Dilogarithm of One Minus Z Plus Dilogarithm of One Minus Reciprocal of Z

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Theorem

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

where:

$\map {\Li_2} z$ is the Dilogarithm function of $z$
$z \in \C$ and $z < 1$.


Proof

From the definition of the dilogarithm function:

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

Taking the derivative of both sides at $-\dfrac z {1 - z}$

\(\ds \frac {\d } {\d z} \map {\Li_2} {-\dfrac z {1 - z} }\) \(=\) \(\ds -\paren {\dfrac {\map \ln {1 - \paren {-\dfrac z {1 - z} } } } {\paren {-\dfrac z {1 - z} } } \paren {-\dfrac 1 {\paren {1 - z}^2} } }\) $x \to -\dfrac z {1 - z}$ and $\rd x \to -\dfrac 1 {\paren {1 - z}^2}$
\(\ds \) \(=\) \(\ds \dfrac {-\map \ln {\dfrac {1 - z} {1 -z} + \dfrac z {1 -z} } } {z \paren {1 - z} }\)
\(\ds \) \(=\) \(\ds \dfrac {-\map \ln {\dfrac 1 {1 -z} } } {z \paren {1 - z} }\)
\(\ds \) \(=\) \(\ds \dfrac {-\paren {\map \ln 1 - \map \ln {1 - z} } } {z \paren {1 - z} }\) Difference of Logarithms
\(\ds \) \(=\) \(\ds \dfrac {-\map \ln 1 } {z \paren {1 - z} } + \dfrac {\map \ln {1 - z} } {z \paren {1 - z} }\)
\(\ds \) \(=\) \(\ds \dfrac {\map \ln {1 - z} } {z \paren {1 - z} }\) Natural Logarithm of 1 is 0
\(\ds \) \(=\) \(\ds \dfrac {\map \ln {1 - z} } {\paren {1 - z} } + \dfrac {\map \ln {1 - z} } z\)


Now integrating both sides with respect to $z$, we obtain:

\(\ds \int_0^z \frac \d {\d z} \map {\Li_2} {-\dfrac z {1 - z} }\) \(=\) \(\ds \int_0^z \paren {\dfrac {\map \ln {1 - z} } {\paren {1 - z} } + \dfrac {\map \ln {1 - z} } z} \rd z\)
\(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {-\dfrac z {1 - z} }\) \(=\) \(\ds \int_0^z \dfrac {\map \ln {1 - z} } {\paren {1 - z} } \rd z + \int_0^z \dfrac {\map \ln {1 - z} } z \rd z\) Fundamental Theorem of Calculus and Linear Combination of Definite Integrals
\(\ds \) \(=\) \(\ds \int_0^z \dfrac {\map \ln {1 - z} } {\paren {1 - z} } \rd z - \map {\Li_2} z\) Definition of Dilogarithm Function
\(\ds \leadsto \ \ \) \(\ds \map {\Li_2} z + \map {\Li_2} {-\dfrac z {1 - z} }\) \(=\) \(\ds \int_0^z \dfrac {\map \ln {1 - z} } {\paren {1 - z} } \rd z\)
\(\ds \) \(=\) \(\ds \int_1^{1 - z} \dfrac {\map \ln u } u \paren {-\rd u}\) $\paren {1 - z} \to u$ and $\rd z \to -\rd u$ U-Substitution
\(\ds \) \(=\) \(\ds -\frac 1 2 \map {\ln^2} {1 - z} + C\) Primitive of Logarithm of x over x

We now solve for $C$ by setting $z = 0$

\(\ds \map {\Li_2} 0 + \map {\Li_2} 0\) \(=\) \(\ds -\frac 1 2 \ln^2 1 + C\)
\(\ds \leadsto \ \ \) \(\ds 0\) \(=\) \(\ds 0 + C\) Dilogarithm of Zero and Natural Logarithm of 1 is 0
\(\ds \leadsto \ \ \) \(\ds 0\) \(=\) \(\ds C\)

Therefore:

\(\ds \map {\Li_2} z + \map {\Li_2} {-\dfrac z {1 - z} }\) \(=\) \(\ds -\frac 1 2 \map {\ln^2} {1 - z}\)
\(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {1 - z} + \map {\Li_2} {1 - \dfrac 1 z}\) \(=\) \(\ds -\frac 1 2 \map {\ln^2} z\) $z \to \paren {1 - z}$

$\blacksquare$


Sources

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