Formelsammlung Mathematik: Unendliche Reihen: Reihen zum Polylogarithmus

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${\text{Li}}_{2}\left({\frac {1}{2}}\right)=\sum _{k=1}^{\infty }{\frac {1}{k^{2}\,2^{k}}}={\frac {\pi ^{2}}{12}}-{\frac {1}{2}}\log ^{2}2$

Beweis

In der Identität ${\text{Li}}_{2}(z)+{\text{Li}}_{2}(1-z)={\frac {\pi ^{2}}{6}}-\log(z)\,\log(1-z)$ setze $z={\frac {1}{2}}$.

 

${\text{Li}}_{3}\left({\frac {1}{2}}\right)=\sum _{k=1}^{\infty }{\frac {1}{k^{3}\,2^{k}}}={\frac {7}{8}}\zeta (3)-{\frac {\pi ^{2}}{12}}\log 2+{\frac {1}{6}}\log ^{3}2$

Beweis

In der Identität ${\text{Li}}_{3}(z)+{\text{Li}}_{3}(1-z)+{\text{Li}}_{3}\left(1-{\frac {1}{z}}\right)=\zeta (3)+{\frac {1}{6}}\log ^{3}z+{\frac {\pi ^{2}}{6}}\log z-{\frac {1}{2}}\log ^{2}(z)\,\log(1-z)$ setze $z={\frac {1}{2}}$.

$2\,{\text{Li}}_{3}\left({\frac {1}{2}}\right)+\underbrace {{\text{Li}}_{3}(-1)} _{-{\frac {3}{4}}\zeta (3)}=\zeta (3)-{\frac {\pi ^{2}}{6}}\log 2+{\frac {1}{3}}\log ^{3}2$

 

${\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)=\sum _{k=1}^{\infty }{\frac {1}{k^{2}\,\phi ^{2k}}}={\frac {\pi ^{2}}{15}}-\log ^{2}\phi$

${\text{Li}}_{2}\left({\frac {1}{\phi }}\right)=\sum _{k=1}^{\infty }{\frac {1}{k^{2}\,\phi ^{k}}}={\frac {\pi ^{2}}{10}}-\log ^{2}\phi$

${\text{Li}}_{2}\left(-{\frac {1}{\phi }}\right)=\sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k^{2}\,\phi ^{k}}}=-{\frac {\pi ^{2}}{15}}+{\frac {1}{2}}\log ^{2}\phi$

Beweis

Bei den Identitäten

$(1)\quad {\text{Li}}_{2}(z)+{\text{Li}}_{2}(-z)={\frac {1}{2}}\,{\text{Li}}_{2}(z^{2})$

$(2)\quad {\text{Li}}_{2}(z)+{\text{Li}}_{2}(1-z)={\frac {\pi ^{2}}{6}}-\log(z)\,\log(1-z)$

$(3)\quad {\text{Li}}_{2}\left(1-{\frac {1}{z}}\right)+{\text{Li}}_{2}(1-z)=-{\frac {1}{2}}\log ^{2}z$

setze jeweils $z={\frac {1}{\phi }}$.

$(1)\quad {\text{Li}}_{2}\left({\frac {1}{\phi }}\right)+{\text{Li}}_{2}\left(-{\frac {1}{\phi }}\right)={\frac {1}{2}}\,{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)$

$(2)\quad {\text{Li}}_{2}\left({\frac {1}{\phi }}\right)+{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)={\frac {\pi ^{2}}{6}}-2\log \phi$

$(3)\quad {\text{Li}}_{2}\left(-{\frac {1}{\phi }}\right)+{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)=-{\frac {1}{2}}\log ^{2}\phi$

Bildet man $(2)+(3)-(1)\,$, so ist

$2\,{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)={\frac {\pi ^{2}}{6}}-{\frac {5}{2}}\log ^{2}\phi -{\frac {1}{2}}{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)\,\Rightarrow \,{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)={\frac {\pi ^{2}}{15}}-\log ^{2}\phi$.

Aus $(2)\,$ folgt damit ${\text{Li}}_{2}\left({\frac {1}{\phi }}\right)={\frac {\pi ^{2}}{6}}-2\log ^{2}\phi -{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)={\frac {\pi ^{2}}{10}}-\log ^{2}\phi$.

Und aus $(3)\,$ folgt ${\text{Li}}_{2}\left(-{\frac {1}{\phi }}\right)=-{\frac {1}{2}}\log ^{2}\phi -{\text{Li}}_{2}\left({\frac {1}{\phi ^{2}}}\right)=-{\frac {\pi ^{2}}{15}}+{\frac {1}{2}}\log ^{2}\phi$.

 

${\text{Li}}_{3}\left({\frac {1}{\phi ^{2}}}\right)=\sum _{k=1}^{\infty }{\frac {1}{k^{3}\,\phi ^{2k}}}={\frac {4}{5}}\zeta (3)+{\frac {2}{3}}\log ^{3}\phi -{\frac {2\pi ^{2}}{15}}\log \phi$

Beweis

Bei der Identität ${\text{Li}}_{3}(z)+{\text{Li}}_{3}(1-z)+{\text{Li}}_{3}\left(1-{\frac {1}{z}}\right)=\zeta (3)+{\frac {1}{6}}\log ^{3}z+{\frac {\pi ^{2}}{6}}\log z-{\frac {1}{2}}\log ^{2}(z)\,\log(1-z)$

setze $z={\frac {1}{\phi }}$, dann ist ${\text{Li}}_{3}\left({\frac {1}{\phi }}\right)+{\text{Li}}_{3}\left({\frac {1}{\phi ^{2}}}\right)+{\text{Li}}_{3}\left(-{\frac {1}{\phi }}\right)=\zeta (3)-{\frac {1}{6}}\log ^{3}\phi -{\frac {\pi ^{2}}{6}}\log \phi +\log ^{3}\phi$.

Wegen ${\text{Li}}_{3}\left({\frac {1}{\phi }}\right)+{\text{Li}}_{3}\left(-{\frac {1}{\phi }}\right)={\frac {1}{4}}{\text{Li}}_{3}\left({\frac {1}{\phi ^{2}}}\right)$ ist damit ${\frac {5}{4}}{\text{Li}}_{3}\left({\frac {1}{\phi ^{2}}}\right)=\zeta (3)+{\frac {5}{6}}\log ^{3}\phi -{\frac {\pi ^{2}}{6}}\log \phi$.