Messing around with the abel-plana formula for $\sum_n \frac 1{n^3}$ I've just discovered the Abel-Plana formula: http://en.wikipedia.org/wiki/Argument_principle
I'm trying to use it to get a closed-form expression for $\sum_{n=1}^\infty \frac 1{n^3}$.
So far, I have the 
$$ \sum_{n=1}^\infty \frac 1{n^3}= 1+2i\int_0^{\infty} \frac{dt}{\bigl(\exp(2\pi t)-1\bigr)(it+1)^3}. $$ This integral has poles at $t=0$ and $t=i$. I know I should use the residue theorem but I'm not sure how to apply it to this integral. Any thoughts would be appreciated.
PS- I know that the answer should be $\zeta(3)$ but I want to know what the Abel-Plana formula has to say about it.

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Following your Wikipedia link, you should write
\begin{align}
\color{#f00}{\sum_{n = 1}^{3}{1 \over n^{3}}} & =
\sum_{n = 0}^{3}{1 \over \pars{1 + n}^{3}} = \int_{0}^{\infty}{\dd x \over \pars{1 + x^{3}}} + \left.\half\,{1 \over \pars{1 + x}^{3}}
\right\vert_{\ x\ =\ 0} -
2\,\Im\int_{0}^{\infty}
{1 \over \pars{1 + \ic x}^{3}}\,{\dd x \over \expo{2\pi x} - 1}
\\[5mm] & =
1 - 2 \Im\int_{0}^{\infty}
{\pars{1 - \ic x}^{3} \over \pars{1 + x^{2}}^{3}}
\,{\dd x \over \expo{2\pi x} - 1} =
\color{#f00}{1 - 2\int_{0}^{\infty}{x\pars{x^{2} - 3} \over \pars{x^{2} + 1}^{3}\pars{\expo{2\pi x} - 1}}\,\dd x}
\end{align}
