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.
Suppose we seek a functional equation for
$$S(x) = \sum_{k\ge 1} \frac{1}{(2k-1)} \frac{1}{\exp(x(2k-1))-\exp(-x(2k-1))}.$$
(The factor of two that is missing is due to the sum that appeared in the post that I linked to in the introduction.)
The sum $S(x)$ is harmonic and may be evaluated by inverting its Mellin transform.
Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have $$\lambda_k = \frac{1}{(2k-1)}, \quad \mu_k = 2k-1 \quad \text{and} \quad g(x) = \frac{1}{\exp(x)-\exp(-x)}.$$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is computed as follows: $$g^*(s) = \int_0^\infty \frac{1}{\exp(x)-\exp(-x)} x^{s-1} dx = \int_0^\infty \frac{\exp(-x)}{1-\exp(-2x)} x^{s-1} dx \\ = \int_0^\infty \sum_{q\ge 0} \exp(-(2q+1)x) x^{s-1} dx = \sum_{q\ge 0} \frac{1}{(2q+1)^s} \Gamma(s) = \left(1-\frac{1}{2^s}\right) \Gamma(s) \zeta(s).$$
Hence the Mellin transform $Q(s)$ of $S(x)$ is given by $$ Q(s) = \left(1-\frac{1}{2^{s+1}}\right) \left(1-\frac{1}{2^s}\right) \Gamma(s) \zeta(s) \zeta(s+1) \\ \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \left(1-\frac{1}{2^{s+1}}\right) \zeta(s+1)$$ where $\Re(s) > 1$.
Intersecting the fundamental strip and the half-plane from the zeta function term we find that the Mellin inversion integral for an expansion about zero is $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ which we evaluate in the left half-plane $\Re(s)<3/2.$
The two zeta function terms cancel the poles of the gamma function term and we are left with just
$$\begin{align} \mathrm{Res}(Q(s)/x^s; s=1) & = \frac{\pi^2}{16x} \quad\text{and}\\ \mathrm{Res}(Q(s)/x^s; s=0) & = -\frac{1}{4} \log 2. \end{align}$$
This shows that $$S(x) = \frac{\pi^2}{16x} -\frac{1}{4} \log 2 + \frac{1}{2\pi i} \int_{-1/2-i\infty}^{-1/2+i\infty} Q(s)/x^s ds.$$
To treat the integral recall the duplication formula of the gamma function: $$\Gamma(s) = \frac{1}{\sqrt\pi} 2^{s-1} \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{s+1}{2}\right).$$
which yields for $Q(s)$
$$\left(1-\frac{1}{2^{s+1}}\right) \left(1-\frac{1}{2^s}\right) \frac{1}{\sqrt\pi} 2^{s-1} \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{s+1}{2}\right) \zeta(s) \zeta(s+1)$$
Furthermore observe the following variant of the functional equation of the Riemann zeta function: $$\Gamma\left(\frac{s}{2}\right)\zeta(s) = \pi^{s-1/2} \Gamma\left(\frac{1-s}{2}\right) \zeta(1-s)$$
which gives for $Q(s)$ $$\left(1-\frac{1}{2^{s+1}}\right) \left(1-\frac{1}{2^s}\right) \frac{1}{\sqrt\pi} 2^{s-1} \pi^{s-1/2} \Gamma\left(\frac{s+1}{2}\right) \Gamma\left(\frac{1-s}{2}\right) \zeta(1-s)\zeta(s+1) \\ = \left(1-\frac{1}{2^{s+1}}\right) \left(1-\frac{1}{2^s}\right) \frac{1}{\sqrt\pi} 2^{s-1} \pi^{s-1/2} \frac{\pi}{\sin(\pi(s+1)/2)} \zeta(1-s)\zeta(s+1) \\ = \left(1-\frac{1}{2^{s+1}}\right) \left(1-\frac{1}{2^s}\right) 2^{s-1} \frac{\pi^s}{\sin(\pi(s+1)/2)} \zeta(1-s)\zeta(s+1).$$
Now put $s=-u$ in the remainder integral to get
$$\frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty} \left(1-\frac{2^u}{2}\right) \left(1-2^u\right) 2^{-u-1} \frac{\pi^{-u}}{\sin(\pi(-u+1)/2)} \zeta(1+u)\zeta(1-u) x^u du \\ = \frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty} \left(1-\frac{2^u}{2}\right) \left(1-2^u\right) 2^{-u-1} \\ \times \frac{\pi^{u}}{\sin(\pi(-u+1)/2)} \zeta(1+u)\zeta(1-u) (x/\pi^2)^u du.$$
We may shift this to $3/2$ as there is no pole at $u=1.$
Now $$\sin(\pi(-u+1)/2) = \sin(\pi(-u-1)/2+\pi) \\ = - \sin(\pi(-u-1)/2) = \sin(\pi(u+1)/2)$$
and furthermore $$\left(1-\frac{2^u}{2}\right) \left(1-2^u\right) 2^{-u-1} = \frac{1}{2} \left(1-\frac{2^u}{2}\right) \left(\frac{1}{2^u}-1\right) = 2^{u-2} \left(\frac{1}{2^{u-1}}-1\right) \left(\frac{1}{2^u}-1\right) \\ = 2^{u-2} \left(1-\frac{1}{2^{u-1}}\right) \left(1-\frac{1}{2^u}\right) \\ = 2^{u-2} \left(1-\frac{1}{2^{u+1}}\right) \left(1-\frac{1}{2^u}\right) - 3\times 2^{u-2} \frac{1}{2^{u+1}} \left(1-\frac{1}{2^u}\right) \\ = \frac{1}{2} 2^{u-1} \left(1-\frac{1}{2^{u+1}}\right) \left(1-\frac{1}{2^u}\right) - \frac{3}{4} 2^{u-1} \left(1-\frac{1}{2^u}\right) \frac{1}{2^{u}}.$$
We have shown that $$S(x) = \frac{\pi^2}{16x} -\frac{1}{4} \log 2 + \frac{1}{2} S(\pi^2/x) \\ - \frac{3}{4} \frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} \left(1-\frac{1}{2^u}\right) \Gamma(u) \zeta(u) \zeta(u+1) (x/\pi^2/2)^u du$$
or alternatively
$$S(x) = \frac{\pi^2}{16x} -\frac{1}{4} \log 2 + \frac{1}{2} S(\pi^2/x) - \frac{3}{4} T(2\pi^2/x)$$
where $$T(x) = \sum_{k\ge 1} \frac{1}{k} \frac{1}{\exp(kx)-\exp(-kx)}$$
with functional equation $$T(x) = \frac{1}{24} x - \frac{1}{2}\log 2 + \frac{\pi^2}{12x} - T(2\pi^2/x).$$
which finally yields $$S(x) = \frac{1}{2} S(\pi^2/x) - \frac{1}{32} x + \frac{1}{8} \log 2 + \frac{3}{4} T(x).$$
Using $\sinh$ with $$S(x) = \sum_{k\ge 1} \frac{1}{(2k-1)} \frac{1}{\sinh((2k-1)x)} \quad\text{and}\quad T(x) = \sum_{k\ge 1} \frac{1}{k} \frac{1}{\sinh(kx)}$$
we obtain the functional equation $$S(x) = \frac{1}{2} S(\pi^2/x) - \frac{1}{16} x + \frac{1}{4} \log 2 + \frac{3}{4} T(x).$$
We also have $$T(\sqrt{2}\pi) = \frac{\sqrt{2}\pi}{24} - \frac{1}{2}\log 2 + \frac{\pi\sqrt{2}}{24} = \frac{\sqrt{2}\pi}{12} - \frac{1}{2}\log 2.$$