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!
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 $\displaystyle b=\frac{3\zeta(2)}{20}=\frac{\pi^2}{40}$.