Poles of Gamma Function

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Theorem

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


Proof

First we examine the location of the poles.

We examine the Weierstrass form of the Gamma function:

$\dfrac 1 {\map \Gamma z} = z e^{\gamma z} \displaystyle \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z / n} }$

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

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

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

This occurs at $z \in \set {-1, -2, \dotsc}$.


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 factor in the product is zero, this excludes the possibility that any other factor is zero.

So the zeros are simple.

Hence:

$\dfrac 1 {\map \Gamma z} = 0 \iff z \in \set {0, -1, -2, -3, \ldots}$

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


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

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


Let the functions $\Gamma_n$ be defined by

$\displaystyle \map {\Gamma_n} z = \int_0^n t^{z - 1} e^{-t} \rd t$

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:

$\displaystyle \cmod {\map \Gamma z - \map {\Gamma_n} Z } \le \int_n^\infty t^{x - 1} e^{-t} \rd t$

where $x = \map \Re z$.

Let $a \in \C$ with $\map \Re a > 0$.

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

By definition of exponential function, we have the expansion:

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

from which we see that:

$\forall \alpha \in \R_{>0}: z^\alpha = \map o {e^z}$

where $o$ denores little-o notation.

In particular:

$\exists c \in \R_{>0}: \forall t \ge 1: e^{-t} \le c t^{-\map \Re a - r - 1}$

Then:

\(\displaystyle \cmod {\map \Gamma z - \map {\Gamma_n} z}\) \(\le\) \(\displaystyle \int_n^\infty t^{x - 1} c t^{-\map \Re a -r - 1} \rd t\)
\(\displaystyle \) \(\le\) \(\displaystyle c\int_n^\infty t^{-2} \rd t\)
\(\displaystyle \) \(\le\) \(\displaystyle \frac c n\)

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

$\blacksquare$


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