Für ${\displaystyle |z|\leq 1}$ ist ${\displaystyle \sum _{k=1}^{\infty }{\frac {z^{k}}{k^{2}}}+\sum _{k=1}^{\infty }{\frac {(-1)^{k}\,z^{k}}{k^{2}}}=\sum _{k=1}^{\infty }{\frac {2\,z^{2k}}{(2k)^{2}}}={\frac {1}{2}}\,\sum _{k=1}^{\infty }{\frac {(z^{2})^{k}}{k^{2}}}}$.

Und für alle anderen ${\displaystyle z\in \mathbb {C} }$ folgt die Gleichung aus der analytischen Fortsetzbarkeit.

${\displaystyle {\text{Li}}_{2}(1-z)=\int _{0}^{1-z}{\frac {-\log(1-t)}{t}}\,dt=\int _{1}^{z}{\frac {\log t}{1-t}}\,dt=\int _{z}^{1}{\frac {-\log t}{1-t}}\,dt}$

${\displaystyle ={\frac {\pi ^{2}}{6}}-\int _{0}^{z}{\frac {-\log t}{1-t}}\,dt={\frac {\pi ^{2}}{6}}-\underbrace {{\Big [}\log(t)\,\log(1-t){\Big ]}_{0}^{z}} _{\log(z)\,\log(1-z)}-\underbrace {\int _{0}^{z}{\frac {-\log(1-t)}{t}}\,dt} _{{\text{Li}}_{2}(z)}}$

Für ${\displaystyle t\neq 0,1\,}$ ist ${\displaystyle \left[{\text{Li}}_{2}\left(1-{\frac {1}{t}}\right)\right]'={\frac {-\log \left({\frac {1}{t}}\right)}{1-{\frac {1}{t}}}}\,{\frac {1}{t^{2}}}={\frac {-\log t}{(1-t)\,t}}={\frac {-\log t}{1-t}}-{\frac {\log t}{t}}}$.

Integriert man nach ${\displaystyle t\,}$ von ${\displaystyle 1\,}$ bis ${\displaystyle z\,}$, so ist ${\displaystyle {\text{Li}}_{2}\left(1-{\frac {1}{z}}\right)=\int _{1}^{z}{\frac {-\log t}{1-t}}\,dt-{\frac {1}{2}}\log ^{2}(z)}$.

Dabei ist ${\displaystyle \int _{1}^{z}{\frac {-\log t}{1-t}}\,dt=\int _{0}^{1-z}{\frac {\log(1-t)}{t}}\,dt=-{\text{Li}}_{2}(1-z)}$.

Für ${\displaystyle z\neq 0,1\,}$ ist ${\displaystyle \left[{\text{Li}}_{3}\left(1-{\frac {1}{z}}\right)\right]'={\frac {{\text{Li}}_{2}\left(1-{\frac {1}{z}}\right)}{1-{\frac {1}{z}}}}\,{\frac {1}{z^{2}}}={\frac {-{\text{Li}}_{2}\left(1-{\frac {1}{z}}\right)}{z\,(1-z)}}}$.

Benutze die Identität ${\displaystyle -{\text{Li}}_{2}\left(1-{\frac {1}{z}}\right)={\frac {1}{2}}\log ^{2}(z)+{\text{Li}}_{2}(1-z)}$ und schreibe ${\displaystyle {\frac {1}{z\,(1-z)}}}$ als ${\displaystyle {\frac {1}{z}}+{\frac {1}{1-z}}}$.

${\displaystyle \left[{\text{Li}}_{3}\left(1-{\frac {1}{z}}\right)\right]'=\underbrace {\frac {{\frac {1}{2}}\log ^{2}z}{z}} _{\left[{\frac {1}{6}}\log ^{3}z\right]'}+{\frac {{\frac {1}{2}}\log ^{2}z}{1-z}}+{\frac {{\text{Li}}_{2}(1-z)}{z}}+\underbrace {\frac {{\text{Li}}_{2}(1-z)}{1-z}} _{{\big [}-{\text{Li}}_{3}(1-z){\big ]}'}}$

Also ist ${\displaystyle \left[{\text{Li}}_{3}(1-z)+{\text{Li}}_{3}\left(1-{\frac {1}{z}}\right)-{\frac {1}{6}}\log ^{3}z\right]'={\frac {{\frac {1}{2}}\log ^{2}z}{1-z}}+{\frac {{\frac {\pi ^{2}}{6}}-\log(z)\,\log(1-z)-{\text{Li}}_{2}(z)}{z}}}$,

und somit ist ${\displaystyle \left[{\text{Li}}_{3}(z)+{\text{Li}}_{3}(1-z)+{\text{Li}}_{3}\left(1-{\frac {1}{z}}\right)-{\frac {1}{6}}\log ^{3}z-{\frac {\pi ^{2}}{6}}\log z\right]'}$

${\displaystyle ={\frac {{\frac {1}{2}}\log ^{2}z}{1-z}}-{\frac {\log(z)\,\log(1-z)}{z}}=-\left[{\frac {1}{2}}\,\log ^{2}(z)\,\log(1-z)\right]'}$.

Damit ist ${\displaystyle {\text{Li}}_{3}(z)+{\text{Li}}_{3}(1-z)+{\text{Li}}_{3}\left(1-{\frac {1}{z}}\right)={\frac {1}{6}}\log ^{3}z+{\frac {\pi ^{2}}{6}}\log z-{\frac {1}{2}}\log ^{2}(z)\,\log(1-z)+C}$.

Für ${\displaystyle z\to 1\,}$ erkennt man, dass die Konstante ${\displaystyle C={\text{Li}}_{3}(1)=\zeta (3)\,}$ sein muss.