Closed form for $\int_0^1\frac{\mathrm{Li}_s(x-x^2)}{x-x^2}\mathrm dx$

Question

I am trying to evaluate
$$F(s)=\sum_{n\geq1}\frac1{n^{s+1}{2n\choose n}}$$
I started off by noting that $$\frac1{n{2n\choose n}}=\frac12\int_0^1\left[x-x^2\right]^{n-1}\mathrm dx$$
So
$$F(s)=\int_0^1\sum_{n\geq1}\frac{\left[x-x^2\right]^{n-1}}{n^s}\mathrm dx$$
And when we recall the definition of the polylogarithm function
$$\mathrm{Li}_s(z)=\sum_{n\geq1}\frac{z^n}{n^s}$$
It becomes apparent that
$$F(s)=\int_0^1\frac{\mathrm{Li}_{s}(x-x^2)}{x-x^2}\mathrm dx$$
Which I do not know how to deal with. Could I have some help evaluating this integral? Thanks.

Answer 1

An expression as a generalized hypergeometric function can be directly obtained by expressing the ratio of two consecutive terms in the series which defines $F(s)$ as a rational fraction in $n$ allowing thus to express the series as a generalized hypergeometric series.

Alternatively, starting from the integral, as remarked by @mrtaurho \begin{align} F(s)&=\int_0^{1/2}\frac{\mathrm{Li}_{s}(x-x^2)}{x-x^2}\,dx+\int_{1/2}^1\frac{\mathrm{Li}_{s}(x-x^2)}{x-x^2}\,dx\\ &=2\int_0^{1/2}\frac{\mathrm{Li}_{s}(x-x^2)}{x-x^2}\,dx\\ &=2\int_0^{1/4}\frac{\mathrm{Li}_{s}(t)}{t}\frac{dt}{\sqrt{1-4t}}\\ &=2\int_0^{1}\frac{\mathrm{Li}_{s}(\frac{u}{4})}{u}\frac{du}{\sqrt{1-u}} \end{align} (we changed $x\to 1-x$ n the second integral of the first expression, then $t=x(1-x)$ and finally $t=u/4$). Now, if $s$ is an integer, we use the representation in terms of the hypergeometric function (here) \begin{equation} \mathrm{Li}_s(z)=z\,_{s+1}F_s\left( 1,1,\ldots,1;2,2,\ldots,2;z \right) \end{equation} to obtain \begin{equation} F(s)=\frac{1}{2}\int_0^{1}\left( 1-u \right)^{-1/2}{}_{s+1}F_s\left( 1,1,\ldots,1;2,2,\ldots,2;\frac{u}{4} \right)\,du \end{equation} Then, from the tabulated integral (DLMF), \begin{equation} {{}_{p+1}F_{q+1}}\left({a_{0},\dots,a_{p}\atop b_{0},\dots,b_{q}};z\right)=% \frac{\Gamma\left(b_{0}\right)}{\Gamma\left(a_{0}\right)\Gamma\left(b_{0}-a_{0% }\right)}\int_{0}^{1}t^{a_{0}-1}(1-t)^{b_{0}-a_{0}-1}{{}_{p}F_{q}}\left({a_{1}% ,\dots,a_{p}\atop b_{1},\dots,b_{q}};zt\right)\mathrm{d}t \end{equation} valid for $\Re b_0>\Re a_0>0$ if $p\le q$ and, additionally, $\left|\mathrm{ph}(1-z)\right|<\pi$ if $p=q+1$. Here $a_0=1,b_0=3/2,z=1/4$, thus \begin{equation} F(s)={}_{s+2}F_{s+1}\left( 1,1,\ldots,1;\frac{3}{2},2,2,\ldots,2;\frac{1}{4} \right) \end{equation}