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

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.

• But there are thousands of possible functions then. I mean, I don´t know how gamma should behave when x is not natural... it doesn´t matter? Oct 25, 2013 at 1:13
• es.wikipedia.org/wiki/Teorema_de_Bohr-Mollerup Oct 25, 2013 at 1:56
• es.wikipedia.org/wiki/Funci%C3%B3n_gamma Oct 25, 2013 at 1:58
• Have you looked into the functional equation for $\Gamma (z)$? Its behavior is similar to the factorial function. The technique of analytic continuation is extremely useful for the extension you seek. Oct 25, 2013 at 19:35

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 $z\neq0,1,2,...$.