# Gamma function can be extended as a meromorphic function

Let $$\Gamma (z)= \int_{0}^{\infty} e^{-t}t^{z-1}dz$$ for $\Re z\gt 0$ Can be extended as a meromorphic function to the entire complex plane with simple poles at non positive integers.

How can I show this statement?

• The simplest way is by taking the product representation of $\Gamma$, and showing that both definitions agree say on $(0,1)$. – Daniel Fischer Jan 6 '14 at 0:01
• Can you write write you said more explicitly please? @DanielFischer thank you. – user315 Jan 6 '14 at 0:06
• I'm on my way to bed, sorry. If nobody did until then, I'll expand tomorrow. – Daniel Fischer Jan 6 '14 at 0:11

The key is the following relation: $$\Gamma(z) =\frac{\Gamma(z+n)}{ z(z+1)...(z+n-1)}$$ for all $z$ such that ${\Re}z>0$ and noticing that $\Gamma(z+n)$ is holomorphic for all $z$ such that ${\Re}z>-n$.
• Since you have asked this question, I assume you are familiar with the formula $\Gamma(x+1)=x\Gamma(x)$. Since you are self-learning, try to use induction to show that that formula implies $\Gamma(z+n)=(z+n-1)(z+n-2)\cdots(z+1)z\Gamma(z)$. Divide both sides by $(z+n-1)(z+n-2)\cdots(z+1)z$. – robjohn Jan 6 '14 at 1:11
• @B11b See robjohn comment. Also, the right hand side function is holomorphic except for $0,-1,...,-n+1$, and it coincides with $\Gamma$ (on $\Gamma$'s domain). Hence its value does not depend on $n$ (due to the identity principle). Taking sufficiently large $n$, we can compute the value of the holomorphic continuation for all complex numbers except $0,-1,-2,...$. – ir7 Jan 6 '14 at 2:39
The long-time standard idea is to integrate by parts, giving $\Gamma(z)=\Gamma(z+1)/z$. This extends $\Gamma(z)$ to the left by unit-width strips, inductively.
Hint: use the functional equation $\Gamma(z+1) = z\Gamma(z)$ to repeatedly extend the Gamma function to a larger and larger domain...