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Is there a closed form for this sum?

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

I have tried using

- How to prove $\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$
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- Accurate identities related to $\sum\limits_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}x^n$ and $\sum\limits_{n=0}^{\infty}\frac{(2n)!}{(n!)^4}x^n$
- How do I prove that $\lim_{n\to+\infty}\frac{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}{\sqrt{n}}=?$
- How to divide natural number N into M nearly equal summands?

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

To solve but got stuck at evaluating

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

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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.

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