Originally Posted by

**angel.white** Back of the book says the answer is zero, but I keep getting infinity (which I think means it is divergent), and I don't know where the discrepancy lies :/

$\displaystyle \int_{-\infty}^\infty xe^{-x^2}$

$\displaystyle =\lim_{N\to\infty}\int_0^N xe^{-x^2} + \lim_{B\to\infty}\int_{-B}^0 xe^{-x^2}$

$\displaystyle =\lim_{N\to\infty}\left[\frac {-1}2e^{-x^2}\right]_0^N + \lim_{B\to\infty}\left[\frac {-1}2e^{-x^2}\right]_{-B}^0$

$\displaystyle =\lim_{N\to\infty}\left[\frac {-1}{2e^{x^2}}\right]_0^N + \lim_{B\to\infty}\left[\frac {-1}{2e^{x^2}}\right]_{-B}^0$

$\displaystyle =\lim_{N\to\infty}\left[\frac{-1}{2e^{N^2}}+\frac{1}{2e^{0^2}}\right] +\lim_{B\to\infty}\left[\frac {-1}{2e^{0^2}}+\frac {1}{2e^{-B^2}}\right]$ Mr F says: And the mistake is in this line ..... When you sub $\displaystyle \, x = -B \, $ into $\displaystyle \, x^2\, $, the result is $\displaystyle \, B^2\, $, not $\displaystyle \, -B^2\, $ .....

$\displaystyle =\lim_{N\to\infty}\left[\frac{-1}{2e^{N^2}}+\frac{1}{2}\right] +\lim_{B\to\infty}\left[\frac {-1}{2}+\frac {e^{B^2}}{2}\right]$

$\displaystyle =\left[\frac{-1}{2e^{\infty^2}}+\frac{1}{2}\right] +\left[\frac {-1}{2}+\frac {e^{\infty^2}}{2}\right]$

$\displaystyle =\frac{-1}{\infty}+\frac{1}{2} + \frac {-1}{2}+\frac {\infty}{2}$

$\displaystyle =\frac{-1}{\infty}+\frac {\infty}{2}$

$\displaystyle =0+\infty$

$\displaystyle =\infty$