Question

`\rm\sum_(n = 1)^ \infty \frac{ \coth(\pi nx) + x \coth \bigg( (\pi n)/(x) \bigg) }{ {n}^(3) } \\`​

Answer 1

Recalling the Mittag-Leffler expansion for the hyperbolic cotangent,

`\displaystyle \coth(\pi z) = -\frac1{\pi z} + \frac{2z^2}\pi \sum_(m=1)^\infty \frac1{z^2+m^2}`

which, by symmetry, is the same as

`\displaystyle \coth(\pi z) = -\frac1{\pi z} + \frac{z^2}\pi \sum_(m\in A) \frac1{z^2+m^2}`

where
`A = \Bbb Z \setminus \{0\}`.

By symmetry again, the given sum is

`\displaystyle S = \sum_(n=1)^\infty \frac{\coth(\pi n x) + x \coth\left(\frac{\pi n}x\right)}{n^3} = \frac12 \sum_(n\in A) \frac{\coth(\pi n x) + x \coth\left(\frac{\pi n}x\right)}{n^3}`

Now rewrite the summand using the M-L expansion.

`\displaystyle S = \frac12 \sum_(n\in A) \frac1{n^3} \left(-\frac1{\pi nx} + \frac{n^2x^2}\pi \sum_(m\in A) \frac1{n^2x^2+m^2} - \frac x{\pi n} + (n^2)/(\pi x^2) \sum_(m\in A) \frac1{(n^2)/(x^2) + m^2}\right) \\\\ = -\frac1{2\pi x} \left[(x^3+1)\sum_(n\in A)\frac1{n^4} - x^3 \sum_((m,n)\in A^2) \frac1n\cdot\frac1{n^2x^2+m^2} - x^2 \sum_((m,n)\in A^2) \frac1n\cdot\frac1{m^2x^2+n^2}\right]`

By symmetry,

`\displaystyle \sum_((m,n)\in A^2) \frac1n \cdot \frac1{m^2x^2+n^2} = \sum_((m,n)\in A^2) \frac1m\cdot\frac1{n^2x^2+m^2}`

and recalling the definition of the Riemann zeta function, this brings us to

`\displaystyle S = -\frac{\zeta(4)}\pi \cdot \frac{x^3+1}x + \frac1{2\pi x} \sum_((m,n)\in A^2) \left(\frac{x^3}n + \frac{x^2}m\right) \frac1{n^2x^2+m^2}`

and that's unfortunately as far as I've gotten...