# Improper integrals

• April 12th 2010, 09:53 PM
monster
Improper integrals
Have been going through evaluating improper integrals by expressing them as limits and have got to the following,

integral from 0 to infintiy of 1/(x^3) with respect to x,

evaluating this from 1 to infinity would seem to make sense to me but from 0 i'm unsure how to proceed.

Thanks for any help in advance
• April 12th 2010, 10:22 PM
Prove It
Quote:

Originally Posted by monster
Have been going through evaluating improper integrals by expressing them as limits and have got to the following,

integral from 0 to infintiy of 1/(x^3) with respect to x,

evaluating this from 1 to infinity would seem to make sense to me but from 0 i'm unsure how to proceed.

Thanks for any help in advance

$\int_0^{\infty}{\frac{1}{x^3}\,dx} = \lim_{a \to 0}\lim_{b \to \infty}\int_a^b{\frac{1}{x^3}\,dx}$

$= \lim_{a \to 0}\lim_{b \to \infty}\left[-\frac{1}{x^4}\right]_a^b$

$= \lim_{a \to 0}\lim_{b \to \infty}\left[\frac{1}{a^4} - \frac{1}{b^4}\right]$

$= \lim_{a \to 0}\left[\frac{1}{a^4} - 0\right]$

$= \lim_{a \to 0}\frac{1}{a^4}$

$= \infty$.

The integral is divergent.