# 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 $t=1-x$ we get

$$\int^z_0 \frac{\log(t)}{1-t} \, dt \,=-\int^{1-z}_{1} \frac{\log(1-x)}{x} \, dx$$

For $0<z <1$

$$\int^{1}_{1-z} \frac{\log(1-x)}{x} \, dx = \int^1_0 \frac{\log(1-x)}{x}\, dx – \int_0^{1-z} \frac{\log(1-x)}{x}\, dx$$

Now it is easy to see that

$$\int^{1}_{1-z} \frac{\log(1-x)}{x} \, dx =-\mathrm{Li}_2(1)+\mathrm{Li}_2(1-z)$$

Which implies that

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

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

Now since $\mathrm{Li}_2(1) = \zeta(2) = \frac{\pi^2}{6}$

$$\mathrm{Li}_2\left(z\right)+\mathrm{Li}_2(1-z)\, = \frac{\pi^2}{6}-\log(z) \log(1-z) \,\,$$

We can easily deduce that for $z=\frac{1}{2}$

$$2\mathrm{Li}_2\left(\frac{1}{2}\right)= \frac{\pi^2}{6}-\log^2\left(\frac{1}{2}\right) \,\,$$

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

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