$${\text{Li}_2(z) = - \int_{0}^{z} \frac{\ln(1-x)}{x} \ dx }$$
$${\text{Li}_2(1-z) = \int_{1}^{z} \frac{\ln(x)}{1-x} \ dx }$$
$${\text{Li}_2(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^2} }$$
$${\text{Li}_s(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^s} }$$
$${\text{Li}_2(z) + \text{Li}_2(-z) = \frac{1}{2} \text{Li}_2(z^2) }$$
$${\text{Li}_2(z) + \text{Li}_2(1-z) = \frac{\pi^2}{6} - \ln(z) \ln(1-z) }$$
$${\text{Li}_2(z) + \text{Li}_2\left(\frac{1}{z}\right) = - \frac{\pi^2}{6} - \frac{\left(\ln(-z) \right)^2}{2} }$$
$${\text{Li}_2(-z) + \text{Li}_2\left(-\frac{1}{z}\right) = - \frac{\pi^2}{6} - \frac{\left(\ln(z) \right)^2}{2} }$$
$${\text{Li}_2(1-z) + \text{Li}_2\left(1 - \frac{1}{z}\right) = - \frac{\left(\ln(z) \right)^2}{2} }$$
$${\text{Li}_{s+1}(z) = \int_{0}^{z} \frac{\text{Li}_s(x)}{x} \ dx }$$
$${ \frac{d}{dz} \left( \text{Li}_{s}(z) \right) = \frac{\text{Li}_{s-1}(z)}{z} }$$
$${\text{Li}_2(1) = \frac{\pi^2}{6} }$$
$${\text{Li}_2(-1) = -\frac{\pi^2}{12} }$$
$${\text{Li}_2(0) = 0 }$$
$${\text{Li}_2 \left(\frac{1}{2} \right) = \frac{\pi^2}{12} - \frac{(\ln 2)^2}{2} }$$
$${\text{Li}_2 \left(\frac{1 - \sqrt{5}}{2} \right) = - \frac{\pi^2}{15} + \frac{1}{2} \left( \ln \frac{\sqrt{5} + 1}{2} \right)^2 }$$
$${\text{Li}_2 \left(- \frac{1 + \sqrt{5}}{2} \right) = - \frac{\pi^2}{10} - \left( \ln \frac{\sqrt{5} + 1}{2} \right)^2 }$$
$${\text{Li}_2 \left(\frac{3 - \sqrt{5}}{2} \right) = \frac{\pi^2}{15} - \left( \ln \frac{\sqrt{5} + 1}{2} \right)^2 }$$
$${\text{Li}_2 \left(\frac{\sqrt{5} - 1}{2} \right) = \frac{\pi^2}{10} - \left( \ln \frac{\sqrt{5} + 1}{2} \right)^2 }$$
$${\text{Li}_2(i) = -\frac{\pi^2}{48} + i G}$$
G = Catalan's constant
$${\text{L}(z) = \text{Li}_2(z) + \frac{1}{2} \ln(z) \ln(1-z) }$$