FANDOM

1,168 Pages

An improper integral (not to be confused with an indefinite integral) is the limit of a definite integral where the endpoints approach a given value. For a given improper integral to exist, it must converge, which it does if the limit exists. An example would be

$\int\limits_1^\infty\dfrac{dx}{x^2}$

This can be solved by taking a limit.

$\lim_{b\to\infty}\int\limits_1^b\dfrac{dx}{x^2}=\lim_{b\to\infty}\left[-\frac{1}{x}\Big|_0^b\right]=\lim_{b\to\infty}\left[-\frac{1}{b}+\frac{1}{1}\right]=1$

Improper integrals result in infinite values when the series is divergent. For instance,

$\int\limits_1^\infty\dfrac{dx}{x}=\lim_{b\to\infty}\int\limits_1^b\frac{dx}{x}=\lim_{b\to\infty}\left[\ln(|x|)\Big|_0^b\right]=\lim_{b\to\infty}\Big[\ln(|b|)-\ln(|1|)\Big]=\infty$

Improper integrals where both endpoints approach infinity can be solved by breaking them into two improper integrals. For example:

$\int\limits_{-\infty}^\infty\dfrac{dx}{1+x^2}=\int\limits_{-\infty}^0\frac{dx}{1+x^2}+\int\limits_0^\infty\frac{dx}{1+x^2}$
$=\lim_{b\to\infty}\int\limits_{-b}^0\dfrac{dx}{1+x^2}+\lim_{b\to\infty}\int\limits_0^b\frac{dx}{1+x^2}=\lim_{b\to\infty}\left[\arctan(x)\Big|_{-b}^0\right]+\lim_{b\to\infty}\left[\arctan(x)\Big|_0^b\right]$
$=\lim_{b\to\infty}\Big[\arctan(0)-\arctan(-b)\Big]+\lim_{b\to\infty}\Big[\arctan(b)-\arctan(0)\Big]=\frac{\pi}{2}+\frac{\pi}{2}=\pi$

Improper integrals can also be used when over a defined area when there is a vertical asymptote. For example,

$\int\limits_0^1\dfrac{dx}{\sqrt{x}}=\lim_{a\to0^+}\int\limits_a^1\frac{dx}{\sqrt{x}}=\lim_{a\to0^+}\left[2\sqrt{x}\Big|^1_a\right]=\lim_{a\to0^+}\Big[2\sqrt1-2\sqrt{a}\Big]=2$

Examples of improper integrals

Community content is available under CC-BY-SA unless otherwise noted.