7
$\begingroup$

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.

$\endgroup$
7
  • 1
    $\begingroup$ Some references here may help... $\endgroup$ – Raymond Manzoni Feb 2 '19 at 18:04
  • 1
    $\begingroup$ 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 $s$? $\endgroup$ – mrtaurho Feb 2 '19 at 18:09
  • $\begingroup$ @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. $\endgroup$ – clathratus Feb 2 '19 at 19:19
  • 2
    $\begingroup$ It can be expressed as a generalized hypergeometric function if $s$ is an integer :$$ F(s)={}_{s+2}F_{s+1}\left( 1,1,\ldots,1;\frac{3}{2},2,2,\ldots,2;\frac{1}{4} \right)$$ $\endgroup$ – Paul Enta Feb 2 '19 at 20:28
  • 3
    $\begingroup$ 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 $\endgroup$ – FDP Feb 4 '19 at 11:05
8
$\begingroup$

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}

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ – clathratus Feb 2 '19 at 22:39
  • 1
    $\begingroup$ 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 $z=1$. $\endgroup$ – Paul Enta Feb 2 '19 at 22:49
  • $\begingroup$ Aah I see... I guess your $\mathrm{ph}(z)$ notation is my $\arg(z)$ notation. Thanks! $\endgroup$ – clathratus Feb 2 '19 at 22:52
  • $\begingroup$ Yes! you are right $\endgroup$ – Paul Enta Feb 2 '19 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.