My problem consists of 3 parts. Let $\alpha,\beta>0$ and $\alpha\beta=\pi^2$
(1) Let $f(\alpha)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)(e^{(2k+1)\alpha}+1)}+\frac{1}{8}\log\alpha$
Then $f(\alpha)=f(\beta)$
(2) $\alpha\sum_{k=0}^{\infty}sech^2{\frac{(2k+1)\alpha}{2}} + \beta\sum_{k=0}^{\infty}sech^2{\frac{(2k+1)\beta}{2}}=1$
(3) $\sum_{k=0}^{\infty}sech^2{\frac{(2k+1)\alpha}{2}}=\frac{1}{2\pi}$
I solved (2) by differentiating (1) in $\alpha$ and solved (3) by putting in $\alpha=\beta=\pi$.
So the difficult part is (1). I tried to use some identities or telescoping, however it is too hard to use the condition that $\alpha\beta=\pi^2$ in infinite summation and proper identities to use. And I thought about contour integrals of a nice functions such that residues are of the form in (1) but also failed to find such function. Finally by integrating both side of (1) and if I show the equality, then I can prove (1) since we can use the fact that when $\alpha=\beta=\pi$, the initial condition is satisfied.
(2) is the result of (1) so it is not useful to show that the differentiated terms are equal.
Is there any method to prove the identity (1) clearly? Thank you for reading my question.
p.s. I considered to use poisson summation formula, but cannot find such suitable function.
I have seen similar nice symmetric relations in several papers by Dixit et al. One of them is
http://arxiv.org/pdf/1312.1232v1.pdf
You may search arXiv.org for other related papers by him.
All of them have nice relation like $F(\alpha)=F(\beta)$ where $\alpha\beta=1$, or $\pi^2$
Hope the references are useful.