Dilogarithm of Minus Z Plus Dilogarithm of Minus Reciprocal of Z

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Theorem

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

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:

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

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

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


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

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

We now solve for $C$ by setting $z = 1$

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

Therefore:

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

$\blacksquare$


Sources

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