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 [latex]{\displaystyle\int }_{a}^{+\infty }f\left(x\right)dx?[/latex] We can integrate [latex]{\displaystyle\int }_{a}^{t}f\left(x\right)dx[/latex] for any value of [latex]t[/latex], so it is reasonable to look at the behavior of this integral as we substitute larger values of [latex]t[/latex]. Figure 1 shows that [latex]{\displaystyle\int }_{a}^{t}f\left(x\right)dx[/latex] may be interpreted as area for various values of [latex]t[/latex]. 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.
Figure 1. To integrate a function over an infinite interval, we consider the limit of the integral as the upper limit increases without bound.
Definition
Let [latex]f\left(x\right)[/latex] be continuous over an interval of the form [latex]\left[a,\text{+}\infty \right)[/latex]. Then
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 [latex]f\left(x\right)[/latex] be continuous over [latex]\left(\text{-}\infty ,\text{+}\infty \right)[/latex]. Then
provided that [latex]{\displaystyle\int }_{\text{-}\infty }^{0}f\left(x\right)dx[/latex] and [latex]{\displaystyle\int }_{0}^{+\infty }f\left(x\right)dx[/latex] both converge. If either of these two integrals diverge, then [latex]{\displaystyle\int }_{\text{-}\infty }^{+\infty }f\left(x\right)dx[/latex] diverges. (It can be shown that, in fact, [latex]{\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[/latex] for any value of [latex]a.[/latex])
In our first example, we return to the question we posed at the start of this section: Is the area between the graph of [latex]f\left(x\right)=\frac{1}{x}[/latex] and the [latex]x[/latex] -axis over the interval [latex]\left[1,\text{+}\infty \right)[/latex] finite or infinite?
Example: Finding an Area
Determine whether the area between the graph of [latex]f\left(x\right)=\frac{1}{x}[/latex] and the x-axis over the interval [latex]\left[1,\text{+}\infty \right)[/latex] is finite or infinite.
Show Solution
We first do a quick sketch of the region in question, as shown in the following graph.
Figure 2. We can find the area between the curve [latex]f\left(x\right)=\frac{1}{x}[/latex] and the x-axis on an infinite interval.
We can see that the area of this region is given by [latex]A={\displaystyle\int }_{1}^{\infty }\frac{1}{x}dx[/latex]. Then we have
[latex]\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}[/latex]
Since the improper integral diverges to [latex]+\infty[/latex], 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 [latex]f\left(x\right)=\frac{1}{x}[/latex] and the x-axis over the interval [latex]\left[1,\text{+}\infty \right)[/latex] about the [latex]x[/latex] -axis.
Show Solution
The solid is shown in Figure 3. Using the disk method, we see that the volume V is
The improper integral converges to [latex]\pi[/latex]. Therefore, the volume of the solid of revolution is [latex]\pi[/latex].
In conclusion, although the area of the region between the x-axis and the graph of [latex]f\left(x\right)=\frac{1}{x}[/latex] over the interval [latex]\left[1,\text{+}\infty \right)[/latex] 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.
Example: Evaluating an Improper Integral over an Infinite Interval
Evaluate [latex]{\displaystyle\int }_{\text{-}\infty }^{0}\frac{1}{{x}^{2}+4}dx[/latex]. State whether the improper integral converges or diverges.
Show Solution
Begin by rewriting [latex]{\displaystyle\int }_{\text{-}\infty }^{0}\frac{1}{{x}^{2}+4}dx[/latex] as a limit using the equation 2 from the definition. Thus,
[latex]\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}[/latex]
The improper integral converges to [latex]\frac{\pi }{4}[/latex].
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 [latex]f[/latex] and [latex]g[/latex] are differentiable functions over an open interval [latex]\left(a, \infty \right)[/latex] for some value of [latex]a[/latex]. If either:
[latex]\underset{x\to \infty}{\lim}f(x)=0[/latex] and [latex]\underset{x\to \infty}{\lim}g(x)=0[/latex]
[latex]\underset{x\to \infty}{\lim}f(x)= \infty[/latex] (or [latex]-\infty[/latex]) and [latex]\underset{x\to \infty}{\lim}g(x)= \infty[/latex] (or [latex]-\infty[/latex]), then
If either [latex]{\displaystyle\int }_{\text{-}\infty }^{0}x{e}^{x}dx[/latex] or [latex]{\displaystyle\int }_{0}^{+\infty }x{e}^{x}dx[/latex] diverges, then [latex]{\displaystyle\int }_{\text{-}\infty }^{+\infty }x{e}^{x}dx[/latex] diverges. Compute each integral separately. For the first integral,
[latex]\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}[/latex]
The first improper integral converges. For the second integral,
Thus, [latex]{\displaystyle\int }_{0}^{+\infty }x{e}^{x}dx[/latex] diverges. Since this integral diverges, [latex]{\displaystyle\int }_{\text{-}\infty }^{+\infty }x{e}^{x}dx[/latex] diverges as well.
try it
Evaluate [latex]{\displaystyle\int }_{-3}^{+\infty }{e}^{\text{-}x}dx[/latex]. State whether the improper integral converges or diverges.
Watch the following video to see the worked solution to the above Try It
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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 [latex]{\displaystyle\int }_{a}^{b}f\left(x\right)dx[/latex], where [latex]f\left(x\right)[/latex] is continuous over [latex]\left[a,b\right)[/latex] and discontinuous at [latex]b[/latex]. Since the function [latex]f\left(x\right)[/latex] is continuous over [latex]\left[a,t\right][/latex] for all values of [latex]t[/latex] satisfying [latex]a
Figure 4. As [latex]t[/latex] approaches b from the left, the value of the area from a to [latex]t[/latex] approaches the area from a to b.
We use a similar approach to define [latex]{\displaystyle\int }_{a}^{b}f\left(x\right)dx[/latex], where [latex]f\left(x\right)[/latex] is continuous over [latex]\left(a,b\right][/latex] and discontinuous at [latex]a[/latex]. We now proceed with a formal definition.
Definition
Let [latex]f\left(x\right)[/latex] be continuous over [latex]\left[a,b\right)[/latex]. Then,
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 [latex]f\left(x\right)[/latex] is continuous over [latex]\left[a,b\right][/latex] except at a point [latex]c[/latex] in [latex]\left(a,b\right)[/latex], then
provided both [latex]{\displaystyle\int }_{a}^{c}f\left(x\right)dx[/latex] and [latex]{\displaystyle\int }_{c}^{b}f\left(x\right)dx[/latex] converge. If either of these integrals diverges, then [latex]{\displaystyle\int }_{a}^{b}f\left(x\right)dx[/latex] diverges.
The following examples demonstrate the application of this definition.
Example: Integrating a Discontinuous Integrand
Evaluate [latex]{\displaystyle\int }_{0}^{4}\frac{1}{\sqrt{4-x}}dx[/latex], if possible. State whether the integral converges or diverges.
Show Solution
The function [latex]f\left(x\right)=\frac{1}{\sqrt{4-x}}[/latex] is continuous over [latex]\left[0,4\right)[/latex] and discontinuous at 4. Using equation 1 from the definition, rewrite [latex]{\displaystyle\int }_{0}^{4}\frac{1}{\sqrt{4-x}}dx[/latex] as a limit:
[latex]\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}[/latex]
The improper integral converges.
Example: Integrating a Discontinuous Integrand
Evaluate [latex]{\displaystyle\int }_{0}^{2}x\text{ln}xdx[/latex]. State whether the integral converges or diverges.
Show Solution
Since [latex]f\left(x\right)=x\ln{x}[/latex] is continuous over [latex]\left(0,2\right][/latex] and is discontinuous at zero, we can rewrite the integral in limit form using equation 2 from the definition:
[latex]\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}[/latex]
The improper integral converges.
Example: Integrating a Discontinuous Integrand
Evaluate [latex]{\displaystyle\int }_{-1}^{1}\frac{1}{{x}^{3}}dx[/latex]. State whether the improper integral converges or diverges.
Show Solution
Since [latex]f\left(x\right)=\frac{1}{{x}^{3}}[/latex] is discontinuous at zero, using equation 3 from the definition, we can write
If either of the two integrals diverges, then the original integral diverges. Begin with [latex]{\displaystyle\int }_{-1}^{0}\frac{1}{{x}^{3}}dx:[/latex]
[latex]\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}[/latex]
Evaluate [latex]{\displaystyle\int }_{0}^{2}\frac{1}{x}dx[/latex]. State whether the integral converges or diverges.
Hint
Write [latex]{\displaystyle\int }_{0}^{2}\frac{1}{x}dx[/latex] in limit form using equation 2 from the definition.
Show Solution
[latex]+\infty[/latex], diverges
Watch the following video to see the worked solution to the above Try It
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