$\displaystyle \int_{-\infty }^{\infty }xe^\frac{-x^2}{a^2}dx$
I tried to solve it by integration by parts, but all messed up.
Hello, roshanhero!
This integration is straight-forward . . .
$\displaystyle \int_{-\infty }^{\infty }xe^{-\frac{x^2}{a^2}}dx$
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?
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 ....
Of related interest: http://www.mathhelpforum.com/math-he...ancel-out.html
$\displaystyle xe^{-\frac{x^2}{a^2}}= - \frac{a^2}{2}\frac{d}{dx}e^{-\frac{x^2}{a^2}}$
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