Dilogarithm Reflection Formula

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Theorem

$\map {\Li_2} z + \map {\Li_2} {1 - z} = \map \zeta 2 - \map \ln z \map \ln {1 - z}$

where:

$\map {\Li_2} z$ is the Dilogarithm function of $z$
$\map \zeta 2$ is the Riemann $\zeta$ function of $2$.


Proof

From the definition of the dilogarithm function:

$\ds \map {\Li_2} z = -\int_0^z \dfrac {\map \ln {1 - x} } x \rd x$

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\ds u\) \(=\) \(\ds \map \ln {1 - x}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds -\frac 1 {1 - x}\) Derivative of $\ln x$, Chain Rule


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \frac 1 x\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \map \ln x\) Primitive of Reciprocal


Then:

\(\ds -\int_0^z \dfrac {\map \ln {1 - x} } x \rd x\) \(=\) \(\ds -\paren {\bigintlimits {\map \ln {1 - x} \map \ln x} 0 z + \int_0^z \frac {\map \ln x} {1 - x} \rd x}\) Integration by Parts
\(\ds \) \(=\) \(\ds -\map \ln z \map \ln {1 - z} - \int_0^z \frac {\map \ln x} {1 - x} \rd x\) simplifying
\(\ds \) \(=\) \(\ds -\map \ln z \map \ln {1 - z} + \int_1^{1 - z} \dfrac {\map \ln {1 - t} } t \rd t\) $\paren {1 - x} \to t$ and $\rd x \to -\rd t$
\(\ds \) \(=\) \(\ds -\map \ln z \map \ln {1 - z} - \paren {\map {\Li_2} {1 - z} - \map {\Li_2} 1}\) Definition of Dilogarithm Function
\(\ds \) \(=\) \(\ds -\map \ln z \map \ln {1 - z} - \map {\Li_2} {1 - z} + \map {\Li_2} 1\)
\(\ds \leadsto \ \ \) \(\ds \map {\Li_2} z + \map {\Li_2} {1 - z}\) \(=\) \(\ds -\map \ln z \map \ln {1 - z} - \map {\Li_2} {1 - z} + \map {\Li_2} 1 + \map {\Li_2} {1 - z}\)
\(\ds \) \(=\) \(\ds \map \zeta 2 - \map \ln z \map \ln {1 - z}\) Dilogarithm of One

$\blacksquare$


Sources

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