Number Theory/Gamma function

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

Definition

In order to fulfill the necessary realation/condition:

${\displaystyle n! = \Gamma(z + 1)}$

we can set the Gamma function to be:

${\displaystyle \Gamma(z) = \displaystyle\int\limits_{0}^{\infty} e^{-t} t^{z - 1}~dt}$

whereby ${\displaystyle \operatorname{Re}(z) > 0}$, because by partial integration we get:

${\displaystyle \displaystyle\int\limits_{a}^{b} e^{-t} t^{n}~dt = \bigg[-e^{-t} t^n\bigg]_a^b + n \displaystyle\int\limits_{a}^{b} e^{-t} t^{n - 1}~dt}$

If ${\displaystyle a \to 0}$ and ${\displaystyle b \to \infty}$ we receive for ${\displaystyle \Gamma(n+1) = n!}$, when ${\displaystyle n \in \mathbb{N}}$.

Theorem: poles of the ${\displaystyle \Gamma-}$function

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

Proof

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

${\displaystyle \Gamma(z) = \int\nolimits_0^{\infty} e^{-t} t^{z - 1}~dt}$

Now, we can split the integral, which results in:

${\displaystyle \Gamma(z) = \int\nolimits_0^{1} e^{-t} t^{z - 1}~dt + \int\nolimits_1^{\infty} e^{-t} t^{z - 1}~dt}$

However, for ${\displaystyle e^{-t}}$ we can substitute its series representation:

${\displaystyle e^{-t} = \displaystyle\sum\limits_{n = 0}^{\infty}~\displaystyle\frac{(-1)^n}{n!}t^n}$

in the first integral, which leads to:

${\displaystyle \Gamma(z) = \int\nolimits_0^{1} \displaystyle\sum\limits_{n = 0}^{\infty} \displaystyle\frac{(-1)^n}{n!}t^n t^{z - 1}~dt + \int\nolimits_1^{\infty} e^{-t} t^{z - 1}~dt}$

Since we have a polynomial in the first integral, we can use the power rule and we get the following:

${\displaystyle \Gamma(z) = \int\nolimits_0^{1} \displaystyle\sum\limits_{n = 0}^{\infty}~\displaystyle\frac{(-1)^n}{n!(n + z)}~dt + \int\nolimits_1^{\infty} e^{-t} t^{z - 1}~dt }$

If we examine the denominator of the series, we can see, that for:

${\displaystyle \begin{array}{|c|c|c|c|c|c|} n & 0 & \hspace{8pt}1 & \hspace{8pt}2 & \hspace{8pt}3 & \dots \\ \hline z & 0 & -1 & -2 & -3 & \dots \\ \end{array} }$

poles occur, with which we can obtain the residue:

${\displaystyle \operatorname{Res}(\Gamma(z)\text{, }-n) = \displaystyle\frac{(-1)^n}{n!}}$

If we use now the residue theorem, we can evalute the following contour integral:

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

As the series:

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

is equal to the sum

${\displaystyle \displaystyle\sum\limits_{n = 0}^{\infty}~\displaystyle\frac{(-1)^n}{n!}t^n}$

with ${\displaystyle t = 1}$, which is ${\displaystyle e^{-1}}$, the integral is equal to:

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