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Some time ago I proved on MSE that the identity $$ \zeta(3)=\frac{5}{2}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\binom{2n}{n}}\tag{1} $$ (crucial for Apery's proof of $\zeta(3)\not\in\mathbb{Q}$) follows from creative telescoping. The proof can be found at page $5$ of my notes, too. Today I realized that $(1)$ is also a consequence of $$ \text{Li}_2\left(\frac{1}{\varphi^2}\right) = \sum_{n\geq 1}\frac{1}{n^2 \varphi^{2n}}=\frac{\pi^2}{15}-\log^2\varphi \tag{2}$$ with $\varphi$ being the golden ratio $\frac{1+\sqrt{5}}{2}$. The proof of such statement, together with some consequences of it, will soon appear at the end of this post. Now my actual question:

Q: Is the implication $(2)\rightarrow(1)$ already known in the literature?

For starters, $(2)$ is a simple consequence of $\varphi^2=\varphi+1$ and the functional identities fulfilled by the dilogarithm function. The same principle easily leads to $$ \text{Li}_3\left(\frac{1}{\varphi^2}\right) = \frac{4}{5}\zeta(3)+\frac{2}{3}\log^3\varphi-\frac{2\pi^2}{15}\log\varphi. \tag{3}$$ Since by partial fraction decomposition we have $$ I\stackrel{\text{def}}{=}\frac{1}{8}\int_{\frac{1}{\varphi^2}}^{1}\frac{(u+1)\log^2(u)}{u(1-u)}\,du = \frac{1}{32}\left[-2\log(u)\log(1-u)+\frac{\log^3(u)}{3}-4\log(u)\text{Li}_2(u)-4\text{Li}_3(u)\right]_{1/\varphi^2}^{1}\tag{4} $$ it follows that $I=\frac{1}{10}\zeta(3)$. If we enforce the substitution $u\mapsto\frac{1-t}{1+t}$ we get: $$ \frac{1}{10}\zeta(3) = \int_{0}^{\frac{1}{\sqrt{5}}}\frac{\text{arctanh}^2(t)}{t(1-t^2)}\,dt\stackrel{t\mapsto \tanh x}{=}\int_{0}^{\log\varphi}x^2\coth(x)\,dx \tag{5}$$ then by enforcing the substitution $x\mapsto\text{arcsinh}(v)$ we get: $$ \frac{1}{10}\zeta(3) = \int_{0}^{1/2}\frac{\text{arcsinh}^2(v)}{v}\,dv\tag{6} $$ and since the Taylor series of $\arcsin^2$ or $\text{arcsinh}^2$ at the origin is well-known, $(1)$ just follows by termwise integration. Since $\coth(z)=\frac{1}{z}+\sum_{n\geq 1}\frac{2z}{z^2+\pi^2 n^2}$, $(5)$ also implies: $$\begin{eqnarray*} \frac{1}{10}\zeta(3)&=&\frac{1}{2}\log^2\varphi+\sum_{n\geq 1}\left[\log^2\varphi-\pi^2 n^2\log\left(1+\frac{\log^2\varphi}{\pi^2 n^2}\right)\right]\\&=&\frac{3}{2}\log^2\varphi-\pi^2\log\left(1+\frac{\log^2\varphi}{\pi^2}\right)+\sum_{m\geq 2}\frac{(-1)^m\log^{2m}(\varphi)}{m \pi^{2m-2}}\left(\zeta(2m-2)-1\right).\tag{7}\end{eqnarray*}$$

Jack D'Aurizio
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  • I can't wait to see what those consequences are :) Maybe we have a closed form for $\zeta(3)$! – Srini Aug 10 '17 at 17:51
  • There are a vast number of dilogarithm and polylogarithm identities known... – Angina Seng Aug 10 '17 at 17:53
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    @LordSharktheUnknown Within that "vast" number of identities, is there one that addresses the question in the OP? – Mark Viola Aug 10 '17 at 18:07
  • There are prepublication notes by Tim Jameson on $\zeta(3)$ and $\zeta(4)$ http://www.maths.lancs.ac.uk/~jameson/polylog.pdf there is also the paper by Yue and Williams which may help you. https://projecteuclid.org/euclid.pjm/1102620561 It would be really interesting if you could work to draw out all the connections in their simplest and clearest form and then publish your work. – James Arathoon Aug 10 '17 at 21:12
  • @JamesArathoon: page $11$ of Jameson's article gives a positive answer to my question. Please turn such comment into an answer, I would be glad to accept it. – Jack D'Aurizio Aug 10 '17 at 21:28
  • Your notes are a treasure trove of useful techniques and interesting results, thank you! One note (though offtopic, here): in exercise 18, there's a small mistake, $\zeta(3)<2\ln\frac32$ is obviously wrong, as the RHS is $<1$. From the preceding identity, it's clear that it should be $\zeta(3)<1+\frac12\ln\frac32=1.202732554054082190989\ldots$, a remarkably sharp estimate. –  Aug 11 '17 at 10:02
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    ah, you've fixed it. thank you for listening! :) –  Aug 11 '17 at 18:29
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    Leonard Lewin, a microwave engineer, who wrote the book "Dilogarithms and Associated Functions" (1958) that helped reignite the modern interest in Polylogarithms, says he was partly inspired as a school child by bizarre integrals involving the Golden Section in the pages of Edward's Calculus. (see e.g. https://projecteuclid.org/download/pdf_1/euclid.bams/1183548696 ). His book was republished in 1981 under the title "Polylogarithms and Associated Functions". He includes many formulae involving the Golden Section in his books. – James Arathoon Aug 13 '17 at 08:47
  • Incidentally Leonard Lewin worked at the STL Labs in Harlow, Essex, UK where they first invented fibre optics; allowing us to communicate quickly and cheaply over the internet today. – James Arathoon Aug 13 '17 at 08:48

3 Answers3

3

There is something about the results that reminds me of work done on Euler double sums by Borwein, Boersma and another author whose name evades me. There was also some work by Peter Jordan on infinite sums of Psi functions that had application to wing theory.

I apologize if that sends you off on the wrong path!

Lysistrata
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  • Any pointer is gladly accepted, don't worry. Do you have some explicit reference (URLs or articles I can find)? – Jack D'Aurizio Aug 10 '17 at 18:29
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    Borwein, Zucker and Boersma The evaluation of character Euler double Sums http://docserver.carma.newcastle.edu.au/255/3/bbz7.pdf

    That paper contains a reference to Peter Jordan's work.

     – Lysistrata Aug 10 '17 at 18:53
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Lengthy comment which may be of some help to you:

The simplest integrals I could find (repeatedly transforming by substitution from the basic $\coth(x)$ integral you give above) for $\zeta(2)$ and $\zeta(3)$ involving the Golden Ratio $\phi$ as one of the limits were

$$\zeta(2)=\frac{10}{3} \int_1^{\phi} \log(x) \left( \frac{1}{x-1} +\frac{1}{x+1}-\frac{1}{x} \right) dx$$

$$\zeta(3)={10} \int_1^{\phi} \log(x)^2 \left( \frac{1}{x-1} +\frac{1}{x+1}-\frac{1}{x} \right) dx$$

from the partial fraction expansion $$\frac{1}{x}\frac{(x^2+1)}{(x^2-1)}=\left( \frac{1}{x-1} +\frac{1}{x+1}-\frac{1}{x} \right)$$

These are not a million miles away from the standard integrals used to calculate to di- and tri-logarithms.

James Arathoon
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2

Comment becomes an answer

There are prepublication notes by Tim Jameson on $\zeta(3)$ and $\zeta(4)$ http://www.maths.lancs.ac.uk/~jameson/polylog.pdf

There is also the paper by Yue and Williams which may help you. https://projecteuclid.org/euclid.pjm/1102620561

It would be really interesting if you could work to draw out all the connections in their simplest and clearest form and then publish your work.

James Arathoon
  • 3,659