# Math Help - Gamma Function Conversion to Polar Coordinates

1. ## Gamma Function Conversion to Polar Coordinates

This is not a homework problem; I'm just making sure I can follow all the logic to this problem.

Hi I have the following part of my book which I am trying to understand but I am stuck on a single part of the conversion from rectangular coordinates to polar coordinates:

The book lists:
$(\Gamma(1/2))^2 = 4 \int_0^{\frac{\pi}{2}} \int_0^\infty e^{-r^2}r dr d\theta = 4\frac{\pi}{2} (\frac{e^{-r^2}}{-2}) \ evaluated \ from \ 0 \ to \ \infty$

When I do a u substitution for the improper integral how do I get evaluate the integral with lim as $t \rightarrow \infty \ for \int_0^t e^u du$ I have set $u = -r^2, du = -2rdr$.

The result is $\pi$
Thank you.

2. Originally Posted by mathisfunforme
This is not a homework problem; I'm just making sure I can follow all the logic to this problem.

Hi I have the following part of my book which I am trying to understand but I am stuck on a single part of the conversion from rectangular coordinates to polar coordinates:

The book lists:
$(\Gamma(1/2))^2 = 4 \int_0^{\frac{\pi}{2}} \int_0^\infty e^{-r^2}r dr d\theta = 4\frac{\pi}{2} (\frac{e^{-r^2}}{-2}) \ evaluated \ from \ 0 \ to \ \infty$

When I do a u substitution for the improper integral how do I get evaluate the integral with lim as $t \rightarrow \infty \ for \int_0^t e^u du$ I have set $u = -r^2, du = -2rdr$.

The result is $\pi$
Thank you.
First note that the integral with respect to $\theta$ can be done straight off as the integrand is independednt of $\theta$

That leaves you with the problem of evaluating:

$\int_0^{\infty} r e^{-r^2}\; dr$

Now don't bother with substitution, you should recognise that:

$\frac{d}{dr}e^{-r^2}=-2re^{-r^2}$

Hence using the fundamental theorem of caculus:

$\int_0^{\infty} r e^{-r^2}\; dr=\left[ -\frac{1}{2}e^{-r^2}\right]_0^{\infty}=\frac{1}{2}$

CB