Dilogarithm of One Minus Reciprocal of Golden Mean

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Theorem

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


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.


Proof

We first note the following:

\(\text {(1)}: \quad\) \(\ds -\frac 1 \phi\) \(=\) \(\ds 1 - \phi\) Reciprocal Form of One Minus Golden Mean
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \frac 1 {\phi^2}\) \(=\) \(\ds 1 - \dfrac 1 \phi\) dividing through by $-\phi$ and rearranging


We now note:

\(\ds \map {\Li_2} z + \map {\Li_2} {-z}\) \(=\) \(\ds \dfrac 1 2 \map {\Li_2} {z^2}\) Dilogarithm of Square
\(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {\frac 1 \phi} + \map {\Li_2} {-\frac 1 \phi}\) \(=\) \(\ds \dfrac 1 2 \map {\Li_2} {\frac 1 {\phi^2} }\) setting $z := \dfrac 1 \phi$
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {\frac 1 \phi} + \map {\Li_2} {1 - \phi} - \dfrac 1 2 \map {\Li_2} {1 - \dfrac 1 \phi}\) \(=\) \(\ds 0\) substituting from $(1)$ and $(2)$ above
\(\ds \map {\Li_2} {1 - z} + \map {\Li_2} {1 - \dfrac 1 z}\) \(=\) \(\ds -\dfrac 1 2 \map {\ln^2} z\) Dilogarithm of One Minus Z Plus Dilogarithm of One Minus Reciprocal of 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 $z := \dfrac 1 \phi$
\(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {1 - \frac 1 \phi} + \map {\Li_2} {1 - \phi}\) \(=\) \(\ds -\dfrac 1 2 \paren {\map \ln 1 - \map \ln \phi}^2\) Difference of Logarithms
\(\text {(4)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {1 - \frac 1 \phi} + \map {\Li_2} {1 - \phi}\) \(=\) \(\ds -\dfrac 1 2 \map {\ln^2} \phi\) Natural Logarithm of 1 is 0
\(\text {(5)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {\frac 1 \phi} - \dfrac 3 2 \map {\Li_2} {1 - \dfrac 1 \phi}\) \(=\) \(\ds \dfrac 1 2 \map {\ln^2} \phi\) $(3) - (4)$
\(\ds \map {\Li_2} z + \map {\Li_2} {1 - z}\) \(=\) \(\ds \map \zeta 2 - \map \ln z \map \ln {1 - z}\) Dilogarithm Reflection Formula
\(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {\frac 1 \phi} + \map {\Li_2} {1 - \frac 1 \phi}\) \(=\) \(\ds \map \zeta 2 - \map \ln {\frac 1 \phi} \map \ln {1 - \frac 1 \phi}\) $z := \dfrac 1 \phi$
\(\ds \) \(=\) \(\ds \map \zeta 2 - \map \ln {\frac 1 \phi} \map \ln {\frac 1 {\phi^2} }\)
\(\ds \) \(=\) \(\ds \map \zeta 2 - \paren {\map \ln 1 - \map \ln \phi} \paren {\map \ln 1 - \map \ln {\phi^2} }\) Difference of Logarithms
\(\ds \) \(=\) \(\ds \map \zeta 2 + \map \ln \phi \paren {-2 \map \ln \phi}\) Natural Logarithm of 1 is 0
\(\ds \) \(=\) \(\ds \map \zeta 2 - 2 \paren {\map \ln \phi}^2\)
\(\text {(6)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {\frac 1 \phi} + \map {\Li_2} {1 - \frac 1 \phi}\) \(=\) \(\ds \map \zeta 2 - 2 \paren {\map \ln \phi}^2\)
\(\ds \leadsto \ \ \) \(\ds -\dfrac 5 2 \map {\Li_2} {1 - \dfrac 1 \phi}\) \(=\) \(\ds \dfrac 1 2 \map {\ln^2} \phi - \paren {\map \zeta 2 - 2 \paren {\map \ln \phi}^2}\) $(5) - (6)$
\(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {1 - \dfrac 1 \phi}\) \(=\) \(\ds \dfrac 2 5 \map \zeta 2 - \paren {\map \ln \phi}^2\) multiplying through by $-\dfrac 2 5$ and rearranging

$\blacksquare$


Sources

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