I am trying to evaluate the integral∫0π2log⁡(eix+e−ix)−log⁡{2}}{e2ix−1}eixdxI am wondering if a complex variable substitution the likes of iu=eix, du=eixdx is justifiable? In my mind. when x=0,u=0 and when x=π2,u=1 which would make the integral−∫01log⁡(iu−iu)−log⁡(2)1+u2,du

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Derick Alvarado · Accepted Answer

With some rearranging the integral is equal to12i∫0π2log⁡cos⁡xsin⁡xdx=12i∫01log⁡t1−t2dtby the substitution t=cos⁡x. Then expand the denominator as a geometric series∑n=0∞12i∫01log⁡t⋅t2ndt=∑n=0∞−12i(2n+1)2=iπ216by integration by parts and then using the sum of the odd terms of the Basel problem.

I am trying to evaluate the integral \int_{0}^{\frac{\pi}{2}} \frac{\log(e^{ix}+e^{-ix})]-\log{2}}{e^{2ix}-1}e^{ix} dx I am

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