calculus - How to evaluate $I=int_0^{frac{pi}{2}}sin^2xln(sin^2(tan x))dx$

$$I=\int_0^{\frac{\pi}{2}}\sin^2x\ln(\sin^2(\tan x))dx \hspace{15mm}(1)$$
Now, using definite integral property of $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$ $$I=\int_0^{\frac{\pi}{2}}\cos^2x\ln(\sin^2(\cot x))dx\hspace{15mm}(2)$$

After $\tan x=t$ substitution in $(1)$ and $\cot x=m$ in $(2)$, to get: $$I=\int_0^{\frac{\pi}{2}}\frac{t^2}{(1+t^2)^2}\ln(\sin^2 t)dt=\int_0^{\frac{\pi}{2}}\frac{m^2}{(1+m^2)^2}\ln(\sin^2 m)dm$$
After variable change and addition, I get:

$$2I=\int_0^{\frac{\pi}{2}}\frac{x^2}{(1+x^2)^2}\ln(\sin^4 x)dx\implies \frac{I}{2}=\int_0^{\frac{\pi}{2}}\frac{x^2}{(1+x^2)^2}\ln(\sin x)dx$$

How could I proceed? Any other solutions which happen to be more efficient/simple?

Answer

$$I=\int_0^{\frac{\pi}{2}}\cos^2x\ln(\sin^2(\tan x))dx$$

Use the substitution $t = \tan(x)$

$$I=2\int_0^{\infty}\frac{\ln|\sin t|}{(1+t^2)^2}dt$$

Then use the fourier expansion of $\log|\sin x|$

$$I=-2\sum\frac{1}{k}\int_0^{\infty}\frac{\cos(2kt)}{(1+t^2)^2}dt-2\int_0^{\infty}\frac{\log(2)}{(1+t^2)^2}dt$$

For the second integral

$$2\int_0^{\infty}\frac{\log(2)}{(1+t^2)^2}$$

We can use the beta function to deduce that

$$2\int_0^{\infty}\frac{\log(2)}{(1+t^2)^2} =\frac{\pi}{2}\log(2)$$

For the second integral

$$\int_0^{\infty}\frac{\cos(2kt)}{(1+t^2)^2}dt = Re\int_0^{\infty}\frac{e^{2kit}}{(1+t^2)^2}dt$$

It can be showing by complex analysis that

$$\int_0^{\infty}\frac{\cos(2kt)}{(1+t^2)^2}dt = \frac{\pi}{4}e^{-2k}(1+2k)$$

Finally we have the integral

$$I=-\frac{\pi}{2}\sum \frac{e^{-2k}(1+2k)}{k } -\frac{\pi}{2}\log(2)$$

The sum can be computed using

$$-\log(1-x) = \sum \frac{x^k}{k}$$

Hence we deduce that

$$I=-\frac{\pi}{2}\left\{\log\left(\frac{2}{e^2-1} \right) +\frac{2e^ 2}{e^2-1}\right\}$$

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ADDENDUM

As required A proof using complex analysis for

$$\int_0^{\infty}\frac{\cos(2kt)}{(1+t^2)^2}dt$$

This can be done by considering

$$F(z) = \frac{e^{2ki z}}{(1+z^2)^2}$$

The poles are $z=\pm i$, then consider half a circle in the upper half plane.

Note there is only one pole of order 2 inside the contour. Hence by the residue theorem

$$\int_C \frac{e^{2ki z}}{(1+z^2)^2}+\int_{-\infty}^0\frac{e^{2ki x}}{(1+x^2)^2} \,dx+ \int^{\infty}_0\frac{e^{2ki x}}{(1+x^2)^2}\,dx = 2\pi i \text{Res} \frac{e^{2ki z}}{(1+z^2)^2}|_{z = i}$$

The integral along the arc converges to 0

$$2\int^{\infty}_0\frac{\cos(2k x)}{(1+x^2)^2} \,dx = 2\pi i \text{Res} \frac{e^{2ki z}}{(1+z^2)^2}|_{z = i}$$

The pole of order 2 can be calculated using

$$\lim_{z \to i} \frac{d}{dz} (z-i)^2 \frac{e^{2ki z}}{(1+z^2)^2}$$

This simplifies to

$$2\int^{\infty}_0\frac{\cos(2k x)}{(1+x^2)^2} \,dx = \frac{\pi}{2}e^{-2k}(1+2k)$$

$$\int^{\infty}_0\frac{\cos(2k x)}{(1+x^2)^2} \,dx = \frac{\pi}{4}e^{-2k}(1+2k)$$

For many other approaches see Calculating the integral $\int_{0}^{\infty} \frac{\cos x}{1+x^2}\mathrm{d}x$ without using complex analysis