Seven years ago, I asked about closed-forms for the binomial sum $$\sum_{n=1}^\infty \frac{1}{n^k\,\binom {2n}n}$$

Some alternative results have been made. Up to a certain $k$, it seems it can be expressed surprisingly by a log sine integral,

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

and zeta function $\zeta(s)$. Hence,

$$\begin{aligned} \frac\pi2\,\rm{Ls}_1\Big(\frac{\pi}3\Big) &=\;3\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} =\zeta(2) \\ \frac\pi2\,\rm{Ls}_2\Big(\frac{\pi}3\Big) &=-\frac34\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} -\zeta(3) =-\frac\pi2\,\rm{Cl}_2\Big(\frac\pi3\Big)\\ \frac{6\pi}{35}\,\rm{Ls}_3\Big(\frac{\pi}3\Big) &=\frac{36}{17}\sum_{n=1}^\infty \frac{1}{n^4\,\binom {2n}n} =\zeta(4)\\ \frac{2^3\pi}{3!}\rm{Ls}_4\Big(\frac{\pi}3\Big) &=-3\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n} -19\zeta(5)-2\zeta(2)\zeta(3) \\ 32\pi\,\rm{Ls}_5\Big(\frac{\pi}3\Big) &=144 \sum_{n=1}^\infty \frac{1}{n^6\,\binom {2n}n} +2029\zeta(6)+192\zeta(3)^2 \\ \frac{2^8\pi}{5!}\rm{Ls}_6\Big(\frac{\pi}3\Big) &=-24 \sum_{n=1}^\infty \frac{1}{n^7\,\binom {2n}n} -493\zeta(7)-48\zeta(2)\zeta(5)-164\zeta(3)\zeta(4) \\ \end{aligned}$$

where $\rm{Cl}_2\big(\tfrac\pi3\big)$ is Gieseking's constant and other $\rm{Ls}_{2n}\big(\tfrac\pi3\big)$ can be found here. I found these using Mathematica's integer relations sub-routine. Unfortunately, either the pattern stops at this point, or some other variables are involved. Note that Borwein and Straub also found,

$$\pi\,\rm{Ls}_7\Big(\frac{\pi}3\Big) =-135\pi\,\rm{Gl}_{6,1}\Big(\frac{\pi}{3}\Big)+\Big(2152-\tfrac{103}{864}\Big)\zeta(8)+45\zeta(2)\zeta(3)^2\quad$$


$$\rm{Gl}_{m,1}\Big(\frac{\pi}3\Big) = \sum_{n=1}^\infty \frac{\sum_{k=1}^{n-1}\frac1k}{n^m}\sin\Big(\frac{n\,\pi}3\Big)= \sum_{n=1}^\infty \frac{H_{n-1}}{n^m}\sin\Big(\frac{n\,\pi}3\Big)$$

with harmonic number $\rm{H}_n$.

Q: Can we bring this table higher and find a relation between the log sine integral $\rm{Ls}_7\big(\frac{\pi}3\big)$ and binomial sums?

$\color{blue}{Update:}$ Given the generalized log sine integral,

$$\rm{Ls}_m^{(k)}(\sigma) = \int_0^{\sigma}x^k\Big(\ln\big(2\sin\tfrac{x}{2}\big)\Big)^{m-1-k}\,dx$$

where the post was just the case $k=0$. If we use $k=1$ instead,

$$\rm{Ls}_m^{(1)}(\sigma) = \int_0^{\sigma} x\,\Big(\ln\big(2\sin\tfrac{x}{2}\big)\Big)^{m-2}\,dx$$

this paper mentions that Borwein et al found,

$$\sum_{n=1}^\infty \frac{1}{n^m\,\binom {2n}n} = \frac{(-2)^{\color{red}{m-2}}}{(m-2)!}\int_0^{\pi/3} x\,\Big(\ln\big(2\sin\tfrac{x}{2}\big)\Big)^{m-2}\rm{dx}$$

Note: The paper made a typo. (Corrected in red.)

  • $\begingroup$ The last formula (attributed here to Borwein et al) can't be correct. For $m=2$ is gives zero. $\endgroup$ – Dr. Wolfgang Hintze May 5 '19 at 16:57
  • $\begingroup$ @ Tito Piezas III I meant m=1. But I repeat that the formula is wrong. Numerically, for m=1..4 we obtain from the formula the values {0., -1.09662, -1.04589, -1.02219} while the sum gives {0.6046, 0.548311, 0.522946, 0.511097}. $\endgroup$ – Dr. Wolfgang Hintze May 5 '19 at 17:17
  • $\begingroup$ Using the usual procedure of replacing the sum by an integral I obtain the following expression for the sum $s(k) = \text{Li}_k\left(\frac{1}{4}\right)+\frac{1}{\Gamma(k)} \int_0^{\frac{\pi }{6}} \frac{2^k u \log ^{k-1}\left(\frac{1}{2 \sin (u)}\right)}{\cos ^2(u)} \, du$. The formula is correct numerically for the first few values of $k$. $\endgroup$ – Dr. Wolfgang Hintze May 5 '19 at 17:22
  • $\begingroup$ @Dr.WolfgangHintze: I checked it. The authors made a small typo. The factor should be $$\frac{(-2)^{m-2}}{(m-2)!}$$ It now works. $\endgroup$ – Tito Piezas III May 5 '19 at 17:28
  • 3
    $\begingroup$ Except for m=1 it is okay now. $\endgroup$ – Dr. Wolfgang Hintze May 5 '19 at 17:47

Persistence pays off! Given the log sine integral,

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

The case $\rm{Ls}_7\big(\frac{\pi}3\big)$ was elusive, but $\rm{Ls}_\color{red}8\big(\frac{\pi}3\big)$ was found. Hence,

$$\frac{2^{10}\cdot9\pi}{7!} \int_0^{\pi/3}\ln^7\big(2\sin\tfrac{x}{2}\big)\,dx+6^3 \sum_{n=1}^\infty \frac{1}{n^9\,\binom {2n}n}\\=-13921\zeta(9)-6^4\zeta(2)\zeta(7)-6087\zeta(3)\zeta(6)-4428\zeta(4)\zeta(5)-192\zeta^3(3) $$

though I don't know why the lower level is more elusive.


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