On closed forms for the binomial sum $\sum_{n=1}^\infty \frac{z^n}{n^p\,\binom {2n}n}$ for general $p$?

Solution 1:

In general, we have $$\begin{align} \operatorname{ls}^{(1)}_k(a) := \int_0^a x \, \ln \left( \frac{\sin x}{\sin a} \right)^{k-1} \, \mathrm{d}x \\&= \sin a \int_0^1 \dfrac{\sin^{-1}( x \, \sin a)}{\sqrt{1-x^2 \sin^2 a}} \, \ln(x)^{k-1}\, \mathrm{d}x \\&= \sin a \int_0^1 \sum_{n=1}^\infty \frac{(2 x \sin a)^{2n-1}}{n \binom{2n}{n}}\, \ln(x)^{k-1}\, \mathrm{d}x \\&= (k-1)! \,(-2)^{-k-1} \sum_{n=1}^\infty \frac{(2 \sin a)^{2n}}{n^{k+1} \binom{2n}{n}} \end{align}$$ If we define $$\operatorname{ls}_k(a) := \int_0^a \ln \left( \frac{\sin x}{\sin a} \right)^{k-1} \, \mathrm{d}x,$$ we may note that your previous post establishes closed forms in term of zeta functions for $$\frac{\pi}{6} \operatorname{ls}_k\left(\frac{\pi}{6}\right)- \operatorname{ls}^{(1)}_k\left(\frac{\pi}{6}\right)$$ for $k \in {1,2,3,4,5,6,8}.$

Solution 2:

Hoping that you enjoy hypergeometric functions, $$ A_p(z)=\sum_{n=1}^\infty \frac{z^n}{n^p\,\binom {2n}n}=\frac{z}{2} \, \, _{p+1}F_p\left(1,\cdots,1;\frac{3}{2},2,\cdots,2;\frac{z}{4}\right)$$ and what you wrote in comments is correct.