Euler's work IV

The strong connection between the $\operatorname{Li}_{2}$ function and $\phi$

Today we show the proof of this result posted by @seriesbot_q \[ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2\binom{2n}{n}} =\frac{\pi^2}{6}-3\ln^2(\phi)\]
Not surprisingly, this series is related to the work of Euler, specially his work on the properties of the $\arctan$ and his work on the beautiful properties that relates the dilogarithm function with the golden ratio.

Proof

In a previous post we showed the proof of the following series expansion:
\[\frac{\arcsin(x)}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{2^{2n}}{\binom{2n}{n}(2n+1)}x^{2n+1} = \sum_{n=0}^{\infty} \frac{2^{2n} n!^2}{(2n+1)!}x^{2n+1}\]

This series expansion is a consequence of the Euler's formula for $\arctan x$.

If we allow x to take complex values we can transform this series in another very useful:

Let $\displaystyle x= -iv$ and recall that $\displaystyle i\arcsin (-ia) = \operatorname{arcsinh}(a)$

Hence \[\frac{\arcsin(-iv)}{\sqrt{1+v^2}} = -i\sum_{n=0}^{\infty} \frac{2^{2n}}{\binom{2n}{n}(2n+1)}(-1)^nv^{2n+1}\] Then, multiplying by $i$ both sides \[\frac{i\arcsin(-iv)}{\sqrt{1+v^2}} = \sum_{n=0}^{\infty} \frac{2^{2n}}{\binom{2n}{n}(2n+1)}(-1)^nv^{2n+1}\] Therefore \[\frac{\operatorname{arcsinh}(v)}{\sqrt{1+v^2}} = \sum_{n=0}^{\infty} \frac{2^{2n}}{\binom{2n}{n}(2n+1)}(-1)^nv^{2n+1}\] If we multiply by $\displaystyle \frac{1}{v}$ \[\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}} = \sum_{n=0}^{\infty} \frac{2^{2n}}{\binom{2n}{n}(2n+1)}(-1)^nv^{2n}\] Integrating from $0$ to $x$ \[\int_{0}^{x}\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}}dv = \sum_{n=0}^{\infty} \frac{2^{2n}}{\binom{2n}{n}(2n+1)^2}(-1)^nx^{2n+1}\] If we let $\displaystyle x = \frac{1}{2}$ \[\int_{0}^{\frac{1}{2}}\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}}dv = \frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^n}{\binom{2n}{n}(2n+1)^2}\] \[\Longrightarrow 2\int_{0}^{\frac{1}{2}}\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}}dv = \sum_{n=0}^{\infty} \frac{(-1)^n}{\binom{2n}{n}(2n+1)^2}\]
The series in the right hand side is the series that we want in a closed-expression form. Therefore we have to find the value of the integral in the left hand side:

Recall that
\[ \operatorname{arcsinh}(a) = \ln\left(a+\sqrt{1+a^2}\right)\] Therefore, if we denote $\displaystyle \phi = \frac{1+\sqrt{5}}{2}$ the golden ratio: \begin{align*} I = 2\int_{0}^{\frac{1}{2}}\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}}dv =& 2\int_{0}^{\frac{1}{2}}\frac{\ln\left(v+\sqrt{1+v^2}\right)}{v\sqrt{1+v^2}}dv\\ =& 4\int_{1}^{\phi }\frac{\ln(w)}{(w+1)(w-1)}dw \quad \left(w \mapsto v+\sqrt{1+v^2} \right)\\ =& \underbrace{2\int_{1}^{\phi} \frac{\ln(w)}{w-1}dw}_{I_{1}} -\underbrace{2\int_{1}^{\phi} \frac{\ln(w)}{w+1}dw}_{I_{2}} \end{align*} Recall the definition of the dilogarithm function: \[ \operatorname{Li}_{2}(z) = \int_{z}^{0} \frac{\ln(1-t)}{t} dt\] Hence \begin{align*} I_{1}= 2\int_{1}^{\phi} \frac{\ln(w)}{w-1}dw =& 2\int_{0}^{\phi-1} \frac{\ln(1+s)}{s}ds \quad (s \mapsto w-1)\\ =& -2\int_{1-\phi}^{0} \frac{\ln(1-r)}{r}ds \quad (s \mapsto -w)\\ =&-2\operatorname{Li}_{2}(1-\phi)\\ =&-2\operatorname{Li}_{2}(-\phi^{-1}) \quad (1-\phi = -\phi^{-1}) \end{align*} For $I_{2}$: \begin{align*} I_{2}= 2\int_{1}^{\phi} \frac{\ln(w)}{w+1}dw \stackrel{IBP}{=} &2\ln(w)\ln(w+1)\Big|_{1}^{\phi} - 2\int_{1}^{\phi}\frac{\ln(w+1)}{w}dw \\ =& 2\ln(\phi)\ln(\phi+1) + 2\int_{-\phi}^{-1}\frac{\ln(1-s)}{s}ds \quad (s\mapsto -w)\\ =& 2\ln(\phi)\ln(\phi+1) + 2\int_{-\phi}^{0}\frac{\ln(1-s)}{s}ds - 2\int_{-1}^{0}\frac{\ln(1-s)}{s}ds\\ =& 2\ln(\phi)\ln(\phi+1) + 2\operatorname{Li}_{2}(-\phi) - 2\operatorname{Li}_{2}(-1)\\ \end{align*} Therefore \[I= 2\int_{0}^{\frac{1}{2}}\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}}dv = -2\operatorname{Li}_{2}(-\phi^{-1}) - 2\ln(\phi)\ln(\phi+1) - 2\operatorname{Li}_{2}(-\phi) + 2\operatorname{Li}_{2}(-1) \]
A famous result for the dilogarithm function states that $\displaystyle \operatorname{Li}_{2}(-1)=-\frac{1}{12}\pi^2$
Now, this is the interesting part: it turned out that the relationship between the dilogarithm function and the golden rate is very strong. Euler proved the following identities that relate both concepts (I will add the proof of this results in an appendix later): \[\operatorname{Li}_{2}(-\phi^{-1}) = -\frac{1}{15}\pi^2 + \frac{1}{2}\ln^2(\phi)\] \[\operatorname{Li}_{2}(-\phi) -\frac{1}{10}\pi^2 -\ln^2(\phi)\] Here you can find more of this identities. Additionally, we can make the following transformation: \begin{align*} \ln(\phi)\ln(1+\phi) =& \ln(\phi)\ln\left(\phi(1+\phi^{-1})\right)\\ =& \ln^2(\phi)+ \ln(\phi)\phi(1+\phi^{-1}) \end{align*} where \[ 1+\phi^{-1} = 1+\frac{2}{1+\sqrt{5}}\frac{1-\sqrt{5}}{1-\sqrt{5}}=1 - \frac{1-\sqrt{5}}{2} = \frac{1+\sqrt{5}}{2} = \phi\] Hence \[ \ln(\phi)\ln(1+\phi) = \ln^2(\phi)+ \ln(\phi)\ln(1+\phi^{-1}) = 2\ln^2(\phi) \] Putting all the ingredients togheter: \begin{align*} I= 2\int_{0}^{\frac{1}{2}}\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}}dv =& \underbrace{-2\operatorname{Li}_{2}(-\phi^{-1})}_{\frac{2}{15}\pi^2-\ln^2(\phi)} \underbrace{-2\ln(\phi)\ln(\phi+1)}_{-4\ln^2(\phi)} \underbrace{-2\operatorname{Li}_{2}(-\phi)}_{\frac{2}{10}\pi^2+2\ln^2(\phi)} + \underbrace{2\operatorname{Li}_{2}(-1)}_{-\frac{\pi^2}{6}}\\ =& \frac{\pi^2}{6}-3\ln^2(\phi) \end{align*} Then we can conclude: \[ \boxed{\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2\binom{2n}{n}} =\frac{\pi^2}{6}-3\ln^2(\phi)}\]