In the accepted answer here, the first two steps in computing the integral are
\begin{align} \mathcal{I} =&\frac{1}{2}\int^\infty_0\ln(1-e^{-2x})\ln\left(\frac{x^2}{\pi^2+x^2}\right)\ {\rm d}x\\ =&-\frac{\partial}{\partial a}\Bigg{|}_{a=0}\sum^\infty_{n=1}\frac{1}{n}\int^\infty_0x^ae^{-2nx}\ {\rm d}x+\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n}\int^\infty_0e^{-2nx}\ln(\pi^2+x^2)\ {\rm d}x \end{align} I am stuck, however, on seeing what was done to show that these two lines are equivalent. Could someone add a couple of steps in between, or give the name of the process?
