# More on the log sine integral $\int_0^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta$

I. In this post, the OP asks about the particular log sine integral, $$\mathrm{Ls}_{7}^{\left ( 3 \right )} =-\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\,\mathrm{d}\theta$$

and gives the monster evaluation,

\small{\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\,\color{red}{\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\,\color{red}{\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 I had to use small fonts to make it fit better.

Update: Based on Przemo's answer, I just realized the multiple polylogarithm $$\mathrm{Li}_{m,1}(z)$$ is just a Nielsen generalized polylogarithm $$S_{n,p}(z)$$ in disguise,

$$\color{red}{\mathrm{Li}_{m ,1}}\left ( z \right )=\sum_{k=1}^{\infty }\frac{z^{k}}{k^{m}}\sum_{j=1}^{k-1}\frac{1}{j} = S_{m-1,\,2}(z)$$

discussed below.

$$S_{n,p}(z) = \frac{(-1)^{n+p-1}}{(n-1)!\,p!}\int_0^1\frac{(\ln t)^{n-1}\big(\ln(1-z\,t)\big)^p}{t}dt$$

Given $$\color{blue}{z=-1}$$, for brevity let $$S_{n,p}(-1) = S_{n,p}$$. Then I found the integral can be evaluated using only 6 $$S_{n,p}$$ with small coefficients,

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

However, since there is a linear relation between these six as,

$$1260 S_{5,2}-1506 S_{4,3} + 1004 S_{3,4} -\tfrac{\pi^2}6\Big(157S_{3,2}-279S_{2,3}+558S_{1,4}\Big) = 0$$

then the number of terms can be reduced to $$5$$. The last piece of the puzzle is to express these as polylogarithms. Note that,

$$32S_{3,2}(-1) = 16\zeta(2)\zeta(3)-29\zeta(5)$$

$$32S_{2,3}(-1) = 16\zeta(2)\zeta(3)-31\zeta(5)+64S_{1,4}(-1)$$

$$30S_{1,4}(-1) = -a^5-5a^3\rm{Li}_2(\tfrac12)-15a^2\rm{Li}_3(\tfrac12) -30a\, \rm{Li}_4(\tfrac12)-30\rm{Li}_5(\tfrac12)+30\zeta(5)$$

where $$a=\ln 2$$, and,

$$128S_{5,2}(-1) = 64\zeta(2)\zeta(5)+112\zeta(3)\zeta(4)-251\zeta(7)$$

I've been trying to do so also for $$S_{3,4}(-1)$$ and $$S_{4,3}(-1)$$ but to no avail. The best I could do was,

$$128S_{3,4}(-1)-192S_{4,3}(-1)=-64\zeta(2)\zeta(5)-160\zeta(3)\zeta(4)+315\zeta(7)$$

III. Question: For $$z=-1$$, can we express $$S_{3,4}(-1)$$ and $$S_{4,3}(-1)$$ in terms of polylogarithms $$\rm{Li}_m(x)$$? In general, can we do so for all $$S_{n,p}(-1)$$?

P.S. I know it can be done when $$n = 1$$.

• @ Tito Piezas III: I remember I was pondering on your problem above some time ago and I was not quite sure if the answer is positive. I will try to brush up on this soon. – Przemo May 22 at 10:34
• @Przemo: You may be interested in this related post – Tito Piezas III Jun 3 at 14:58