TITLE: Evaluate $\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta $
QUESTION [24 upvotes]: Evaluate

$$\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\,\mathrm{d}\theta $$

Several days ago,I found this interesting integral from a paper about generalized log-sine integrals,but I can't remember the title of it. The answer of the integral is

\begin{align*}
-\mathrm{Ls}_{7}^{\left ( 3 \right )}\left ( \pi  \right)&=\frac{9}{35}\log^72+\frac{4}{5}\pi ^{2} \log^52+9\zeta \left ( 3 \right )\log^42-\frac{31}{30}\pi ^{4}\log^32\\
&-\left [ 72\mathrm{Li}_5\left ( \frac{1}{2} \right )-\frac{9}{8}\zeta \left ( 5 \right )-\frac{51}{4}\pi ^{2}\zeta \left ( 3 \right ) \right ]\log^22\\
&+\left [ 72\mathrm{Li}_{5,1}\left ( \frac{1}{2} \right )-216\mathrm{Li}_6\left ( \frac{1}{2} \right )+36\pi ^{2}\mathrm{Li}_4\left ( \frac{1}{2} \right ) \right ]\log2+72\mathrm{Li}_{6,1}\left ( \frac{1}{2} \right )\\
&-216\mathrm{Li}_7\left ( \frac{1}{2} \right )+36\pi ^{2}\mathrm{Li}_5\left ( \frac{1}{2} \right )-\frac{1161}{32}\zeta \left ( 7 \right )-\frac{375}{32}\pi ^{2}\zeta \left ( 5 \right )+\frac{1}{10}\pi ^{4}\zeta \left ( 3 \right )
\end{align*}
  where 
  $$\mathrm{Ls}_n^{\left ( k \right )}\left ( \alpha  \right ):=-\int_{0}^{\alpha }\theta ^{k}\log^{n-1-k}\left | 2\sin\frac{\theta }{2} \right |\mathrm{d}\theta $$
  is the generalized log-sine integral and
  $$\mathrm{Li}_{\lambda ,1}\left ( z \right )=\sum_{k=1}^{\infty }\frac{z^{k}}{k^{\lambda }}\sum_{j=1}^{k-1}\frac{1}{j}$$
  is the multiple polylogarithm.


I found a beautiful way to solve the integrals below
$$\int_{0}^{\frac{\pi }{2}}t^{2n}\log^{m}\left ( 2\cos t  \right )\mathrm{d}t $$
Let's consider
$$\mathcal{I}\left ( x,y \right )=\int_{0}^{\frac{\pi }{2}}\cos\left ( xt \right )\left ( 2\cos t \right )^{y}\mathrm{d}t$$
By using Gamma function,the integral become
$$\mathcal{I}\left ( x,y \right )=\frac{\pi \, \Gamma \left ( y+1 \right )}{2\Gamma \left ( \dfrac{x+y+2}{2} \right )\Gamma \left ( \dfrac{y-x+2}{2} \right )}$$ 
Then we can get
$$\mathcal{I}\left ( x,y \right )=\frac{\pi }{2}\exp\left ( \sum_{k=2}^{\infty }\frac{\left ( -1 \right )^{k}}{k\cdot 2^{k}}\zeta \left ( k \right )\left [ \left ( 2y \right )^{k}-\left ( y-x \right )^{k}-\left ( x+y \right )^{k} \right ] \right )$$
On the other hand,using taylor series 
$$\mathcal{I}\left ( x,y \right )=\sum_{n=0}^{\infty }\frac{\left ( -1 \right )^{n}}{\left ( 2n \right )!}x^{2n}\sum_{m=0}^{\infty }\frac{y^{m}}{m!}\int_{0}^{\frac{\pi }{2}}t^{2n}\log^m\left ( 2\cos t \right )\mathrm{d}t$$
So,the comparison of coefficient shows the answer.For example
$$\int_{0}^{\frac{\pi }{2}}t^{2}\log^2\left ( 2\cos t \right )\mathrm{d}t=4\cdot \frac{\pi }{2}\left [ \frac{12}{4\cdot 16} \zeta \left ( 4 \right )+\frac{1}{2}\frac{8}{2^{2}\cdot 4^{2}}\zeta \left ( 2 \right )^{2}\right ]=\frac{11}{1440}\pi ^{5}$$

I wonder can we use the same way to prove the integral in the beginning,if not,is there another way to handle it?

REPLY [3 votes]: In general, the log sine integral for general $a,b$,
$$\int_0^{\pi}x^a\ln^b\left(2\sin\tfrac{x}2\right) dx$$
and at the special upper limit $\sigma = \pi$, can be concisely expressed in terms of the Nielsen generalized polylogarithm,
$$S_{n,p}(z) = \frac{(-1)^{n+p-1}}{(n-1)!\,p!}\int_0^1\frac{(\ln t)^{n-1}\big(\ln(1-z\,t)\big)^p}{t}dt$$
This reduces to the usual polylogarithm when $p=1$,
$$S_{n-1,1}(z)=\mathrm{Li}_n(z)$$
For brevity, since we will use only $\color{blue}{z=-1}$, let,
$$S_{n,p}(-1) = S_{n,p}$$
The OP's integral then has the compact form of only $6$ terms,

$$\large{\frac1{18}\int_0^{\pi}x^3\ln^3\left(2\sin\tfrac{x}2\right)dx \\=-10S_{5,2}+14S_{4,3}-8S_{3,4}+\frac{\pi^2}6\Big(4S_{3,2}-9S_{2,3}+6S_{1,4}\Big)\\ =\, 0.3341049\dots}$$


P.S. Note also that,
$$32S_{3,2}(-1) = 16\zeta(2)\zeta(3)-29\zeta(5)$$
$$32S_{2,3}(-1) = 16\zeta(2)\zeta(3)-31\zeta(5)+64S_{1,4}(-1)$$
$$128S_{5,2}(-1) = 64\zeta(2)\zeta(5)+112\zeta(3)\zeta(4)-251\zeta(7)$$
$$30S_{1,4}(-1) = -\ln^4(2)\, \rm{Li}_1(\tfrac12)-5\ln^3(2)\, \rm{Li}_2(\tfrac12)-15\ln^2(2) \,\rm{Li}_3(\tfrac12)\\ -30 \ln(2)\, \rm{Li}_4(\tfrac12)-30\rm{Li}_5(\tfrac12)+30\zeta(5)$$
Perhaps $S_{3,4}(-1)$ and $S_{4,3}(-1)$ can also be similarly expressed though it is not sure. If so, then the OP's integral can be evaluated without exotic functions by using only the polylogarithm $\mathrm{Li}_n(z)$, since $\ln(z)$ and $\zeta(n)$ are just special cases of it.