# 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.

• Thanks for Te(fi)xing that up for me :) I am studying for an exam right now and just thought of it on the fly.
– joe
Nov 16, 2012 at 7:53

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\,\mathcal{#1}} \newcommand{\mrm}[1]{\,\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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}