# 3.7 Improper integrals  (Page 3/6)

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

${e}^{3},$ converges

## 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 ${\int }_{a}^{b}f\left(x\right)dx,$ 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 $a the integral ${\int }_{a}^{t}f\left(x\right)dx$ is defined for all such values of $t.$ Thus, it makes sense to consider the values of ${\int }_{a}^{t}f\left(x\right)dx$ as $t$ approaches $b$ for $a That is, we define ${\int }_{a}^{b}f\left(x\right)dx=\underset{t\to {b}^{-}}{\text{lim}}{\int }_{a}^{t}f\left(x\right)dx,$ provided this limit exists. [link] illustrates ${\int }_{a}^{t}f\left(x\right)dx$ as areas of regions for values of $t$ approaching $b.$

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

## Definition

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

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.
3. If $f\left(x\right)$ is continuous over $\left[a,b\right]$ except at a point $c$ in $\left(a,b\right),$ then
${\int }_{a}^{b}f\left(x\right)dx={\int }_{a}^{c}f\left(x\right)dx+{\int }_{c}^{b}f\left(x\right)dx,$

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

The following examples demonstrate the application of this definition.

## Integrating a discontinuous integrand

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

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

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

The improper integral converges.

## Integrating a discontinuous integrand

Evaluate ${\int }_{0}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx.$ State whether the integral converges or diverges.

Since $f\left(x\right)=x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x$ is continuous over $\left(0,2\right]$ and is discontinuous at zero, we can rewrite the integral in limit form using [link] :

$\begin{array}{ccccc}\hfill {\int }_{0}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx& =\underset{t\to {0}^{+}}{\text{lim}}{\int }_{t}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\underset{t\to {0}^{+}}{\text{lim}}\left(\frac{1}{2}{x}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}x-\frac{1}{4}{x}^{2}\right)|{}_{\begin{array}{c}\\ t\end{array}}^{\begin{array}{c}2\\ \end{array}}\hfill & & & \begin{array}{c}\text{Evaluate}\phantom{\rule{0.2em}{0ex}}{\int }^{\text{​}}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}\text{using integration by parts}\hfill \\ \text{with}\phantom{\rule{0.2em}{0ex}}u=\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}dv=x.\hfill \end{array}\hfill \\ & =\underset{t\to {0}^{+}}{\text{lim}}\left(2\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2-1-\frac{1}{2}{t}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}t+\frac{1}{4}{t}^{2}\right).\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =2\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2-1.\hfill & & & \begin{array}{c}\text{Evaluate the limit.}\phantom{\rule{0.2em}{0ex}}\underset{t\to {0}^{+}}{\text{lim}}{t}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.2em}{0ex}}\text{is indeterminate.}\hfill \\ \text{To evaluate it, rewrite as a quotient and apply}\hfill \\ \text{L’Hôpital’s rule.}\hfill \end{array}\hfill \end{array}$

The improper integral converges.

## Integrating a discontinuous integrand

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

Since $f\left(x\right)=1\text{/}{x}^{3}$ is discontinuous at zero, using [link] , we can write

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

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

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

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

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