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

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 :)

• How have you defined $\Gamma$? – user61527 Feb 6 '14 at 16:03
• @T. Bongers I first defined for $Re(z)>0$ and then defined the other part of the complex plane as follows. Let $z$ be a complex number such that $Re(z)<0$ and nonnegative. Then there exists a unique $N\in\mathbb{Z}^+$ such that $-N<Re(z)<-N+1$. Then define $\Gamma\triangleq \frac{\Gamma(z+N)}{z(z+1)\cdots (z+(N-1))}$ – John. p Feb 6 '14 at 16:08
• @T. Bonger I defined it as a Euler's second integral – John. p Feb 6 '14 at 16:29

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:

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