# Tough integral

• Apr 29th 2010, 08:37 AM
ankuoli
Tough integral
$\int x^2*e^(-x^2/\theta)$ the bound are from 0 to infinity but i mostly would like to know how to go about this integral.

Thank you
• Apr 29th 2010, 08:47 AM
Chris L T521
Quote:

Originally Posted by ankuoli
$\int x^2*e^(-x^2/\theta)$ the bound are from 0 to infinity but i mostly would like to know how to go about this integral.

Thank you

Hint: Make the substitution $u=\tfrac{1}{\theta}x^2$. Then apply the definition of the Gamma function.

Can you take it from here? Post back if you get stuck! :)
• Apr 29th 2010, 08:59 AM
kompik
Quote:

Originally Posted by ankuoli
$\int x^2*e^(-x^2/\theta)$ 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 $\theta=1$.
Integration by parts gives:
$\int x. xe^{-x^2} dx= -\frac12e^{-x^2}+\frac12\int e^{-x^2}dx$

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.)
• Apr 29th 2010, 09:02 AM
Chris L T521
Quote:

Originally Posted by kompik
To make things simpler, I will assume $\theta=1$.
Integration by parts gives:
$\int x. xe^{-x^2} dx= -\frac12e^{-x^2}+\frac12\int e^{-x^2}dx$

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.)

You can find a closed form since the bounds of integration are from 0 to $\infty$!
• Apr 29th 2010, 09:34 AM
kompik
Quote:

Originally Posted by kompik
(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 $\theta=1$.
Integration by parts gives:
$\int_0^\infty x. xe^{-x^2} dx= \left[-\frac12 xe^{-x^2}\right]_0^\infty+\frac12\int_0^\infty e^{-x^2}dx=\frac14\sqrt\pi$

The result obtained Gamma function is the same: $\frac12\Gamma\left(\frac32\right)=\frac14\sqrt{\pi }$
• Apr 29th 2010, 09:37 AM
ankuoli
i have not done integration by parts or substitution in a very long time, so i'm a bit stuck. But i do like the idea of the gamma function. Any help on substitution to get started would be great.

thanks
• Apr 29th 2010, 09:40 AM
kompik
Quote:

Originally Posted by ankuoli
i have not done integration by parts or substitution in a very long time, so i'm a bit stuck. But i do like the idea of the gamma function. Any help on substitution to get started would be great.

thanks

If $u=\frac1\theta x^2$ then $du=\frac2\theta xdx$. Can you continue from there?
• Apr 29th 2010, 10:01 AM
ankuoli
Do i then have dv= $x^2 dx$ and v= $1/3*x^3$

and then do uv- $\int v du$ ???

Also to add to the fun of this problem:

I know that the distribution of $x^2$ is an $exp(\theta)$

and the whole integral is $(2x^2/\theta) * e^(-x^2/\theta)$.

Ideally I would like this integral to equal 1/2 but i have a feeling it's not going to happen.
• Apr 29th 2010, 10:07 AM
kompik
Quote:

Originally Posted by ankuoli
Do i then have dv= $x^2 dx$ and v= $1/3*x^3$

and then do uv- $\int v du$ ???

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 $u=x$ and $v'=xe^{-x^2/2}$. (I do not know, what notation are you used to.)
• Apr 29th 2010, 10:21 AM
ankuoli
so how exactly do i use u in the gamma function? sorry i'm very confused
• Apr 29th 2010, 10:23 AM
kompik
Quote:

Originally Posted by ankuoli
so how exactly do i use u in the gamma function? sorry i'm very confused