Objectives
- Distinguish between proper and improper integrals, and apply the appropriate technique in order to determine if the improper integral converges or diverges
Section 7.8 Improper Integrals
Subsection 7.8.1 Before Class
https://mymedia.ou.edu/media/7.8-1/1_dlr6ol1whttps://mymedia.ou.edu/media/7.8-2/1_3v7uc3qlSubsubsection 7.8.1.1 Type 1 Integrals
Example 7.8.1.
Fill out the following table of values for \(\ds \int_0^a e^{-x}\, dx\text{:}\)
| a | Exact Value | Approximate Value |
| 1 | ||
| 100 | ||
| 1000 | ||
| 10000 |
Based on the table, do you expect the integral to settle on a specific value? If so, what is the value (to 3 decimal places)?
Solution.
| a | Exact Value | Approximate Value |
| 1 | \(1-\dfrac{1}{e}\) | \(0.6321205588\) |
| 100 | \(1-\dfrac{1}{e^{100}}\) | 1 |
| 1000 | \(1-\dfrac{1}{e^{1000}}\) | 1 |
| 10000 | \(1-\dfrac{1}{e^{10000}}\) | 1 |
1
Example 7.8.2.
- Compute \(\ds \int_0^a e^{-x}\, dx\) in general
- Take the limit as \(a\to \infty\text{;}\) what is the interpretation of your answer?
Solution.
- \(\displaystyle \ds \int_0^a e^{-x}\, dx = 1-e^{-a}\)
-
\(\ds \lim_{a\to \infty} (1-e^{-a}) = 1\)As the upper bound goes to \(\infty\text{,}\) the area under \(e^{-x}\) from 0 to a goes to 1.
Definition 7.8.3. Improper Integral (Type 1).
An improper integral of type 1 is an integral of the form
\begin{equation*}
\ds \int_a^\infty f(x)\, dx \qquad \int_{-\infty}^a f(x)\, dx \qquad \int_{-\infty}^\infty f(x)\, dx
\end{equation*}
provided that \(f(x)\) is continuous on the domain of integration.
Definition 7.8.4. Convergence/Divergence (Type 1).
An improper integral of type 1 is said to converge if the limits below exist:
\begin{equation*}
\ds \lim_{t\to\infty} \int_a^t f(x)\, dx \qquad \lim_{t\to -\infty} \int_t^a f(x)\, dx
\end{equation*}
If the limit does not exist or is infinite, then we say the integral is divergent
Example 7.8.5.
Determine if \(\ds \int_1^\infty \dfrac{1}{x^2}\, dx\) converges or diverges. If it converges, give its exact value.
Solution.
The integral converges to 1
Example 7.8.6.
Determine if \(\ds \int_1^\infty \dfrac{1}{x^3}\, dx\) converges or diverges. If it converges, give its exact value.
Solution.
The integral converges to \(\dfrac{1}{2}\)
Example 7.8.7.
Evaluate \(\ds \int_{\infty}^0 xe^x\, dx\)
Solution.
\(\ds \int_{-\infty}^0 xe^x\, dx = -1\)
Example 7.8.8.
Show that \(\ds \int_{-\infty}^{\infty} \dfrac{1}{1+x^2}\, dx = \pi\)
Solution.
Split the integral into two pieces:
\begin{equation*}
\ds \int_{-\infty}^\infty \dfrac{1}{1+x^2}\, dx = \int_{-\infty}^0 \dfrac{1}{1+x^2}\, dx + \int_0^\infty \dfrac{1}{1+x^2}\, dx
\end{equation*}
For the first piece, we compute:
\begin{align*}
\ds \int_{-\infty}^0 \dfrac{1}{1+x^2}\, dx \amp = \lim_{t\to-\infty} \int_t^0 \dfrac{1}{1+x^2}\, dx\\
\ds \lim_{t\to-\infty} \int_t^0 \dfrac{1}{1+x^2}\, dx \amp = \lim_{t\to -\infty} \lrpar{\inv{\tan}(x)}\bigg\rvert_t^0\\
\ds \lim_{t\to -\infty} \lrpar{\inv{\tan}(x)}\bigg\rvert_t^0 \amp = \lim_{t\to -\infty} \lrpar{\inv{\tan}(0)-\inv{\tan}(t)}\\
\amp = \dfrac{\pi}{2}
\end{align*}
For the second piece, we have a similar computation:
\begin{align*}
\ds \int_0^\infty \dfrac{1}{1+x^2}\, dx \amp = \lim_{t\to\infty} \int_0^t \dfrac{1}{1+x^2}\, dx\\
\ds \lim_{t\to\infty} \int_0^t \dfrac{1}{1+x^2}\, dx \amp = \lim_{t\to \infty} \lrpar{\inv{\tan}(x)}\bigg\rvert_0^t\\
\ds \lim_{t\to \infty} \lrpar{\inv{\tan}(x)}\bigg\rvert_0^t \amp = \lim_{t\to \infty} \lrpar{\inv{\tan}(t)-\inv{\tan}(0)}\\
\amp = \dfrac{\pi}{2}
\end{align*}
Summing the two pieces gives the answer of \(\pi\)
Example 7.8.9.
Does \(\ds \int_{-\infty}^0 \dfrac{1}{2-4x}\, dx\) converge or diverge? Why?
Solution.
The intgral diverges since the antiderivative involves a logarithm. The end behavior of the log function is unbounded, so the integral diverges.
Subsection 7.8.2 Pre-Class Activities
Example 7.8.10.
Use this space to write any questions you might have from the videos.
Solution.
Answers vary
Example 7.8.11.
Evaluate \(\ds \int_2^\infty e^{-5p}\, dp\)
Solution.
\(\ds \int_2^\infty e^{-5p}\, dp = \dfrac{1}{5}e^{-10}\)
Example 7.8.12.
Determine if \(\ds \int_{-\infty}^{\infty} xe^{-x^2}\, dx\) converges or diverges
Solution.
The integral converges to 0
Example 7.8.13.
Determine if \(\ds \int_{-\infty}^0 \dfrac{z}{z^4 + 4}\, dz\) converges or diverges.
Solution.
The integral converges to \(-\dfrac{\pi}{8}\)
Example 7.8.14.
Evaluate \(\ds \int_e^\infty \dfrac{1}{x(\ln x)^2}\, dx\)
Solution.
\(\ds \int_e^\infty \dfrac{1}{x(\ln x)^2}\, dx = 1\)
Example 7.8.15.
The integral \(\ds \int_0^\infty \sin^2\alpha\, d\alpha\) diverges. Compute the integral to see why. How could we come to the same conclusion without any computation?
Solution.
\(\ds \int \sin^2\alpha\, d\alpha = -\dfrac{1}{4}\sin 2\alpha + \dfrac{1}{2}\alpha\text{,}\) so when we take the limit of the linear term, we will always get \(\infty\) and a divergent integral (even though the limit of the trig term doesn’t exist).
Without integrating, we could also see that the integral would diverge because \(\sin^2\alpha \geq 0 \) so the area is always positive and only accumulates positive values.
Subsection 7.8.3 In Class
Example 7.8.16.
For what values of \(p\) does \(\ds \int_1^\infty \dfrac{1}{x^p}\, dx\) converge?
Solution.
It converges to \(\dfrac{1}{p-1}\) for \(p \gt 1\)
Example 7.8.17.
Evaluate \(\ds \int_0^\infty \dfrac{x^2}{\sqrt{1+x^3}}\, dx\)
Solution.
The integral diverges
Example 7.8.18.
Compute \(\ds \int_{-\infty}^0 2^r\, dr\)
Solution.
\(\ds \int_{-\infty}^0 2^r\, dr = \dfrac{1}{\ln 2}\)
Subsubsection 7.8.3.1 Type 2 Integrals
Definition 7.8.19. Improper Integral (Type 2).
An improper integral of type 2 is an integral of the form
\begin{equation*}
\ds \int_a^b f(x)\, dx
\end{equation*}
where \(f(x)\) has a discontinuity at either \(a\) or \(b\) or in \([a,b]\text{.}\)
Definition 7.8.20. Convergence/Divergence (Type 2).
An improper integral of type 2 is said to converge if the limits below exist:
\begin{equation*}
\ds \lim_{t\to a^+} \int_t^b f(x)\, dx \qquad \lim_{t\to b^-} \int_a^t f(x)\, dx
\end{equation*}
If the limit does not exist or is infinite, then we say the integral is divergent.
Example 7.8.21.
Find \(\ds \int_1^4 \dfrac{1}{\sqrt{x-1}}\, dx\)
Solution.
\(\ds \int_1^4 \dfrac{1}{\sqrt{x-1}}\, dx = 2\sqrt{3}\)
Example 7.8.22.
Determine if \(\ds \int_0^{\pi/2} \csc \theta \, d\theta\) converges or diverges.
Solution.
The integral diverges
Example 7.8.23.
A classmate claims that \(\ds \int_0^5 \dfrac{1}{x-3}\, dx = \ln 2 - \ln 3\text{.}\) Are they correct or incorrect? Why?
Solution.
They are incorrect. If the integral is computed using FTC 2 (disregarding the discontinuity), then one would get this value. However, accounting for the discontinuity shows that the integral diverges.
Example 7.8.24.
Compute \(\ds \int_0^1 \ln x\, dx\)
Solution.
The integral diverges
Example 7.8.25.
If \(f(t)\) is continuous on \((0,\infty)\text{,}\) we define the Laplace transform \(\mathcal{L}\) of \(f\) to be the function \(F\) defined as
\begin{equation*}
\mathcal{L}[f(t)] = F(s) = \int_0^\infty f(t)e^{-st}\, dt
\end{equation*}
Compute \(\mathcal{L}[1]\) and \(\mathcal{L}[t]\text{.}\)
Solution.
\(\mathcal{L}[1] = \dfrac{1}{s}\) and \(\mathcal{L}[t]=\dfrac{1}{s^2}\)
Example 7.8.26.
Evaluate \(\ds \int_0^1 \dfrac{dx}{\sqrt{1-x^2}}\)
Solution.
\(\ds \int_0^1 \dfrac{dx}{\sqrt{1-x^2}} = \dfrac{\pi}{2}\)
Subsection 7.8.4 After Class Activities
Example 7.8.27.
Determine if the following converge or diverge:
- \(\displaystyle \ds \int_0^\infty \dfrac{1}{x^2 + x}\, dx\)
- \(\displaystyle \ds \int_1^\infty \dfrac{\ln x}{x^2}\, dx\)
- \(\displaystyle \ds \int_2^\infty \dfrac{dv}{v^2 + 2v-3}\)
- \(\displaystyle \ds \int_{-1}^2 \dfrac{x}{(x+1)^2}\, dx\)
- \(\displaystyle \ds \int_{-2}^3 \dfrac{1}{x^4}\, dx\)
- \(\displaystyle \ds \int_0^4 \dfrac{1}{x^2-x-2}\, dx\)
Solution.
- \(\ds \int_0^\infty \dfrac{1}{x^2 + x}\, dx\) diverges
- \(\displaystyle \ds \int_1^\infty \dfrac{\ln x}{x^2}\, dx=1\)
- \(\displaystyle \ds \int_2^\infty \dfrac{dv}{v^2 + 2v-3} = \dfrac{1}{4}\ln 5\)
- \(\ds \int_{-1}^2 \dfrac{x}{(x+1)^2}\, dx\) diverges
- \(\ds \int_{-2}^3 \dfrac{1}{x^4}\, dx\) diverges
- \(\ds \int_0^4 \dfrac{1}{x^2-x-2}\, dx =\) diverges
