite interval (in this case the entire real line $\Reals = (-\infty,\infty)$). Define this type of improper integral as follows:

Definition 5.8

For a continuous function $f$ and a real number $a$, define the improper integral of $f$ over $\lival{a}{\infty}$ by

\[ \int_{a}^{\infty} f(x)~\dx ~=~ \lim_{b \to \infty}~\int_{a}^{b} f(x)~\dx ~, \]

define the improper integral of $f$ over $\rival{-\infty}{a}$ by

\[ \int_{-\infty}^{a} f(x)~\dx ~=~ \lim_{b \to -\infty}~\int_{b}^{a} f(x)~\dx ~, \]

and define the improper integral of $f$ over $(-\infty,\infty)$ by

\[ \int_{-\infty}^{\infty} f(x)~\dx ~=~ \int_{-\infty}^{c} f(x)~\dx ~+~ \int_{c}^{\infty} f(x)~\dx ~, \]

for any real number $c$ (typically $c=0$). If the given limit exists (i.e. is a real number) then the improper integral is convergent; otherwise it is divergent.

The limits in the above definitions are always taken after evaluating the integral inside the limit. Just as for “proper” definite integrals, improper integrals can be interpreted as representing the area under a curve.

Example 5.28

Evaluate $~\displaystyle\int_1^{\infty}\,\dfrac{\dx}{x}~$. [r]

Solution: For all real numbers $b > 1$,

\begin{align*} \int_1^{\infty}\,\dfrac{\dx}{x} ~&=~ \lim_{b \to \infty}~\int_{1}^{b} \dfrac{\dx}{x} ~=~ \lim_{b \to \infty}~\left(\ln\,x~\Biggr|_1^{b}\right)\\[6pt] &=~ \lim_{b \to \infty}~ (\ln\,b ~-~ \ln\,1) ~=~ \lim_{b \to \infty}~ b ~=~ \infty \end{align*}

and so the integral is divergent. This means that the area under the curve $y=1/x$ over the interval $\lival{1}{\infty}$—as shown in the graph above—is infinite.

Example 5.29

Evaluate $~\displaystyle\int_1^{\infty}\,\dfrac{\dx}{x^2}~$. [r]

Solution: For all real numbers $b > 1$,

\begin{align*} \int_1^{\infty}\,\dfrac{\dx}{x^2} ~&=~ \lim_{b \to \infty}~\int_{1}^{b} \dfrac{\dx}{x^2} ~=~ \lim_{b \to \infty}~\left(-\frac{1}{x}~\Biggr|_1^{b}\right)\\[6pt] &=~ \lim_{b \to \infty}~ \left(-\frac{1}{b} ~-~ \left(-\frac{1}{1}\right)\right) ~=~ \lim_{b \to \infty}~ \left(1 ~-~ \frac{1}{b}\right) ~=~ 1 ~-~ 0 ~=~ 1 ~. \end{align*}

This means that the area under the curve $y=1/x^2$ over the interval $\lival{1}{\infty}$—as shown in the graph above—equals 1. Thus, an infinite region can have a finite area. Length and area are different and not necessarily related concepts, as this example illustrates. Notice that $y=1/x^2$ approaches the $x$-axis asymptote much faster than $y=1/x$ does—fast enough to make the integral convergent.

Example 5.30

Evaluate $~\displaystyle\int_{-\infty}^0\,e^x~\dx~$. [r] tikzpicture[>=latex,every node/.style=font=] [fill=fillcolor,domain=-3.95:0,variable= ] (-3.95,0) -- plot ( , 1.5*exp( )) -- (0,0) -- cycle; scope[color=linecolor,line width=1.5pt] -3.95,-3.94,...,0.00 1.5*exp( ) stroke scope [<->,black!60,line width=1pt,anchor=base] (0,2) node[above] y |- (-4.3,0) node[left] x; [below] at (0,-0.1) 0; [above left] at (-1.5,0.7) y = e^x; tikzpicture

Solution: For all real numbers $b < 0$,

\begin{align*} \int_{-\infty}^0\,e^x~\dx ~&=~ \lim_{b \to -\infty}~\int_{b}^0\,e^x~\dx ~=~ \lim_{b \to -\infty}~\left(e^x~\Biggr|_b^{0}\right)\\[6pt] &=~ \lim_{b \to -\infty}~ (1 ~-~ e^b) ~=~ 1 ~-~ 0 ~=~ 1 ~. \end{align*}

This means that the area under the curve $y=e^x$ over the interval $\rival{-\infty}{0}$—as shown in the graph above—equals 1.

Example 5.31

Evaluate $~\displaystyle\int_{0}^{\infty}\,\sin\,x~\dx~$. [r]

Solution: Since

\begin{align*} \int_{0}^{\infty}\,\sin\,x~\dx ~&=~ \lim_{b \to \infty}~\int_{0}^{b}\,\sin\,x~\dx\\[6pt] &=~ \lim_{b \to \infty}~\left(-\cos\,x~\Biggr|_0^{b}\right) ~=~ \lim_{b \to \infty}~(-\cos\,b ~+~ 1) \end{align*}

then the integral is divergent, since $\lim_{b \to \infty}~\cos\,b$ does not exist ($\cos b$ oscillates between 1 and -1). This means that the net area over $\lival{0}{\infty}$—counted as positive above the $x$-axis and negative below—is indeterminate.

Example 5.32

Evaluate $~\displaystyle\int_{-\infty}^{\infty}\,\frac{\dx}{1 + x^2}~$. [r]

Solution: Split the integral at $x=0$:

\begin{align*} \int_{-\infty}^{\infty}\,\frac{\dx}{1 + x^2} ~&=~ \int_{-\infty}^{0}\,\frac{\dx}{1 + x^2} ~+~ \int_{0}^{\infty}\,\frac{\dx}{1 + x^2}\\[6pt] &=~ \left(\lim_{b \to -\infty}~\int_{b}^{0}\,\frac{\dx}{1 + x^2}\right) ~+~ \left(\lim_{b \to \infty}~\int_{0}^{b}\,\frac{\dx}{1 + x^2}\right)\\[6pt] &=~ \left(\lim_{b \to -\infty}~\tan^{-1} x~\Biggr|_b^{0}\right) ~+~ \left(\lim_{b \to \infty}~\tan^{-1} x~\Biggr|_0^{b}\right)\\[6pt] &=~ \lim_{b \to -\infty}~ (\tan^{-1} 0 ~-~ \tan^{-1} b) ~+~ \lim_{b \to \infty}~ (\tan^{-1} b ~-~ \tan^{-1} 0)\\[4pt] &=~ (0 - (-\pi/2)) ~+~ (\pi/2 - 0) ~=~ \pi \end{align*}

This means that the area under the curve $y=\frac{1}{1+x^2}$ over the entire real line $(-\infty,\infty)$—as shown in the graph above—equals $\pi$. Note that if the integral were split at any number $c$ then the answer would be the same. Another way to evaluate the integral would have been to use the symmetry around the $y$-axis—as $f(x)=\frac{1}{1+x^2}$ is an even function—so that

\[ \int_{-\infty}^{\infty}\,\frac{\dx}{1 + x^2} ~=~ 2\,\int_{0}^{\infty}\,\frac{\dx}{1 + x^2} ~=~ \cdots ~=~ 2 (\pi/2 - 0) ~=~ \pi ~. \]

Since the integrand is continuous over $\Reals$, a common way of evaluating the integral—especially among students—is to simply use $\pm\infty$ as actual limits of integration, thus avoiding the need to take a limit:

\[ \int_{-\infty}^{\infty}\,\frac{\dx}{1 + x^2} ~=~ \tan^{-1} x~\Biggr|_{-\infty}^{\infty} ~=~ \tan^{-1} (\infty) ~-~ \tan^{-1} (-\infty) ~=~ \frac{\pi}{2} ~-~ \frac{-\pi}{2} ~=~ \pi \]

This type of shortcut is fine as long as you are aware of what plugging $x=\pm\infty$ into $\tan^{-1} x$ actually means, and that there are no numbers for which the integrand is undefined (which would yield an improper integral of a different type, to be discussed shortly).

The second type of improper integral is of a function not continuous or not bounded over its interval of integration. For example, the integral in property (3) of the Dirac delta function is of that type, since $\delta$ is discontinuous at $x=0$. Define this type of improper integral as follows:

Definition 5.9

For a function $f$ that is continuous on $\lival{a}{b}$ but has either a discontinuity or vertical asymptote at $x=b$, define the improper integral of $f$ over $\lival{a}{b}$ by

\[ \int_{a}^{b} f(x)~\dx ~=~ \lim_{c \to b-}~\int_{a}^{c} f(x)~\dx ~. \]

Likewise, if $f$ is continuous on $\rival{a}{b}$ but has either a discontinuity or vertical asymptote at $x=a$, then define the improper integral of $f$ over $\rival{a}{b}$ by

\[ \int_{a}^{b} f(x)~\dx ~=~ \lim_{c \to a+}~\int_{c}^{b} f(x)~\dx ~. \]

If $f$ is continuous on $\ival{a}{b}$ but has either a discontinuity or vertical asymptote at $x=c$ for $a < c < b$, then define the improper integral of $f$ over $\ival{a}{b}$ by

\[ \int_{a}^{b} f(x)~\dx ~=~ \int_{a}^{c} f(x)~\dx ~+~ \int_{c}^{b} f(x)~\dx ~, \]

where the integrals on the right are evaluated as in the first two definitions. If the given limit exists (i.e. is a real number) then the improper integral is convergent; otherwise it is divergent.
Adjust these definitions accordingly for infinite intervals—e.g. $\lival{a}{\infty}$, $\rival{-\infty}{b}$, or $(-\infty,\infty)$—to be consistent with the definitions of improper integrals of that type.

Example 5.33

Evaluate $~\displaystyle\int_{0}^{1}\,\frac{\dx}{x}~$. [r]

Solution: Since $x=0$ is a vertical asymptote for $y = \frac{1}{x}$,

\begin{align*} \int_{0}^{1}\,\frac{\dx}{x} ~&=~ \lim_{c \to 0+}~\int_{c}^{1}\,\frac{\dx}{x} ~=~ \lim_{c \to 0+}~\left(\ln\,x~\Biggr|_c^{1}\right)\\[6pt] &=~ \lim_{c \to 0+}~(\ln\,1 ~-~ \ln\,c) ~=~ 0 ~-~ (-\infty) ~=~ \infty \end{align*}

and so the integral is divergent. This means that the area under the curve $y=1/x$ over the interval $\rival{0}{1}$—as shown in the graph above—is infinite. The region is infinite in the $y$ direction.

Example 5.34

Evaluate $~\displaystyle\int_{0}^{1}\,\frac{\dx}{\sqrt{x}}~$. [r]

Solution: Since $x=0$ is a vertical asymptote for $y = \frac{1}{\sqrt{x}}$,

\begin{align*} \int_{0}^{1}\,\frac{\dx}{\sqrt{x}} ~&=~ \lim_{c \to 0+}~\int_{c}^{1}\,\frac{\dx}{\sqrt{x}} ~=~ \lim_{c \to 0+}~\left(2\,\sqrt{x}~\Biggr|_c^{1}\right)\\[6pt] &=~ \lim_{c \to 0+}~(2 ~-~ 2\,\sqrt{c}\,) ~=~ 2 ~-~ 0 ~=~ 2 ~. \end{align*}

This means that the area under the curve $y=1/\sqrt{x}$ over the interval $\rival{0}{1}$—as shown in the graph above—equals 2. The region is infinite in the $y$ direction.

Example 5.35

Evaluate $~\displaystyle\int_{1}^{3}\,\lfloor x \rfloor~\dx~$. [r]

Solution: Recall from Example 3.22 in Section 3.3 that the floor function $y=\lfloor x \rfloor$ has jump discontinuities at each integer value of $x$, as shown in the graph on the right. The integral $\int_{1}^{3}\,\lfloor x \rfloor\;\dx$ is thus an improper integral over the interval $\lival{1}{3}$, which needs to be split at the point of discontinuity $x=2$ within that interval:

\begin{align*} \int_{1}^{3}\,\lfloor x \rfloor~\dx ~&=~ \int_{1}^{2}\,\lfloor x \rfloor~\dx ~+~ \int_{2}^{3}\,\lfloor x \rfloor~\dx\\[6pt] &=~ \left(\lim_{b \to 2-}~\int_{1}^{b}\,\lfloor x \rfloor~\dx\right) ~+~ \left(\lim_{c \to 3-}~\int_{2}^{c}\,\lfloor x \rfloor~\dx\right)\\[6pt] &=~ \left(\lim_{b \to 2-}~\int_{1}^{b}\,1~\dx\right) ~+~ \left(\lim_{c \to 3-}~\int_{2}^{c}\,2~\dx\right)\\[6pt] &=~ \lim_{b \to 2-}~\left(x~\Biggr|_1^{b}\right) ~+~ \lim_{c \to 3-}~\left(2x~\Biggr|_2^{c}\right)\\[6pt] &=~ \lim_{b \to 2-}~(b - 1) ~+~ \lim_{c \to 3-}~(2c - 4) ~=~ (2-1) ~+~ (6-4) ~=~ 3 \end{align*}

Similar to some of the above examples, the following result is easy to prove (see the exercises):

Theorem 5.19

For any real number $a > 0$, the improper integral

\[ \int_{a}^{\infty}\,\frac{\dx}{x^p} \]

is convergent if $p > 1$, and divergent if $0 < p \le 1$.

The following test for convergence or divergence is sometimes helpful:

Theorem 5.20

Comparison Test for Improper Integrals:

If $\abs{f(x)} \le g(x)$ for all $x$ in $\lival{a}{\infty}$, and if $\int_a^{\infty} g(x)\,\dx$ is convergent, then $\int_a^{\infty} f(x)\,\dx$ is convergent.
If $f(x) \ge g(x) \ge 0$ for all $x$ in $\lival{a}{\infty}$, and if $\int_a^{\infty} g(x)\,\dx$ is divergent, then $\int_a^{\infty} f(x)\,\dx$ is divergent.

The idea behind part (a) is that if $-g(x) \le f(x) \le g(x)$ over $\lival{a}{\infty}$, then—thinking of improper integrals as areas—the integral of $f$ is “squeezed” between the two finite integrals for $\pm g$. There are, however, some subtle issues to prove about the limit in the integral of $f$—finite bounds might not necessarily mean the limit exists.^[2]

Example 5.36

Show that $~\displaystyle\int_{1}^{\infty}\,\dfrac{\sin\,x}{x^2}~\dx~$ is convergent. [r] $tikzpicture[>=latex,every node/.style=font=] [fill=fillcolor,domain=1:4,samples=400,variable= ] (1,0) -- plot ( , sin(9* r)/( * )) -- (4,0) -- cycle; [dashed] (1,1) -- (1,-1); [->,black!60,line width=1pt,anchor=base] (0,-1.1) -- (0,1.1) node[above] y; [->,black!60,line width=1pt,anchor=base] (0,0) -- (4.3,0) node[right] x; scope[dotted] 1.00,1.01,...,4.00 1.1/( * ) 1.00,1.01,...,4.00 -1.1/( * ) stroke scope scope[color=linecolor,line width=1.5pt] 1.00,1.01,...,4.00 sin(9* r)/( * ) stroke scope [left] at (0,0) 0; [below left] at (1,-0.1) 1; [right] at (1.1,1) y = 1x^2; [right] at (1,-1) y = - 1x^2; [color=linecolor,line width=1.5pt] (2.9,0.8) -- (3.6,0.8); [right] at (3.6,0.8) \,xx^2; tikzpicture$

Solution: By Example 5.29, the integral $\int_{1}^{\infty} \frac{1}{x^2}\,\dx$ is convergent. So since $\abs{\sin\,x} \le 1$ for all $x$, then

\[ \ABS{\frac{\sin\,x}{x^2}} ~\le~ \frac{1}{x^2} \]

for all $x$ in $\lival{1}{\infty}$. Thus, by the Comparison Test, $\int_{1}^{\infty} \frac{\sin\,x}{x^2}\,\dx$ is convergent. The graph on the right shows how the curve $y=\tfrac{\sin\,x}{x^2}$ is bounded between the curves $y=\pm\tfrac{1}{x^2}$.

The rules and properties from Section 5.3 concerning definite integrals still apply to improper integrals, provided the improper integrals are convergent. For example, suppose a function $f$ has a discontinuity or vertical asymptote at $x=c$. If both improper integrals $\int_a^{c} f(x)\,\dx$ and $\int_c^{b} f(x)\,\dx$ are convergent, then the improper integral $\int_a^{b} f(x)\,\dx$ is convergent and

\[ \int_a^{b} f(x)~\dx ~=~ \int_a^{c} f(x)~\dx ~+~ \int_c^{b} f(x)~\dx ~. \]

Likewise, if $\int_a^{c} f(x)\,\dx$ and $\int_c^{\infty} f(x)\,\dx$ are convergent, then so is $\int_a^{\infty} f(x)\,\dx$, with

\[ \int_a^{\infty} f(x)~\dx ~=~ \int_a^{c} f(x)~\dx ~+~ \int_c^{\infty} f(x)~\dx ~. \]

Exercises

For Exercises 1-15, evaluate the given improper integral.

$\displaystyle\int_{1}^{\infty} \frac{\dx}{x^3}\vphantom{\displaystyle\int_{0}^{1}\frac{\dx}{\sqrt[3]{x}}}$
$\displaystyle\int_{0}^{1} \frac{\dx}{\sqrt[3]{x}}$
$\displaystyle\int_{0}^{\infty} e^{-x} ~\dx\vphantom{\displaystyle\int_{0}^{1}\frac{\dx}{\sqrt[3]{x}}}$
$\displaystyle\int_{0}^{\infty} e^{-2x} ~\dx\vphantom{\displaystyle\int_{0}^{1}\frac{\dx}{\sqrt[3]{x}}}$
$\displaystyle\int_{-1}^{1} \frac{\dx}{x}\vphantom{\displaystyle\int_{0}^{1}\frac{\dx}{\sqrt[3]{x}}}$

5
$\displaystyle\int_{0}^{\infty} x e^{-x^2}~\dx\vphantom{\displaystyle\int_{0}^{\pi/2}}$
$\displaystyle\int_{-\infty}^{0} 2^x ~\dx\vphantom{\displaystyle\int_{0}^{\pi/2}}$
$\displaystyle\int_{0}^{\pi/2} \tan x ~\dx$
$\displaystyle\int_{0}^{1} \frac{\ln\,x}{x}~\dx\vphantom{\displaystyle\int_{0}^{\pi/2}}$
$\displaystyle\int_{-1}^{1} \frac{\dx}{\sqrt{1 - x^2}}\vphantom{\displaystyle\int_{0}^{\pi/2}}$

5
$\displaystyle\int_{0}^{3} \lceil x \rceil ~\dx\vphantom{\displaystyle\int_{0}^{1} \frac{\dx}{(x - 1)^3}}$
$\displaystyle\int_{-\infty}^{\infty} \frac{\dx}{x^2 ~+~ 4}\vphantom{\displaystyle\int_{0}^{1} \frac{\dx}{(x - 1)^3}}$
$\displaystyle\int_{0}^{1} \frac{\dx}{(x - 1)^3}$
$\displaystyle\int_{2}^{\infty} \frac{\dx}{x\, \ln x}\vphantom{\displaystyle\int_{0}^{1} \frac{\dx}{(x - 1)^3}}$
$\displaystyle\int_{1}^{\infty} \frac{\dx}{x\,\sqrt{x^2 - 1}}\vphantom{\displaystyle\int_{0}^{1} \frac{\dx}{(x - 1)^3}}$
In a standby system of two non-identical components, the normal operating component A has a failure rate of $\lambda_A > 0$ failures per unit time, while the standby component B—which takes over when A fails—has a failure rate $\lambda_B > 0$ (with $\lambda_A \ne \lambda_B$).
1. Find the standby system’s reliability $R(t)$ beyond time $t \ge 0$, where
  
  \[ R(t) ~=~ \int_t^{\infty} \frac{\lambda_A \lambda_B}{\lambda_A - \lambda_B} \left(e^{-\lambda_B x} ~-~ e^{-\lambda_A x}\right)~\dx ~. \]
2. Show that the system’s mean time to failure (MTTF) $m$, where $m = \int_0^{\infty} R(t)\,\dt~$, is $m = \frac{1}{\lambda_A} + \frac{1}{\lambda_B}$.
Show that for all $a > 0$, $~\displaystyle\int_{a}^{\infty}\,\frac{\dx}{x^p}~$ is convergent if $p > 1$, and divergent if $0 < p \le 1$.
Show that for all $a > 0$, $~\displaystyle\int_{0}^{a}\,\frac{\dx}{x^p}~$ is convergent if $0 < p < 1$, and divergent if $p \ge 1$. 2
Is $~\displaystyle\int_{1}^{\infty} \frac{\dx}{x + x^4}~$ convergent? Explain.
Is $~\displaystyle\int_{2}^{\infty} \frac{\dx}{x - \sqrt{x}}~$ convergent? Explain.

Example 5.31 showed that $\int_0^{\infty}\sin x\,\dx$ is divergent. What is the flaw in the argument that the integral must be 0 since each “hump” of $\sin x$ above the $x$-axis is canceled by one below the $x$-axis?
This exercise concerns the subtraction rule $\int_a^{\infty} (f(x) - g(x))\,\dx = \int_a^{\infty} f(x)\,\dx \;-\; \int_a^{\infty} g(x)\,\dx$.
1. Show that $\frac{1}{x (x+1)} = \frac{1}{x} - \frac{1}{x+1}$ for all $x$ except 0 and -1
2. Show that $\int_1^{\infty} \frac{\dx}{x (x+1)}$ is convergent.
3. Show that both $\int_1^{\infty} \frac{\dx}{x}$ and $\int_1^{\infty} \frac{\dx}{x+1}$ are divergent.
4. Does part (c) contradict parts(a)-(b) and the subtraction rule? Explain.
[r]
The improper integral $\int_{-\infty}^{\infty} \delta(x)\,\dx = 1$ is one of the notable “improprieties” of the Dirac delta function $\delta$. One way to think of that integral is by approximating $\delta$ by triangular “pulse” functions $D_n$ (for $n \ge 1$), as in the picture on the right.
1. Write a formula for each $D_n(x)$ over all of $\Reals$.
2. Show that $\int_{-\infty}^{\infty} D_n(x)\,\dx = 1\,$ for all integers $n \ge 1$.
3. Show that $\lim_{n \to \infty} D_n(0) = \infty = \delta(0)$.
4. Do the $D_n$ functions begin to resemble $\delta$ as $n \to \infty$?

Created by the physicist P.A.M. Dirac (1902-1984), who won a Nobel Prize in physics in 1933. The function is neither real-valued nor continuous at $x=0$. The “graph” in Figure 5.5.1 is perhaps misleading, as $\infty$ is not an actual point on the $y$-axis. One interpretation is that $\delta$ is an abstraction of an instantaneous pulse or burst of something, preceded and followed by nothing. To learn more about this fascinating and useful function, see § 15 in Dirac, P.A.M., The Principles of Quantum Mechanics, 4th ed., Oxford, UK: Oxford University Press, 1958. ↩
See pp.140-141 in Buck, R.C., Advanced Calculus, 2nd ed., New York: McGraw-Hill Book Co., 1965. ↩