An interesting Integral = (2π)^2(ln2)

[Tony1]

Prove that,

$$\int_{0}^{\pi/2}\ln^2\left(\ln^2 (\sin x)\over \pi^2+\ln^2 (\sin x)\right){\mathrm dx\over \tan x}=\color{blue}{(2\pi)^2\ln 2}$$

 

[jvicens]

First, we need to use the substitution $u=\ln(\sin x)$ to simplify the integral:

$$\int_{0}^{\pi/2}\ln^2\left(\ln^2 (\sin x)\over \pi^2+\ln^2 (\sin x)\right){\mathrm dx\over \tan x}=\int_{-\infty}^{\ln(\frac{\pi}{2})}\ln^2\left(\frac{u^2}{\pi^2+u^2}\right){\mathrm du}$$

Next, we can use the property of logarithms to simplify the integrand:

$$\int_{-\infty}^{\ln(\frac{\pi}{2})}\ln^2\left(\frac{u^2}{\pi^2+u^2}\right){\mathrm du}=\int_{-\infty}^{\ln(\frac{\pi}{2})}\ln^2(u^2)-\ln^2(\pi^2+u^2){\mathrm du}$$

Now, we can use integration by parts to evaluate the integral on the right side:

$$\int_{-\infty}^{\ln(\frac{\pi}{2})}\ln^2(u^2)-\ln^2(\pi^2+u^2){\mathrm du}=\left[u^2\ln^2(u^2)-2u\ln(u^2)+2u\right]_{-\infty}^{\ln(\frac{\pi}{2})}-\int_{-\infty}^{\ln(\frac{\pi}{2})}2u\left(\frac{2u}{u^2+\pi^2}\right){\mathrm du}$$

Simplifying and evaluating the limits, we get:

$$\left[u^2\ln^2(u^2)-2u\ln(u^2)+2u\right]_{-\infty}^{\ln(\frac{\pi}{2})}-\int_{-\infty}^{\ln(\frac{\pi}{2})}\frac{4u^2}{u^2+\pi^2}{\mathrm du}=\ln^2(\ln^2(\pi)-\frac{\pi}{2})+2\ln(\ln(\frac{\pi}{2}))+2\ln(\frac{\pi