<s>[INST] 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!
 [/INST] 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 $- I$.
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