Calculating Riemann zeta function of a complex number given the complex contour integral Can you please demonstrate how one would calculate the Riemann Zeta function of any complex number, given that the Riemann Zeta function is equal to the following (shown in http://arxiv.org/pdf/1208.3429v1.pdf): 
If you utilize a technique in complex analysis (Such as Cauchy's Integral Formula), may you please explain the process step-by-step.
Thank you,
Best Regards,
J.M
 A: Sorry for the late response, I have been busy with finals and such.
Just for an example, let's calculate $\zeta(-1)$.
$$ \zeta(-1)=\frac{\Gamma(2)}{2\pi i}\oint_{\gamma}\frac{u^{-2}}{e^{-u}-1}du  $$
Where $\gamma$ is the Hankel contour.  Let's just focus on the integrand right now and expand it into a series.  The full derivation on how this can be expanded is found in the link I provided in the comments.  Thus:
$$ \frac{u^{-2}}{e^{-u}-1}=-\frac{1}{u^3}-\frac{1}{2u^2}-\frac{1}{12u}+\dots $$
Now, we can integrate this series term by term.  Essentially, we are coming in from $-\infty$and going around the unit circle and then back out again to $-\infty$.  The two paths coming in and out will cancel each other, as the only pole is at $u=0$.  Thus, we can just focus on the unit circle contour.  In general, the contour integral around the unit circle of the function $1/z^n$ is zero unless $n=1$. Thus,
$$\oint_{\gamma} \frac{u^{-2}}{e^{-u}-1}du=-\frac{1}{12}\oint_{\gamma}\frac{du}{u} $$
Then, by the Cauchy integral formula or the Residue theorem,
$$\oint_{\gamma}\frac{du}{u}=2\pi i  $$
So,
$$ \zeta{(-1)}=\frac{-\Gamma(2)}{12(2\pi i)}2\pi i=-\frac1{12} $$
