# Dilogarithm difference formula proof

$$\mathrm{Li}_2(z) + \mathrm{Li}_2 \left(\frac{z}{z-1} \right) = – \frac{1}{2} \log^2 (1-z) \,\,\,\, \, z<1$$

$$\textit{proof}$$

Start by the following

$$\mathrm{Li}_2 \left(\frac{z}{z-1} \right) = -\int^{\frac{z}{z-1}}_0 \frac{ \log(1-t)}{t}\, dt$$

Differentiate both sides with respect to $z$

$$\frac{d}{dz}\mathrm{Li}_2 \left(\frac{z}{z-1} \right) = \frac{1}{(z-1)^2}\left( \frac{ \log \left(1-\frac{z}{z-1}\right)}{\frac{z}{z-1}} \right)$$

Upon simplification we obtain

$$\frac{d}{dz}\mathrm{Li}_2 \left(\frac{z}{z-1} \right) = \frac{- \log(1-z)}{z(z-1)}$$

Using partial fractions decomposition

$$\frac{d}{dz}\mathrm{Li}_2 \left(\frac{z}{z-1} \right) = \frac{\log(1-z)}{1-z}+ \frac{\log(1-z)}{z}$$

Integrate both sides with respect to $z$

$$\mathrm{Li}_2 \left(\frac{z}{z-1} \right) = -\frac{1}{2} \log^2(1-z) – \mathrm{Li}_2(z) +C$$

Put  $z=-1$ to find the constant

$$\mathrm{Li}_2 \left(\frac{1}{2} \right) = -\frac{1}{2} \log^2(2) – \mathrm{Li}_2(-1) +C$$

Remember that

$$\mathrm{Li}_2 \left(\frac{1}{2} \right) = \frac{\pi^2}{12}-\frac{1}{2} \log^2\left(\frac{1}{2} \right)\,\, , \, \, \mathrm{Li}_2(-1) = -\frac{\pi^2}{12}$$

Hence we deduce that $C=0$

$$\mathrm{Li}_2 \left(\frac{z}{z-1} \right) = -\frac{1}{2} \log^2(1-z) – \mathrm{Li}_2(z)$$

Which can be written as

$$\mathrm{Li}_2 \left(\frac{z}{z-1} \right)+\mathrm{Li}_2(z) = -\frac{1}{2} \log^2(1-z) \, \,$$

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