Question Number 2930 by Filup last updated on 30/Nov/15
$$\mathrm{Prove}\:\mathrm{that}\:\Gamma\left({i}\right)=−{i}\left({i}\right)! \\ $$$$\mathrm{where}\:{i}=\sqrt{−\mathrm{1}} \\ $$
Commented by 123456 last updated on 02/Dec/15
$$\Gamma\left({z}\right)\Gamma\left(\mathrm{1}−{z}\right)=\frac{\pi}{\mathrm{sin}\:\pi{z}} \\ $$$$\Gamma\left({z}\right)\Gamma\left(−{z}\right)=−\frac{\pi}{{z}\:\mathrm{sin}\:\pi{z}} \\ $$$$\Gamma\left(\bar {{z}}\right)=\bar {\Gamma}\left({z}\right) \\ $$$${z}={x}\imath \\ $$$$\mid\Gamma\left({x}\imath\right)\mid^{\mathrm{2}} =\frac{\pi}{{x}\:\mathrm{sinh}\:\pi{x}} \\ $$
Answered by 123456 last updated on 02/Dec/15
$${x}!=\Gamma\left({x}+\mathrm{1}\right) \\ $$$$\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right) \\ $$$$\mathrm{so} \\ $$$${x}!={x}\Gamma\left({x}\right) \\ $$$${x}={i} \\ $$$${i}!={i} \Gamma\left({i}\right) \\ $$$$\Gamma\left({i}\right)=\frac{{i}!}{{i}}=\frac{{i}!{i}}{{i}^{\mathrm{2}} }=−{i}!{i} \\ $$