Gamma Function

Gamma function is an integral

$$\Gamma (x+1)=x\Gamma (x)$$

Euler found that the integral $\Gamma (n+1)=\int \_0^1(-\ln x{)}^n,dx$ has the property $\Gamma (n+1)=n\Gamma (n)$ when he takes the integral by parts:

$$\begin{array}{rl}\int (-\ln x{)}^{n}dx& =x(-\ln x{)}^n-\int xd(-\ln x{)}^n\\ & =x(-\ln x{)}^n+n\int (-\ln x{)}^{n-1}dx\\ ⟹\int \_0^1(-\ln x{)}^n,dx& =n\int \_0^1(-\ln x{)}^{n-1},dx\end{array}$$

let $u=-\ln x$, and define

$$\begin{array}{rl}\Gamma (n+1)& =\int \_{\infty }^0{u}^n,d(e^{-u})\\ & =\int \_0^{\infty }{u}^n{e}^{-u},du\end{array}\int \_0^{\infty }{u}^{n-1}{e}^{-u}du$$

redefine $x$ as $n-1$, and we have:

$$\Gamma (x)=\int \_0^{\infty }{u}^{x-1}{e}^{-u},du$$

Tips

$n$ here is a real number. I use the real number definition of power function here.

Properties

Integrate by parts:

\begin{align} \Gamma (x+1) & = \int\_{0}^{\infty} u^{x} e^{-u} , du \\ & = \int\_{0}^{\infty} u^{x} (-e^{-u})' d u \\ & = \left\[ u^{x}(-e^{-u}) \right]_{0}^{\infty} - \int_{0}^{\infty} (-e^{-u}) (x u ^{x-1}) , du \\ & = - \lim\_{ u \to \infty } \frac{u^{x}}{e^{u}} + x \int\_{0}^{\infty} e^{-u}u^{x-1} , du \\ & = x\Gamma(x) \end{align}

$\Gamma (1)=1$

\Gamma(1) = \int\_{0}^{\infty} e^{-u} , du \stackrel{s(u)=-u}= - \int\_{0}^{-\infty} e^s , ds = - \left\[ e^{s} \right]\_{0}^{-\infty} = 1

$\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }$

let $u(t)=t^2$, and you would find that it’s Gaussian Integral

$$\begin{array}{rl}\Gamma (1/2)& =\int \_0^{\infty }{u}^{-1/2}{e}^{-u}du\\ & =\int \_0^{\infty }{t}^{-1}{e}^{-t^2}d{t}^2\\ & =2\int \_0^{\infty }{e}^{-x^2}dx\\ & =\int \_{-\infty }^{\infty }{e}^{-x^2},dx=\sqrt{\pi }\end{array}$$