TITLE: Definite integration $\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx$
QUESTION [11 upvotes]: How do I integrate $$\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx\quad ?$$
The actual integral that I encountered is:
$$\int_{-\infty}^\infty dx \left(\frac{N}{\cosh(\frac{\pi }{c}(x-1))}+\frac{1}{\cosh(\frac{\pi}{c}x)} \right) 2 \tan^{-1}\left(\frac{2x-2}{c} \right)$$ where c is a constant with $$\Re c>0$$ Not sure if these two terms makes it easier.
I was trying to solve just the last term, but I couldn't make any progress. Numerical integration gives $\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx= -1.01334 $. Any hint on how to do it analytically?

REPLY [5 votes]: Since we have the trivial representations $\displaystyle \int_0^{\infty }\left(\int_0^{\infty }  \sin ((2 x-2) y) e^{-y (1+z)}\textrm{d}z\right)\textrm{d}y=\arctan(2x-2)$ and then $\displaystyle \int_{-\infty }^{\infty } \frac{\sin ((2 x-2) y)}{\cosh (\pi  x)} \textrm{d}x=-\frac{\sin(2 y)}{\cosh(y)}$, we arrive at
$$\mathcal{I}=-\int_0^{\infty}\left(\int_0^{\infty}\frac{\sin(2 y)}{\cosh(y)} e^{-(1+z)y} \textrm{d}y\right)\textrm{d}z,$$
where expanding $\operatorname{sech}(y)$ in series, integrating with respect to $y$ and identifying the polygammas with a complex argument, we have
$$\mathcal{I}= \int_0^{\infty}\left(-\frac{1}{4} i \psi ^{(0)}\left(1+\frac{i}{2}+\frac{z}{4}\right)+\frac{1}{4} i \psi ^{(0)}\left(\frac{1}{2}+\frac{i}{2}+\frac{z}{4}\right)+\frac{1}{4} i \psi ^{(0)}\left(1-\frac{i}{2}+\frac{z}{4}\right)-\frac{1}{4} i \psi ^{(0)}\left(\frac{1}{2}-\frac{i}{2}+\frac{z}{4}\right)\right)\textrm{d}z=i \log \left(\frac{\displaystyle\Gamma \left(\frac{1}{2}-\frac{i}{2}\right) \Gamma \left(1+\frac{i}{2}\right)}{\displaystyle \Gamma \left(\frac{1}{2}+\frac{i}{2}\right) \Gamma \left(1-\frac{i}{2}\right)}\right),$$
where the last integration is trivially developed by using negapolygamma function.
End of story