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?
 A: $$I=\int^{\frac{\pi}{2}}_0 x \log|\sin(x )|  \, dx =\int^{\frac{\pi}{2}}_0 x \log|2\sin(x )| -\frac{\pi^2}{8}\log(2)$$
We relate the integral to the Clausen function 
\begin{align} 
\int^{\frac{\pi}{2}}_0 x \log|2\sin(x )| dx &=\frac{1}{2} \int^{\frac{\pi}{2}}_0 \mathrm{Cl}_2(2\theta)\, d\theta\\ &=\frac{1}{2} \int^{\frac{\pi}{2}}_0 \sum_{n=1}^{\infty}\frac{\sin(2n\theta)}{n^2}\, d\theta \\ &=-\frac{1}{4}\sum_{n=1}\frac{(-1)^n}{n^3}+\frac{1}{4}\sum_{n=1}\frac{1}{n^3}\\ 
&=\frac{7}{16}\zeta(3) 
\end{align}
Collecting that together we have 
$$I=\frac{7}{16}\zeta(3)-\frac{\pi^2}{8}\log(2)$$
I considered a more general case in this thread 
A: 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
$$ \int_0^\frac{\pi}{2}x\ln(\sin x)~dx= \frac7{16}\zeta(3)-\frac{\pi^2}8\ln2$$
A: The integral from $0$ to $\pi/2$ of $\ln(\sin x)$ is one of Euler's integrals - look here for some information.
This paper explains why
$$\int_{0}^{\pi/2}x\ln(\sin x)\ dx=\frac7{16}\zeta(3)-\frac{\pi^2}8\ln2.
$$
A: Hint: Let $t=\sin x$, then integrate by parts, and you will be left with evaluating 
$\displaystyle\int_0^1\frac{\arcsin^2t}tdt$, at which point you will make use of the formula $\displaystyle\sum_{n=1}^\infty\frac{(2t)^{2n}}{n^2\displaystyle{2n\choose n}}$ 
$=2\arcsin^2t,$ in conjunction with Apery's alternative series for $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{n^3\displaystyle{2n\choose n}}=$ 
$=-\dfrac25\zeta(3)$, by switching the order of summation $\&$ integration. $\big($In case you're 
wondering where the formula for $\arcsin^2t$ comes form, start with the expression 
$\arcsin t=\displaystyle\int_0^t\frac{du}{\sqrt{1-u^2}}$, expand the integrand using the binomial series, then 
switch the order of summation and integration, and square it using the Cauchy  product$\big)$.
