A functional equation relating two harmonic sums. Introduction. I computed two Mellin transforms while browsing / working on the problem at this MSE link. No solution was found, but some interesting auxiliary results appeared. I am writing to ask for independent confirmation of these results, not necessarily using Mellin transforms.
Problem statement. Introduce 
$$S(x) = \sum_{k\ge 1} \frac{1}{(2k-1)} 
\frac{1}{\sinh((2k-1)x)}
\quad\text{and}\quad
T(x) = \sum_{k\ge 1} \frac{1}{k} 
\frac{1}{\sinh(kx)}$$
Prove the functional equation
$$S(x) = \frac{1}{2} S(\pi^2/x)
- \frac{1}{16} x
+ \frac{1}{4} \log 2
+ \frac{3}{4} T(x).$$ 
Evaluate $T(x)$ at $x=\sqrt{2}\pi$ and prove that
$$T(\sqrt{2}\pi) = \frac{\sqrt{2}\pi}{12} - \frac{1}{2}\log 2.$$
Remark. It is hoped that these two problems might reward investigation, perhaps using several different methods. I do ask that possible details of the computations be included.
 A: Let $q = e^{-x}$ so that $$T(x) = \sum_{n = 1}^{\infty}\frac{1}{n\sinh nx} = 2\sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{2n})} = f(q)\text{ (say)}\tag{1}$$ and then we have
\begin{align}
T(x) &= f(q)\notag\\
&= 2\sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{2n})}\notag\\
&= 2\sum_{n = 1}^{\infty}\frac{q^{n}}{n}\sum_{m = 0}^{\infty}q^{2mn}\notag\\
&= 2\sum_{m = 0}^{\infty}\sum_{n = 1}^{\infty}\frac{q^{(2m + 1)n}}{n}\notag\\
&= -2\sum_{m = 0}^{\infty}\log(1 - q^{2m + 1})\notag\\
&= -2\log\prod_{m = 1}^{\infty}(1 - q^{2m - 1})\notag\\
&= -2\log\left(2^{1/4}q^{1/24}\left(\frac{2k}{k'^{2}}\right)^{-1/12}\right)\tag{2}\\
&= -\frac{\log 2}{2} - \frac{\log q}{12} + \frac{1}{6}\log\frac{2k}{k'^{2}}\tag{3}
\end{align}
where $k$ is the elliptic modulus corresponding to nome $q$ and $k' = \sqrt{1 - k^{2}}$.
The product $\prod(1 - q^{2m - 1})$ is essentially Ramanujan's class invariant $g_{n}, g(q)$ and we have accordingly given its representation $(2)$ in terms of nome $q$ and modulus $k$. Also note that $\log q = -\pi K'/K$ where $K, K'$ are complete elliptic integrals of first kind with modulus $k$ and $k' = \sqrt{1 - k^{2}}$.
By the theory of modular equations (given in the link on Ramanujan's class invariant mentioned earlier) it is known that if $K'/K = \sqrt{r}$ where $r$ is a positive rational number then the value of modulus $k$ is an algebraic number and such value of $k$ is called a singular modulus and denoted by $k_{r}$. Here in the current notation we have $\log q = -x$ and hence if $x = \pi\sqrt{r}$ then the value of $k$ is an algebraic number and it is possible to have a closed form for $T(x)$ in terms of logarithm of an algebraic number plus $x/12$.
If we set $x = \pi\sqrt{2}$ then the corresponding value of singular modulus $k$ is $k = k_{2} = \sqrt{2} - 1$. And hence $2k/(1 - k^{2}) = 1$ and it follows from $(3)$ that $$T(\sqrt{2}\pi) = \frac{\sqrt{2}\pi}{12} - \frac{1}{2}\log 2\tag{4}$$ Next we discuss the sum $S(x)$ given by $$S(x) = \sum_{n\text{ odd}}\frac{1}{n\sinh nx} = \sum_{n = 1}^{\infty}\frac{1}{n\sinh nx} - \sum_{n\text{ even}}\frac{1}{n\sinh nx} = T(x) - \frac{T(2x)}{2}\tag{5}$$ Note that when $x$ is replaced by $2x$ then $q = e^{-x}$ is replaced by $q^{2}$ and hence $T(2x) = f(q^{2})$. By Landen's transformation replacing $q$ with $q^{2}$ leads to replacing $k$ with $(1 - k')/(1 + k')$ and hence
\begin{align}
T(2x) &= f(q^{2})\notag\\
&= -\frac{\log 2}{2} - \frac{\log q^{2}}{12} + \frac{1}{6}\log\dfrac{2\cdot\dfrac{1 - k'}{1 + k'}}{1 - \left(\dfrac{1 - k'}{1 + k'}\right)^{2}}\notag\\
&= -\frac{\log 2}{2} - \frac{\log q}{6} + \frac{1}{6}\log\frac{1 - k'^{2}}{2k'}\notag\\
&= -\frac{\log 2}{2} - \frac{\log q}{6} + \frac{1}{6}\log\frac{k^{2}}{2k'}\tag{6}
\end{align}
and hence it follows from $(3), (5)$ and $(6)$ that $$S(x) = -\frac{\log 2}{4} + \frac{1}{12}\log\frac{8}{k'^{3}}\tag{7}$$ Note that the above equation shows that for $x = \pi\sqrt{r}, r\in\mathbb{Q}^{+}$ the value of $S(x)$ is the logarithm of an algebraic number.
Next we deal with the transformation from $x$ to $\pi^{2}/x$. This changes $q = e^{-x}$ to $q' = e^{-\pi^{2}/x}$ and the effect of this is to swap $k$ and $k'$ so that $$S(\pi^{2}/x) = -\frac{\log 2}{4} + \frac{1}{12}\log\frac{8}{k^{3}}\tag{8}$$ and from $(7)$ and $(8)$ we get $$S(x) - (1/2)S(\pi^{2}/x) = -\frac{\log 2}{8} + \frac{1}{8}\log \frac{2k}{k'^{2}}\tag{9}$$ and looking at equations $(3)$ and $(9)$ we get $$S(x) = \frac{1}{2}S\left(\frac{\pi^{2}}{x}\right) - \frac{1}{16}x + \frac{1}{4}\log 2 + \frac{3}{4}T(x)\tag{10}$$ which is the desired functional equation connecting $S(x), S(\pi^{2}/x)$ and $T(x)$.

Like most of my answers dealing with sums involving hyperbolic functions this answer also requires a good understanding of the theory of elliptic integrals and their link with theta functions and other related topics.
Incidentally Ramanujan delved very deep into these topics and he had a very good understanding of both Mellin transform (favorite tool used by OP, see his answers) and elliptic/theta functions and moreover he somehow had the real-analysis equivalent of Mellin transform methods (Ramanujan's Master Theorem) so he could achieve his results without any use of complex analysis.
