Find : $\int_0^{\pi/4}x\ln(\sin x)\mathrm dx$ I'm try to find this integral 
$$\int_0^{\pi/4}x\ln(\sin x)\mathrm dx$$
My try use : 
$\ln(\sin x)=-\ln2-\sum\limits_{n=1}^{\infty}\frac{\cos (2nx)}{n}$
But I don't know how to complete summation ...  
I will happy if someone help me
Thanks!
 A: Your approach works perfectly well:
We can use the Fourier series and integrate by parts to obtain
$$ I \equiv \int \limits_0^{\pi/4} x [- \ln(\sin(x))] \, \mathrm{d} x = \frac{\pi^2}{32} \ln(2) + \frac{1}{4} \sum \limits_{n=1}^\infty \frac{1}{n^2} \left[\frac{\pi}{2} \sin\left(\frac{\pi}{2} n \right) - \frac{1}{n} \left(1 - \cos\left(\frac{\pi}{2} n \right)\right)\right] \, . $$
$\sin\left(\frac{\pi}{2} n \right)$ is non-zero and alternating for odd $n$, while $\cos\left(\frac{\pi}{2} n \right)$ is non-zero and alternating for even $n$. Therefore,
$$ I = \frac{\pi^2}{32} \ln(2) + \frac{\pi}{8} \sum \limits_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} - \frac{1}{4} \sum \limits_{n=1}^\infty \frac{1}{n^3} - \frac{1}{32} \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^3} \, .$$
The first series is Catalan's constant $\mathrm{G}$, the second one is $\zeta(3)$ and the third one is $\eta(3) = \frac{3}{4} \zeta(3)$ (with the Riemann zeta function $\zeta$ and the Dirichlet eta function $\eta$), so we obtain
$$ I = \frac{\pi^2}{32} \ln(2) + \frac{\pi}{8} \mathrm{G} - \frac{35}{128} \zeta(3) $$
and your integral is $- I$.
A: Another way to attack this integral is via Integration By Parts with the help of the Clausen Function $\operatorname{Cl}_2(z)$(and its relatives). The natural choice here is $u=x$ and $\mathrm dv=\log(\sin x)$. The aforementioned Clausen Function allows us to express the anti-derivative of the chosen $\mathrm dv$. Eventually we will get
\begin{align*}
\int_0^\frac\pi4x\log(\sin x)\mathrm dx&=\left[x\left(-\frac12\operatorname{Cl}_2(2x)-x\log(2)\right)\right]_0^{\frac\pi4}+\int_0^\frac\pi4\frac12\operatorname{Cl}_2(2x)+x\log(2)\mathrm dx\\
&=-\frac\pi8\operatorname{Cl}_2\left(\frac\pi2\right)-\frac{\pi^2}{16}\log(2)+\frac{\pi^2}{32}\log(2)+\frac12\int_0^\frac\pi4\operatorname{Cl}_2(2x)\mathrm dx\\
&=-\frac\pi8G-\frac{\pi^2}{32}\log(2)+\frac14\int_0^{\frac\pi2}\operatorname{Cl}_2(x)\mathrm dx\\
&=-\frac\pi8G-\frac{\pi^2}{32}\log(2)+\frac14\left[\zeta(3)-\operatorname{Cl}_3\left(\frac\pi2\right)\right]\\
&=-\frac\pi8G-\frac{\pi^2}{32}\log(2)+\frac14\left[\zeta(3)+\frac18\eta(3)\right]\\
&=-\frac\pi8G-\frac{\pi^2}{32}\log(2)+\frac{35}{128}\zeta(3)
\end{align*}

$$\therefore~\int_0^\frac\pi4x\log(\sin x)\mathrm dx~=~-\frac\pi8\text{G}-\frac{\pi^2}{32}\log(2)+\frac{35}{128}\zeta(3)$$

Here we used several properties of the Clausen Function which overall are quite simple to prove utilizing the integral representation and the series representation of this function. Not only the result coincides with the one given by ComplexYetTrivial, the underlying method is afterall more or less the same. For those who are familiar with the Clausen Function it is rather obvious that the here described method is nothing more than a more conventient way $-$ at least in my opinion $-$ to deal with the occuring Fourier Series. However, from my experience the Clausen Function is quite helpful in order to deal with similiar integrals to the examined and therefore I wanted to share this approach aswell.
