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

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 $\sum^\infty_{n=1}\frac{H_n}{n^q}$

If you need only the closed form:

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

I could have sworn when I began my answer there was a square in the integrand and not a 6 power on the 2x-1 term. Anyway, this is for $(2x-1)^{2}$. But, the same idea can be used on the 6 power, but it will be very long and messy. It may even make tech grunt and groan.

One way to go about it is to use the identity:

$\displaystyle \int_{0}^{\frac{\pi}{2}}x\cos^{p-1}(x)\sin(ax)dx=\frac{\pi}{2^{p+1}}\Gamma(p)\left[\frac{\psi(\frac{p+a+1}{2})-\psi(\frac{p-a+1}{2})}{\Gamma(\frac{p+a+1}{2})\Gamma(\frac{p-a+1}{2})}\right]$

Since you have the 2 inside your log term, you could split your integrand up and write as

$\displaystyle \int_{0}^{1/2}(2x-1)^{2}\left[\ln^{2}(2)+2\ln(2)\ln(\sin(\pi x))+\ln^{2}(\sin(\pi x))\right]dx$

Then, using the above identity, do each separately.

The first integration is straightforward. Take the middle one:

$\displaystyle 2\ln(2)\int_{0}^{1/2}(2x-1)^{2}\ln(\sin(\pi x))dx$

Let $\displaystyle u=2x-1, \;\ dx=1/2du$

$\displaystyle \ln(2)\int_{-1}^{0}u^{2}\ln(\cos(\pi u/2))du$

Now, let $\displaystyle t=\frac{\pi u}{2}, \;\ du=\frac{2}{\pi}dt$.

Due to symmetry, we can write the limits as:

$\displaystyle \left(\frac{2}{\pi}\right)^{3}\ln(2)\int_{0}^{\pi/2}t^{2}\ln(\cos(t))dt$

Now, using the identity I mentioned at the top, diff once w.r.t p, then once w.r.t a.

Then, let p=1 and a=0. As I said, diffing the right side of the above identity may prove messy. So, use some sort of tech. For the last integral on the end, you can make the same subs to get the pi/2 limit, but then diff twice w.r.t p in order to get the squared log term.

So, for $\displaystyle \int_{0}^{\frac{\pi}{2}}x^{2}\ln(\cos(x))dx=\frac{-\pi}{4}\zeta(3)-\frac{\ln(2)}{24}\pi^{3}$

Using this method with that 6 will be brutal...even with tech. So, how about keeping the 2?. As Norbert alluded, 'getting off the horse in the middle of the stream' can be a little annoying.

BTW, that link that Norbert provided is very nice. It outlines how to handle log trig integrals using contours. You do not see this in many places. Check it out.