Improper Integrals: Learn It 1

Giselle Miller

Improper Integrals: Learn It 1

Calculate integrals over infinite intervals
Find integrals when there’s an infinite discontinuity inside your interval
Use the comparison theorem to determine if an improper integral converges

Improper Integrals

Is the area between the graph of [latex]f(x) = \frac{1}{x}[/latex] and the [latex]x[/latex]-axis over the interval [latex][1, +\infty)[/latex] finite or infinite? What about the volume if we revolve this region around the [latex]x[/latex]-axis?

The answer might surprise you: the area is infinite, but the volume is finite. This paradox introduces us to improper integrals—integrals that extend over infinite intervals or involve functions with discontinuities.

improper integrals

Integrals that either have infinite limits of integration or involve functions with discontinuities within the interval of integration.

In this section, we’ll explore how to evaluate these special integrals using limits, discovering when they converge to finite values and when they diverge to infinity.

Integrating over an Infinite Interval

How do we make sense of an integral like [latex]\int_a^{+\infty} f(x) dx[/latex]? We can’t simply plug in infinity as our upper limit.

Instead, we use a limit approach:

First, integrate from [latex]a[/latex] to some finite value [latex]t[/latex]
Then examine what happens as [latex]t[/latex] approaches infinity

Think of it this way: We’re asking “What happens to the area under the curve as we extend our region farther and farther to the right?

Figure 1 below illustrates this concept. As [latex]t[/latex] increases, the shaded region under [latex]f(x)[/latex] gets wider. The improper integral [latex]\int_a^{+\infty} f(x) dx[/latex] represents the limit of these areas as [latex]t \to +\infty[/latex].

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. — Figure 1. To integrate a function over an infinite interval, we consider the limit of the integral as the upper limit increases without bound.

improper integrals with infinite limits

Type 1: Upper limit approaches [latex]+\infty[/latex]
Let [latex]f(x)[/latex] be continuous over [latex][a, +\infty)[/latex]. Then:

[latex]\int_a^{+\infty} f(x) dx = \lim_{t \to +\infty} \int_a^t f(x) dx[/latex]
Type 2: Lower limit approaches [latex]-\infty[/latex]

Let [latex]f(x)[/latex] be continuous over [latex](-\infty, b][/latex]. Then:

[latex]\int_{-\infty}^b f(x) dx = \lim_{t \to -\infty} \int_t^b f(x) dx[/latex]
Type 3: Both limits are infinite
Let [latex]f(x)[/latex] be continuous over [latex](-\infty, +\infty)[/latex]. Then:

[latex]\int_{-\infty}^{+\infty} f(x) dx = \int_{-\infty}^0 f(x) dx + \int_0^{+\infty} f(x) dx[/latex]

Note: You can split at any point [latex]a[/latex], not just [latex]0[/latex].

[latex]\\[/latex]

Convergence vs. Divergence

Converges: The limit exists and equals a finite number
Diverges: The limit doesn’t exist or approaches [latex]\pm\infty[/latex]

For Type 3 integrals, both parts must converge for the whole integral to converge.

Now let’s tackle the question from our introduction: Is the area between [latex]f(x) = \frac{1}{x}[/latex] and the [latex]x[/latex]-axis over [latex][1, +\infty)[/latex] finite or infinite?

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.

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.

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

[latex]\underset{x\to \infty}{\lim}\dfrac{f(x)}{g(x)}=\underset{x\to \infty}{\lim}\dfrac{f^{\prime}(x)}{g^{\prime}(x)}[/latex]

assuming the limit on the right exists or is [latex]\infty[/latex] or [latex]−\infty[/latex].

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

Start by splitting up the integral:

[latex]{\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[/latex].

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,

[latex]\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}[/latex]

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.