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
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,
$$ \zeta{(-1)}=\frac{-\Gamma(2)}{12(2\pi i)}2\pi i=-\frac1{12} $$