Is there a closed form for this sum?

$$\sum _{n=1}^{\infty }\frac{\coth (xn)}{n^3}$$

I have tried using

$$x\coth \left(xn\right)=\frac{1}{n}+2n\sum _{k=1}^{\infty }\frac{1}{n^2+\left(\frac{\pi }{x}\right)^2k^2}$$

To solve but got stuck at evaluating

$$\sum _{n=1}^{ \infty}\sum _{k=1}^{\infty }\frac{1}{k^2\left(n^2+\left(\frac{\pi }{x}\right)^2k^2\right)}$$

  • $\begingroup$ @DougM: Assuming $x\neq0$, $\coth{nx}\to\pm1$ as $n\to\infty$. $\endgroup$ – John Bentin Aug 4 '16 at 16:31
  • $\begingroup$ Sorry, I had $\cosh x$ in my mind. $\endgroup$ – Doug M Aug 4 '16 at 16:37

We may consider that our series equals $$ S(x)=\frac{d}{dx}\sum_{n\geq 1}\frac{\log(\sinh(nx))}{n^4}=\frac{\zeta(4)}{x}+\frac{d}{dx}\sum_{n\geq 1}\frac{1}{n^4}\sum_{m\geq 1}\log\left(1+\frac{n^2 x^2}{\pi^2 m^2}\right) \tag{1}$$ where in the second step we exploited the Weierstrass product for $\frac{\sinh(z)}{z}$. We get: $$\begin{eqnarray*} S(x)&=&\frac{\zeta(4)}{x}+2x\sum_{n,m\geq 1}\frac{1}{m^2 n^2\pi^2+n^4 x^2}\\&=&\frac{\zeta(4)}{x}+2x\left(\frac{5\pi^2+x^2}{180}-\frac{x}{2\pi^2}\sum_{m\geq 1}\frac{1}{m^3}\,\coth\left(\frac{m\pi^2}{x}\right)\right)\tag{2}\end{eqnarray*} $$ and the same can be achieved by applying the Poisson summation formula, but I am not expecting a great simplification in the last sum (on $m\geq 1$) unless $x$ is a particular integer multiple of $\pi$.
However, it is interesting to notice that the above formula leads to: $$ S(\pi) = \sum_{n\geq 1}\frac{\coth(\pi n)}{n^3}=\color{red}{\frac{7\pi^3}{180}},\tag{3} $$ for instance.

  • 1
    $\begingroup$ why on earth people didn't upvote more on such nice answers...(+1) $\endgroup$ – tired Aug 4 '16 at 23:35
  • $\begingroup$ I up vote because I got a question on the general of this question. $\endgroup$ – gymbvghjkgkjkhgfkl Aug 5 '16 at 22:37

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