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,
$$\pi\,\rm{Ls}_7\Big(\frac{\pi}3\Big) =-135\pi\,\rm{Gl}_{6,1}\Big(\frac{\pi}{3}\Big)+\Big(2152-\tfrac{103}{864}\Big)\zeta(8)+45\zeta(2)\zeta(3)^2\quad$$
where,
$$\rm{Gl}_{m,1}\Big(\frac{\pi}3\Big) = \sum_{n=1}^\infty \frac{\sum_{k=1}^{n-1}\frac1k}{n^m}\sin\Big(\frac{n\,\pi}3\Big)= \sum_{n=1}^\infty \frac{H_{n-1}}{n^m}\sin\Big(\frac{n\,\pi}3\Big)$$
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.)
Best Answer
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.