Closed form of $\displaystyle\mathscr{R}=\int_0^{\frac{\pi}{2}}\sin^2x\,\ln\big(\sin^2(\tan x)\big)\,\,dx$

Inspired by Mr. Olivier Oloa in this question. Does the following integral admit a closed form?

\mathscr{R}=\int_0^{\Large\frac{\pi}{2}}\sin^2x\,\ln\big(\sin^2(\tan x)\big)\,\,dx

It will be my last question before I take a long break from my activity on Mathematics StackExchange. So, please be nice. No more downvotes for no reason because this is a challenge problem. BBML. 👋(>‿◠)♥

Edit :

I am also interested in knowing the numerical value of $\mathscr{R}$ to the precision of at least $50$ digits. If you use Mathematica to find its numerical value, please share your method & the code. Thank you.

Solutions Collecting From Web of "Closed form of $\displaystyle\mathscr{R}=\int_0^{\frac{\pi}{2}}\sin^2x\,\ln\big(\sin^2(\tan x)\big)\,\,dx$"

Here’s a slightly different approach. Using the power reduction formula and getting rid of the second exponent of the sine using the properties of the logarithm the integral becomes:

$$I=\int_0^{\frac{\pi}{2}} (1-\cos 2x) \log (\sin (\tan x))dx$$

Using the substitution $2x=u$ we have:

$$I = \frac{1}{2}\int_0^\pi (1-\cos u) \log\bigg(\sin \bigg(\tan \bigg(\frac{u}{2}\bigg)\bigg)\bigg)du$$

We have the integral ready for a Weierstrass substitution, after which it becomes:

$$I = \int_0^\infty\bigg(1-\frac{1-t^2}{1+t^2}\bigg) \log(\sin (t)) \frac{1}{1+t^2}dt$$


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

From now on I’ll use $x$ again. The Fourier series of $\log (\sin x)$ is well known and it is:

$$\log(\sin x)= -\log 2 -\sum_{n=1}^\infty \frac{\cos(2nx)}{n}$$

So the integral, exchanging integration and summation, becomes:

$$I=-2\log 2 \int_0^\infty \frac{x^2}{(1+x^2)^2}dx-2\sum_{n=1}^\infty \frac{1}{n} \int_0^\infty \frac{x^2\cos(2nx)}{(1+x^2)^2}$$

These are both easy integrals from the point of view of residue calculus. The final result is:

$$I=-\frac{\pi \log 2}{2} -\frac{\pi}{2}\sum_{n=1}^\infty \frac{e^{-2n}}{n}+\pi \sum_{n=1}^\infty e^{-2n}$$

These sums can be evaluated using the geometric series and its integral.
So we have:

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


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

The answer is
\mathscr{R}=\frac{\pi }{2} \left(\log \left(\frac{e^2-1}{2} \right)-\frac{2
As Kirill proved we have
Now, the function
is $\pi$-periodic and even function. It is not difficult to calculate its Fourier cosine coefficients $a_n$ such that
F(u)=\frac{a_0}{2}+\sum_{n=1}a_n\cos(2n u)
$$a_n=\frac{2}{\pi}\int_0^\pi F(u)\cos(2n u)du
=\frac{2}{\pi}\int_{-\infty}^\infty \frac{u^2}{(1+u^2)^2}\cos(2n u)du= e^{-2 n} (1-2 n)$$
The last equality is obtained by a simple residue calculus.

On the other hand it is easy and well-known that
\log(\sin^2u)=-2\log 2-\sum_{n=1}^\infty\frac{2}{n}\cos(2nu)
So using Parseval’s formula we get
\mathscr{R}=\frac{\pi}{2}\left(- \log 2-\frac{1}{2}\sum_{n=1}^\infty\frac{2}{n}e^{-2 n} (1-2 n)\right)
and this simplifies easily to the announced closed form.$\qquad\square$

Write $\tan x = u$ to get the integral
$$ \int_0^\infty \frac{u^2\,du}{ (1+u^2)^2} \log(\sin^2 u) $$
and split over periods of length $\pi$ by writing $u=\pi k+s$, $0<s<\pi$ so that the integral is
$$ 2\int_0^\pi \sum_{k\geq0} \frac{u^2}{(1+u^2)^2}\log\sin s\,ds, \qquad u = \pi k+s. $$
The sum can be done explicitly in terms of polygamma functions:
$$ \frac{1}{2\pi}\Im\psi\left(\frac{i+s}{\pi}\right) + \frac{1}{2\pi^2}\Re\psi_1\left(\frac{i+s}{\pi}\right), $$
so the integral is equal to
$$ \frac1\pi \int_0^\pi \log(\sin s)\left(\Im\psi\left(\frac{i+s}{\pi}\right) + \frac{1}{\pi}\Re\psi_1\left(\frac{i+s}{\pi}\right) \right)\,ds$$
which is numerically
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