dilogarithm property.

Question

Prove that $\mathrm{Li}_{2}(-z)+\mathrm{Li}_{2}\left(\frac{z}{1+z}\right)=-\frac{1}{2} \ln ^{2}(1+z)$ I tried to paint in the rows, but I did not succeed. I don't have any more ideas.

Answer 1

By definition, if $ z\in\mathbb{R} $, then $$ \mathrm{Li}_{2}\left(-z\right)=-\int_{0}^{-z}{\frac{\ln{\left(1-x\right)}}{x}\,\mathrm{d}x}=-\int_{0}^{z}{\frac{\ln{\left(1+y\right)}}{y}\,\mathrm{d}y} $$ (We substituted $ x=-y$)

Using the same definition, then substituting $ \small\left\lbrace\begin{aligned}x&=\frac{y}{1+y}\\ \mathrm{d}x&=\frac{\mathrm{d}y}{\left(1+y\right)^{2}}\end{aligned}\right. $, gives that : $$ \mathrm{Li}_{2}\left(\frac{z}{1+z}\right)=-\int_{0}^{\frac{z}{1+z}}{\frac{\ln{\left(1-x\right)}}{x}\,\mathrm{d}x}=-\int_{0}^{z}{\frac{\ln{\left(1-\frac{y}{1+y}\right)}}{y\left(1+y\right)}\,\mathrm{d}y}=\int_{0}^{z}{\frac{\ln{\left(1+y\right)}}{y\left(1+y\right)}\,\mathrm{d}y} $$

Thus : \begin{aligned}\mathrm{Li}_{2}\left(-z\right)+\mathrm{Li}_{2}\left(\frac{z}{1+z}\right)&=\int_{0}^{z}{\frac{\ln{\left(1+y\right)}}{y\left(1+y\right)}\,\mathrm{d}y}-\int_{0}^{z}{\frac{\ln{\left(1+y\right)}}{y}\,\mathrm{d}y}\\ &=-\int_{0}^{z}{\ln{\left(1+y\right)}\left(\frac{1}{y}-\frac{1}{y\left(1+y\right)}\right)\mathrm{d}y}\\ &=-\int_{0}^{z}{\frac{\ln{\left(1+y\right)}}{1+y}\,\mathrm{d}y}\\ &=-\left[\frac{\ln^{2}{\left(1+y\right)}}{2}\right]_{0}^{z}\\ \mathrm{Li}_{2}\left(-z\right)+\mathrm{Li}_{2}\left(\frac{z}{1+z}\right)&=-\frac{1}{2}\ln^{2}{\left(1+z\right)}\end{aligned}