Sahan Manodya
The gamma function is a complex valued function which generalizes the factorial function. Throughout the article we'll use $C$ for the complex half plane $\{z\in\mathbb{C}:\mathrm{Re}(z) > 0\}$. For $z\in\mathbb{C}$ with $\mathrm{Re}(z) > 0$ it is defined by the improper integral
\begin{equation}
\Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} \mathrm{d}t
\end{equation}
This integral converges to complex differentiable function for every $z$ in the half plane $C$ and the convergence is absolute. So the gamma function is holomorphic on $C$. In this article, we will talk about the holomorphic continuation of the gamma function to a maximal domain on complex plane, and explore some of its properties.
Properties of the Gamma Function
Before we discuss the holomorphic continuation of the gamma function, let's see some of the important properties of the gamma function on $C$. One of the most important properties of the gamma function is it satisfies the functional equation
\begin{equation}
\Gamma(z+1) = z\Gamma(z),
\end{equation}
for every $z\in\mathbb{C}$ with $\mathrm{Re}(z) > 0$ which can be proven easily using the integral definition of the Gamma function. This functional equation gives a way to obtain holomorphic extension of $\Gamma$. Alos, from this property we can easily see that the Gamma function satisfies a shifted factorial function on positive integers. That is,
\begin{equation}
\Gamma(n) = (n-1)! \quad \text{for } n \in \mathbb{N}.
\end{equation}
which indicates that Gamma function is unbounded.
Meromorphic Extension
To extend the gamma function to the entire complex plane, we use the functional equation $f(z+1)=z f(z)$ which is already satisfied by the gamma function on $C$. First consider the set $B_0=\{z\in\mathbb{C}:-1<\mathrm{Re}(z)\}-\{0\}$. Now lets going to construct a function $f$ on a subset the complex plane in an inductive manner. First define $f$ on $B_0$ by
\begin{equation}
f(z) = \frac{\Gamma(z+1)}z\text{ for all }z\in B_0
\end{equation}
According to this definition, it is clear that $f$ is homolorphic in $B_0$ and $f$ satisfies $f(z)=\Gamma(z)$ for all $z\in C$. So $f$ is a holomorphic extension of $\Gamma$ to the $B_0$. Now for each negative integer $n$ take $B_n=\{z\in\mathbb{C}:n-1<\mathrm{Re}(z)\leq n\}-\{n\}$. Then for each negative integer $n$, inductively extend $f$ to $B_{n}$ by defining
\begin{equation}
f(z) = \frac{f(z+1)}z\text{ for all }z\in B_{n}\setminus B_{n+1}
\end{equation}
Since $\frac{f(z+1)}z=f(z)$ on $B_{n+1}$, by the way we define $f$, we have $\frac{f(z+1)}z=f(z)$ for every $z\in B_n$ and $f$ is holomorphic in the open set $B_n$ for every negative integer $n$. So in this way we obtain a holomorphic extension of the Gamma function to the union of all $B_n$'s which is $\mathbb{C}\setminus \mathbb{Z}_{\leq 0}$. Here $\mathbb{Z}_{\leq 0}$ is the set of all non-positive integers. Therefore $f$ is a holomorphic extension of the Gamma function to the open set $\mathbb{C}\setminus \mathbb{Z}_{\leq 0}$ and by the Identity theorem it's the unique extension of the Gamma function to that open set. Note that $f$ also satisfies the functional equation $f(z+1)=zf(z)$ for every $z\in \mathbb{C}\setminus \mathbb{Z}_{\leq 0}$.
Now since we have singularities of $f$ at non-positive integers, in order to obtain a maximal set where Gamma function can be extended, we have to find whether the singularities are removable or not. Since $f$ is continuous at $1$ and $f(1)=\Gamma(1)=1$, we have \begin{equation} \lim_{z\to 0}zf(z) = \lim_{z\to 0}f(z+1)=1
\end{equation}
Since the limit exists and it is non-zero, $f$ has a simple pole at $0$ so the singularity is non-removable. Also for every non-positive integer $n$, if $\lim_{z\to n}(z-n)f(z)=L$ then we have
\begin{equation}\begin{split}
\lim_{z\to n-1}(z-(n-1))f(z)&=\lim_{z\to n}(z-n)f(z-1)\\&=\lim_{z\to n}\frac{(z-n)f(z)}{z-1}\\&=\frac L{n-1}
\end{split}\end{equation}
Therefore by mathematical induction, for non-negative integer $n$,
\begin{equation}
\lim_{z\to -n}(z+n)f(z)=\frac{(-1)^n}{n!}
\end{equation}
Since the limit exists and it's non-zero $f$ has simple poles at every non-positive integer, so the singularities are non-removable and $\mathbb{C}\setminus \mathbb{Z}_{\leq 0}$ is a maximal domain which Gamma function can be extended holomorphically. Since $f$ is the unique holomorphic extension of the Gamma function, by the identity theorem, we'll define it to be the Gamma function. So $\Gamma$ function has simple poles at every non-positive integer and $\mathbb{Z}_{\leq 0}$ has no accumulation points in complex plane, so the Gamma function is meromorphic in $\mathbb{C}$.
The important thing is, the functional equation $\Gamma(z+1) = z\Gamma(z)$ is still satisfied by the Gamma function, but now for every $z\in \mathbb{C}\setminus \mathbb{Z}_{\leq 0}$.
Also by above results, we can find the Residue of Gamma function at each pole at $-n$ where $n$ is non-negative integer as
\begin{equation}
\mathrm{Res}[\Gamma,-n] = \lim_{z\to -n}(z+n)\Gamma(z)=\frac{(-1)^n}{n!}
\end{equation}
One of another thing we can obsereve from above results is, since $\Gamma$ has no zeros and it has simple poles at non-positive integers, $\frac1\Gamma$ has a holomorphic extension to the complex plain and that extension is called the reciprocal Gamma function which is entire and has simple zeros at non-positive integers.
Euler Reflection Formula
Finally we will see an important property of the gamma function, which is the famous Euler reflection formula given by
\begin{equation}
\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}\text{ for all }z\in \mathbb{C}\setminus \mathbb{Z}
\end{equation}
To prove this, consider the function $g(z)=\Gamma(z)\Gamma(1-z)\sin(\pi z)$ defined on $\mathbb{C}\setminus \mathbb{Z}$. Clearly $g$ is holomorphic. For $x\in\mathbb{R}$ with $0\in(0,1)$, since both $x$ and $1-x$ lies in $(0,1)$, using the integral representation of Gamma function we can show that
$$ \Gamma(x)\Gamma(1-x)=\int_0^\infty\frac{t^{x-1}}{1+t}\mathrm{d}t=\frac\pi{\sin(\pi x)}$$
Therefore $g$ takes the constant value $\pi$ on $(0,1)$ and since $(0,1)$ has limit points in $\mathbb{C}\setminus \mathbb{Z}$ which is connected, by the identity theorem $g=\pi$, which proves the Euler reflection formula. We can also use this formula to show that $\Gamma(\frac12)=\sqrt{\pi}$, by substituting $z=\frac12$ in the Euler reflection formula and showing that $\Gamma(\frac12)\geq0$. From the integral formula of the Gamma function it can be shown that $\Gamma(\frac12)=\int_{-\infty}^\infty e^{-x^2}\mathrm{d}x$, so by above result we can prove the important integral formula
\begin{equation}
\int_{-\infty}^\infty e^{-x^2}\mathrm{d}x=\sqrt{\pi}.
\end{equation}
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