Poles of 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:


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

The terms of the product clearly do not tend to zero.

So the product is zero only if one of the terms is zero.

This occurs when $1 + \dfrac z n = 0$.

This occurs at $z \in \left\{{-1,-2, \dots }\right\}$.

We also have the expression outside the product to consider.

From Exponential Tends to Zero and Infinity, the exponential function is never $0$.

So 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:
 * $\dfrac 1 {\Gamma \left({z}\right)} = 0 \iff z \in \left\{{0, -1, -2, -3, \ldots}\right\}$

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

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

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

Let the functions $\Gamma_n$ be defined by


 * $\displaystyle \Gamma_n \left({z}\right) = \int_0^n t^{z-1} e^{-t} \, \mathrm d t$

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 Modulus of Complex Integral:


 * $\displaystyle |\Gamma \left({z}\right) - \Gamma_n \left({z}\right)| \le \int_n^\infty t^{x-1} e^{-t} \, \mathrm d t$

where $x = \Re \left({z}\right)$.

Let $a \in \C$ with $\Re \left({a}\right) > 0$.

Let $D$ be an open disk of radius $r$ about $a$.

We have the expansion:


 * $e^z = 1 + z + \dfrac {z^2} 2 + \dotsb$

from which we see that:
 * $\forall \alpha \in \R_{>0}: z^\alpha = o \left({e^z}\right)$

where $o$ denores little-o notation.

In particular:
 * $\exists c \in \R_{>0}: \forall t \ge 1: e^{-t} \le c t^{-\Re \left({a}\right) - r - 1}$

Then:

So $\Gamma_n \to \Gamma$ uniformly in $D$, and the proof is complete.