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?

  • $\begingroup$ $$x^a = e^{a \log{x}}$$ $\endgroup$
    – Ron Gordon
    Dec 19, 2014 at 22:16
  • $\begingroup$ @RonGordon, I suppose I should clarify - I ask for the intermediate steps because I am especially curious to know how the summation and differentiation arose. $\endgroup$
    – Doubt
    Dec 22, 2014 at 19:45


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