Gamma function can be extended as a meromorphic function

Question

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?

Answer 1

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 ${\Re}z>-n$.

Answer 2

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.

Historically, Euler's (and others') discussion of this extension of "factorial" often used the ("Hadamard") product expansion of the Gamma function, but it is not the simplest thing to prove that the integral has the appropriate product expansion, etc. True, it is interesting to understand this, probably best in the light of (post-Weierstrass) Hadamard products, rather than the (to my taste weird, immemorable) explicit manipulations of Whittaker-and-Watson. But the quickest more-or-less airtight proof is just integration by parts.

Answer 3

Hint: use the functional equation $\Gamma(z+1) = z\Gamma(z)$ to repeatedly extend the Gamma function to a larger and larger domain...