$\frac{1}{2}!$ aka $\Gamma(\frac{3}{2})$ I know it's $\frac{\sqrt{\pi}}{2}$ but how can this be evaluated by hand? (Or can it not?)
For quick reference: $$n!=\Gamma(n+1)$$
$$\Gamma(n)=(n-1)!$$
$$\Gamma(n)=\int_0^\infty x^{n-1} e^{-x}\,dx$$
Thanks. And if it's not too much trouble, maybe some guidelines in solving general Gamma functions.
 A: $$\Gamma(3/2) = \Gamma(1/2)/2 = \frac{1}{2}\int_0^{\infty} x^{-1/2} e^{-x} \, dx $$
Change variables to $y=x^{1/2}$, so $dy=dx/(2x^{1/2})$, and we have the integral
$$ I = \int_0^{\infty} e^{-y^2} \, dy. $$
You may or may not have met this one before. Basically the easiest idea for evaluating this is to change coordinates to polars after squaring it.
$$ I^2 = \int_0^{\infty} \int_0^{\infty} e^{-(x^2+y^2)} \, dx \, dy $$
Set $x=r\cos{\theta}$, $y=r\sin{\theta}$, then $ dx \, dy = r \, dr \, d\theta $. The integral is over the first quadrant, so $0<\theta<\pi/2$.
$$ I^2 = \int_0^{\infty} \int_0^{\pi/2} re^{-r^2} \, d\theta \, dr = \frac{\pi}{2} \int_0^{\infty}  re^{-r^2} \, dr $$
The remaining integral has an elementary antiderivative $e^{-r^2}/2$, so the whole thing evaluates to
$$ I^2 = \frac{\pi}{4}, $$
and you take the positive root for obvious reasons.
A: By definition of the Gamma function, we have
$$\Gamma\left(\frac{1}{2}\right)=\intop_{0}^{\infty}\frac{e^{-t}}{\sqrt{t}}\ dt.$$
Now let $u=\sqrt{t}$, so that $du=\frac{1}{2\sqrt{t}}\  dt$. Then we have
$$\Gamma\left(\frac{1}{2}\right)=\int_{0}^{\infty}2e^{-u^{2}}du=\intop_{-\infty}^{+\infty}e^{-u^{2}}du.$$
This last integral is the Gaussian integral, whose value can be found many different ways. Several are included in the link.
