Zurück zu Identitäten
Für | z | ≤ 1 {\displaystyle |z|\leq 1} ist ∑ k = 1 ∞ z k k 2 + ∑ k = 1 ∞ ( − 1 ) k z k k 2 = ∑ k = 1 ∞ 2 z 2 k ( 2 k ) 2 = 1 2 ∑ k = 1 ∞ ( z 2 ) k k 2 {\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 z ∈ C {\displaystyle z\in \mathbb {C} } folgt die Gleichung aus der analytischen Fortsetzbarkeit.
Li 2 ( 1 − z ) = ∫ 0 1 − z − log ( 1 − t ) t d t = ∫ 1 z log t 1 − t d t = ∫ z 1 − log t 1 − t d t {\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} = π 2 6 − ∫ 0 z − log t 1 − t d t = π 2 6 − [ log ( t ) log ( 1 − t ) ] 0 z ⏟ log ( z ) log ( 1 − z ) − ∫ 0 z − log ( 1 − t ) t d t ⏟ Li 2 ( z ) {\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 t ≠ 0 , 1 {\displaystyle t\neq 0,1\,} ist [ Li 2 ( 1 − 1 t ) ] ′ = − log ( 1 t ) 1 − 1 t 1 t 2 = − log t ( 1 − t ) t = − log t 1 − t − log t t {\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 t {\displaystyle t\,} von 1 {\displaystyle 1\,} bis z {\displaystyle z\,} , so ist Li 2 ( 1 − 1 z ) = ∫ 1 z − log t 1 − t d t − 1 2 log 2 ( z ) {\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 ∫ 1 z − log t 1 − t d t = ∫ 0 1 − z log ( 1 − t ) t d t = − Li 2 ( 1 − z ) {\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 z ≠ 0 , 1 {\displaystyle z\neq 0,1\,} ist [ Li 3 ( 1 − 1 z ) ] ′ = Li 2 ( 1 − 1 z ) 1 − 1 z 1 z 2 = − Li 2 ( 1 − 1 z ) z ( 1 − z ) {\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 − Li 2 ( 1 − 1 z ) = 1 2 log 2 ( z ) + Li 2 ( 1 − z ) {\displaystyle -{\text{Li}}_{2}\left(1-{\frac {1}{z}}\right)={\frac {1}{2}}\log ^{2}(z)+{\text{Li}}_{2}(1-z)} und schreibe 1 z ( 1 − z ) {\displaystyle {\frac {1}{z\,(1-z)}}} als 1 z + 1 1 − z {\displaystyle {\frac {1}{z}}+{\frac {1}{1-z}}} . [ Li 3 ( 1 − 1 z ) ] ′ = 1 2 log 2 z z ⏟ [ 1 6 log 3 z ] ′ + 1 2 log 2 z 1 − z + Li 2 ( 1 − z ) z + Li 2 ( 1 − z ) 1 − z ⏟ [ − Li 3 ( 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 [ Li 3 ( 1 − z ) + Li 3 ( 1 − 1 z ) − 1 6 log 3 z ] ′ = 1 2 log 2 z 1 − z + π 2 6 − log ( z ) log ( 1 − z ) − Li 2 ( z ) z {\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 [ Li 3 ( z ) + Li 3 ( 1 − z ) + Li 3 ( 1 − 1 z ) − 1 6 log 3 z − π 2 6 log z ] ′ {\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]'} = 1 2 log 2 z 1 − z − log ( z ) log ( 1 − z ) z = − [ 1 2 log 2 ( z ) log ( 1 − z ) ] ′ {\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 Li 3 ( z ) + Li 3 ( 1 − z ) + Li 3 ( 1 − 1 z ) = 1 6 log 3 z + π 2 6 log z − 1 2 log 2 ( z ) log ( 1 − z ) + C {\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 z → 1 {\displaystyle z\to 1\,} erkennt man, dass die Konstante C = Li 3 ( 1 ) = ζ ( 3 ) {\displaystyle C={\text{Li}}_{3}(1)=\zeta (3)\,} sein muss.
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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e954f0db099d3a1a4ad24ed63a75222b4336785c)
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und schreibe 
als 
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![{\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 ]}'}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c262bd773f30d8f0f4098d7cca48604620294d21)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1d2cd8129f3aa45fada24a20d88aee391650358)
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![{\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]'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f48ca57f34fbea37ab2a7bf6ee4dcffb2a38b7c2)
![{\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]'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61adb6bb942843201cb896e3db351485a6a8b7b3)
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erkennt man, dass die Konstante 
sein muss.
![{\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)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b19f37cb82660ec47b4186d2a19152a269c92ca)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ef96ab191811af478182325e141037e3cd75f0e)
