Dilogarithm of Minus Reciprocal of Golden Mean

Theorem

$\map {\Li_2} {-\dfrac 1 \phi} = -\dfrac 2 5 \map \zeta 2 + \dfrac 1 2 \paren {\map \ln \phi}^2$

where:

$\map {\Li_2} x$ is the dilogarithm function of $x$

$\map \zeta 2$ is the Riemann $\zeta$ function of $2$

$\phi$ denotes the golden mean.

We now note:

$\ds \map {\Li_2} {1 - z} + \map {\Li_2} {1 - \dfrac 1 z}$

$=$

$\ds -\dfrac 1 2 \map {\ln^2} z$

$\ds \leadsto \ \ $

$\ds \map {\Li_2} {1 - \frac 1 \phi} + \map {\Li_2} {1 - \dfrac 1 {\frac 1 \phi} }$

$=$

$\ds -\dfrac 1 2 \map {\ln^2} {\frac 1 \phi}$

setting $z := \dfrac 1 \phi$

$\ds \leadsto \ \ $

$\ds \map {\Li_2} {1 - \frac 1 \phi} + \map {\Li_2} {1 - \phi}$

$=$

$\ds -\dfrac 1 2 \paren {\map \ln 1 - \map \ln \phi}^2$

$\ds \leadsto \ \ $

$\ds \map {\Li_2} {1 - \frac 1 \phi} + \map {\Li_2} {-\frac 1 \phi}$

$=$

$\ds -\dfrac 1 2 \map {\ln^2} \phi$

$\ds \leadsto \ \ $

$\ds \paren {\dfrac 2 5 \map \zeta 2 - \paren {\map \ln \phi}^2} + \map {\Li_2} {-\frac 1 \phi}$

$=$

$\ds -\dfrac 1 2 \map {\ln^2} \phi$

$\ds \leadsto \ \ $

$\ds \map {\Li_2} {-\dfrac 1 \phi}$

$=$

$\ds -\dfrac 2 5 \map \zeta 2 + \dfrac 1 2 \paren {\map \ln \phi}^2$

rearranging

$\blacksquare$