TITLE: Log Sine: $\int_0^\pi \theta^2 \ln^2\big(2\sin\frac{\theta}{2}\big)d \theta.$
QUESTION [12 upvotes]: Hi I am trying to calculate
$$
I:=\int_0^\pi \theta^2 \ln^2\big(2\sin\frac{\theta}{2}\big)d \theta.
$$
Here is a related  Integral $\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta$..  This paper may also be of interest to people here : http://www.math.uwo.ca/~dborwein/cv/zeta4.pdf.
We can expand the log in the integral to obtain three interals, one trivial, the other 2 are not so easy, any ideas? I tried doing the following
$$
\left( \ln 2 +\ln \sin \frac{\theta}{2} \right)^2=\ln^2(2)+\ln^2\sin\frac{\theta}{2}+2\ln (2)\ln \sin\big(\frac{\theta}{2}\big).
$$
We can write I as 
$$
I=\ln^2(2)\int_0^\pi \theta^2d\theta  +\int_0^\pi\theta^2 \ln^2 \sin \frac{\theta}{2}d\theta+2\ln 2 \int_0^\pi\theta^2 \ln \sin{\frac{\theta}{2}}d\theta.
$$
Change of variables $x=\theta/2$ and performing the trivial integral we obtain
$$
I=\frac{\pi^3\ln^2 2}{3}+8\int_0^{\pi/2} x^2 \ln^2 \sin x\, dx+16\ln 2\int_0^{\pi/2} x^2 \ln \sin x \, dx.
$$
I am stuck at this point, I was trying to somehow work these two integrals into the form of $$
\int_0^{\pi/2} \ln \sin x dx= \frac{-\pi\ln(2)}{2}\approx -1.08879
$$
but couldn't do so.  Thanks.

REPLY [4 votes]: Hint:
$$\int_0^\pi\theta^2\ln^2\left(2\sin\dfrac{\theta}{2}\right)d\theta$$
$$=\int_\pi^0(\pi-\theta)^2\ln^2\left(2\sin\dfrac{\pi-\theta}{2}\right)d(\pi-\theta)$$
$$=\int_0^\pi\theta^2\ln^2\left(2\cos\dfrac{\theta}{2}\right)d\theta-2\pi\int_0^\pi\theta\ln^2\left(2\cos\dfrac{\theta}{2}\right)d\theta+\pi^2\int_0^\pi\ln^2\left(2\cos\dfrac{\theta}{2}\right)d\theta$$
Which may use the result in Integral $\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta$.