$\frac{1}{2}!$ aka $\Gamma(\frac{3}{2})$ [duplicate]

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.

marked as duplicate by Jack D'Aurizio, Jonas Meyer, Michael Hardy, Adriano, Grigory MApr 6 '15 at 9:29

• $\Gamma(\frac12)=\sqrt{\pi}$ is a classical result, and the fuctional equation $x\Gamma(x)=\Gamma(x+1)$ gives you the result. Is this the sort of thing you are looking for? – Olivier Bégassat Apr 6 '15 at 0:57
• Thanks, but not really. I'm not understanding where $\sqrt{\pi}$ or $\frac{\sqrt{\pi}}{2}$ come from. – Jacob Wheeler Apr 6 '15 at 1:05

$$\Gamma(3/2) = \Gamma(1/2)/2 = \frac{1}{2}\int_0^{\infty} x^{-1/2} e^{-x} \, dx$$ 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.
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
$$\Gamma\left(\frac{1}{2}\right)=\int_{0}^{\infty}2e^{-u^{2}}du=\intop_{-\infty}^{+\infty}e^{-u^{2}}du.$$