Relations between Dilogarithms and The Golden Ratio

Today I came across the following relation in this Twitter post by @infseriesbot. We will show today how to proof this Interesting relations.

$\begin{aligned} \operatorname{Li}_{2}\left(\frac{1}{2}\right) &=\frac{\pi^{2}}{12}-\frac{\ln ^{2} 2}{2} \\ \mathrm{Li}_{2}\left(\frac{1}{\phi}\right) &=\frac{\pi^{2}}{10}-\ln ^{2} \phi \\ \mathrm{Li}_{2}\left(-\frac{1}{\phi}\right) &=\frac{\ln ^{2} \phi}{2}-\frac{\pi^{2}}{15} \\ \mathrm{Li}_{2}\left(\frac{1}{\phi^{2}}\right) &=\frac{\pi^{2}}{15}-\ln ^{2} \phi \end{aligned}$

Consider the following three Dilogarithm relations

$Li_{2}\left(x\right)+Li_{2}\left(-x\right)=\frac{Li_2\left(x^2\right)}{2}$ (1)

$Li_{2}\left(x\right)+Li_{2}\left(1-x\right)=\zeta(2)-\log\left(x\right)\log\left(1-x\right)$ (2)

$Li_{2}\left(1-x\right)+Li_{2}\left(1-\frac{1}{x}\right)=-\frac{\log^2\left(x\right)}{2}$ (3)

(1) is proved in the end of the post (equation (A.5)). The proof for (2) and (3) can be found in this previous post.

The first relation can be proved automatically just by letting $x=\frac{1}{2}$ in (2)

$Li_{2}\left(\frac{1}{2}\right)+Li_{2}\left(\frac{1}{2}\right)=\zeta(2)-\log\left(\frac{1}{2}\right)\log\left(\frac{1}{2}\right)$

$\boxed{\operatorname{Li}_{2}\left(\frac{1}{2}\right) &=\frac{\pi^{2}}{12}-\frac{\ln ^{2} 2}{2}}$

For the other three relations, recall the Golden ratio

$\phi=\frac{1+\sqrt{5}}{2}$

it´s easy to verify that

$\phi=\frac{1}{\phi}+1 , \qquad\phi^2=\phi+1 ,$

$\phi^{-2}+\phi^{-1}=1 \, \Rightarrow \frac{1}{\phi^2}=1-\frac{1}{\phi}$

letting $x=\frac{1}{\phi}$ in (1), (2) and (3) we obtain

$Li_{2}\left(\frac{1}{\phi}\right)+Li_{2}\left(-\frac{1}{\phi}\right)=\frac{Li_2\left(\frac{1}{\phi^2}\right)}{2}$ (4)

$Li_{2}\left(\frac{1}{\phi}\right)+Li_{2}\left(\frac{1}{\phi^2}\right)=\zeta(2)-2\log^2\left(\phi\right)$ (5)

$Li_{2}\left(\phi^{-2}\right)+Li_{2}\left(-\phi^{-1}\right)=-\frac{\log^2\left(\phi\right)}{2}$ (6)

Subtracting (6) from (4) we get

$Li_{2}\left(-\phi^{-1}\right)-\frac{3}{2}Li_{2}\left(\phi^{-2}\right)=\frac{\log^2\left(\phi\right)}{2}$ (7)

Now Subtracting (7) - (5) gives

$\frac{5}{2}Li_{2}\left(\phi^{-2}\right)=\zeta(2)-\frac{5}{2}\log^2\left(\phi\right)$

$Li_{2}\left(\phi^{-2}\right)=\frac{2}{5}\zeta(2)-\log^2\left(\phi\right)$

which gives us

$\boxed{Li_{2}\left(\frac{1}{\phi^{2}}\right)=\frac{\pi^2}{15}-\log^2\left(\phi\right)}$ (8)

Plugging (8) in (5) gives

$Li_{2}\left(\frac{1}{\phi}\right)+\frac{2}{5}\zeta(2)-\log^2\left(\phi\right)=\zeta(2)-2\log^2\left(\phi\right)$

$Li_{2}\left(\frac{1}{\phi}\right)=\frac{3}{5}\zeta(2)-\log^2\left(\phi\right)$

$\boxed{Li_{2}\left(\frac{1}{\phi}\right)=\frac{\pi^2}{10}-\log^2\left(\phi\right)}$ (9)

Plugging (8) in (6) gives

$\boxed{Li_{2}\left(-\frac{1}{\phi}\right)=\frac{\log^2\left(\phi\right)}{2}-\frac{\pi^2}{15}}$ (10)

Appendix

Recall the following representation of the Gamma function

$\frac{\Gamma(x)}{s^x}=\int_{0}^{\infty}e^{-st}t^{x-1}dt$

let $t=-\ln u$

$\frac{\Gamma(x)}{s^x}=\int_{1}^{0}e^{s \ln(u)}(-\ln(u))^{x-1}\frac{(-du)}{u}$

$\frac{\Gamma(x)}{s^x}=(-1)^{x-1}\int_{0}^{1}u^{s-1}\ln^{x-1}(u)du$

$\boxed{\frac{1}{s^x}=\frac{(-1)^{x-1}}{\Gamma(x)}\int_{0}^{1}u^{s-1}\ln^{x-1}(u)du}$ (A.1)

Let s=k and x=n

$\frac{1}{k^n}=\frac{(-1)^{n-1}}{\Gamma(n)}\int_{0}^{1}u^{k-1}\ln^{n-1}(u)du$ (A.2)

now consider the definition of polylogarithm function

$Li_{n}(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^n}$ (A.3)

Plugging (A.2) in (A.3) we get

$Li_{n}(x)=\sum_{k=1}^{\infty}x^k\frac{(-1)^{n-1}}{\Gamma(n)}\int_{0}^{1}u^{k-1}\ln^{n-1}(u)du$

$Li_{n}(x)=\frac{(-1)^{n-1}}{\Gamma(n)}\int_{0}^{1}\frac{\ln^{n-1}(u)}{u}\sum_{k=1}^{\infty}(ux)^kdu$

$Li_{n}(x)=\frac{(-1)^{n-1}}{\Gamma(n)}\int_{0}^{1}\frac{\ln^{n-1}(u)}{u}\cdot \frac{ux}{1-ux}du$

$\boxed{Li_{n}(x)=\frac{(-1)^{n-1}}{\Gamma(n)}\int_{0}^{1}\frac{x\ln^{n-1}(u)}{1-ux} du}$ (A.4)

Now consider the folowing

$Li_{n}(x)+Li_{n}(-x)=\frac{(-1)^{n-1}}{\Gamma(n)}\int_{0}^{1}\ln^{n-1}(u)\left\{\frac{x}{1-ux}-\frac{x}{1+ux}\right\} du$

by (A.4)

$Li_{n}(x)+Li_{n}(-x)=\frac{(-1)^{n-1}}{\Gamma(n)}\int_{0}^{1}\ln^{n-1}(u)\left\{\frac{2ux^2}{1-u^2x^2}\right\} du$

Now let $u^2=y \, \Rightarrow \, 2udu=dy$

$Li_{n}(x)+Li_{n}(-x)=\frac{(-1)^{n-1}}{\Gamma(n)}\int_{0}^{1}\ln^{n-1}(y^{1/2})\left\{\frac{yx^2}{1-yx^2}\right\} dy$

$Li_{n}(x)+Li_{n}(-x)=2^{1-n}\underbrace{\frac{(-1)^{n-1}}{\Gamma(n)}\int_{0}^{1}\frac{x^2\ln^{n-1}(y)}{1-yx^2} dy}__{=Li_n(x^2) \,\,\, \text{by \,\,(4)}}$

$\boxed{Li_{n}(x)+Li_{n}(-x)=2^{1-n}Li_n(x^2)}$ (A.5)

Reference

POLYLOGARITHMS, MULTIPLE ZETA VALUES, AND THE SERIES OF HJORTNAES AND COMTET Notes by Tim Jameson (December 2009)

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