TITLE: Rewrite $\Gamma(-z)$ in terms of $\Gamma(z)$
QUESTION [7 upvotes]: Is it possible to rewrite $\Gamma(-z)$ in terms of $\Gamma(z)$ where $\Gamma(z) = \int^\infty_0 t^{z-1}e^{-t}dt$?

REPLY [4 votes]: We have that $\Gamma(1+z)=z\Gamma(z)$, or $\Gamma(1-z)=-z\Gamma(-z)$.
We also have that $\Gamma(1-z)\Gamma(z)=\dfrac{\pi}{\sin(\pi z)}$.
Putting these two together yields $$\Gamma(z)\Gamma(-z)=-\dfrac{\pi}{z\sin(\pi z)}$$