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

esbagoar7kh

esbagoar7kh

Answered question

2022-04-29

I am trying to evaluate the integral
0π2log(eix+eix)log{2}}{e2ix1}eixdx
I 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(iuiu)log(2)1+u2,du

Answer & Explanation

Derick Alvarado

Derick Alvarado

Beginner2022-04-30Added 10 answers

With some rearranging the integral is equal to
12i0π2logcosxsinxdx=12i01logt1t2dt
by the substitution t=cosx. Then expand the denominator as a geometric series
n=012i01logtt2ndt=n=012i(2n+1)2=iπ216
by integration by parts and then using the sum of the odd terms of the Basel problem.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?