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,

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

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

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(n) = (n-1)!. }

z=1/2

For z=1/2,

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) = \int_{0}^{\infty}t^{-1/2}e^{-t}dt }

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

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

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 =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 the function 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. }