A friend of mine send me the problem to integrate
$$\operatorname{PV}\int_0^{\infty}\frac{x\operatorname{tan}(\pi x)}{(1+x^2)^2}dx$$ where $\operatorname{PV}$ is Cauchy principle value.
I'm getting $\frac{1}{2}\psi^{(1)}\left(\frac{1}{2}+i\right)$ which is complex trigamma argument however, he has got answer in real closed form $\frac{\pi^2}{(e^{-\pi}+e^{\pi})^2.}$.
My work
I evaluted integral as follows
Recalling the result to due Weiestrass factorization theorem $$ \operatorname{cos}(\pi x)=\prod_{n\geq 0}\left(1-\frac{4 x}{(2n+1)^2}\right)$$ taking $\log$ and differentiating with rest to $x$ we have $$\operatorname{tan}(\pi x)=\frac{8}{\pi}\sum_{n\geq 0}\frac{x}{(2n+1)^2-4x^2}$$ and thus subbing this result the integral, we have $$\frac{8}{\pi}\sum_{n\geq 0}\int_0^{\infty}\frac{x^2dx}{((2n+1)^2-4x^2)(1+x^2)^2}$$
Making partial fraction and integration gives us $$\sum_{n\geq 0}\frac{2}{(2n+2i+1)^2}=\frac{1}{2}\psi^{1}\left(\frac{1}{2}+i\right)$$
As the per WA $$\Re\frac{1}{2}\left(\psi^{(1)}\left(\frac{1}{2}+i\right)\right)=\frac{\pi^2}{(e^{-\pi }+e^{\pi})^2}\tag{1}\label{mainfm}$$
My question is, How do i prove the relation \eqref{mainfm}?
I tried for the references of relationship between hyperbolic function and trigamma function however, I cannot get any such relationship.
Any sort of help/ reference or different approches will be appreciated. Thank you.
Interestingly, making change of $\operatorname{tan}(\pi x)$ as $\operatorname{tanh}(\pi x)$ I came up with the following closed form
$$\int_0^{\infty}\frac{x\tanh(\pi x)}{(1+x^2)^2}dx=\frac{\pi^2}{4}-2$$
$(1)$ follows immediately from the "conjugation property" and the reflection formula for $\psi^{(1)}$: $$\psi^{(1)}(\bar{z})=\overline{\psi^{(1)}(z)};\qquad\psi^{(1)}(z)+\psi^{(1)}(1-z)=\frac{\pi^2}{\sin^2\pi z}.$$