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

I tried to solve it by integration by parts, but all messed up.

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- Apr 4th 2010, 03:13 AMroshanheroDifficult integration
$\displaystyle \int_{-\infty }^{\infty }xe^\frac{-x^2}{a^2}dx$

I tried to solve it by integration by parts, but all messed up. - Apr 4th 2010, 03:47 AMMoo
Hello,

Note that the function $\displaystyle f(x)=xe^{-x^2/a^2}$ is an odd function. And since it's integrated over an interval that is symmetric to 0, the value of the integral is just 0 ! - Apr 4th 2010, 03:49 AMSoroban
Hello, roshanhero!

This integration is straight-forward . . .

Quote:

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

We have: .$\displaystyle \int e^{-\frac{x^2}{a^2}}\left(x\,dx\right) $

$\displaystyle \text{Let: }\:u \,=\,-\frac{x^2}{a^2} \quad\Rightarrow\quad du \,=\,-\frac{2x}{a^2}\,dx \quad\Rightarrow\quad x\,dx \,=\,-\frac{a^2}{2}\,du $

$\displaystyle \text{Substitute: }\;\int e^u\left(-\frac{a^2}{2}\,du\right) \;=\;-\frac{a^2}{2}\int e^u\,du $

Got it?

- Apr 4th 2010, 04:06 AMmr fantastic
No. The reason this integral is zero is not because the function is odd. If that were the case, then the Cauchy distribution would have a mean equal to zero .... (Surprised)

Of related interest: http://www.mathhelpforum.com/math-he...ancel-out.html - Apr 4th 2010, 04:17 AMKrizalid
in order to say it's zero, we need to prove convergence.

- Apr 4th 2010, 11:57 PMCaptainBlack
$\displaystyle xe^{-\frac{x^2}{a^2}}= - \frac{a^2}{2}\frac{d}{dx}e^{-\frac{x^2}{a^2}}$

Then the fundamental theorem of calculus will allow you to show that

$\displaystyle \lim_{a,b \to \infty}\int_{-a }^{b }xe^\frac{-x^2}{a^2}dx=0$

Alternatively recognise that you are being asked to find a multiple of of the mean of a zero mean normally distributed RV, which is therefore 0.

CB