Calculus/Improper integrals

In a definite integral ($\underset{a}{\overset{b}{\int }}f(x)dx$) the function has defined intervals and the function itself is continuous. In this section, we deal with integrals of functions where the interval is infinite (type I) or the function has infinite discontinuity in the intervals [a,b] (type II).

Type I: Infinite Integrals
An integral with infinite region includes ±$\mathrm{\infty }$ included in the interval such as $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)dx$. We cannot simply find the antiderivative and plug in $\mathrm{\infty }$. We can however rewrite the integral using a limit. Let $\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{1}{x^2}dx=\underset{b\to \mathrm{\infty }}{\lim }\underset{1}{\overset{b}{\int }}\frac{1}{x^2}dx$
Now this represents a definite integral so we can find the antiderivative and see if the integral converges. $\underset{b\to \mathrm{\infty }}{\lim }\underset{1}{\overset{b}{\int }}\frac{1}{x^2}dx=\underset{b\to \mathrm{\infty }}{\lim }[-\frac{1}{x}{]}_1^b=\underset{b\to \mathrm{\infty }}{\lim }-\frac{1}{b}+1=1$
We can now define the the type 1 integral:
(a) If there is some value $b$ where $b\ge a$ and $\underset{a}{\overset{b}{\int }}f(x)dx$ exists, then
$\underset{a}{\overset{\mathrm{\infty }}{\int }}f(x)dx=\underset{b\to \mathrm{\infty }}{\lim }\underset{a}{\overset{b}{\int }}f(x)dx$

(b) If there is some value $a$ where $a\le b$ and $\underset{a}{\overset{b}{\int }}f(x)dx$ exists, then
$\underset{-\mathrm{\infty }}{\overset{b}{\int }}f(x)dx=\underset{a\to -\mathrm{\infty }}{\lim }\underset{a}{\overset{b}{\int }}f(x)dx$

(c) We can also define $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)dx$ as

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)dx=\underset{-\mathrm{\infty }}{\overset{a}{\int }}f(x)dx+\underset{a}{\overset{\mathrm{\infty }}{\int }}f(x)dx$ assuming that both integrals converge.

**note that if the limits fail to exist, we say that the integral diverges and if the improper integrals yield a finite solution, the integral converges.

Lets look at an example: Evaluate the integral if it converges. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x{e}^{x^{-2}}dx$
$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x{e}^{x^{-2}}dx=\underset{a\to -\mathrm{\infty }}{\lim }\underset{a}{\overset{0}{\int }}x{e}^{x^{-2}}dx+\underset{b\to \mathrm{\infty }}{\lim }\underset{0}{\overset{b}{\int }}x{e}^{x^{-2}}dx$ Use the chain rule to find the antiderivative with $u=-x^2,dx=-\frac{du}{2x}$

$\underset{a\to \mathrm{\infty }}{\lim }\underset{a}{\overset{0}{\int }}x{e}^{x^{-2}}dx=\frac{-e^{(0{)}^{-2}}}{2}-\underset{a\to -\mathrm{\infty }}{\lim }\frac{-e^{(a{)}^{-2}}}{2}=(-\frac{1}{2})-(0)$

$\underset{b\to \mathrm{\infty }}{\lim }\underset{0}{\overset{b}{\int }}x{e}^{x^{-2}}dx=\underset{b\to \mathrm{\infty }}{\lim }\frac{-e^{(b{)}^{-2}}}{2}-\frac{-e^{(0{)}^{-2}}}{2}=(0)+\frac{1}{2}$

$\frac{1}{2}-\frac{1}{2}=0$ This shows that the integral converges to 0.

Type II: Infinite Discontinuity
Integrating a function that contains a vertical asymptote. Template:Calculus:TOC