<s>[INST] Tough definite integral: $\int_0^\frac{\pi}{2}x\ln^2(\sin x)~dx$ Any ideas on evaluating the definite integral
$$\int_0^\frac{\pi}{2}x\ln^2(\sin x)\ dx$$
The best numerical approximation I could get is $0.2796245358$.
Is there even a closed form solution?
 [/INST] Here is an elementary approach. Denote $ K= \int_0^\infty \frac{\ln t\ln(1+t^2) \tan^{-1}t}{1+t^2}dt $
\begin{align}
I= &\int_0^\frac{\pi}{2}x\ln^2(\sin x)\ dx
\overset{t=\tan x}= \frac14 \int_0^\infty \frac{\tan^{-1}t \ln^2\frac{t^2}{1+t^2}}{1+t^2}dt\\
=& \int_0^\infty \frac{\tan^{-1}t\ln^2t}{1+t^2}\overset{t\to 1/t}{dt} +\frac14
\int_0^\infty \frac{\tan^{-1}t\ln^2(1+t^2)}{1+t^2}\overset{t\to 1/t}{dt}
-K\\
=& \frac\pi8 \int_0^\infty \frac{\ln^2t}{1+t^2}dt
 + \frac\pi{16}\int_0^\infty \frac{\ln^2(1+t^2)}{1+t^2}dt -I -K\\
 =& \frac\pi8 \cdot\frac{\pi^3}8
+\frac\pi{16}\left(\frac{\pi^3}6+2\pi\ln^22\right)  -I-K
=\frac{5\pi^4}{192}+\frac{\pi^2}8\ln^22-\frac12K\tag1
\end{align}
Note that, with the substitution $x=\frac{(1+t^2)y}{1-y}$
\begin{align}
\int_0^1 \frac{2t\ln\frac{1-y}y}{1+t^2y^2}dy
= \int_0^\infty \frac{2t\ln\frac{1+t^2}x}{(1+x)^2+t^2}dx=\ln(1+t^2)\tan^{-1}t
\end{align}
Then, integrate $K$ as follows
\begin{align}
K=& \int_0^\infty \frac{\ln t}{1+t^2}\int_0^1 \frac{2t\ln\frac{1-y}y}{1+t^2y^2}dy \>\overset{t^2\to t}{dt}\\
= &\frac12 \int_0^1 \ln\frac{1-y}y\int_0^\infty 
\frac{\ln t}{(1+t)(1+y^2t)}dt\>dy
=\int_0^1 \ln\frac{1-y}y
\frac{\ln^2y}{1-y^2}dy\\
=&\frac{\pi^4}{16}+\frac12\int_0^1\frac{\ln^2y\ln(1-y)}{1-y}dy
+ \frac12\int_0^1\frac{\ln^2y\ln(1-y)}{1+y}dy
\end{align}
The pair of integrals above are known and given below
\begin{align}
&\int_0^1\frac{\ln^2y\ln(1-y)}{1-y}dy= -\frac{\pi^4}{180}\\
&\int_0^1\frac{\ln^2y\ln(1-y)}{1+y}dy= 
-4Li_4\left(\frac12\right) +\frac{\pi^4}{90}+\frac{\pi^2}6\ln^22-\frac16\ln^42\\
\end{align}
Plug them into $K$ and then into (1) to obtain
$$\int_0^\frac{\pi}{2}x\ln^2(\sin x)\ dx
= Li_4\left(\frac12\right)-\frac{19\pi^4}{2880}+\frac{\pi^2}{12}\ln^22+\frac1{24}\ln^42
$$
</s>