Tag Archives: dilogarithm

Square difference formula for polylgoarithm proof

$$\mathrm{Li}_{\,n}(-z) + \mathrm{Li}_{\,n}(z) = 2^{1-n} \,\mathrm{Li}_{\,n}(z^2) $$ $$\textit{proof}$$ As usual we write the series representation of the LHS $$\sum_{k=1}^\infty \frac{z^k}{k^n}+\sum_{k=1}^\infty \frac{(-z)^k}{k^n}$$ Listing the first few terms $$z+\frac{z^2}{2^n}+\frac{z^3}{3^n}+\cdots +\left(-z+\frac{z^2}{2^n}-\frac{z^3}{3^n}+\cdots \right) $$ The odd terms will cancel $$2\frac{z^2}{2^n}+2\frac{z^4}{4^n}+2\frac{z^6}{6^n}+\cdots $$ Take \( 2^{1-n} … Continue reading

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Dilogarithm functional equation proof

$$\mathrm{Li}_2(z) + \mathrm{Li}_{2}(1-z) = \frac{\pi^2}{6}-\log(z) \log(1-z) \,\,\,\, ,\,0<z<1$$ $$\textit{proof}$$ Start by the following $$\mathrm{Li}_2\left(z\right) = -\int^{z}_0 \frac{\log(1-t)}{t} \, dt $$ Now integrate by parts to obtain $$\mathrm{Li}_2\left(z\right)= -\int^z_0 \frac{\log(t)}{1-t} \, dt -\log(z) \log(1-z) $$ By the change of variable \(t=1-x … Continue reading

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