# Number Theory/Gamma function

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The Gamma function is a result of the efforts to expand the factorials $n!$ onto the complex plane.

## Definition

In order to fulfill the necessary realation/condition:

$n! = \Gamma(z + 1)$ we can set the Gamma function to be:

$\Gamma(z) = \displaystyle\int\limits_{0}^{\infty} e^{-t} t^{z - 1}~dt$ whereby $\operatorname{Re}(z) > 0$ , because by partial integration we get:

$\displaystyle\int\limits_{a}^{b} e^{-t} t^{n}~dt = \bigg[-e^{-t} t^n\bigg]_a^b + n \displaystyle\int\limits_{a}^{b} e^{-t} t^{n - 1}~dt$ If $a \to 0$ and $b \to \infty$ we receive for $\Gamma(n+1) = n!$ , when $n \in \mathbb{N}$ .

## Theorem: poles of the $\Gamma-$ function

The poles of the $\Gamma-$ function appear at $-n$ , whereby $n \in \mathbb{N} \cup 0$ ; the residues are at $\operatorname{Res}(\Gamma(z)\text{, }-n) = \frac{(-1)^n}{n!}$ and its contour integral is equal to $\frac{2 \pi i}{e}$ .

#### Proof

The Euler definition of Legendre for the $\Gamma-$ function is:

$\Gamma(z) = \int\nolimits_0^{\infty} e^{-t} t^{z - 1}~dt$ Now, we can split the integral, which results in:

$\Gamma(z) = \int\nolimits_0^{1} e^{-t} t^{z - 1}~dt + \int\nolimits_1^{\infty} e^{-t} t^{z - 1}~dt$ However, for $e^{-t}$ we can substitute its series representation:

$e^{-t} = \displaystyle\sum\limits_{n = 0}^{\infty}~\displaystyle\frac{(-1)^n}{n!}t^n$ in the first integral, which leads to:

$\Gamma(z) = \int\nolimits_0^{1} \displaystyle\sum\limits_{n = 0}^{\infty} \displaystyle\frac{(-1)^n}{n!}t^n t^{z - 1}~dt + \int\nolimits_1^{\infty} e^{-t} t^{z - 1}~dt$ Since we have a polynomial in the first integral, we can use the power rule and we get the following:

$\Gamma(z) = \int\nolimits_0^{1} \displaystyle\sum\limits_{n = 0}^{\infty}~\displaystyle\frac{(-1)^n}{n!(n + z)}~dt + \int\nolimits_1^{\infty} e^{-t} t^{z - 1}~dt$ If we examine the denominator of the series, we can see, that for:

$\begin{array}{|c|c|c|c|c|c|} n & 0 & \hspace{8pt}1 & \hspace{8pt}2 & \hspace{8pt}3 & \dots \\ \hline z & 0 & -1 & -2 & -3 & \dots \\ \end{array}$ poles occur, with which we can obtain the residue:

$\operatorname{Res}(\Gamma(z)\text{, }-n) = \displaystyle\frac{(-1)^n}{n!}$ If we use now the residue theorem, we can evalute the following contour integral:

$\int\nolimits_{C}\Gamma(z)~dz = 2\pi i \displaystyle\sum\limits_{n \geq 0} \displaystyle\frac{(-1)^n}{n!} = \displaystyle\frac{2 \pi i}{e}$ As the series:

$\displaystyle\sum\limits_{n \geq 0}~\displaystyle\frac{(-1)^n}{n!}$ is equal to the sum

$\displaystyle\sum\limits_{n = 0}^{\infty}~\displaystyle\frac{(-1)^n}{n!}t^n$ with $t = 1$ , which is $e^{-1}$ , the integral is equal to:

$\displaystyle\frac{2 \pi i}{e}$ 