Improper Integral $\int\limits_0^{\frac12}(2x - 1)^6\log^2(2\sin\pi x)\,dx$

How can I find a closed form for the following integral $$\int_0^{\frac12}(2x - 1)^6\log^2(2\sin\pi x)\,dx?$$

Solution 1:

Here is a proof of Cleo's claim that \begin{align} \int^\frac{1}{2}_0(2x-1)^6\ln^2(2\sin(\pi x))\ {\rm d}x=\boxed{\displaystyle\frac{11\pi^2}{360}+\frac{60}{\pi^4}\zeta^2(3)-\frac{720}{\pi^6}\zeta(3)\zeta(5)} \end{align} It doesn't take much to show that the integral is equivalent to $$\mathscr{J}=\frac{64}{\pi^7}\int^\frac{\pi}{2}_0x^6\ln^2(2\cos{x})\ {\rm d}x$$ Use the identity $\displaystyle\ln^2(2\cos{x})={\rm Re}\ln^2(1+e^{i2x})+x^2$ to get \begin{align} \color{red}{\Large{\mathscr{J}}} &=\frac{\pi^2}{72}+\frac{64}{\pi^7}{\rm Re}\int^\frac{\pi}{2}_0x^6\ln^2(1+e^{i2x})\ {\rm d}x\\ &=\frac{\pi^2}{72}+\frac{1}{2\pi^7}{\rm Im}\int^{-1}_1\frac{\ln^{6}{z}\ln^2(1+z)}{z}{\rm d}z\\ &=\frac{\pi^2}{72}-\frac{3}{\pi^6}\int^1_0\frac{\ln^5{z}\ln^2(1-z)}{z}{\rm d}z+\frac{10}{\pi^4}\int^1_0\frac{\ln^3{z}\ln^2(1-z)}{z}{\rm d}z-\frac{3}{\pi^2}\int^1_0\frac{\ln{z}\ln^2(1-z)}{z}{\rm d}z\\ &=\frac{\pi^2}{72}+\frac{720}{\pi^6}\sum^\infty_{n=1}\frac{H_n}{(n+1)^7}-\frac{120}{\pi^4}\sum^\infty_{n=1}\frac{H_n}{(n+1)^5}+\frac{6}{\pi^2}\sum^\infty_{n=1}\frac{H_n}{(n+1)^3}\\ &=\frac{\pi^2}{72}+\frac{720}{\pi^6}\left(\frac{\pi^8}{7560}-\zeta(3)\zeta(5)\right)-\frac{120}{\pi^4}\left(\frac{\pi^6}{1260}-\frac{1}{2}\zeta^2(3)\right)+\frac{6}{\pi^2}\left(\frac{\pi^4}{360}\right)\\ &=\boxed{\displaystyle\color{red}{\Large{\frac{11\pi^2}{360}+\frac{60}{\pi^4}\zeta^2(3)-\frac{720}{\pi^6}\zeta(3)\zeta(5)}}} \end{align} Generalized Euler sum $\sum^\infty_{n=1}\frac{H_n}{n^q}$

Solution 2:

If you need only the closed form:

$$\frac{11}{60}\zeta(2)+\frac{60}{\pi^4}\zeta^2(3)-\frac{720}{\pi^6}\zeta(3)\zeta(5)$$