Simple Poles of Gamma Function
Theorem
The domain of the Gamma function $\Gamma$ as a complex function is as follows: $$ \mathbb{C} \setminus \left( \mathbb{Z} \setminus \mathbb{N} \right) = \mathbb{C} \setminus \left\{ 0 , -1, -2, \cdots \right\} $$ Moreover, the set of singularities of $\Gamma$, $\left( \mathbb{Z} \setminus \mathbb{N} \right)$, is a set of simple poles.
- $\mathbb{N}$ represents the set of natural numbers, $\mathbb{Z}$ represents the set of integers, and $\mathbb{C}$ represents the set of complex numbers.
Explanation
Visualization
The above figure is a graph of only the real part of the Gamma function on the real axis, and it can be seen that the function values diverge at non-positive integers.
Code
The following is Julia code for visualization.
using SpecialFunctions, LaTeXStrings
z = -5:0.001:5
Γz = gamma.(Complex.(z))
plot(z, real.(Γz), lw = 2, color = :black, xticks = -5:1:5,
xlims = [-5, 5], ylims = [-5, 5], size = [400, 400])
hline!([0], color = :black)
vline!(-5:0, style = :dash, color = :red)
xlabel!(L"\Re(z)")
ylabel!(L"\Re(\Gamma (z))")
Proof
Euler’s Reflection Formula: $$ {\Gamma (1-p) \Gamma ( p )} = { {\pi} \over {\sin \pi p } } $$
In $\Re (z) > 0$, the Gamma function $\Gamma (z) = \int_{0}^{\infty} t^{z-1} e^{t} dt$ is well defined for values other than $0$. Dividing both sides of Euler’s reflection formula by $\Gamma (z)$, we get: $$ \Gamma (1-z) = {\frac{ \pi }{ \Gamma (z) \sin \pi z }} $$ Substituting $z = 1$, we get: $$ \Gamma (0) = \Gamma (1-1) = {\frac{ \pi }{ \Gamma (1) \cdot \sin 0 \pi }} = {\frac{ \pi }{ 0! \cdot 0 }} = \infty $$ And substituting $z = 2$, we get: $$ \Gamma (-1) = \Gamma (1-2) = {\frac{ \pi }{ \Gamma (2) \cdot \sin 1 \pi }} = {\frac{ \pi }{ 1! \cdot 0 }} = \infty $$ Since this divergence is the same for all $z \in \left\{ 0, 1, 2, \dots \right\}$, $\Gamma$ has singularities at all non-positive integers, and all these singularities are simple poles because there is only one $\sin \pi z$ in the denominator.
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