<s>[INST] Evaluate integral $\int_0^\frac{\pi}{2}x\ln(\sin x)~dx$ I've corrected typing error in the integral.
I apologize for my mistake.
Reedited question:
Can anybody solve integral:
$$\int_0^\frac{\pi}{2}x\ln(\sin x)~dx$$
I'm just trying to guess some simple formula for $\zeta(3)$. My "strategy" is simple: Find some conjectures and check them numerically.
Value of similar integral is well known:
$$\int_0^\pi x\ln(\sin x)~dx=-\dfrac{\pi^2}{2}\ln(2) $$
Is there any idea for antiderivative of $x\ln(\sin x)$? It's too difficult to solve it for me.
Any ideas?
 [/INST] Note
\begin{align}\int_0^\frac{\pi}{2}x\ln(2\sin x)~dx
=\frac12 \int_0^\frac{\pi}{2}x\ln(2\sin 2x)~dx
 + \frac12 \int_0^\frac{\pi}{2}x\ln\frac{\sin x}{\cos x}~dx\tag1\\
\end{align}
where
\begin{align}
&\int_0^\frac{\pi}{2} \overset{2x\to x} {x\ln(2\sin 2x)}~dx 
=\frac14 \int_0^{\pi} \overset{x\to\pi-x} {x\ln(2\sin x)}~{dx}
=\frac\pi8 \int_0^{\pi}\ln(2\sin x)~dx =0
\end{align}
and
\begin{align}
&\int_0^\frac{\pi}{2}x\ln\frac{\sin x}{\cos x} ~dx 
\overset{t=\tan x}= \int_0^\infty \frac{\ln t\tan^{-1}t}{1+t^2} dt \\
=& \int_0^\infty \frac{\ln t}{1+t^2} \int_0^1 \frac t{1+ t^2 y^2}dy ~dt
\overset{t^2\to t}=\frac14\int_0^1 \int_0^\infty \frac{\ln t}{(1+t)(1+ t y^2)} \overset{t\to 1/(y^2t)}{dt ~dy}\\
=& \frac14\int_0^1 \int_0^\infty \frac{-\ln t -\ln y^2}{(1+t)(1+ t y^2)}dt ~dy = -\frac14\int_0^1 \int_0^\infty \frac{\ln y}{(1+t)(1+ t y^2)}dt ~dy\\
 =& \frac12\int_0^1 \frac{\ln^2 y}{1-y^2}dy
= \frac12\int_0^1 \frac{\ln^2 y}{1-y}dy - \frac12\int_0^1 \frac{y \ln^2 y}{1-y^2}\overset{y^2\to y}{dy}\\
=& \frac7{16}\int_0^1 \frac{\ln^2 y}{1-y}dy
= \frac7{16}\cdot 2\zeta(3)=\frac 78\zeta(3)
\end{align}
Substitute above results into (1) to get
$\int_0^\frac{\pi}{2}x\ln(2\sin x)~dx= \frac7{16}\zeta(3)$, which leads to
$$ \int_0^\frac{\pi}{2}x\ln(\sin x)~dx= \frac7{16}\zeta(3)-\frac{\pi^2}8\ln2$$
</s>