$\Gamma(x)=\lim_{n\to\infty} \frac{n^x n!}{x(x+1)\cdots (x+n)}$?

Question

I'm now reading Artin's Gamma Function.

$\Gamma(x)=\lim_{n\to\infty} \frac{n^x n!}{x(x+1)\cdots (x+n)}$?

He proved the above equality when $x$ is real using the fact $\Gamma$ is log-convex.

How do i extend this to complex plane?

I don't know analytic continuation so please give me a relatively elementary proof if it is possible. Thank you :)

Answer 1

I'll prove it for the case $\Re(z)>0$, WLOG.

Note that $\forall n\in\mathbb{Z}^+, \int_0^n t^{z-1} (1-\frac{t}{n})^n dt = \frac{n^z n!}{z(z+1)\cdots(z+n)}$.

Fix $n\in \mathbb{Z}^+$ and $\mu$ be the Lebesgue measure.

Then, there exists $F_n\in L^1(\mu$) such that $F_n\upharpoonright (0,n] = t^{z-1}(1-\frac{t}{n})^n$ and $F_n=F_n\chi_{(0,n]}$.

Note that $|F_n(t)| \leq t^{z-1} e^{-t}$ and $\lim_{n\to\infty} F_n(t) = t^{z-1}e^{-t}$.

By applying one of Monotone Convergence or Dominated Convergence Theorem:

$\lim_{n\to\infty} \frac{n^z n!}{z(z+1)\cdots(z+n)}=\int_0^\infty t^{z-1}e^{-t}$