$\displaystyle \int x^2*e^(-x^2/\theta)$ the bound are from 0 to infinity but i mostly would like to know how to go about this integral.
Thank you
To make things simpler, I will assume $\displaystyle \theta=1$.
Integration by parts gives:
$\displaystyle \int x. xe^{-x^2} dx= -\frac12e^{-x^2}+\frac12\int e^{-x^2}dx$
So I think, that with some effort you can express it using Phi-function or some of the related functions, but you cannot expect a closed form using only elementary functions.
(I hope I haven't made some embarrassing error in the calculation - but plenty of attentive eyes are here, ready to spot any mistake.)
Indeed, there was an error - forgotten x.
I still assume $\displaystyle \theta=1$.
Integration by parts gives:
$\displaystyle \int_0^\infty x. xe^{-x^2} dx= \left[-\frac12 xe^{-x^2}\right]_0^\infty+\frac12\int_0^\infty e^{-x^2}dx=\frac14\sqrt\pi$
The result obtained Gamma function is the same: $\displaystyle \frac12\Gamma\left(\frac32\right)=\frac14\sqrt{\pi }$
Do i then have dv=$\displaystyle x^2 dx$ and v=$\displaystyle 1/3*x^3$
and then do uv-$\displaystyle \int v du$ ???
Also to add to the fun of this problem:
I know that the distribution of $\displaystyle x^2$ is an $\displaystyle exp(\theta)$
and the whole integral is $\displaystyle (2x^2/\theta) * e^(-x^2/\theta)$.
Ideally I would like this integral to equal 1/2 but i have a feeling it's not going to happen.
Sorry if I confused you, but when I mentioned u, I was not talking about integration by parts but about Chris' suggestion with Gamma-function.
In the integration by parts I took $\displaystyle u=x$ and $\displaystyle v'=xe^{-x^2/2}$. (I do not know, what notation are you used to.)