Gamma function

The Gamma funciton is defined as

$\Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}dt$

Relations and values

Factorial

Using integration by parts,

$\Gamma (z)=\int _{0}^{\infty }t^{z-1}d(e^{-t}dt)$

$=\left[t^{z-1}(-e^{-t})\right]_{0}^{\infty }-\int _{0}^{\infty }(z-1)t^{(z-2)}(-e^{-t})dt.$

At $t=0$, $t^{z-1}(-e^{-t})$ goes to 0. Using L'Hopital's rule, it's easy to show that $\lim _{t\to \infty }{\frac {t^{z-1}}{e^{t}}}=0$.

So,

$=(z-1)\int _{0}^{\infty }t^{(z-1)-1}e^{-t}dt$

$=(z-1)\Gamma (z-1).$

Thus,

$\Gamma (z)=(z-1)\Gamma (z-1).$

Using this, and the the fact that $\Gamma (1)=1$, then we can get the factorial function. For $n$ a positive integer,

$\Gamma (n)=(n-1)!.$

z=1/2

For z=1/2,

$\Gamma (1/2)=\int _{0}^{\infty }t^{-1/2}e^{-t}dt$

Substituting $x=t^{1/2}$,

$=\int _{0}^{\infty }({\frac {1}{x}})e^{-x^{2}}(2xdx)$

$=2\int _{0}^{\infty }e^{-x^{2}}dx$

$=\int _{-\infty }^{\infty }e^{-x^{2}}dx$

where in the last line we used the fact that $e^{-x^{2}}$ is an even function. The integral is called the Gaussian integral and has a well known value of ${\sqrt {\pi }}$.

Thus,

$\Gamma (1/2)={\sqrt {\pi }}.$