I am trying to evaluate the integral∫0π2log(eix+e-ix)]-log2e2ix-1eixdxI am wondering if a complex variable substitution the likes of&nbsp;iu=eix,&nbsp;du=eix&nbsp;dx&nbsp;&nbsp;is justifiable? In my mind. when x=0,u=0 and when&nbsp;x=π2,u=1&nbsp;which would make the integral−∫01log⁡(iu−iu)−log⁡(2)1+u2,du

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I am trying to evaluate the integral∫0π2log(eix+e-ix)]-log2e2ix-1eixdxI am wondering if a complex variable substitution the likes of&amp;nbsp;iu=eix,&amp;nbsp;du=eix&amp;nbsp;dx&amp;nbsp;&amp;nbsp;is justifiable? In my mind. when x=0,u=0 and when&amp;nbsp;x=π2,u=1&amp;nbsp;which would make the integral−∫01log⁡(iu−iu)−log⁡(2)1+u2,du

Alexzander Evans · 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 \(\displaystyle{\int_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}{\frac{{{\log{{\left({e}^{{{i}{x}}}+{e}^{{-{i}{x}}}\right)}}}}}{-}}{\log{{\left\lbrace{2}\right\rbrace}}}\rbrace{\left\lbrace{e}^{{{2}{i}{x}}}-{1}\right\rbrace}{e}^{{{i}{x}}}{\left.{d}{x}\right.}\) I

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