0

I'm reading this proof of $\Gamma(\frac12)=\sqrt{\pi}$, but I just can't understand this part.

$$\left[{\Gamma(\frac12)}\right]^2=4\int_0^\infty\int_0^\infty{e^{-(u^2+v^2)}}dvdu=4\int_0^{\frac{\pi}{2}}\int_0^\infty{e^{-r^2}rdrd{\theta}}$$

I think it is something about polar coordinates but unclear. I would really appreciate if someone could give me a clear explanation. Thank you :)

Dimen
  • 415
  • Exactly: make a change of variables $u = r\cos\theta$, $v =r\sin \theta$. – Pedro Jul 01 '17 at 03:29
  • I thought so but I'm still not sure how I should change the range when I make a change of variable in double integrals. Also I'm not sure how to deal with $d(rsin{\theta})d(rcos{\theta})$ – Dimen Jul 01 '17 at 03:32
  • And note that the region covered is the first quadrant. In rectangular coordinates, this is $(x,y)\in (0,\infty)\times(0,\infty)$, but in polar coordinates it is $(r,\theta)\in(0,\infty)\times(0,\pi/2)$. – MPW Jul 01 '17 at 03:33
  • One has $dA=dx ; dy =r ; dr; d\theta$ – MPW Jul 01 '17 at 03:34
  • $\theta$ ranges in $(0,\pi/2)$ and $r$ in $(0,\infty)$, as in your formula. The Jacobian of the change of variables I wrote above is $r$, hence the result. If this is new to you, you should read about the change of variables formula. – Pedro Jul 01 '17 at 03:35
  • @Dimen https://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables – spaceisdarkgreen Jul 01 '17 at 03:35
  • It meant $\iint_{\mathbb{R}^2} f(\sqrt{x^2+y^2})dxdy= \int_0^\infty 2\pi r f(r)dr$ – reuns Jul 01 '17 at 03:36
  • @Dimen Instead of deleting the post, you can post a solution to your own question if you understood what to do. This has the advantage that it will help future visitors to the site. – Pedro Jul 01 '17 at 05:10
  • (Given your reputation score I am not sure if you can just yet answer posts, but I have given you an upvote in good faith, so that you eventually can.) – Pedro Jul 01 '17 at 05:11
  • 3
    Possible duplicate of Why is $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ ? – StubbornAtom Jul 01 '17 at 06:22

0 Answers0