# Gamma Function

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The Gamma function can be thought of as an extension of the definition of the factorial \begin{align} n!=n(n-1)(n-2) \dots 2 \cdot 1 \end{align} What is the factorial of $1/2$ or the complex number $-1/2+i$? We set the stage by calculating Eulers Integral of the second kind. \begin{align} I_n:=\int_0^{\infty} t^n e^{-t} dt \end{align} Integration by parts gives us \begin{align} I_n=\underbrace{ \left[ -t^n e^{-t} \right]_0^{\infty}}_{=0} + n \int_0^{\infty} t^{n-1} e^{-t} dt=n \cdot I_{n-1} \end{align} another integration will give $I_n=n(n-1)I_{n-2}$ and so on. The last integral gives \begin{align} I_0= n \int_0^{\infty} e^{-t} dt=1 \end{align} so we have for $n \in \mathbb{N}_0$ \begin{align} \int_0^{\infty} t^n e^{-t} dt=n! \end{align} But we can also allow other values for $n$, thus

we define the Gamma function $\Gamma$ for positive arguments $x \in \mathbb{R}_{\geq 0}$ \begin{align} \Gamma (x) := \int_0^{\infty} t^{x-1} e^{-t} dt \end{align} with properties \begin{align} \Gamma (x+1) &=x \Gamma (x) \hspace{1.45cm} \text{for } x \in \mathbb{C}\\ \Gamma (n+1) &=n! \hspace{2cm} \text{for } n \in \mathbb{N}_0\\ \Gamma \left( \frac{1}{2} \right) &=\sqrt{\pi} \end{align}

The Gamma function diverges for negative integers values on the real axis of $x$ and $x \rightarrow 0$.

Furhter Reading: Tilo Arens $\&$ Co - Mathematik (Spektrum Verlag)