How to evaluate $\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$.

Question

How can i evaluate

$$\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$$ $$=\frac{\pi ^3}{6}-\frac{\pi ^3}{6\sqrt{2}}-4\pi +6\pi \ln \left(2\right)-\frac{\pi }{2\sqrt{2}}\ln ^2\left(2\sqrt{2}+3\right)-\sqrt{2}\pi \operatorname{Li}_2\left(2\sqrt{2}-3\right)$$

This is what I've done thus far. \begin{align*} &\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx\\ &=\frac{\pi }{2}\int _0^{\pi }\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx\\[2mm] &=\frac{\pi }{4}\int _0^{\pi }\sin \left(\frac{x}{2}\right)\operatorname{Li}_2\left(\cos \left(x\right)\right)\:dx+\frac{\pi }{4}\int _{\pi }^{2\pi }\sin \left(\frac{x}{2}\right)\operatorname{Li}_2\left(\cos \left(x\right)\right)\:dx\\[2mm] &=\frac{\pi }{2}\int _0^{\pi }\sqrt{\frac{1+\cos \left(x\right)}{2}}\operatorname{Li}_2\left(-\cos \left(x\right)\right)\:dx=\pi \int _0^{\infty }\frac{\operatorname{Li}_2\left(\frac{t^2-1}{1+t^2}\right)}{\left(1+t^2\right)\sqrt{1+t^2}}\:dt\\[2mm] &=\frac{\pi }{2}\int _1^{\infty }\frac{\operatorname{Li}_2\left(\frac{x-2}{x}\right)}{x\sqrt{x}\sqrt{x-1}}\:dx=\frac{\pi }{2}\int _0^1\frac{\operatorname{Li}_2\left(1-2x\right)}{\sqrt{1-x}}\:dx\\[2mm] &=\pi \zeta \left(2\right)+\frac{\pi }{\sqrt{2}}\int _{-1}^1\frac{\sqrt{1+x}\ln \left(1-x\right)}{x}\:dx \end{align*} But I'm not sure how to proceed with either that polylogarithmic integral on the $5$th line nor the last one.

I'd appreciate any hints or ideas, thanks.

Answer 1

$$\small \int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)dx=\frac{\pi^3}{6}+6\pi \ln 2-4\pi +\frac{\pi}{\sqrt 2}\left(\operatorname{Li}_2(-(3+2\sqrt 2))-\operatorname{Li}_2(-(3-2\sqrt 2))\right)$$ $$\small\int_{-1}^1\frac{\sqrt{1+x}\ln(1-x)}{x}dx=6\sqrt 2 \ln 2 -4\sqrt 2+\operatorname{Li}_2(-(3+2\sqrt 2))-\operatorname{Li}_2(-(3-2\sqrt 2))$$

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To show this result we can continue with the integral left at the end of the question body. $$\int_{-1}^1\frac{\sqrt{1+x}\ln(1-x)}{x}dx=\int_{-1}^1 \frac{\ln(1-x)}{\sqrt{1+x}}dx+\int_{-1}^1 \frac{\ln(1-x)}{x\sqrt{1+x}}dx=\mathcal I+\mathcal J$$ $$\mathcal I=\int_{-1}^1 \frac{\ln(1-x)}{\sqrt{1+x}}dx \overset{1+x=t^2}=2\int_0^\sqrt 2\ln(2-x^2)dx=2\int_0^\sqrt 2 (x-\sqrt 2)'\ln(2-x^2)dx$$ $$\overset{IBP}=2\sqrt 2\ln 2 -4\int_0^\sqrt 2 \frac{x}{x+\sqrt 2}dx=6\sqrt 2 \ln 2 -4\sqrt 2$$

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$$\mathcal J(t)=\int_{-1}^1 \frac{\ln(1-tx)}{x\sqrt{1+x}}dx\Rightarrow \mathcal J'(t)=-\int_{-1}^1 \frac{1}{(1-tx)\sqrt{1+x}}dx$$ $$\overset{1+x\to x^2}=-\frac{2}{t}\int_0^\sqrt 2\frac{1}{\frac{1+t}{t}-x^2}dx=-\frac{2\operatorname{arctanh}\left(\sqrt 2 \sqrt{\frac{t}{1+t}}\right)}{\sqrt t\sqrt{1+t}}$$ $$\mathcal J(0)=0\Rightarrow \mathcal J=-2\int_0^1 \frac{\operatorname{arctanh}\left(\sqrt 2 \sqrt{\frac{t}{1+t}}\right)}{\sqrt t\sqrt{1+t}}dt\overset{\frac{t}{1+t}=x^2}=-4\int_0^\frac{1}{\sqrt 2}\frac{\operatorname{arctanh}(\sqrt 2 x)}{1-x^2}dx$$ $$\overset{\sqrt 2 x\to x}=-4\sqrt 2 \int_0^1 \frac{\operatorname{arctanh} x}{2-x^2}dx\overset{x=\frac{1-t}{1+t}}=\int_0^1 \frac{\ln t}{3-2\sqrt 2+t}dt-\int_0^1\frac{\ln t}{3+2\sqrt 2 +t}dt$$ $$=\operatorname{Li}_2(-(3+2\sqrt 2))-\operatorname{Li}_2(-(3-2\sqrt 2))$$ Putting this togheter gives the announced result.