Proof that the gamma function is an extension of the factorial function

Question

I've already proved that $$\Gamma (n)= (n-1)!$$ but I don´t really know what else to do to verify that $\Gamma$ is an extension of the factorial function for real numbers (positive) Thank you! And I´m sorry for my language, I am Spanish, so thank you again for trying to understand me.

Answer 1

Consider $$\Gamma (z)=\int_0^\infty e^{-t}t^{z}{dt\over t}.$$ We can rewrite this as $$\Gamma (z)=\int_0^1 e^{-t}t^{z}{dt\over t}+\int_1^\infty e^{-t}t^{z}{dt\over t}.$$ In the first term of this sum we see that the power series representation of $e^{-t}$ converges uniformly which implies that the series can be integrated term by term. So, $$\Gamma (z)=\int_0^1\sum_{n=0}^\infty {(-1)^n\over n!}t^{z+n}{dt\over t}+\int_1^\infty e^{-t}t^{z}{dt\over t}=\sum_{n=0}^\infty {(-1)^n\over n!(z+n)}+\int_1^\infty e^{-t}t^{z}{dt\over t}.$$ We can see that the series converges for $z\neq 0, -1, -2,...$ which is a meromorphic function. Its poles are simple poles at the non-positive integers. The residue at $-n$ is $(-1)^n\over n!$. The last integral extends as an entire function of z. Thus $\Gamma (z)$ has been analytically continued to the entire complex plane except for $z\neq0,1,2,...$.

Answer 2

It is a result in Ahlfohrs, Complex Analysis, actually exercise 1 on page 196, that the factorial function can be extended in any way we like at a few non-integer points, and an entire holomorphic function can be constructed to fit those points. In particular, if real valued, we can make any real-analytic extension that we want. So, the fact that the gamma function is analytic is not the big restriction. It is a simpler:

this is the only log-convex extension of the factorial.

http://en.wikipedia.org/wiki/Logarithmic_convexity

http://en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem

http://en.wikipedia.org/wiki/Gamma_function