Improper integrals:

**^_^Engineer_Adam^_^** · September 5th 2007, 07:09 AM

Im stuck with this problem:
$\int_{-\infty}^{+\infty} xe^{-x^2}dx$
kindly help

**Krizalid** · September 5th 2007, 07:23 AM

Use that $\int_{ - \infty }^\infty {f(x)\,dx} = \mathop {\lim }\limits_{a \to - \infty } \int_a^c {f(x)\,dx} + \mathop {\lim }\limits_{b \to \infty } \int_c^b {f(x)\,dx}$

**topsquark** · September 5th 2007, 07:33 AM

Originally Posted by ^_^Engineer_Adam^_^

Im stuck with this problem:
$\int_{-\infty}^{+\infty} xe^{-x^2}dx$
kindly help

Another hint. Let $y = x^2$ .

-Dan

**^_^Engineer_Adam^_^** · September 5th 2007, 07:36 AM

$-\frac{e^{-x^2}}{2}\big|_{-\infty}^{0} +(- \frac{e^{-x^2}}{2})\big|_{0}^{\infty}$

the answer is zero?

**red_dog** · September 5th 2007, 08:56 AM

$\displaystyle\int_{-\infty}^{\infty}xe^{-x^2}dx=\lim_{a\to\infty}\int_{-a}^axe^{-x^2}dx$
$f(x)=xe^{-x^2}$ is an odd function and $[-a,a]$ is a symmetric interval, so the integral is 0, thus the limit is 0.

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