Improper Integrals
Improper Integrals with Infinite Integration Limits
1. If $f(x)$ is continuous on $[a, \infty)$, then
$
\int_a^{\infty} f(x) d x=\lim _{b \rightarrow \infty} \int_a^b f(x) d x
$
2. If $f(x)$ is continuous on $(-\infty, b]$, then
$
\int_{-\infty}^b f(x) d x=\lim _{a \rightarrow-\infty} \int_a^b f(x) d x
$
3. If $f(x)$ is continuous on $(-\infty, \infty)$, then
$
\int_{-\infty}^{\infty} f(x) d x=\int_{-\infty}^c f(x) d x+\int_c^{\infty} f(x) d x
$
where $c$ is any real number.
Improper Integrals with Infinite Discontinuities
1. If $f(x)$ is continuous on $[a, b)$ and has an infinite discontinuity at $b$, then
$
\int_a^b f(x) d x=\lim _{c \rightarrow b^{-}} \int_a^c f(x) d x
$
2. If $f(x)$ is continuous on $(a, b]$ and has an infinite discontinuity at $a$, then
$
\int_a^b f(x) d x=\lim _{c \rightarrow a^{+}} \int_c^b f(x) d x
$
3. If $f(x)$ is continuous on $[a, b]$, except for some number $c$ in $(a, b)$ at which $f$ has an infinite discontinuity, then
$
\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x
$
where $c$ is any real number.
In each case, if the limit is finite we say that the improper integral converges and that the limit is the value of the improper integral. If the limit fails to exists, the improper integral diverges.
Example 1
- Evaluate $\int_0^{\infty} x e^{-x^2} d x$
▶️Answer/Explanation
Solution
$
\begin{aligned}
& \int_0^{\infty} x e^{-x^2} d x \\
& =\lim _{b \rightarrow \infty} \int_0^b x e^{-x^2} d x=\lim _{b \rightarrow \infty}\left[-\frac{1}{2} e^{-x^2}\right]_0^b \\
& =-\frac{1}{2} \lim _{b \rightarrow \infty}\left[e^{-b^2}-e^0\right]=-\frac{1}{2}(0-1)=\frac{1}{2}
\end{aligned}
$
Example2
- Evaluate $\int_{-\infty}^{\infty} \frac{d x}{1+x^2}$
▶️Answer/Explanation
Solution
$\begin{aligned} & \int_{-\infty}^{\infty} \frac{d x}{1+x^2}=\int_{-\infty}^0 \frac{d x}{1+x^2}+\int_0^{\infty} \frac{d x}{1+x^2} \\ & =\lim _{a \rightarrow-\infty} \int_a^0 \frac{d x}{1+x^2}+\lim _{b \rightarrow \infty} \int_0^b \frac{d x}{1+x^2} \\ & =\lim _{a \rightarrow-\infty}\left[\tan ^{-1} x\right]_a^0+\lim _{b \rightarrow \infty}\left[\tan ^{-1} x\right]_0^b \\ & =\lim _{a \rightarrow-\infty}\left(\tan ^{-1} 0-\tan ^{-1} a\right)+\lim _{b \rightarrow \infty}\left(\tan ^{-1} b-\tan ^{-1} 0\right) \\ & =\left(0-\left(-\frac{\pi}{2}\right)\right)+\left(\frac{\pi}{2}-0\right)=\pi\end{aligned}$
Example3
- Evaluate $\int_1^5 \frac{d x}{\sqrt{x-1}}$
▶️Answer/Explanation
Solution
$\begin{aligned} & \int_1^5 \frac{d x}{\sqrt{x-1}} \\ & =\lim _{b \rightarrow 1^{+}} \int_b^5 \frac{d x}{\sqrt{x-1}}=\lim _{b \rightarrow 1^{+}}[2 \sqrt{x-1}]_b^5 \\ & =\lim _{b \rightarrow 1^{+}}[2 \sqrt{4}-2 \sqrt{b-1}] \\ & =4\end{aligned}$
Example 5
Find $\int_0^1 \frac{d x}{1-x}$.
▶️Answer/Explanation
Solution
$\begin{aligned} & \int_0^1 \frac{d x}{1-x} \\ & =\lim _{b \rightarrow 1^{-}} \int_0^b \frac{d x}{1-x}=\lim _{b \rightarrow 1^{-}}[-\ln |1-x|]_0^b \\ & =\lim _{b \rightarrow 1^{-}}[-\ln |1-b|+\ln 1] \\ & =\infty\end{aligned}$
Example6
$
\int_2^{\infty} \frac{1}{\sqrt{x-1}} d x=
$
(A) $-\infty$
(B) -2
(C) 1
(D) $\infty$
▶️Answer/Explanation
Ans:D
Example7
- $
\int_0^{\infty} \frac{1}{(x+3)(x+4)} d x=
$
(A) $-\ln \frac{4}{3}$
(B) $-\ln \frac{3}{4}$
(C) 0
(D) $\ln 4$
▶️Answer/Explanation
Ans:B
Example8
- $
\int_0^{\infty} x^2 e^{-x^3}=
$
(A) $\frac{1}{3}$
(B) $\frac{1}{2}$
(C) 1
(D) divergent
▶️Answer/Explanation
Ans:C