While trying to work on an old post, I managed after some manipulations to show that the nasty improper tan integral $$ I=\int_0^{\pi} \left( \frac{\pi}{2} - x \right) \frac{\tan x}{x} \, {\rm d}x\approx 2.138967 $$ could be represented as $$ I=\sum_{k=1}^\infty\frac{\ln \left(\frac{2 k+1}{2 k-1}\right)}{2 k-1}- \sum_{k=1}^\infty\frac{\ln \left(\frac{2 k-1}{2 k+1}\right)}{2 k+1}\ .\qquad(\star) $$ The purpose of the original post was to find a closed-form solution for $I$. Clearly the problem would be solved if closed-form solutions for the (equally nasty) logarithmic sums in $(\star)$ could be found. I have worked a bit on those, but didn't achieve much. My question is: can the sums in $(\star)$ be evaluated in closed form? Many thanks, folks.



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