1. Oct 15, 2008

### ahmedxx

hiiii

can any body help me in solving this problem

http://www.up3.cc [Broken]

thanks alot

Moderation Note: Edited irritating title.

Last edited by a moderator: May 3, 2017
2. Oct 15, 2008

### cristo

Staff Emeritus
Re: help me pleeeeeeeeeeease

You need to show some work before we can help you. What happened to the homework posting template?

3. Oct 16, 2008

### Alexitron

http://blue-whitegt.com/covers/Integral.JPG [Broken]

Last edited by a moderator: May 3, 2017
4. Oct 16, 2008

### ahmedxx

thanks Alexitron

you help me alot

I'm now reading on Gamma function.

regards.....

5. Oct 16, 2008

### HallsofIvy

Staff Emeritus
To do this specific integral, you don't need the Gamma function.

If
$$I= \int_{-\infty}^\infty e^{-\frac{t^2}{2}} dt$$
then, because of the symmetry,
$$\frac{I}{2}= \int_0^\infty e^{-\frac{t^2}{2}}dt$$
which also means that
$$\frac{I}{2}= \int_0^\infty e^{-\frac{x^2}{2}}dx$$
and
$$\frac{I}{2}= \int_0^\infty e^{-\frac{y^2}{2}}dy$$
so that
$$\frac{I^2}{4}= \left(\int_0^\infty e^{-\frac{x^2}{2}}dx\right)\left( \int_0^\infty e^{-\frac{y^2}{2}}dy\right)$$
That can be interpreted as a double integral over the first quadrant of the plane and changing to polart coordinates gives an easy integral.