# Thread: Tough integral

1. ## Tough integral

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

Thank you

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

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

Can you take it from here? Post back if you get stuck!

3. Originally Posted by ankuoli
$\displaystyle \int x^2*e^(-x^2/\theta)$ 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.)

4. Originally Posted by kompik
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.)
You can find a closed form since the bounds of integration are from 0 to $\displaystyle \infty$!

5. 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 $\displaystyle \theta=1$.
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: $\displaystyle \frac12\Gamma\left(\frac32\right)=\frac14\sqrt{\pi }$

6. 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

7. 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 $\displaystyle u=\frac1\theta x^2$ then $\displaystyle du=\frac2\theta xdx$. Can you continue from there?

8. 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.

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

and then do uv-$\displaystyle \int v du$ ???
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 $\displaystyle u=x$ and $\displaystyle v'=xe^{-x^2/2}$. (I do not know, what notation are you used to.)

10. so how exactly do i use u in the gamma function? sorry i'm very confused

11. Originally Posted by ankuoli
so how exactly do i use u in the gamma function? sorry i'm very confused
Good start to learn the basics about this function could be wikipedia.