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](https://drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view), 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?
> 
> 
> 

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For starters, $(2)$ is a simple consequence of $\varphi^2=\varphi+1$ and the functional identities fulfilled by the [dilogarithm](http://mathworld.wolfram.com/Dilogarithm.html) 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](https://math.stackexchange.com/questions/878477/a-closed-form-of-sum-k-0-infty-frac-1k1k-gamma2-left-frack2), $(1)$ 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:
$$\begin{eqnarray\*} \frac{1}{10}\zeta(3)&=&\frac{1}{2}\log^2\varphi+\sum\_{n\geq 1}\left[\log^2\varphi-\pi^2 n^2\log\left(1+\frac{\log^2\varphi}{\pi^2 n^2}\right)\right]\\&=&\frac{3}{2}\log^2\varphi-\pi^2\log\left(1+\frac{\log^2\varphi}{\pi^2}\right)+\sum\_{m\geq 2}\frac{(-1)^m\log^{2m}(\varphi)}{m \pi^{2m-2}}\left(\zeta(2m-2)-1\right).\tag{7}\end{eqnarray\*}$$