Substitution of Complex Variable in integral

Question

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 wondering if a complex variable substitution the likes of $iu=e^{ix}$, $\,du=e^{ix}\,dx$ is justifiable? In my mind. when $x=0, u=0$ and when $x=\pi/2, u=1$ which would make the integral $$-\int_{0}^{1} \frac{\log(iu-i/u)-\log(2)}{1+u^2}\,du$$

but it is not numerically equivalent. Is it possible to introduce a complex variable substitution to solve the integral, and if so how do I go about it?

Answer 1

With some rearranging the integral is equal to

$$\frac{1}{2i}\int_0^{\frac{\pi}{2}}\frac{\log\cos x}{\sin x}dx = \frac{1}{2i}\int_0^1\frac{\log t}{1-t^2}dt$$

by the substitution $t = \cos x$. Then expand the denominator as a geometric series

$$\sum_{n=0}^\infty \frac{1}{2i} \int_0^1 \log t \cdot t^{2n}dt=\sum_{n=0}^\infty \frac{-1}{2i(2n+1)^2} = \frac{i\pi^2}{16}$$

by integration by parts and then using the sum of the odd terms of the Basel problem.

Answer 2

I think you are on the wrong track. I would look at $$ \log(e^{ix}+e^{-ix})-\log2 =\log\left(\frac{e^{ix}+e^{-ix}}2\right) =\log(\sin x) $$ and $$ \frac{e^{ix}}{e^{2ix}-1} =\frac1{e^{ix}-e^{-ix}} =\frac1{2\cos x} $$ so your integral turns into $$ \int_0^{\pi/2}\frac{\log(\sin x)}{2\cos x}\;dx. $$ Can you work on this?