Closed form for $f_m=\sum _{n=0} ^{\infty} \frac{4^{n}}{(2n+1)^{m}\binom{2n}{n}}$

Question

I am interested in a non hypergeometric closed form of a family of series involving the inverse of the central binomial coefficient $\binom{2n}{n}$, similar to the ones in this question $$f(m)=\sum _{n=0} ^{\infty} \frac{4^{n}}{(2n+1)^{m}\binom{2n}{n}}, \quad m \geq 2 $$

The only ones I know are: $$\begin{eqnarray} f(2) &=& 2 \rm G,\\ f(3) &=& \frac{3}{16} \pi ^3 + \frac{1}{4} \pi \log ^2 2 -4 \Im \operatorname{Li} _3 (1+i),\\ f(4) &=& -\frac{3}{16} \pi ^3 \log2 - \frac{1}{12} \pi \log ^3 2 +8 \Im \operatorname{Li} _4 (1+i)-\frac{1}{384} \left( \psi^{(3)}\left(\frac14\right) - \psi^{(3)}\left(\frac34\right) \right) \end{eqnarray} $$

where $\rm G$ is Catalan's constant, $\operatorname{Li}_n(x)$ is the polylogarithm, and $\psi ^{(n)}(x)$ is the polygamma function. It feels like there's a certain pattern to them and that the values of $f(m)$ lie in an algebra of $\rm G, \pi, \log 2$ and certain polylog and polygamma values.

More generally, we can consider the functions $$f_m(x)=\sum _{n=0} ^{\infty} \frac{(2x)^{2n}}{(2n+1)^{m}\binom{2n}{n}}, \quad m \in \mathbb{Z} $$

and notice that $$f_{m+1}(t) = \frac{1}{t} \int _0 ^t f_m(x) dx$$

For $m=1$ the expression reduces to: $$f_1(x) = \frac{\operatorname{arcsin}(x)}{x\sqrt{1-x^2}}$$

For the cases $m=2$ and $m=3$ the substitution $x\rightarrow \sin \phi$ expresses the integrals in the form considered in this difficult to read but wonderful article but I couldn't proceed to higher values of $m$.

Answer 1

$f(m)$ can express as rational linear combination of multiple zeta values of level 2. If you know multiple zeta values of level 2 such that

\begin{align} t(\boldsymbol k)=\sum_{0<n_1<\dots<n_r}\frac1{(2n_1-1)^{k_1}\dots(2n_r-1)^{k_r}} \end{align}

then you can expressed as the series

\begin{align*} f(m)=\sum_{0\leq n}\frac{2^{2n}}{(2n+1)^m\binom{2n}n}=\sum_{\substack{\operatorname{wt}(\boldsymbol k)=m\\k_1\geq 2}}2^jt(k_1,\dots,k_{j-1},\bar{k_j}). \end{align*}

It can be proved by integral substitution. (You can also prove this only series transformation.)

Answer 2

This is my own attempt at a solution. Consider the generalized log-sine integrals defined by $$\operatorname{Ls}^{(k)} _{~n} (\sigma) = -\int _0 ^\sigma \theta^k \log^{n-k-1} \left( 2\sin\frac{\theta}{2} \right) d\theta, \quad 0 \leq \theta \leq 2\pi$$ Now, let's look at the case $\sigma = \pi$. Consider the exponential generating function: $$-\sum _{n \geq0, ~k\geq0} \operatorname{Ls}^{(k)} _{~n+k+1} (\pi) \frac{\lambda^n}{n!}\frac{(i \mu)^k}{k!} =\\= \int _0 ^\pi \sum _{n \geq 0} \frac{\lambda^n \log ^n\left( 2\sin \frac{\theta}{2} \right) }{n!} \sum_{k\geq0} \frac{(i\mu\theta)^k}{k!} d\theta = \int _0 ^\pi \left( 2\sin \frac{\theta}{2} \right)^\lambda e^{i\mu\theta}d\theta $$ Following J. Borwein and A. Straub, the right hand integral could be further simplified: $$-\sum _{n \geq0, ~k\geq0} \operatorname{Ls}^{(k)} _{~n+k+1} (\pi) \frac{\lambda^n}{n!}\frac{(i \mu)^k}{k!} = i \sum_{n\geq0} \binom{\lambda}{n} \frac{(-1)^n e^{i\pi\frac{\lambda}{2}}-e^{i\pi\mu}}{\mu-\frac{\lambda}{2}+n}\tag{*}$$

This allows us to find the values of $\operatorname{Ls}^{(k)} _{~n} (\pi)$ by differentiating the right hand sum and setting $\lambda =0, \mu = 0$.

Now, let's define, and I am very sorry about this notation, the cosecant log-sine integrals by $$\operatorname{cscLs}^{(k)} _{~n,~m} (\sigma) = -\int _0 ^\sigma \frac{\theta^{k+m}}{\left( 2\sin \frac{\theta}{2} \right) ^m} \log^{n-k-1} \left( 2\sin \frac{\theta}{2} \right) d\theta$$ These integrals are connected to the integrals in question by a binomial expansion of the $\log ^{n-k-1} \left (2 \sin \frac{\theta}{2} \right)$, like in the answer to this question by Raymond Manzoni. So, let's consider the latter funcitons first. We wish to find $\operatorname{cscLs}^{(0)} _{~n,~1} (\pi)$. We can study them in the same way, by considering their EGF: $$-\sum _{n \geq0, ~k\geq0} \operatorname{cscLs}^{(k)} _{~n+k+1,~m} (\pi) \frac{\lambda^n}{n!}\frac{(i \mu)^k}{k!} =\\= \int _0 ^\pi \frac{\theta^m}{\left( 2\sin \frac{\theta}{2} \right)^m}\sum _{n \geq 0} \frac{\lambda^n \log ^n \left( 2\sin \frac{\theta}{2} \right) }{n!} \sum_{k\geq0} \frac{(i\mu\theta)^k}{k!} d\theta =\\= \int _0 ^\pi \frac{\theta^m}{\left( 2\sin \frac{\theta}{2} \right)^m} \left( 2\sin \frac{\theta}{2} \right)^\lambda e^{i\mu\theta}d\theta =(-i)^m \frac{\partial^m}{\partial\mu^m}\int _0 ^\pi \left( 2\sin \frac{\theta}{2} \right)^{\lambda-m} e^{i\mu\theta}d\theta$$ which means that from $(*)$, by setting $\lambda \rightarrow \lambda -m$, we get: $$-\sum _{n \geq0, ~k\geq0} \operatorname{cscLs}^{(k)} _{~n+k+1,~m} (\pi) \frac{\lambda^n}{n!}\frac{(i \mu)^k}{k!} =(-i)^{m-1} \frac{\partial^m}{\partial\mu^m} \sum_{n\geq0} \binom{\lambda-m}{n} \frac{(-1)^n e^{i\pi\frac{\lambda-m}{2}}-e^{i\pi\mu}}{\mu-\frac{\lambda-m}{2}+n}$$ Again,we can now find the values of $\operatorname{cscLs}^{(k)} _{~n, ~m} (\pi)$ by differentiating the right hand sum with respect to $\mu$ and $\lambda$ and then setting $\lambda =0, \mu = 0$.