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.
Best Answer
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: $$\begin{eqnarray*}\sum_{k\geq 0}\frac{1}{(2k+1)\sinh(\pi(2k+1))}&=&\frac{1}{\pi}\sum_{k\geq 0}\frac{1}{(2k+1)^2}+\frac{2}{\pi}\sum_{k\geq 0}\sum_{n\geq 1}\frac{(-1)^n}{n^2+(2k+1)^2}\\&=&-\frac{\pi}{8}+\frac{2}{\pi}\sum_{h\geq 1}\frac{r_2(h)\cdot\eta(h)}{h}\tag{1}\end{eqnarray*}$$ 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: $$\begin{eqnarray*} \#\{(a,b)\in\mathbb{Z}:a^2+b^2=n\} &=& 4\left(\chi_4 * 1\right)(n)\\&=& 4\left(d_{1(4)}(n)-d_{3(4)}(n)\right) \tag{3} \end{eqnarray*}$$ 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.