LogSine Integral $I=-\int_0^{\pi/3} \ln^2\big(2\cos \frac{\theta}{2}\big) d\theta$ These are known as LogSine integrals at $2\pi/3$, so I will call the integral Ls as this is common in the literature.  I am trying to prove
$$
Ls=-\int_0^{\pi/3} \ln^2\big(2\cos \frac{\theta}{2}\big) d\theta=-\frac{13\pi^3}{162}-2Gl_{2,1}\big(\frac{2\pi}{3}\big)
$$
where $Gl_{2,1}$ can be reduced to one-dimensional polylogarithmic constants.  I know we can write
$$
\ln^2\big(2\cos \frac{\theta}{2}\big) =\big(\ln 2+\ln \cos\frac{\theta}{2}\big)^2=\ln^2 2+\ln^2 \cos \frac{\theta}{2} +2\ln 2 \ln \cos \frac{\theta}{2},
$$
but am totally stuck at this point.  Thanks
 A: Using the principal brach of $\log z$,
$$\log(1+e^{2ix}) = \log(e^{i x}(e^{-ix}+ e^{i x})) = \log(e^{ix})+ \log(2 \cos x) = ix + \log(2 \cos x) .$$
Squaring both sides,
$$ \int_{0}^{\pi /6} \log^{2}(1+e^{2ix}) \ dx = \int_{0}^{\pi /6} \Big( ix + \log(2 \cos x) \Big)^2 \ dx .$$
Then equating the real parts on both sides of the equation and rearranging,
$$ \begin{align} \int_{0}^{\pi/3} \log^{2} \left( 2 \cos \frac{x}{2}\right) \ dx &= 2 \int_{0}^{\pi /6} \log^{2}(2 \cos x) \ dx \\ &= 2 \int_{0}^{\pi /6} x^{2} \ dx + 2 \ \text{Re} \int_{0}^{\pi /6} \log^{2}(1+e^{2ix}) \ dx \\ &= \frac{\pi^{3}}{324} + 2 \ \text{Re} \int_{0}^{\pi /6} \log^{2}(1+e^{2ix}) \ dx . \end{align}$$
Now make the substitution $z = e^{2ix}$.
Then
$$\int_{0}^{\pi/3} \log^{2} \left( 2 \cos \frac{x}{2} \right) \ dx  = \frac{\pi^{3}}{324} + \text{Re} \frac{1}{i} \int_{C} \frac{\log^{2}(1+z)}{z} \ dz$$
where $C$ is a portion of the unit circle in the first quadrant of the complex plane.
But since we're using the principal branch of $\log z$, $\log(1+z)$ is analytic on the complex plane for $\text{Re}(z) > -1$.
So the path doesn't matter.
And therefore
$$ \int_{0}^{\pi /3} \log^{2}\left( 2 \cos \frac{x}{2} \right) \ dx = \frac{\pi^{3}}{324} + \text{Re} \frac{1}{i} \int_{0}^{e^{i \pi/3}} \frac{\log^{2}(1+z)}{z} \ dz .$$ 
You can find an antiderivative of the integrand in terms of polylogarithms by integrating by parts twice.
$$ \begin{align} \int \frac{\log^{2}(1+z)}{z} \ dz &= \log^{2}(1+z)\log(-z) - 2 \int \frac{\log(1+z) \log(-z)}{z} \ dz \\ &= \log^{2}(1+z) \log(-z) + 2 \text{Li}_{2}(1+z) \log(1+z) - 2 \int \frac{\text{Li}_{2}(1+z)}{1+z} \ dz \\ &= \log^{2}(1+z) \log(-z) + 2 \text{Li}_{2}(1+z) \log(1+z) - 2 \text{Li}_{3}(1+z) + C \end{align} $$
Evaluating the integral at the limits and then simplifying a bit,
$$ \int_{0}^{\pi /3} \log^{2}\left( 2 \cos \frac{x}{2} \right) \ dx = \frac{7 \pi^{3}}{324} - \frac{\pi}{6} \log^{2}(3) + \log(3) \text{Im} \  \text{Li}_{2}(1+e^{i \pi /3}) + \frac{\pi}{3} \text{Re} \  \text{Li}_{2}(1+e^{i \pi /3}) $$
$$ - 2 \ \text{Im} \  \text{Li}_{3}(1+e^{i \pi /3}) \approx 0.439089177455491 .$$
