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Given the log sine integral,

$$\rm{Ls}_m\Big(\frac{\pi}3\Big) = \int_0^{\pi/3}\Big(\ln\big(2\sin\tfrac{\theta}{2}\big)\Big)^{m-1}\,d\theta$$

we have in this post,

$$\begin{aligned} \frac\pi{2!}\,\rm{Ls}_2\Big(\frac{\pi}3\Big) &=-\frac34\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} -\zeta(3)\\ \frac{2^3\pi}{3!}\rm{Ls}_4\Big(\frac{\pi}3\Big) &=-3\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n} -19\zeta(5)-2\zeta(2)\zeta(3) \\ \frac{2^8\pi}{5!}\rm{Ls}_6\Big(\frac{\pi}3\Big) &=-24 \sum_{n=1}^\infty \frac{1}{n^7\,\binom {2n}n} -493\zeta(7)-48\zeta(2)\zeta(5)-164\zeta(3)\zeta(4) \\ \frac{2^{10}\cdot9\pi}{7!}\rm{Ls}_8\Big(\frac{\pi}3\Big) &=-6^3 \sum_{n=1}^\infty \frac{1}{n^9\,\binom {2n}n}-13921\zeta(9)-6^4\zeta(2)\zeta(7)-6087\zeta(3)\zeta(6)\\ &-4428\zeta(4)\zeta(5)-192\zeta^3(3)\end{aligned}$$

while in this post,

$$\begin{aligned} \rm{Ls}_2\Big(\frac\pi3\Big) &= -\rm{Cl}_2\Big(\frac\pi3\Big)\\ \rm{Ls}_4\Big(\frac\pi3\Big) &= -\tfrac92\rm{Cl}_4\Big(\frac\pi3\Big)-\tfrac12\pi\,\zeta(3)\\ \rm{Ls}_6\Big(\frac\pi3\Big) &= -\tfrac{135}2\rm{Cl}_6\Big(\frac\pi3\Big)-\tfrac{35}{36}\pi^3\,\zeta(3)-\tfrac{15}{2}\pi\,\zeta(5) \end{aligned}$$

with the Clausen function for even $m$,

$$\rm{Cl}_m(x) =\sum_{k=1}^\infty \frac{\sin(kx)}{k^m}$$

These identities imply a triple relation between $\rm{Ls}_m\big(\frac\pi3\big)$, $\rm{Cl}_m\big(\frac\pi3\big)$, and $\sum_{n=1}^\infty \frac{1}{n^{m+1}\,\binom {2n}n}$

Q: However, the triple relation for $m=8$ is missing. Is there also for $m=8$?

Tito Piezas III
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  • Yes, I already tried. Either Mathematica gave me the wrong digits (see this post), or I used the wrong zeta constants, or there really is no triple relation for $m=8$. – Tito Piezas III Jun 10 '19 at 06:20
  • You want a relation between $\text{Ls}_8(\pi/3)$ and $\text{Cl}_8(\pi/3)$? – FDP Jun 11 '19 at 14:38
  • @FDP: A relation between any of the three: $$\rm{Ls}8\big(\frac\pi3\big),;\rm{Cl}_8\big(\frac\pi3\big),;\sum{n=1}^\infty \frac{1}{n^9,\binom {2n}n}$$ that is not in the list above will be quite nice to add. – Tito Piezas III Jun 11 '19 at 14:56
  • Using GP PARI: clau(x)={imag(polylog(8,exp(Ix)))} logsin(s)={intnum(x=0,s,log(2sin(x/2))^7)} lindep([clau(Pi/3),logsin(Pi/3),Pizeta(7),Pi^3zeta(5),Pi^5*zeta(3)]) no relation found. – FDP Jun 11 '19 at 15:06
  • @FDP: Thanks, I already tried with an old version of Mathematica, but there's always the uncertainty that my version may not be giving the correct digits. – Tito Piezas III Jun 11 '19 at 15:11
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    $$\sum _{n=1}^{\infty } \frac{1}{n^8 \binom{2 n}{n}}=-4 \pi \text{Gl}\left({6,1},\frac{\pi }{3}\right)+36 \text{Gl}\left({7,1},\frac{\pi }{3}\right)+\frac{\pi ^2 \zeta (3)^2}{9}+4 \zeta (5) \zeta (3)-\frac{266677 \pi ^8}{440899200}$$ Here $\text{Gl}$ denotes level six Multiple Zeta Values. It is highly likely that these MZVs are irreducible. – Infiniticism Sep 30 '20 at 11:27

0 Answers0