I was trying to evaluate the following integral which I found on Quora.
$$I =\int\limits_0^{\pi/2}\ln\left(\frac{\ln^2(\sin x)}{\pi^2+\ln^2(\sin x)}\right)\frac{\ln \cos x}{\tan x}\, dx $$
I used the substitution $\sin x = e^{-t}$ to get
$$I = \int\limits_0^\infty \ln\left(\frac{t^2}{\pi^2+t^2}\right)\ln\sqrt{1-e^{-2t}}\, dt$$
As I noticed,
$$\ln \left(\frac{t^2}{\pi^2+t^2}\right) = -2 \int\limits_0^\pi \frac{a}{a^2+t^2}\, da$$ And thus $$I = -\int\limits_0^\infty\int\limits_0^\pi \frac{a\ln(1-e^{-2t})}{a^2+t^2}\, da\, dt$$ If interchanging of the order of integration is allowed, then we get $$I = -\int\limits_0^\pi a\int\limits_0^\infty \frac{\ln(1-e^{-2t})}{t^2+a^2}\, dt\, da $$ Question :
- Is the interchange of integration is allowed?
- If yes, how can I evaluate the integral?
I tried to use the fact that $\int\limits_0^\infty e^{-au}\cos(ut)\, du = \frac{a}{a^2+t^2}$, but that made things more complicated. Please help.
