\int_0^{\pi/2}x\ln^2\left(2 \sin(x) \right)\,dx

Lets show the following result:


\int_0^{\pi/2}x\ln^2\left(2 \sin(x) \right)\,dx=-\frac{19}{32}\zeta(4)+\text{Li}_4\left(\frac12\right)+\frac{7\ln(2)\zeta(3)}{8}+\frac{\ln^4(2)}{24} -\frac{\ln^2(2)\pi^2}{24}


Recall (see here)

\ln^2\left(2 \sin(x) \right)=\frac{\pi^2}4-\pi x+x^2+2 \sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}\cos(2nx)}{n}

Than


\begin{aligned} \int_0^{\pi/2}x\ln^2\left(2 \sin(x) \right)\,dx&=\frac{\pi^2}4\int_0^{\pi/2}x\,dx-\pi\int_0^{\pi/2}x^2\,dx+\int_0^{\pi/2}x^3\,dx+2 \sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n}\int_0^{\pi/2}x\cos(2nx)\,dx\\ &=\frac{\pi^4}{32}-\frac{\pi^4}{24}+\frac{\pi^4}{64}+2 \sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n}\left(-\frac{1}{2n}\int_0^{\pi/2}\sin(2nx)\,dx \right)\\ &=\frac{\pi^4}{32}-\frac{\pi^4}{24}+\frac{\pi^4}{64}+\frac12 \sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^3}\left((-1)^n-1 \right)\\ &=\frac{\pi^4}{192}+\frac12 \sum_{n=1}^\infty \frac{\operatorname{H}_{n}}{n^3}\left((-1)^n-1 \right)-\frac12 \sum_{n=1}^\infty \frac{1}{n^4}\left((-1)^n-1 \right)\\ &=\frac{\pi^4}{192}+\frac12 \sum_{n=1}^\infty \frac{(-1)^n\operatorname{H}_{n}}{n^3}-\frac12 \sum_{n=1}^\infty \frac{\operatorname{H}_{n}}{n^3}+\frac12 \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^4}+\frac12 \sum_{n=1}^\infty \frac{1}{n^4}\\ &=\frac{15}{32}\zeta(4)-\frac{5}{8}\zeta(4)+\frac{7}{16}\zeta(4)+\frac12\zeta(4)+\frac12\left(-\frac{11}{4}\zeta(4)+2\text{Li}_4\left(\frac12\right)+\frac{7\ln(2)\zeta(3)}{4}+\frac{\ln^4(2)}{12} -\frac{\ln^2(2)\pi^2}{12} \right)\\ &=-\frac{19}{32}\zeta(4)+\text{Li}_4\left(\frac12\right)+\frac{7\ln(2)\zeta(3)}{8}+\frac{\ln^4(2)}{24} -\frac{\ln^2(2)\pi^2}{24} \qquad \blacksquare \end{aligned}

Where we used the results:

Proved here

\sum_{n=1}^\infty \frac{\operatorname{H}_{n}}{n^3}=\frac{5}{4}\zeta(4)

Proved here


\sum_{n=1}^\infty \frac{(-1)^n\operatorname{H}_{n}}{n^3}=-\frac{11 \pi^4}{360}+2\text{Li}_4\left(\frac12\right)+\frac{7\ln(2)\zeta(3)}{4}+\frac{\ln^4(2)}{12} -\frac{\ln^2(2)\pi^2}{12}

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