A very odd-looking statement about $\zeta(3)$ and $\text{Li}_2\left(\frac{1}{\varphi^2}\right)$ 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:
$$\begin{eqnarray*} \frac{1}{10}\zeta(3)&=&\frac{1}{2}\log^2\varphi+\sum_{n\geq 1}\left[\log^2\varphi-\pi^2 n^2\log\left(1+\frac{\log^2\varphi}{\pi^2 n^2}\right)\right]\\&=&\frac{3}{2}\log^2\varphi-\pi^2\log\left(1+\frac{\log^2\varphi}{\pi^2}\right)+\sum_{m\geq 2}\frac{(-1)^m\log^{2m}(\varphi)}{m \pi^{2m-2}}\left(\zeta(2m-2)-1\right).\tag{7}\end{eqnarray*}$$
 A: 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!
A: 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.
A: 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.
