TITLE: Integral $\int_0^{\infty} \frac{\log x}{\cosh^2x} \ \mathrm{d}x = \log\frac {\pi}4- \gamma$
QUESTION [25 upvotes]: Inspired by the user @Integrals, I thought I'd find some nice integrals! Especially interesting are those involving $\log \pi$. From Borwein and Devlin's "The Computer as Crucible", pg. 58 - show that
$$\displaystyle \int_0^{\infty} \frac{\log x}{\cosh^2x} \ \mathrm{d}x = \log\frac {\pi}4 - \gamma,$$
where $\gamma$ is the Euler-Mascheroni constant.

REPLY [14 votes]: Largely speaking, we have  
$$\begin{align}
\int_0^{\infty}  \frac{\log x}{\cosh^2x}\ \mathrm{d}x &=\lim_{a\to0} \lim_{b\to2}\frac{\partial}{ \partial a \partial b}\left(-4\int_0^{ \infty}\frac{x^{a-1}}{1+e^{b x}} \ dx\right)\\&=
\lim_{a\to0} \lim_{b\to2}\frac{\partial}{\partial a \partial b}\left(b^{-a} (2^{3-a}-4) \Gamma(a) \zeta(a)\right)\\&=\log\left(\frac{\pi}{4}\right)-\gamma\end{align}$$
Done.