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
$$T(\sqrt{2}\pi) =
\frac{\sqrt{2}\pi}{24} - \frac{1}{2}\log 2
+ \frac{\pi\sqrt{2}}{24}
= \frac{\sqrt{2}\pi}{12} - \frac{1}{2}\log 2.$$
