Dilogarithm of One Half

Theorem

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

where:

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

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

Proof

\(\ds \map {\Li_2} z + \map {\Li_2} {1 - z}\)

\(=\)

\(\ds \map \zeta 2 - \map \ln z \map \ln {1 - z}\)

Dilogarithm Reflection Formula

\(\ds \leadsto \ \ \)

\(\ds \map {\Li_2} {\frac 1 2} + \map {\Li_2} {\frac 1 2}\)

\(=\)

\(\ds \map \zeta 2 - \map \ln {\frac 1 2} \map \ln {\frac 1 2}\)

$z := \dfrac 1 2$

\(\ds \leadsto \ \ \)

\(\ds 2 \map {\Li_2} {\frac 1 2}\)

\(=\)

\(\ds \map \zeta 2 - \paren {\map \ln {\frac 1 2} }^2\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 2 \paren {\map \zeta 2 - \paren {-\map \ln 2}^2}\)

Logarithm of Reciprocal and dividing by $2$

\(\ds \)

\(=\)

\(\ds \dfrac 1 2 \paren {\map \zeta 2 - \paren {\map \ln 2}^2}\)

$\blacksquare$