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.

II. The Nielsen generalized polylogarithm,

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

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,


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 $n = 1$.

  • $\begingroup$ @ 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. $\endgroup$ – Przemo May 22 '19 at 10:34
  • $\begingroup$ @Przemo: You may be interested in this related post $\endgroup$ – Tito Piezas III Jun 3 '19 at 14:58

By shifting contour we know that this integral essentially reduces to the calculation of $$\int_0^1 \frac{\log ^3(x) \log ^3(x+1)}{x} \, dx$$ And some lower weight analogues. All weight 7 log integrals are expressible via elementary terms, polylog terms and at most 3 irreducible alternate Euler-like constants. See my article here for the reduction of the original integral to this log integral, and here for the systematic evaluation of high weight log integrals including this one.

  • $\begingroup$ For weight seven, at most $2$ irreducible constants, not $3$. $\endgroup$ – pisco Dec 2 '19 at 12:22
  • 1
    $\begingroup$ @pisco I added the one first appeared in weight 6. : ) $\endgroup$ – Display name Dec 2 '19 at 12:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.