Poles of the Gamma Function

Theorem

The gamma function $\Gamma :\C \to \C$ is analytic throughout the complex plane except at $\left\{{0, -1, -2, -3, \ldots}\right\}$ where it has simple poles.

Proof

First we examine the location of the poles.

We examine the Weierstrass form of the Gamma function:

$\displaystyle \frac 1 {\Gamma(z)} = ze^{\gamma z} \prod_{n=1}^\infty \left({\left({1 + \frac z n}\right) e^{\frac{-z} n} }\right)$

The terms of the product clearly do not tend to zero, and so the product is only zero if one of the terms is zero; this occurs when $1 + \dfrac z n = 0$, which occurs at $z \in \left\{{-1,-2, \dots }\right\}$.

We also have the expression outside the product to consider; since the exponential function is never $0$ (although tending towards it, this expression is zero whenever $z=0$.

Furthermore, if some term in the product is zero, this excludes the possibility that any other term is zero, so the zeros are simple.

Hence $\displaystyle \frac 1 {\Gamma(z)} = 0$ when, and only when, $z \in \left\{{0, -1, -2, -3, \ldots}\right\}$.

Therefore $\Gamma$ has simple poles at these points.

Next we show that $\Gamma$ is analytic on $\Re(z) > 0$.

By Gamma Difference Equation this proves that $\Gamma$ is analytic on $\C \backslash \left\{{0, -1, -2, -3, \ldots}\right\}$.

Let the functions $\Gamma_n$ be defined by

$\displaystyle \Gamma_n(z) = \int_0^n t^{z-1} e^{-t}\ dt$

Clearly each $\Gamma_n$ is analytic, so by Uniform Limit of Analytic Functions is Analytic it is sufficient to show that $\Gamma_n \to \Gamma$ locally uniformly.

By Absolute Value of Complex Integral we have

$\displaystyle |\Gamma(z) - \Gamma_n(z)| \leq \int_n^\infty t^{x-1} e^{-t}\ dt$

where $x = \Re(z)$.

Let $a \in \C$ with $\Re(a) > 0$, and let $D$ be an open disk of radius $r$ about $a$.

We have the expansion

$\displaystyle e^z = 1 + z + \frac {z^2}2 + \cdots $

from which we see that $z^\alpha = o(e^z)$ for all $\alpha > 0$.

In particular, there exists $c > 0$ such that $e^{-t} \leq c t^{-\Re(a)-r-1}$ for all $t \geq 1$.

Then we have

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \vert\Gamma(z) - \Gamma_n(z)\vert\)	\(\leq\)	\(\displaystyle \int_n^\infty t^{x-1} c t^{-\Re(a)-r-1}\ dt\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle c\int_n^\infty t^{-2} dt\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \frac c n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So $\Gamma_n \to \Gamma$ uniformly in $D$, and we are done.

$\blacksquare$