TITLE: Why is $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?
QUESTION [75 upvotes]: It seems as if no one has asked this here before, unless I don't know how to search.
The Gamma function is
$$
\Gamma(\alpha)=\int_0^\infty x^{\alpha-1} e^{-x}\,dx.
$$
Why is
$$
\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\text{ ?}
$$
(I'll post my own answer, but I know there are many ways to show this, so post your own!)

REPLY [3 votes]: By the Bohr-Mollerup theorem,
$$\Gamma(1/2)=\lim_{n\to\infty}\frac{\sqrt n(n!)}{(1/2)(-1/2)\dots(1/2-n)}=\lim_{n\to\infty}\frac{\sqrt n4^{n+1}(n!)^2}{(2n)!(n+\frac12)}$$
Apply the Stirling approximation and watch almost everything simplify!
$$\Gamma(1/2)=\sqrt\pi\lim_{n\to\infty}\frac n{n+\frac12}=\sqrt\pi$$