LogSine Integral $\int_0^{\pi/3}\ln^n\big(2\sin\frac{\theta}{2}\big)\mathrm d\theta$ I am trying to integrate a special case of the log sine integral $\rm{Ls}_n(\sigma)$ at $\sigma=\pi/3$ :
$$
\rm{Ls}_{n}\big(\tfrac{\pi}{3}\big)=-\int_0^{\pi/3}\bigg[\ln\big(2\sin\tfrac{\theta}{2}\big)\bigg]^{n-1}\mathrm d\theta
$$
where $n$ is a non-negative integer.
This problem is strongly related to the hypergeometric form of the Log Sine integral.
The closed form is rather simple, although I am having trouble computing it. We can use standard log rules on the inside expression, although I am not sure how this will help us...Thanks
 A: 
I. Special case

The log sine integral,
$$\rm{Ls}_{n}\big(\tfrac{\pi}{3}\big)=\int_0^{\pi/3}\bigg[\ln\big(2\sin\tfrac{x}{2}\big)\bigg]^{n-1}\mathrm dx$$
can be given a nice closed-form in terms of $\pi$, the zeta function $\zeta(s)$, and Clausen function $\rm{Cl}_n(x)$ for the first few even $n$. Let,
$$\rm{Cl}_n(\alpha) =\sum_{m=1}^\infty \frac{\sin(m\alpha)}{m^n},\quad \text{even}\;n$$
$$\rm{Cl}_n(\alpha) =\sum_{m=1}^\infty \frac{\cos(m\alpha)}{m^n},\quad \text{odd}\;n$$
Then,
$$\begin{aligned}
\rm{Ls}_2\big(\tfrac\pi3\big) &= -\rm{Cl}_2(\tfrac\pi3\big)\\
\rm{Ls}_4\big(\tfrac\pi3\big) &= -\tfrac92\rm{Cl}_4(\tfrac\pi3\big)-\tfrac12\pi\,\zeta(3)\\
\rm{Ls}_6\big(\tfrac\pi3\big) &= -\tfrac{135}2\rm{Cl}_6(\tfrac\pi3\big)-\tfrac{35}{36}\pi^3\,\zeta(3)-\tfrac{15}{2}\pi\,\zeta(5)
\end{aligned}$$
See Borwein's "Special Values of Generalized Log-sine Integrals". Note that the special case $n=2$ is Gieseking's constant.
Unfortunately, the simple pattern seems to stop at $n=6$. I tried to find $n=8$ using $\rm{Cl}_8(\tfrac\pi3\big)$ and analogous products of $\pi$ and $\zeta(s)$, but an integer relations algorithm couldn't find it. (Presumable a new function comes to play at $n=8$.)

II. General case

More generally, we have,
$$\rm{Ls}_n^{(k)}(z) = \int_0^{z}x^k \Big(\ln\big(2\sin\tfrac{x}{2}\big)\Big)^{n-1-k}\,dx$$
If we focus on the case $n-1-k=1$, or $n=k+2$,
$$\rm{Ls}_{k+2}^{(k)}(z) = \int_0^{z}x^k \ln\big(2\sin\tfrac{x}{2}\big)\,dx$$
then for all $k$ we have the concise formula in terms of the Clausen function and zeta function,
$$\frac{1}{k!}\int_0^{z}x^k \ln\big(2\sin\tfrac{x}{2}\big)\,dx = P(k)+\sum_{j=2}^{k+2} \frac{(-1)^{j+\lfloor(j+1)/2\rfloor }}{(k+2-j)!} \operatorname{Cl}_j(z)\,z^{k+2-j}$$
with floor function $\lfloor n\rfloor$ and where,
$$P(k)=\frac{1-(-1)^k}{2}\,\zeta(k+2)\,i^{k-1}$$
with $i = \sqrt{-1}$, and $|z| < 2\pi$. 
P.S. Note that for even $k$, then $P(k) = 0$, while for odd $k$, it is a real number.
A: $$\text{Consider }\int_0^\frac{\pi}{3}\left(2\sin\frac{\theta}{2}\right)^a~d\theta~,$$
$$\int_0^\frac{\pi}{3}\left(2\sin\frac{\theta}{2}\right)^a~d\theta$$
$$=2^a\int_0^\frac{\pi}{3}\sin^a\frac{\theta}{2}d\theta$$
$$=2^{a+1}\int_0^\frac{\pi}{6}\sin^a\theta~d\theta$$
$$=2^{a+1}\int_0^\frac{1}{2}x^a~d(\sin^{-1}x)$$
$$=2^{a+1}\int_0^\frac{1}{2}\dfrac{x^a}{\sqrt{1-x^2}}dx$$
$$=2^{a+1}\int_0^\frac{1}{4}\dfrac{x^\frac{a}{2}}{\sqrt{1-x}}d\left(x^\frac{1}{2}\right)$$
$$=2^a\int_0^\frac{1}{4}\dfrac{x^\frac{a-1}{2}}{\sqrt{1-x}}dx$$
$$=2^aB\left(\dfrac{1}{4};\dfrac{a+1}{2},\dfrac{1}{2}\right)$$
$$\therefore\int_0^\frac{\pi}{3}\ln^n\left(2\sin\frac{\theta}{2}\right)d\theta=\dfrac{d^n}{da^n}\left(2^aB\left(\dfrac{1}{4};\dfrac{a+1}{2},\dfrac{1}{2}\right)\right)(a=0)$$
