Calculate $\int_0^{\pi/2}\frac{\sin(x)\log{\sin{(x)}}}{x}\,dx$ Inspired by a question I saw these days, I try to calculate in closed form
$$\int_0^{\pi/2}\frac{\sin(x)\log{\sin{(x)}}}{x}\,dx$$
So far no fruitful idea that is worth sharing. What way would you propose? Note I prefer ways suggested,  not necessarily solutions, but I have nothing against any of the options you prefer.
 A: It's not a closed form, but I hope can be useful. Using $$\log\left(\sin\left(x\right)\right)=-\log\left(2\right)-\sum_{n\geq1}\frac{\cos\left(2nx\right)}{n}$$
 we have$$\int_{0}^{\pi/2}\frac{\sin\left(x\right)\log\left(\sin\left(x\right)\right)}{x}=-\log\left(2\right)\textrm{Si}\left(\frac{\pi}{2}\right)-\sum_{n\geq1}\frac{1}{n}\int_{0}^{\pi/2}\frac{\sin\left(x\right)\cos\left(2nx\right)}{x}dx.$$
 Now we use the identity $$\sin\left(x\right)\cos\left(2nx\right)=\frac{1}{2}\left(\sin\left(x-2nx\right)+\sin\left(2nx+x\right)\right)$$
 to obtain $$\int_{0}^{\pi/2}\frac{\sin\left(x\right)\log\left(\sin\left(x\right)\right)}{x}=-\log\left(2\right)\textrm{Si}\left(\frac{\pi}{2}\right)+\frac{1}{2}\sum_{n\geq1}\frac{\textrm{Si}\left(\frac{\pi}{2}\left(2n+1\right)\right)-\textrm{Si}\left(\frac{\pi}{2}\left(2n-1\right)\right)}{n}.$$
 Now, we have that $$\frac{\textrm{Si}\left(\frac{\pi}{2}\left(2n+1\right)\right)-\textrm{Si}\left(\frac{\pi}{2}\left(2n-1\right)\right)}{n}=O\left(\frac{1}{\pi n^{2}}\right)$$
 at $n\rightarrow\infty$
  so we have the approximation $$\int_{0}^{\pi/2}\frac{\sin\left(x\right)\log\left(\sin\left(x\right)\right)}{x}\simeq-\log\left(2\right)\textrm{Si}\left(\frac{\pi}{2}\right)-\frac{\zeta\left(2\right)}{2\pi}.$$ Note that numerically the integral is $-1.05585$ and my result is $-1.21193...$
