Gamma function

The Gamma funciton is defined as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}dt}

Relations and values

Factorial

Using integration by parts,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(z) = \int_{0}^{\infty}t^{z-1}d(e^{-t}dt)}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \left[t^{z-1}(-e^{-t}) \right]_0^{\infty}- \int_{0}^{\infty}(z-1)t^{(z-2)}(-e^{-t})dt.}

At Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=0} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t^{z-1}(-e^{-t})} goes to 0. Using L'Hopital's rule, it's easy to show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 $x=t^{1/2}$,

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

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \int_{-\infty}^{\infty} e^{-x^2}dx }

where in the last line we used the fact that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-x^2}} is an even function. The integral is called the Gaussian integral and has a well known value of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt\pi} .

Thus,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(1/2) = \sqrt\pi. }