Evaluating Improper Integrals

Learning Outcomes

Evaluate an integral over an infinite interval
Evaluate an integral over a closed interval with an infinite discontinuity within the interval

Integrating over an Infinite Interval

How should we go about defining an integral of the type ${\displaystyle\int }_{a}^{+\infty }f\left(x\right)dx?$ We can integrate ${\displaystyle\int }_{a}^{t}f\left(x\right)dx$ for any value of $t$ , so it is reasonable to look at the behavior of this integral as we substitute larger values of $t$ . Figure 1 shows that ${\displaystyle\int }_{a}^{t}f\left(x\right)dx$ may be interpreted as area for various values of $t$ . In other words, we may define an improper integral as a limit, taken as one of the limits of integration increases or decreases without bound.

This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and decreasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. The region in the first curve is small, and progressively gets wider under the second and third graph as t moves further to the right away from a on the x-axis.

Definition

Let $f\left(x\right)$ be continuous over an interval of the form $\left[a,\text{+}\infty \right)$ . Then

${\displaystyle\int }_{a}^{+\infty }f\left(x\right)dx=\underset{t\to \text{+}\infty }{\text{lim}}{\displaystyle\int }_{a}^{t}f\left(x\right)dx$ ,

provided this limit exists.
Let $f\left(x\right)$ be continuous over an interval of the form $\left(\text{-}\infty ,b\right]$ . Then

${\displaystyle\int }_{\text{-}\infty }^{b}f\left(x\right)dx=\underset{t\to \text{-}\infty }{\text{lim}}{\displaystyle\int }_{t}^{b}f\left(x\right)dx$ ,

provided this limit exists.

In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.
Let $f\left(x\right)$ be continuous over $\left(\text{-}\infty ,\text{+}\infty \right)$ . Then

${\displaystyle\int }_{\text{-}\infty }^{+\infty }f\left(x\right)dx={\displaystyle\int }_{\text{-}\infty }^{0}f\left(x\right)dx+{\displaystyle\int }_{0}^{+\infty }f\left(x\right)dx$ ,

provided that ${\displaystyle\int }_{\text{-}\infty }^{0}f\left(x\right)dx$ and ${\displaystyle\int }_{0}^{+\infty }f\left(x\right)dx$ both converge. If either of these two integrals diverge, then ${\displaystyle\int }_{\text{-}\infty }^{+\infty }f\left(x\right)dx$ diverges. (It can be shown that, in fact, ${\displaystyle\int }_{\text{-}\infty }^{+\infty }f\left(x\right)dx={\displaystyle\int }_{\text{-}\infty }^{a}f\left(x\right)dx+{\displaystyle\int }_{a}^{+\infty }f\left(x\right)dx$ for any value of $a.$ )

In our first example, we return to the question we posed at the start of this section: Is the area between the graph of $f\left(x\right)=\frac{1}{x}$ and the $x$ -axis over the interval $\left[1,\text{+}\infty \right)$ finite or infinite?

Example: Finding an Area

Determine whether the area between the graph of $f\left(x\right)=\frac{1}{x}$ and the x-axis over the interval $\left[1,\text{+}\infty \right)$ is finite or infinite.

Show Solution

We first do a quick sketch of the region in question, as shown in the following graph.

This figure is the graph of the function y = 1/x. It is a decreasing function with a vertical asymptote at the y-axis. In the first quadrant there is a shaded region under the curve bounded by x = 1 and x = 4.

We can see that the area of this region is given by $A={\displaystyle\int }_{1}^{\infty }\frac{1}{x}dx$ . Then we have

\begin{array}{ccccc} A & = \int_{1}^{\infty} \frac{1}{x} d x \\ = lim_{t \to + \infty} \int_{1}^{t} \frac{1}{x} d x & Rewrite the improper integral as a limit. \\ = lim_{t \to + \infty} ln | x | |_{\begin{matrix} 1 \end{matrix}}^{\begin{matrix} t \end{matrix}} & Find the antiderivative. \\ = lim_{t \to + \infty} (ln | t | - ln 1) & Evaluate the antiderivative. \\ = + \infty . & Evaluate the limit. \end{array}

$\begin{array}{ccccc}\hfill A& ={\displaystyle\int }_{1}^{\infty }\frac{1}{x}dx\hfill & & & \\ & =\underset{t\to \text{+}\infty }{\text{lim}}{\displaystyle\int }_{1}^{t}\frac{1}{x}dx\hfill & & & \text{Rewrite the improper integral as a limit.}\hfill \\ & =\underset{t\to \text{+}\infty }{\text{lim}}\text{ln}|x||{}_{\begin{array}{c}\\ 1\end{array}}^{\begin{array}{c}t\\ \end{array}}\hfill & & & \text{Find the antiderivative.}\hfill \\ & =\underset{t\to \text{+}\infty }{\text{lim}}\left(\text{ln}|t|-\text{ln}1\right)\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =\text{+}\infty .\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$

Since the improper integral diverges to $+\infty$ , the area of the region is infinite.

Example: Finding a Volume

Find the volume of the solid obtained by revolving the region bounded by the graph of $f\left(x\right)=\frac{1}{x}$ and the x-axis over the interval $\left[1,\text{+}\infty \right)$ about the $x$ -axis.

Show Solution

The solid is shown in Figure 3. Using the disk method, we see that the volume V is

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

$V=\pi {\displaystyle\int }_{1}^{+\infty }\frac{1}{{x}^{2}}dx$ .

This figure is the graph of the function y = 1/x. It is a decreasing function with a vertical asymptote at the y-axis. The graph shows a solid that has been generated by rotating the curve in the first quadrant around the x-axis.

Then we have

\begin{array}{ccccc} V & = π \int_{1}^{+ \infty} \frac{1}{x^{2}} d x \\ = π lim_{t \to + \infty} \int_{1}^{t} \frac{1}{x^{2}} d x & Rewrite as a limit. \\ = π lim_{t \to + \infty} - \frac{1}{x} |_{\begin{matrix} 1 \end{matrix}}^{\begin{matrix} t \end{matrix}} & Find the antiderivative. \\ = π lim_{t \to + \infty} (- \frac{1}{t} + 1) & Evaluate the antiderivative. \\ = π . \end{array}

$\begin{array}{ccccc}\hfill V& =\pi {\displaystyle\int }_{1}^{+\infty }\frac{1}{{x}^{2}}dx\hfill & & & \\ & =\pi \underset{t\to \text{+}\infty }{\text{lim}}{\displaystyle\int }_{1}^{t}\frac{1}{{x}^{2}}dx\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\pi \underset{t\to \text{+}\infty }{\text{lim}}-\frac{1}{x}|{}_{\begin{array}{c}\\ 1\end{array}}^{\begin{array}{c}t\\ \end{array}}\hfill & & & \text{Find the antiderivative.}\hfill \\ & =\pi \underset{t\to \text{+}\infty }{\text{lim}}\left(\text{-}\frac{1}{t}+1\right)\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =\pi .\hfill & & & \end{array}$

The improper integral converges to $\pi$ . Therefore, the volume of the solid of revolution is $\pi$ .

In conclusion, although the area of the region between the x-axis and the graph of $f\left(x\right)=\frac{1}{x}$ over the interval $\left[1,\text{+}\infty \right)$ is infinite, the volume of the solid generated by revolving this region about the x-axis is finite. The solid generated is known as Gabriel’s Horn.

Interactive

Visit this website to read more about Gabriel’s Horn.

Example: Evaluating an Improper Integral over an Infinite Interval

Evaluate ${\displaystyle\int }_{\text{-}\infty }^{0}\frac{1}{{x}^{2}+4}dx$ . State whether the improper integral converges or diverges.

Show Solution

Begin by rewriting ${\displaystyle\int }_{\text{-}\infty }^{0}\frac{1}{{x}^{2}+4}dx$ as a limit using the equation 2 from the definition. Thus,

\begin{array}{ccccc} \int_{- \infty}^{0} \frac{1}{x^{2} + 4} d x & = lim_{x \to - \infty} \int_{t}^{0} \frac{1}{x^{2} + 4} d x & Rewrite as a limit. \\ = lim_{t \to - \infty} \frac{1}{2} \tan^{- 1} \frac{x}{2} |_{\begin{matrix} t \end{matrix}}^{\begin{matrix} 0 \end{matrix}} & Find the antiderivative. \\ = \frac{1}{2} lim_{t \to - \infty} (\tan^{- 1} 0 - \tan^{- 1} \frac{t}{2}) & Evaluate the antiderivative. \\ = \frac{π}{4} . & Evaluate the limit and simplify. \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int }_{\text{-}\infty }^{0}\frac{1}{{x}^{2}+4}dx& =\underset{x\to \text{-}\infty }{\text{lim}}{\displaystyle\int }_{t}^{0}\frac{1}{{x}^{2}+4}dx\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\underset{t\to \text{-}\infty }{\text{lim}}\frac{1}{2}{\tan}^{-1}\frac{x}{2}|{}_{\begin{array}{c}\\ t\end{array}}^{\begin{array}{c}0\\ \end{array}}\hfill & & & \text{Find the antiderivative.}\hfill \\ & =\frac{1}{2}\underset{t\to \text{-}\infty }{\text{lim}}\left({\tan}^{-1}0-{\tan}^{-1}\frac{t}{2}\right)\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =\frac{\pi }{4}.\hfill & & & \text{Evaluate the limit and simplify.}\hfill \end{array}$

The improper integral converges to $\frac{\pi }{4}$ .

Because improper integrals require evaluating limits at infinity, at times we may be required to use L’Hôpital’s Rule to evaluate a limit.

Recall: L’Hôpital’s Rule

Suppose $f$ and $g$ are differentiable functions over an open interval $\left(a, \infty \right)$ for some value of $a$ . If either:

$\underset{x\to \infty}{\lim}f(x)=0$ and $\underset{x\to \infty}{\lim}g(x)=0$
$\underset{x\to \infty}{\lim}f(x)= \infty$ (or $-\infty$ ) and $\underset{x\to \infty}{\lim}g(x)= \infty$ (or $-\infty$ ), then

lim_{x \to \infty} \frac{f (x)}{g (x)} = lim_{x \to \infty} \frac{f^{'} (x)}{g^{'} (x)}

$\underset{x\to \infty}{\lim}\dfrac{f(x)}{g(x)}=\underset{x\to \infty}{\lim}\dfrac{f^{\prime}(x)}{g^{\prime}(x)}$

assuming the limit on the right exists or is $\infty$ or $−\infty$ .

Example: Evaluating an Improper Integral on $\left(-\infty ,+\infty \right)$

Evaluate ${\displaystyle\int }_{-\infty }^{+\infty }x{e}^{x}dx$ . State whether the improper integral converges or diverges.

Show Solution

Start by splitting up the integral:

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

${\displaystyle\int }_{\text{-}\infty }^{+\infty }x{e}^{x}dx={\displaystyle\int }_{\text{-}\infty }^{0}x{e}^{x}dx+{\displaystyle\int }_{0}^{+\infty }x{e}^{x}dx$ .

If either ${\displaystyle\int }_{\text{-}\infty }^{0}x{e}^{x}dx$ or ${\displaystyle\int }_{0}^{+\infty }x{e}^{x}dx$ diverges, then ${\displaystyle\int }_{\text{-}\infty }^{+\infty }x{e}^{x}dx$ diverges. Compute each integral separately. For the first integral,

\begin{array}{ccccc} \int_{- \infty}^{0} x e^{x} d x & = lim_{t \to - \infty} \int_{t}^{0} x e^{x} d x & Rewrite as a limit. \\ = lim_{t \to - \infty} (x e^{x} - e^{x}) |_{t}^{0} & \begin{matrix} Use integration by parts to find the \\ antiderivative. (Here u = x and d v = e_{d v}^{x} .) \end{matrix} \\ = lim_{t \to - \infty} (- 1 - t e^{t} + e^{t}) & Evaluate the antiderivative. \\ = - 1. & \begin{matrix} Evaluate the limit. Note: lim_{t \to - \infty} t e^{t} is \\ indeterminate of the form 0 \cdot \infty . Thus, \\ lim_{t \to - \infty} t e^{t} = lim_{t \to - \infty} \frac{t}{e^{- t}} = lim_{t \to - \infty} \frac{- 1}{e^{- t}} = lim_{t \to - \infty} - e^{t} = 0 by \\ L'H \hat{o} pital's Rule. \end{matrix} \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int _{-\infty }^{0}x{e}^{x}dx}& ={\underset{t\to -\infty}\lim}{\displaystyle\int _{t}^{0}x{e}^{x}dx}\hfill & & & \text{Rewrite as a limit.}\hfill \\ & ={\underset{t\to -\infty}\lim}\left(x{e}^{x}-{e}^{x}\right)\Biggr|_{t}^{0} \hfill & & & \begin{array}{c}\text{Use integration by parts to find the} \hfill \\ \text{antiderivative. (Here } u=x \text{ and } dv=e_{dv}^{x}\text{.)}\end{array} \\ & ={\underset{t\to -\infty}\lim}\left(-1-t{e}^{t}+{e}^{t}\right)\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =-1.\hfill & & & \begin{array}{c}\text{Evaluate the limit.}\mathit{\text{Note:}} {\underset{t\to -\infty}\lim}t{e}^{t}\text{is}\hfill \\ \text{indeterminate of the form}0\cdot \infty .\text{Thus,}\hfill \\ {\underset{t\to -\infty}\lim}t{e}^{t}={\underset{t\to -\infty}\lim}\frac{t}{{e}^{\text{-}t}}={\underset{t\to -\infty}\lim}\frac{-1}{{e}^{-t}}={\underset{t\to -\infty}\lim}-{e}^{t}=0\text{by}\hfill \\ \text{L'H}\hat{o}\text{pital's Rule.}\hfill \end{array}\hfill \end{array}$

The first improper integral converges. For the second integral,

\begin{array}{ccccc} \int_{0}^{+ \infty} x e^{x} d x & = lim_{t \to + \infty} \int_{0}^{t} x e^{x} d x & Rewrite as a limit. \\ = lim_{t \to + \infty} (x e^{x} - e^{x}) |_{t}^{0} & Find the antiderivative. \\ = lim_{t \to + \infty} (t e^{t} - e^{t} + 1) & Evaluate the antiderivative. \\ = lim_{t \to + \infty} ((t - 1) e^{t} + 1) & Rewrite. (t e^{t} - e^{t} is indeterminate.) \\ = + \infty . & Evaluate the limit. \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int _{0}^{+\infty }x{e}^{x}dx}& =\underset{t\to +\infty }{\text{lim}}{\displaystyle\int _{0}^{t}x{e}^{x}dx}\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\underset{t\to \text{+}\infty }{\text{lim}}\left(x{e}^{x}-{e}^{x}\right)\Biggr|_{t}^{0} \hfill & & & \text{Find the antiderivative.}\hfill \\ & =\underset{t\to \text{+}\infty }{\text{lim}}\left(t{e}^{t}-{e}^{t}+1\right)\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =\underset{t\to \text{+}\infty }{\text{lim}}\left(\left(t - 1\right){e}^{t}+1\right)\hfill & & & \text{Rewrite.} (t e^t-e^t \text{ is indeterminate.)} \hfill \\ & =\text{+}\infty .\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$

Thus, ${\displaystyle\int }_{0}^{+\infty }x{e}^{x}dx$ diverges. Since this integral diverges, ${\displaystyle\int }_{\text{-}\infty }^{+\infty }x{e}^{x}dx$ diverges as well.

try it

Evaluate ${\displaystyle\int }_{-3}^{+\infty }{e}^{\text{-}x}dx$ . State whether the improper integral converges or diverges.

Hint

${\displaystyle\int }_{-3}^{+\infty }{e}^{\text{-}x}dx=\underset{t\to \text{+}\infty }{\text{lim}}{\displaystyle\int }_{-3}^{t}{e}^{\text{-}x}dx$

Show Solution

${e}^{3}$ , converges

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Try It

Integrating a Discontinuous Integrand

Now let’s examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form ${\displaystyle\int }_{a}^{b}f\left(x\right)dx$ , where $f\left(x\right)$ is continuous over $\left[a,b\right)$ and discontinuous at $b$ . Since the function $f\left(x\right)$ is continuous over $\left[a,t\right]$ for all values of $t$ satisfying [latex]a

This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and increasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. There is also a vertical asymptote at x = b. The region in the first curve is small, and progressively gets wider under the second and third graph as t gets further from a, and closer to b on the x-axis.

We use a similar approach to define ${\displaystyle\int }_{a}^{b}f\left(x\right)dx$ , where $f\left(x\right)$ is continuous over $\left(a,b\right]$ and discontinuous at $a$ . We now proceed with a formal definition.

Definition

Let $f\left(x\right)$ be continuous over $\left[a,b\right)$ . Then,

${\displaystyle\int }_{a}^{b}f\left(x\right)dx=\underset{t\to {b}^{-}}{\text{lim}}{\displaystyle\int }_{a}^{t}f\left(x\right)dx$ .
Let $f\left(x\right)$ be continuous over $\left(a,b\right]$ . Then,

${\displaystyle\int }_{a}^{b}f\left(x\right)dx=\underset{t\to {a}^{+}}{\text{lim}}{\displaystyle\int }_{t}^{b}f\left(x\right)dx$ .

In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.
If $f\left(x\right)$ is continuous over $\left[a,b\right]$ except at a point $c$ in $\left(a,b\right)$ , then

${\displaystyle\int }_{a}^{b}f\left(x\right)dx={\displaystyle\int }_{a}^{c}f\left(x\right)dx+{\displaystyle\int }_{c}^{b}f\left(x\right)dx$ ,

provided both ${\displaystyle\int }_{a}^{c}f\left(x\right)dx$ and ${\displaystyle\int }_{c}^{b}f\left(x\right)dx$ converge. If either of these integrals diverges, then ${\displaystyle\int }_{a}^{b}f\left(x\right)dx$ diverges.

The following examples demonstrate the application of this definition.

Example: Integrating a Discontinuous Integrand

Evaluate ${\displaystyle\int }_{0}^{4}\frac{1}{\sqrt{4-x}}dx$ , if possible. State whether the integral converges or diverges.

Show Solution

The function $f\left(x\right)=\frac{1}{\sqrt{4-x}}$ is continuous over $\left[0,4\right)$ and discontinuous at 4. Using equation 1 from the definition, rewrite ${\displaystyle\int }_{0}^{4}\frac{1}{\sqrt{4-x}}dx$ as a limit:

\begin{array}{ccccc} \int_{0}^{4} \frac{1}{\sqrt{4 - x}} d x & = lim_{t \to 4^{-}} \int_{0}^{t} \frac{1}{\sqrt{4 - x}} d x & Rewrite as a limit. \\ = lim_{t \to 4^{-}} (- 2 \sqrt{4 - x}) |_{\begin{matrix} 0 \end{matrix}}^{\begin{matrix} t \end{matrix}} & Find the antiderivative. \\ = lim_{t \to 4^{-}} (- 2 \sqrt{4 - t} + 4) & Evaluate the antiderivative. \\ = 4. & Evaluate the limit. \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int }_{0}^{4}\frac{1}{\sqrt{4-x}}dx& =\underset{t\to {4}^{-}}{\text{lim}}{\displaystyle\int }_{0}^{t}\frac{1}{\sqrt{4-x}}dx\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\underset{t\to {4}^{-}}{\text{lim}}\left(-2\sqrt{4-x}\right)|{}_{\begin{array}{c}\\ 0\end{array}}^{\begin{array}{c}t\\ \end{array}}\hfill & & & \text{Find the antiderivative.}\hfill \\ & =\underset{t\to {4}^{-}}{\text{lim}}\left(-2\sqrt{4-t}+4\right)\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =4.\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$

The improper integral converges.

Example: Integrating a Discontinuous Integrand

Evaluate ${\displaystyle\int }_{0}^{2}x\text{ln}xdx$ . State whether the integral converges or diverges.

Show Solution

Since $f\left(x\right)=x\ln{x}$ is continuous over $\left(0,2\right]$ and is discontinuous at zero, we can rewrite the integral in limit form using equation 2 from the definition:

\begin{array}{ccccc} \int_{0}^{2} x ln x d x & = lim_{t \to 0^{+}} \int_{t}^{2} x ln x d x & Rewrite as a limit. \\ = lim_{t \to 0^{+}} (\frac{1}{2} x^{2} ln x - \frac{1}{4} x^{2}) |_{\begin{matrix} t \end{matrix}}^{\begin{matrix} 2 \end{matrix}} & \begin{matrix} Evaluate \int x \ln x d x using integration by parts \\ with u = ln x and d v = x . \end{matrix} \\ = lim_{t \to 0^{+}} (2 ln 2 - 1 - \frac{1}{2} t^{2} ln t + \frac{1}{4} t^{2}) . & Evaluate the antiderivative. \\ = 2 ln 2 - 1. & \begin{matrix} Evaluate the limit. lim_{t \to 0^{+}} t^{2} \ln t is indeterminate. \\ To evaluate it, rewrite as a quotient and apply \\ L'h \hat{o} pital's rule. \end{matrix} \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int }_{0}^{2}x\text{ln}xdx& =\underset{t\to {0}^{+}}{\text{lim}}{\displaystyle\int }_{t}^{2}x\text{ln}xdx\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\underset{t\to {0}^{+}}{\text{lim}}\left(\frac{1}{2}{x}^{2}\text{ln}x-\frac{1}{4}{x}^{2}\right)|{}_{\begin{array}{c}\\ t\end{array}}^{\begin{array}{c}2\\ \end{array}}\hfill & & & \begin{array}{c}\text{Evaluate}{\displaystyle\int}x\ln{x}dx\text{ using integration by parts}\hfill \\ \text{with }u=\text{ln}x\text{ and }dv=x.\hfill \end{array}\hfill \\ & =\underset{t\to {0}^{+}}{\text{lim}}\left(2\text{ln}2 - 1-\frac{1}{2}{t}^{2}\text{ln}t+\frac{1}{4}{t}^{2}\right).\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =2\text{ln}2 - 1.\hfill & & & \begin{array}{c}\text{Evaluate the limit.}\underset{t\to {0}^{+}}{\text{lim}}{t}^{2}\ln{t}\text{ is indeterminate.}\hfill \\ \text{To evaluate it, rewrite as a quotient and apply}\hfill \\ \text{L'h}\hat{o}\text{pital's rule.}\hfill \end{array}\hfill \end{array}$

The improper integral converges.

Example: Integrating a Discontinuous Integrand

Evaluate ${\displaystyle\int }_{-1}^{1}\frac{1}{{x}^{3}}dx$ . State whether the improper integral converges or diverges.

Show Solution

Since $f\left(x\right)=\frac{1}{{x}^{3}}$ is discontinuous at zero, using equation 3 from the definition, we can write

\int_{- 1}^{1} \frac{1}{x^{3}} d x = \int_{- 1}^{0} \frac{1}{x^{3}} d x + \int_{0}^{1} \frac{1}{x^{3}} d x

${\displaystyle\int }_{-1}^{1}\frac{1}{{x}^{3}}dx={\displaystyle\int }_{-1}^{0}\frac{1}{{x}^{3}}dx+{\displaystyle\int }_{0}^{1}\frac{1}{{x}^{3}}dx$ .

If either of the two integrals diverges, then the original integral diverges. Begin with ${\displaystyle\int }_{-1}^{0}\frac{1}{{x}^{3}}dx:$

\begin{array}{ccccc} \int_{- 1}^{0} \frac{1}{x^{3}} d x & = lim_{t \to 0^{-}} \int_{- 1}^{t} \frac{1}{x^{3}} d x & Rewrite as a limit. \\ = lim_{t \to 0^{-}} (- \frac{1}{2 x^{2}}) |_{\begin{matrix} - 1 \end{matrix}}^{\begin{matrix} t \end{matrix}} & Find the antiderivative. \\ = lim_{t \to 0^{-}} (- \frac{1}{2 t^{2}} + \frac{1}{2}) & Evaluate the antiderivative. \\ = + \infty . & Evaluate the limit. \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int }_{-1}^{0}\frac{1}{{x}^{3}}dx& =\underset{t\to {0}^{-}}{\text{lim}}{\displaystyle\int }_{-1}^{t}\frac{1}{{x}^{3}}dx\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\underset{t\to {0}^{-}}{\text{lim}}\left(-\frac{1}{2{x}^{2}}\right)|{}_{\begin{array}{c}\\ -1\end{array}}^{\begin{array}{c}t\\ \end{array}}\hfill & & & \text{Find the antiderivative.}\hfill \\ & =\underset{t\to {0}^{-}}{\text{lim}}\left(-\frac{1}{2{t}^{2}}+\frac{1}{2}\right)\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =\text{+}\infty .\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$

Therefore, ${\displaystyle\int }_{-1}^{0}\frac{1}{{x}^{3}}dx$ diverges. Since ${\displaystyle\int }_{-1}^{0}\frac{1}{{x}^{3}}dx$ diverges, ${\displaystyle\int }_{-1}^{1}\frac{1}{{x}^{3}}dx$ diverges.

try it

Evaluate ${\displaystyle\int }_{0}^{2}\frac{1}{x}dx$ . State whether the integral converges or diverges.

Hint

Write ${\displaystyle\int }_{0}^{2}\frac{1}{x}dx$ in limit form using equation 2 from the definition.

Show Solution

$+\infty$ , diverges

Watch the following video to see the worked solution to the above Try It

You can view the transcript for this segmented clip of “3.7 Improper Integrals” here (opens in new window).

Module 3: Techniques of Integration