Recall that the series:
n^s \underset{def}{=} e^{x\ln(n)}e^{i\ln(n)y}
\Gamma(s)\zeta(s)=\sum_{n=1}^\infty \frac{\left( \int z^{s-1}e^{-z}dz \right)}
=\sum_{n=1}^{\infty}\int_{0}^\infty z^{s-1}e^{-nz}dz
=\int_{0}^\infty \frac{z^{s-1}e^{-z}}{1-e^{-z}}dz=\int_{0}^\infty \frac{z^{s-1}}{e^z-1}dz
\zeta(s)= \frac{1}{(e^{2\pi i s}-1)\Gamma(s)}\int_{0}^\infty \frac{z^{s-1}}{e^z-1}dz
\zeta(1-n)=\frac{(-1)^{n-1}(n-1)!}{2\pi i} \int_{C} \frac{z^{-n}}{e^{-z}-1}dz
\frac{z}{e^{-z}-1}= -\frac{z}{2} +\sum_{{n=0}}^\infty \frac{b_{2n}}{(2n)!}z^
\zeta(1-s)=\frac{\zeta(s)}{2(2\pi)^{s-1}\Gamma(1-s)\sin\left( \frac{\pi s}{2} \right)} \hspace{1in} Re(s)<0