Dilogarithme

${\displaystyle \operatorname {Li} _{2}(z)+\operatorname {Li} _{2}(-z)={\frac {1}{2}}\operatorname {Li} _{2}(z^{2})}$[1]

${\displaystyle \operatorname {Li} _{2}(1-z)+\operatorname {Li} _{2}\left(1-{\frac {1}{z}}\right)=-{\frac {\ln ^{2}z}{2}}}$[2]

${\displaystyle \operatorname {Li} _{2}(z)+\operatorname {Li} _{2}(1-z)={\frac {\pi ^{2}}{6}}-\ln z\cdot \ln(1-z)}$

${\displaystyle \operatorname {Li} _{2}(-z)-\operatorname {Li} _{2}(1-z)+{\frac {1}{2}}\operatorname {Li} _{2}(1-z^{2})=-{\frac {\pi ^{2}}{12}}-\ln z\ln(z+1)}$

${\displaystyle \operatorname {Li} _{2}(z)+\operatorname {Li} _{2}\left({\frac {1}{z}}\right)=-{\frac {\pi ^{2}}{6}}-{\frac {1}{2}}\ln ^{2}(-z)}$

${\displaystyle \operatorname {Li} _{2}\left({\frac {1}{3}}\right)-{\frac {1}{6}}\operatorname {Li} _{2}\left({\frac {1}{9}}\right)={\frac {\pi ^{2}}{18}}-{\frac {\ln ^{2}3}{6}}}$

${\displaystyle \operatorname {Li} _{2}\left(-{\frac {1}{2}}\right)+{\frac {1}{6}}\operatorname {Li} _{2}\left({\frac {1}{9}}\right)=-{\frac {\pi ^{2}}{18}}+\ln 2\ln 3-{\frac {\ln ^{2}2}{2}}-{\frac {\ln ^{2}3}{3}}}$

${\displaystyle \operatorname {Li} _{2}\left({\frac {1}{4}}\right)+{\frac {1}{3}}\operatorname {Li} _{2}\left({\frac {1}{9}}\right)={\frac {\pi ^{2}}{18}}+2\ln 2\ln 3-2\ln ^{2}2-{\frac {2}{3}}\ln ^{2}3}$

${\displaystyle \operatorname {Li} _{2}\left(-{\frac {1}{3}}\right)-{\frac {1}{3}}\operatorname {Li} _{2}\left({\frac {1}{9}}\right)=-{\frac {\pi ^{2}}{18}}+{\frac {1}{6}}\ln ^{2}3}$

${\displaystyle \operatorname {Li} _{2}\left(-{\frac {1}{8}}\right)+\operatorname {Li} _{2}\left({\frac {1}{9}}\right)=-{\frac {1}{2}}\ln ^{2}{\frac {9}{8}}}$

${\displaystyle 36\operatorname {Li} _{2}\left({\frac {1}{2}}\right)-36\operatorname {Li} _{2}\left({\frac {1}{4}}\right)-12\operatorname {Li} _{2}\left({\frac {1}{8}}\right)+6\operatorname {Li} _{2}\left({\frac {1}{64}}\right)=\pi ^{2}}$

${\displaystyle \operatorname {Li} _{2}(-1)=-{\frac {\pi ^{2}}{12}}}$

${\displaystyle \operatorname {Li} _{2}(0)=0}$

${\displaystyle \operatorname {Li} _{2}\left({\frac {1}{2}}\right)={\frac {\pi ^{2}}{12}}-{\frac {\ln ^{2}2}{2}}}$

${\displaystyle \operatorname {Li} _{2}(1)={\frac {\pi ^{2}}{6}}}$

${\displaystyle \operatorname {Li} _{2}(2)={\frac {\pi ^{2}}{4}}-\mathrm {i} \pi \ln 2}$

${\displaystyle \operatorname {Li} _{2}\left(-{\frac {{\sqrt {5}}-1}{2}}\right)=-{\frac {\pi ^{2}}{15}}+{\frac {1}{2}}\ln ^{2}{\frac {{\sqrt {5}}-1}{2}}}$

${\displaystyle \operatorname {Li} _{2}\left(-{\frac {{\sqrt {5}}+1}{2}}\right)=-{\frac {\pi ^{2}}{10}}-\ln ^{2}{\frac {{\sqrt {5}}+1}{2}}}$

${\displaystyle \operatorname {Li} _{2}\left({\frac {3-{\sqrt {5}}}{2}}\right)={\frac {\pi ^{2}}{15}}-\ln ^{2}{\frac {{\sqrt {5}}-1}{2}}}$

${\displaystyle \operatorname {Li} _{2}\left({\frac {{\sqrt {5}}-1}{2}}\right)={\frac {\pi ^{2}}{10}}-\ln ^{2}{\frac {{\sqrt {5}}-1}{2}}}$

On rencontre couramment la fonction de Spence en physique des particules, dans le calcul des corrections radiatives. Dans ce contexte, la fonction est souvent définie avec une valeur absolue à l'intérieur du logarithme :

${\displaystyle \operatorname {\Phi } (x)=-\int _{0}^{x}{\frac {\ln |1-u|}{u}}\,\mathrm {d} u={\begin{cases}\operatorname {Li} _{2}(x),&x\leq 1~;\\{\frac {\pi ^{2}}{3}}-{\frac {1}{2}}\ln ^{2}(x)-\operatorname {Li} _{2}({\frac {1}{x}}),&x>1.\end{cases}}}$