TITLE: Prove: $\int_{-\infty}^{\infty} \frac{\tan^{-1}e^{-\pi x}}{\cosh\frac{3x\zeta(2)}{20}} dx = 10$
QUESTION [6 upvotes]: I want to solve the following integral:
$$\int_{-\infty}^{\infty} \frac{\tan^{-1}e^{-\pi x}}{\cosh\frac{3x\zeta(2)}{20}} dx = 10$$
Have not tried it yet, but it may be tough. All I know is that  ζ(2) is  π²/6. I don't know what to do. Any help would be appreciated. Thanks!

REPLY [10 votes]: Consider the integral 
\begin{align}
I(a,b)&=\int_{-\infty}^{\infty}\frac{\arctan e^{-ax}}{\cosh b x}dx=\\
&=\int_{0}^{\infty}\frac{\arctan e^{-ax}}{\cosh b x}dx+\int_{0}^{\infty}\frac{\arctan e^{ax}}{\cosh b x}dx=\\
&=\frac{\pi}{2}\int_{0}^{\infty}\frac{dx}{\cosh b x}=\\
&=\frac{\pi}{4b}\int_{-\infty}^{\infty}\frac{dx}{\cosh  x}=\\
&=\frac{\pi^2}{4b}.
\end{align}
Now set $\displaystyle b=\frac{3\zeta(2)}{20}=\frac{\pi^2}{40}$.