mellin Transform $\sum _{k=1}^{\infty } k^2 \text{sech}(k x)=\frac{\pi ^3}{8 x^3}$

Question

I try to calculate $$\sum _{k=1}^{\infty } k^2 \text{sech}(k x)=\frac{\pi ^3}{8 x^3}$$ using mellin transform but it seem something miss?? i calculate first the mellin transform $$2^{1-2 s} \zeta (s) \left(\zeta \left(s,\frac{1}{4}\right)-\zeta \left(s,\frac{3}{4}\right)\right) \Gamma (s)$$ but numerically it is an approximant but t is not exactly

Answer 1

$$\begin{eqnarray*}\sum_{n\geq 1}\frac{n^2}{\cosh(nx)} = \sum_{n\geq 1}\frac{2n^2 e^{-nx}}{1+e^{-2nx}}&=&\sum_{n\geq 1}\sum_{m\geq 0}2n^2(-1)^m e^{-(2m+1)nx}\\&=&\frac{1}{2}\sum_{m\geq 0}(-1)^m \frac{\cosh((m+1/2)x)}{\sinh^3((m+1/2)x)}.\end{eqnarray*}$$ Now the function $f(x)=\frac{\cosh((m+1/2)x)}{\sinh^3((m+1/2)x)}$ has a pole of order $3$ at $x=0$, and its Laurent series is given by: $$ f(x) = \frac{4}{(2m+1)^3 x^3} -\frac{2m+1}{60}\,x +\ldots $$ hence: $$ \sum_{n\geq 1}\frac{n^2}{\cosh(nx)}\approx \sum_{m\geq 0}\frac{2(-1)^m}{(2m+1)^3 x^3} = \frac{\pi^3}{8 x^3} $$ (for a proof of the last identity, have a look at this question) as wanted, but in the above line we are clearly neglecting the contributes given by the other singularities of $\frac{\cosh(x)}{\sinh^3(x)}$.

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It is interesting to point out that the given approximation is essentially equivalent to: $$ \sum_{n\in\mathbb{Z}}\frac{n^2}{\cosh(n)}\approx \int_{\mathbb{R}}\frac{x^2}{\cosh x}\,dx = \frac{\pi^3}{4}, $$ where the fact that the limit form of the Euler-MacLaurin formula fails to give an exact equality can be seen as a consequence of the Abel-Plana summation formula.

Answer 2

Working a little hard it possible to get at x=Pi $$\sum _{k=1}^{\infty } k^2 \text{sech}(\pi k)=\frac{1}{8}-\frac{i \left(\psi _{e^{\pi }}^{(2)}\left(1-\frac{i}{2}\right)-\psi _{e^{\pi }}^{(2)}\left(1+\frac{i}{2}\right)\right)}{4 \pi ^3}$$