When looking at the functional equation for the Riemann zeta function, I came across the statement:
For $s$ an even positive integer, the product $\sin{(\frac{\pi s}{2})}\Gamma({1-s})$ is regular.
How come this does not evaluate to zero at positive even values of $s$, due to the sine function, or remain undefined due to the poles in the Gamma function? Is it to do with the series expansions by any chance?
We know that the functional equation of the Riemann zeta function $$\zeta\left(s\right)=2^{s}\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma\left(1-s\right)\zeta\left(1-s\right) $$ holds on the whole complex plane except for $s=1 $. In particular, it is possible to calculate your product at $s=2n $, with $n=1,2,\dots $. We have $$\sin\left(n\pi\right)\Gamma\left(1-2n\right)=\frac{\zeta\left(2n\right)}{2^{s}\pi^{s-1}\zeta\left(1-2n\right)} $$ (we are sure that $\zeta\left(1-2n\right)\neq0 $ because $1-2n$ is a negative odd integer) and we have closed form for zeta in these particular cases $$\zeta\left(2n\right)=\left(-1\right)^{n+1}\frac{B_{2n}\left(2\pi\right)^{2n}}{2\left(2n\right)!} $$ and $$\zeta\left(1-2n\right)=\zeta\left(-\left(2n-1\right)\right)=-\frac{B_{2n}}{2n}$$ where $B_{n} $ are Bernoulli numbers. So $$\sin\left(n\pi\right)\Gamma\left(1-2n\right)=\left(-1\right)^{n}\frac{\pi}{2\left(2n-1\right)!}. $$