Dilogarithm of Square
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Theorem
- $\map {\Li_2} z + \map {\Li_2} {-z} = \dfrac 1 2 \map {\Li_2} {z^2}$
where $\Li_2$ denotes the dilogarithm function.
Proof
| \(\ds \map {\Li_2} z + \map {\Li_2} {-z}\) | \(=\) | \(\ds -\paren {\int_0^z \frac {\map \ln {1 - t} } t \rd t + \int_0^z \frac {\map \ln {1 + t} } t \rd t}\) | Definition of Dilogarithm Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\int_0^z \frac {\map \ln {\paren {1 - t} \paren {1 + t} } } t \rd t\) | Linear Combination of Definite Integrals, Sum of Logarithms | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\int_0^z \frac {\map \ln {1 - t^2} } t \rd t\) | Difference of Two Squares | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\int_0^{z^2} \frac {\map \ln {1 - t^2} } t \frac {\map \d {t^2} } {2 t}\) | substituting $t \to t^2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac 1 2 \int_0^{z^2} \frac {\map \ln {1 - t^2} } {t^2} \map \rd {t^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2 \map {\Li_2} {z^2}\) | Definition of Dilogarithm Function |
$\blacksquare$