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

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.

• Some references here may help... – Raymond Manzoni Feb 2 '19 at 18:04
• If I am not mistaken integral can be reduced to $$2\int_0^{1/4} \frac{\operatorname{Li}_s(x)}{x}\frac{\mathrm dx}{\sqrt{1-4x}}$$ Maybe this different form is of help for someone. For clarification: what is $s$? – mrtaurho Feb 2 '19 at 18:09
• @mrtaurho Really I'm interested in the $s\in\Bbb N$ cases, but I would be happy to extend it to $s\in\Bbb C$ if possible, although I doubt that's the case. – clathratus Feb 2 '19 at 19:19
• It can be expressed as a generalized hypergeometric function if $s$ is an integer :$$F(s)={}_{s+2}F_{s+1}\left( 1,1,\ldots,1;\frac{3}{2},2,2,\ldots,2;\frac{1}{4} \right)$$ – Paul Enta Feb 2 '19 at 20:28
• it can be proven that, for $m\geq 0$, integer: \begin{align}\sum_{n=1}^\infty \frac{1}{n^{m+2}\binom{2n}{n}}=\frac{(-1)^m2^m}{m!}\int_0^{\frac{\pi}{3}}\theta\ln^m \left(2\sin\left(\frac{\theta}{2}\right)\right)\,d\theta\end{align} (see: p273, Values of the Riemann zeta function and integrals involving $\ln\left(2\sinh\left(\frac{\theta}{2}\right)\right)$and $\ln\left(2\sin\left(\frac{\theta}{2}\right)\right)$, Zhang Nan-Hue and K.S Williams, Pacific journal of mathematics vol 168,number 2 cf: people.math.carleton.ca/~williams/papers/pdf/197.pdf – FDP Feb 4 '19 at 11:05

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) $$$$\mathrm{Li}_s(z)=z\,_{s+1}F_s\left( 1,1,\ldots,1;2,2,\ldots,2;z \right)$$$$ to obtain $$$$F(s)=\frac{1}{2}\int_0^{1}\left( 1-u \right)^{-1/2}{}_{s+1}F_s\left( 1,1,\ldots,1;2,2,\ldots,2;\frac{u}{4} \right)\,du$$$$ Then, from the tabulated integral (DLMF), $$$${{}_{p+1}F_{q+1}}\left({a_{0},\dots,a_{p}\atop b_{0},\dots,b_{q}};z\right)=% \frac{\Gamma\left(b_{0}\right)}{\Gamma\left(a_{0}\right)\Gamma\left(b_{0}-a_{0% }\right)}\int_{0}^{1}t^{a_{0}-1}(1-t)^{b_{0}-a_{0}-1}{{}_{p}F_{q}}\left({a_{1}% ,\dots,a_{p}\atop b_{1},\dots,b_{q}};zt\right)\mathrm{d}t$$$$ valid for $$\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$$, thus $$$$F(s)={}_{s+2}F_{s+1}\left( 1,1,\ldots,1;\frac{3}{2},2,2,\ldots,2;\frac{1}{4} \right)$$$$
• Nice work (+1). Question: What is this $\left|\mathrm{ph}(1-z)\right|<\pi$ referring to? I have never seen this $\mathrm{ph}$ notation before. – clathratus Feb 2 '19 at 22:39
• Thanks. This condition expresses that the complex number $1-z=\left|1-z\right|\exp(i\phi)$ is such that $-\pi<\phi<\pi$, as this hypergeometric function has a branch point at $z=1$. – Paul Enta Feb 2 '19 at 22:49
• Aah I see... I guess your $\mathrm{ph}(z)$ notation is my $\arg(z)$ notation. Thanks! – clathratus Feb 2 '19 at 22:52