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In der Identität Li2(z)+Li2(1−z)=π26−log(z)log(1−z){\displaystyle {\text{Li}}_{2}(z)+{\text{Li}}_{2}(1-z)={\frac {\pi ^{2}}{6}}-\log(z)\,\log(1-z)} setze z=12{\displaystyle z={\frac {1}{2}}}.
In der Identität Li3(z)+Li3(1−z)+Li3(1−1z)=ζ(3)+16log3z+π26logz−12log2(z)log(1−z){\displaystyle {\text{Li}}_{3}(z)+{\text{Li}}_{3}(1-z)+{\text{Li}}_{3}\left(1-{\frac {1}{z}}\right)=\zeta (3)+{\frac {1}{6}}\log ^{3}z+{\frac {\pi ^{2}}{6}}\log z-{\frac {1}{2}}\log ^{2}(z)\,\log(1-z)} setze z=12{\displaystyle z={\frac {1}{2}}}. 2Li3(12)+Li3(−1)⏟−34ζ(3)=ζ(3)−π26log2+13log32{\displaystyle 2\,{\text{Li}}_{3}\left({\frac {1}{2}}\right)+\underbrace {{\text{Li}}_{3}(-1)} _{-{\frac {3}{4}}\zeta (3)}=\zeta (3)-{\frac {\pi ^{2}}{6}}\log 2+{\frac {1}{3}}\log ^{3}2}
Bei den Identitäten (1)Li2(z)+Li2(−z)=12Li2(z2){\displaystyle (1)\quad {\text{Li}}_{2}(z)+{\text{Li}}_{2}(-z)={\frac {1}{2}}\,{\text{Li}}_{2}(z^{2})} (2)Li2(z)+Li2(1−z)=π26−log(z)log(1−z){\displaystyle (2)\quad {\text{Li}}_{2}(z)+{\text{Li}}_{2}(1-z)={\frac {\pi ^{2}}{6}}-\log(z)\,\log(1-z)} (3)Li2(1−1z)+Li2(1−z)=−12log2z{\displaystyle (3)\quad {\text{Li}}_{2}\left(1-{\frac {1}{z}}\right)+{\text{Li}}_{2}(1-z)=-{\frac {1}{2}}\log ^{2}z} setze jeweils z=1ϕ{\displaystyle z={\frac {1}{\phi }}}. (1)Li2(1ϕ)+Li2(−1ϕ)=12Li2(1ϕ2){\displaystyle (1)\quad {\text{Li}}_{2}\left({\frac {1}{\phi }}\right)+{\text{Li}}_{2}\left(-{\frac {1}{\phi }}\right)={\frac {1}{2}}\,{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)} (2)Li2(1ϕ)+Li2(1ϕ2)=π26−2logϕ{\displaystyle (2)\quad {\text{Li}}_{2}\left({\frac {1}{\phi }}\right)+{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)={\frac {\pi ^{2}}{6}}-2\log \phi } (3)Li2(−1ϕ)+Li2(1ϕ2)=−12log2ϕ{\displaystyle (3)\quad {\text{Li}}_{2}\left(-{\frac {1}{\phi }}\right)+{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)=-{\frac {1}{2}}\log ^{2}\phi } Bildet man (2)+(3)−(1){\displaystyle (2)+(3)-(1)\,}, so ist 2Li2(1ϕ2)=π26−52log2ϕ−12Li2(1ϕ2)⇒Li2(1ϕ2)=π215−log2ϕ{\displaystyle 2\,{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)={\frac {\pi ^{2}}{6}}-{\frac {5}{2}}\log ^{2}\phi -{\frac {1}{2}}{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)\,\Rightarrow \,{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)={\frac {\pi ^{2}}{15}}-\log ^{2}\phi }. Aus (2){\displaystyle (2)\,} folgt damit Li2(1ϕ)=π26−2log2ϕ−Li2(1ϕ2)=π210−log2ϕ{\displaystyle {\text{Li}}_{2}\left({\frac {1}{\phi }}\right)={\frac {\pi ^{2}}{6}}-2\log ^{2}\phi -{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)={\frac {\pi ^{2}}{10}}-\log ^{2}\phi }. Und aus (3){\displaystyle (3)\,} folgt Li2(−1ϕ)=−12log2ϕ−Li2(1ϕ2)=−π215+12log2ϕ{\displaystyle {\text{Li}}_{2}\left(-{\frac {1}{\phi }}\right)=-{\frac {1}{2}}\log ^{2}\phi -{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)=-{\frac {\pi ^{2}}{15}}+{\frac {1}{2}}\log ^{2}\phi }.
Bei der Identität Li3(z)+Li3(1−z)+Li3(1−1z)=ζ(3)+16log3z+π26logz−12log2(z)log(1−z){\displaystyle {\text{Li}}_{3}(z)+{\text{Li}}_{3}(1-z)+{\text{Li}}_{3}\left(1-{\frac {1}{z}}\right)=\zeta (3)+{\frac {1}{6}}\log ^{3}z+{\frac {\pi ^{2}}{6}}\log z-{\frac {1}{2}}\log ^{2}(z)\,\log(1-z)} setze z=1ϕ{\displaystyle z={\frac {1}{\phi }}}, dann ist Li3(1ϕ)+Li3(1ϕ2)+Li3(−1ϕ)=ζ(3)−16log3ϕ−π26logϕ+log3ϕ{\displaystyle {\text{Li}}_{3}\left({\frac {1}{\phi }}\right)+{\text{Li}}_{3}\left({\frac {1}{\phi ^{2}}}\right)+{\text{Li}}_{3}\left(-{\frac {1}{\phi }}\right)=\zeta (3)-{\frac {1}{6}}\log ^{3}\phi -{\frac {\pi ^{2}}{6}}\log \phi +\log ^{3}\phi }. Wegen Li3(1ϕ)+Li3(−1ϕ)=14Li3(1ϕ2){\displaystyle {\text{Li}}_{3}\left({\frac {1}{\phi }}\right)+{\text{Li}}_{3}\left(-{\frac {1}{\phi }}\right)={\frac {1}{4}}{\text{Li}}_{3}\left({\frac {1}{\phi ^{2}}}\right)} ist damit 54Li3(1ϕ2)=ζ(3)+56log3ϕ−π26logϕ{\displaystyle {\frac {5}{4}}{\text{Li}}_{3}\left({\frac {1}{\phi ^{2}}}\right)=\zeta (3)+{\frac {5}{6}}\log ^{3}\phi -{\frac {\pi ^{2}}{6}}\log \phi }.