How to compute $\sum_{n\text{ odd}}\frac{1}{n\sinh n\pi\sqrt 3}$? I came across an old question asking to show that $$\displaystyle\sum_{n\text{ odd}}\frac{1}{n\sinh n\pi}=\frac{\ln 2}{8}.\tag{1}$$ 
Although I have managed to prove this formula, my proof uses various theta functional relations and looks like an overkill. On the other hand, it suggests a few more identities, for example
$$\displaystyle\sum_{n\text{ odd}}\frac{1}{n\sinh n\pi\sqrt3}=\frac{\ln \left(8-4\sqrt{3}\right)}{4}.\tag{2}$$



Question: can one prove (1) and (2) in a more elementary way? ideally, the proof should make clear further generalizations.
The identities:
$$\sum_{n\geq 1}\frac{(-1)^n}{n^2+m^2} = -\frac{1}{2m^2}+\frac{\pi}{2m\sinh(m\pi)},\tag{A}$$
$$ \frac{1}{m^2+n^2}=\int_{0}^{+\infty}\frac{\sin(nx)}{n}e^{-mx}\,dx,\tag{B}$$
$$ \sum_{n\geq 1}(-1)^n\frac{\sin(nx)}{x}=-\frac{x}{2}+\pi\left\lfloor\frac{x+\pi}{2\pi}\right\rfloor\tag{C} $$
give a wide range of possibilities to evaluate our series. For instance, $(A)$ gives:
$$\begin{eqnarray*}\sum_{k\geq 0}\frac{1}{(2k+1)\sinh(\pi(2k+1))}&amp;=&amp;\frac{1}{\pi}\sum_{k\geq 0}\frac{1}{(2k+1)^2}+\frac{2}{\pi}\sum_{k\geq 0}\sum_{n\geq 1}\frac{(-1)^n}{n^2+(2k+1)^2}\\&amp;=&amp;-\frac{\pi}{8}+\frac{2}{\pi}\sum_{h\geq 1}\frac{r_2(h)\cdot\eta(h)}{h}\tag{1}\end{eqnarray*}$$
where $\eta(h)$ equals $-1$ if $h\equiv 2\pmod{4}$, $1$ if $h\equiv 1\pmod{4}$, zero otherwise, and:
$$ r_2(h) = \#\{(n,k)\in\mathbb{N}^2: h=n^2+(2k+1)^2\}.\tag{2} $$
Now it is well-known that $a^2+b^2$ is the only reduced binary quadratic form of discriminant $-4$, hence:
$$\begin{eqnarray*} \#\{(a,b)\in\mathbb{Z}:a^2+b^2=n\} &amp;=&amp; 4\left(\chi_4 * 1\right)(n)\\&amp;=&amp; 4\left(d_{1(4)}(n)-d_{3(4)}(n)\right) \tag{3}
\end{eqnarray*}$$
so that we can evaluate the RHS of $(1)$ through Dirichlet convolution.

Since we have class number one also in the case $a^2+3b^2$, the situation is almost the same for the other series.

The Mellin transform gives another chance. See, for instance, this related problem.
Don't forget that sinh is an odd function, so we can use
$$
n \, sinh(n \pi) = -n \, sinh(-n \pi)
$$
to reduce the summation to
$$
\sum_{n\in2\Bbb Z+1}\frac{1}{n\,sinh(n\pi)} = \sum_{n\in2\Bbb N+1}\frac{1}{n\,sinh(n\pi)}
$$
which is more directly manipulated (without having to play with summation indices) if written anyway as
$$
\sum_{n=1}^{\infty}\frac{1}{(2n+1)\,sinh((2n+1)\pi)}
$$

Not really sure how to simplify to your end, from here, but perhaps this is helpful anyway.

-Cheers