Dilogarithmus

Für die folgenden Zahlen lassen sich ${\displaystyle z}$ und ${\displaystyle \operatorname {Li} _{2}(z)}$ in geschlossener Form darstellen:

${\displaystyle \operatorname {Li} _{2}(-1)=-{\frac {{\pi }^{2}}{12}},\qquad \operatorname {Li} _{2}(0)=0,}$

${\displaystyle \operatorname {Li} _{2}{\biggl (}{\frac {1}{2}}{\biggr )}={\frac {{\pi }^{2}}{12}}-{\frac {1}{2}}\ln ^{2}(2),\qquad \operatorname {Li} _{2}(1)={\frac {{\pi }^{2}}{6}}}$,

${\displaystyle \operatorname {Li} _{2}(-\Phi ^{-1})=-{\frac {{\pi }^{2}}{15}}+{\frac {1}{2}}\ln ^{2}(\Phi ),\qquad \operatorname {Li} _{2}(-\Phi )=-{\frac {{\pi }^{2}}{10}}-\ln ^{2}(\Phi ),}$

${\displaystyle \operatorname {Li} _{2}(\Phi ^{-2})={\frac {{\pi }^{2}}{15}}-\ln ^{2}(\Phi ),\qquad \operatorname {Li} _{2}(\Phi ^{-1})={\frac {{\pi }^{2}}{10}}-\ln ^{2}(\Phi )}$.

Mit dem Kürzel Φ wird hierbei die Zahl des Goldenen Schnittes ausgedrückt: ${\displaystyle \Phi =({\sqrt {5}}+1)/2}$

Mit der sechsten Einheitswurzel ${\displaystyle \omega ={\frac {1}{2}}+{\frac {\sqrt {3}}{2}}i}$ und der Gieseking-Konstante ${\displaystyle V_{0}=1{,}0149\ldots }$ hat man außerdem

${\displaystyle \operatorname {Li} _{2}(\omega )={\frac {\pi ^{2}}{36}}+V_{0}i,\qquad \operatorname {Li} _{2}(\omega ^{2})=-{\frac {\pi ^{2}}{18}}+{\frac {2}{3}}V_{0}i,}$

${\displaystyle \operatorname {Li} _{2}(1+\omega )={\frac {\pi ^{2}}{9}}+\left({\frac {2}{3}}V_{0}+{\frac {1}{3}}\ln(3)\pi \right)i,\qquad \operatorname {Li} _{2}\left({\frac {1}{1+\omega }}\right)={\frac {5\pi ^{2}}{72}}-{\frac {1}{8}}\ln(3)+\left(-{\frac {2}{3}}V_{0}+{\frac {1}{12}}\ln(3)\pi \right)i}$.

Werte des Bloch-Wigner-Dilogarithmus können bisher nur numerisch berechnet werden und man kennt nur wenige algebraische Relationen zwischen Werten des Bloch-Wigner-Dilogarithmus. Eine Vermutung von John Milnor besagt für ${\displaystyle N\geq 3}$:

die Zahlen ${\displaystyle D_{2}(e^{2\pi i{\frac {j}{N}}})}$ für ${\displaystyle 0<j<{\frac {N}{2}}}$ und ${\displaystyle \operatorname {ggT} (j,N)=1}$ sind linear unabhängig über ${\displaystyle \mathbb {Q} }$.

Es gibt zahlreiche algebraische Identitäten zwischen Werten von ${\displaystyle L}$ in rationalen oder algebraischen Argumenten. Beispiele spezieller Werte sind

${\displaystyle L(0)=0,\quad L\left({\frac {1}{2}}\right)={\frac {1}{2}},\quad L(1)=1,\quad L\left({\frac {3-{\sqrt {5}}}{2}}\right)={\frac {2}{5}},\quad L\left({\frac {{\sqrt {5}}-1}{2}}\right)={\frac {3}{5}}}$.

Mit der sechsten Einheitswurzel ${\displaystyle \omega ={\frac {1}{2}}+{\frac {\sqrt {3}}{2}}i}$ und der Gieseking-Konstante ${\displaystyle V_{0}=1,0149...}$ hat man

${\displaystyle R(\omega )=-{\frac {\pi ^{2}}{12}}+V_{0}i,\qquad R(\omega ^{2})=-{\frac {\pi ^{2}}{6}}+\left({\frac {2}{3}}V_{0}+{\frac {1}{6}}\ln(3)\pi \right)i,}$

${\displaystyle R(1+\omega )=\left({\frac {2}{3}}V_{0}+{\frac {1}{6}}\ln(3)\pi \right)i,\qquad R\left({\frac {1}{1+\omega }}\right)=-{\frac {\pi ^{2}}{12}}-{\frac {1}{8}}\ln(3)+{\frac {1}{8}}\ln ^{2}(3)+\left(-{\frac {2}{3}}V_{0}+{\frac {1}{12}}\ln(3)\pi \right)i}$

Der Beweis des Wertes vom Dilogarithmus von Eins wird im sogenannten Basler Problem behandelt. Dieser Beweis kann auf folgende Weise absolviert werden:

Folgende Funktion hat folgende Ableitung:

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}{\biggl [}2\operatorname {Li} _{2}{\biggl (}{\frac {x}{{\sqrt {x^{2}+1}}+1}}{\biggr )}-{\frac {1}{2}}\operatorname {Li} _{2}{\biggl (}{\frac {{\sqrt {x^{2}+1}}-1}{{\sqrt {x^{2}+1}}+1}}{\biggr )}{\biggr ]}={\frac {\operatorname {arsinh} (x)}{x{\sqrt {x^{2}+1}}}}}$

Deswegen gilt folgendes Integral:

${\displaystyle \int _{0}^{\infty }{\frac {\operatorname {arsinh} (x)}{x{\sqrt {x^{2}+1}}}}\,\mathrm {d} x={\biggl [}2\operatorname {Li} _{2}{\biggl (}{\frac {x}{{\sqrt {x^{2}+1}}+1}}{\biggr )}-{\frac {1}{2}}\operatorname {Li} _{2}{\biggl (}{\frac {{\sqrt {x^{2}+1}}-1}{{\sqrt {x^{2}+1}}+1}}{\biggr )}{\biggr ]}_{x=0}^{x=\infty }={\frac {3}{2}}\,{\text{Li}}_{2}(1)}$

Der Satz von Fubini liefert diesen Zusammenhang:

${\displaystyle \int _{0}^{\infty }{\frac {\operatorname {arsinh} (x)}{x{\sqrt {x^{2}+1}}}}\mathrm {d} x=\int _{0}^{\infty }\int _{0}^{1}{\frac {1}{-x^{2}y^{2}+x^{2}+1}}\,\mathrm {d} y\,\mathrm {d} x=\int _{0}^{1}\int _{0}^{\infty }{\frac {1}{-x^{2}y^{2}+x^{2}+1}}\,\mathrm {d} x\,\mathrm {d} y=\int _{0}^{1}{\frac {\pi }{2{\sqrt {1-y^{2}}}}}\,\mathrm {d} y={\frac {\pi ^{2}}{4}}}$

Durch Gleichsetzung der beiden zuletzt genannten Formeln erhält man jenes Resultat:

${\displaystyle {\frac {3}{2}}\,{\text{Li}}_{2}(1)={\frac {\pi ^{2}}{4}}}$

Aufgelöst entsteht der genannte Wert:

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

Exakt dieser Wert ist somit auch die unendliche Summe der Kehrwerte der Quadratzahlen:

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}}$

Diese Tatsache geht direkt aus der Maclaurinschen Reihe vom Dilogarithmus hervor.

Der klassische Dilogarithmus genügt zahlreichen Funktionalgleichungen, zum Beispiel

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

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

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

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

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

${\displaystyle \operatorname {Li} _{2}(z)-{\frac {1}{4}}\operatorname {Li} _{2}(z^{2})=\int _{0}^{1}{\frac {\operatorname {arcsin} (xz)}{\sqrt {1-x^{2}}}}\,\mathrm {d} x}$. Daraus folgt ebenso: ${\displaystyle \operatorname {Li} _{2}(1)={\frac {{\pi }^{2}}{6}}}$.

Der Bloch-Wigner-Dilogarithmus genügt den Identitäten

${\displaystyle \operatorname {D} _{2}(z)=\operatorname {D} _{2}\left(1-{\frac {1}{z}}\right)=\operatorname {D} _{2}\left({\frac {1}{1-z}}\right)=-\operatorname {D} _{2}\left({\frac {1}{z}}\right)=-\operatorname {D} _{2}(1-z)=-\operatorname {D} _{2}\left({\frac {-z}{1-z}}\right)}$

und der 5-Term-Relation

${\displaystyle \operatorname {D} _{2}(x)+\operatorname {D} _{2}(y)+\operatorname {D} _{2}\left({\frac {1-x}{1-xy}}\right)+\operatorname {D} _{2}(1-xy)+\operatorname {D} _{2}\left({\frac {1-y}{1-xy}}\right)=0}$.

Der Rogers-Dilogarithmus erfüllt die Beziehung

${\displaystyle L(x)+L(1-x)=1}$

und Abels Funktionalgleichung

${\displaystyle L(x)+L(y)=L(xy)+L\left({\frac {x(1-y)}{1-xy}}\right)+L\left({\frac {y(1-x)}{1-xy}}\right)}$.

Für ${\displaystyle R}$ hat man

${\displaystyle R(x)+R(1-x)=-{\frac {\pi ^{2}}{6}}}$

und die 5-Term-Relation

${\displaystyle R(x)-R(y)+R\left({\frac {y}{x}}\right)-R\left({\frac {1-x^{-1}}{1-y^{-1}}}\right)+R\left({\frac {1-x}{1-y}}\right)=0}$,

insbesondere ist ${\displaystyle R}$ eine wohldefinierte Funktion auf der Bloch-Gruppe.

Folgende Gleichung gilt für den Fall ${\displaystyle t>0\,\cap \,v>0\,\cap \,tw-uv<0}$:

${\displaystyle {\frac {\ln(tx+u)}{vx+w}}={\frac {\mathrm {d} }{\mathrm {d} x}}{\biggl [}{\frac {1}{v}}\ln {\biggl (}{\frac {uv-tw}{v}}{\biggr )}\ln(vx+w)-{\frac {1}{v}}\operatorname {Li} _{2}{\biggl (}-t{\frac {vx+w}{uv-tw}}{\biggr )}{\biggr ]}}$

Beispiel:

${\displaystyle {\begin{aligned}\int _{0}^{5}{\frac {\ln(x+2)}{3x+4}}\mathrm {d} x&={\biggl [}{\frac {1}{3}}\ln {\biggl (}{\frac {2}{3}}{\biggr )}\ln(3x+4)-{\frac {1}{3}}\operatorname {Li} _{2}{\biggl (}-{\frac {3}{2}}x-2{\biggr )}{\biggr ]}_{x=0}^{x=5}=\\&=-{\frac {1}{3}}\ln {\biggl (}{\frac {3}{2}}{\biggr )}\ln {\biggl (}{\frac {19}{4}}{\biggr )}-{\frac {1}{3}}\operatorname {Li} _{2}{\biggl (}-{\frac {19}{2}}{\biggr )}+{\frac {1}{3}}\operatorname {Li} _{2}(-2)\\\end{aligned}}}$

Folgende Gleichung gilt für den Fall ${\displaystyle t>0\,\cap \,v>0\,\cap \,tw-uv>0}$:

${\displaystyle {\frac {\ln(tx+u)}{vx+w}}={\frac {\mathrm {d} }{\mathrm {d} x}}{\biggl [}{\frac {1}{v}}\ln {\biggl (}t\,{\frac {vx+w}{tw-uv}}{\biggr )}\ln(tx+u)+{\frac {1}{v}}\operatorname {Li} _{2}{\biggl (}-v\,{\frac {tx+u}{tw-uv}}{\biggr )}{\biggr ]}}$

Beispiel:

${\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {\ln(5x+6)}{3x+4}}\mathrm {d} x&={\biggl [}{\frac {1}{3}}\ln {\biggl (}{\frac {15}{2}}x+10{\biggr )}\ln(5x+6)+{\frac {1}{3}}\operatorname {Li} _{2}{\biggl (}-{\frac {15}{2}}x-9{\biggr )}{\biggr ]}_{x=0}^{x=1}=\\&={\frac {1}{3}}\ln {\biggl (}{\frac {35}{2}}{\biggr )}\ln(11)-{\frac {1}{3}}\ln(10)\ln(6)+{\frac {1}{3}}\operatorname {Li} _{2}{\biggl (}-{\frac {33}{2}}{\biggr )}-{\frac {1}{3}}\operatorname {Li} _{2}(-9)\\\end{aligned}}}$

Weitere Funktionen lassen sich mit dem Dilogarithmus integrieren:

Areatangens-Hyperbolicus-Funktionen:

${\displaystyle {\frac {\operatorname {artanh} (x)}{x}}={\frac {\mathrm {d} }{\mathrm {d} x}}{\biggl [}\operatorname {Li} _{2}(x)-{\frac {1}{4}}\operatorname {Li} _{2}(x^{2}){\biggr ]}}$

${\displaystyle {\frac {\operatorname {artanh} (x)}{x{\sqrt {1-x^{2}}}}}={\frac {\mathrm {d} }{\mathrm {d} x}}{\biggl [}2\operatorname {Li} _{2}{\biggl (}{\frac {x}{1+{\sqrt {1-x^{2}}}}}{\biggr )}-{\frac {1}{2}}\operatorname {Li} _{2}{\biggl (}{\frac {1-{\sqrt {1-x^{2}}}}{1+{\sqrt {1-x^{2}}}}}{\biggr )}{\biggr ]}}$

${\displaystyle {\frac {\operatorname {artanh} (x)}{x(1-x^{2})}}={\frac {\mathrm {d} }{\mathrm {d} x}}{\biggl [}{\frac {1}{2}}\operatorname {Li} _{2}{\biggl (}{\frac {2x}{x+1}}{\biggr )}+{\frac {1}{2}}\operatorname {artanh} (x)^{2}{\biggr ]}}$

Areasinus-Hyperbolicus-Funktionen:

${\displaystyle {\frac {\operatorname {arsinh} (x)}{x}}={\frac {\mathrm {d} }{\mathrm {d} x}}{\biggl \{}{\frac {1}{2}}\operatorname {Li} _{2}{\bigl [}1-{\bigl (}{\sqrt {x^{2}+1}}-x{\bigr )}^{2}{\bigr ]}+{\frac {1}{2}}\operatorname {arsinh} (x)^{2}{\biggr \}}}$

Deswegen gilt: ${\displaystyle \int _{0}^{1/2}{\frac {\operatorname {arsinh} (x)}{x}}\mathrm {d} x={\frac {1}{2}}\operatorname {Li} _{2}{\biggl (}{\frac {1}{2}}{\sqrt {5}}-{\frac {1}{2}}{\biggr )}+{\frac {1}{2}}\operatorname {arsinh} {\biggl (}{\frac {1}{2}}{\biggr )}^{2}={\frac {\pi ^{2}}{20}}}$

Daraus folgt: ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(2k)!}{16^{k}(2k+1)^{2}(k!)^{2}}}={\frac {\pi ^{2}}{10}}}$

${\displaystyle {\frac {\operatorname {arsinh} (x)}{x{\sqrt {x^{2}+1}}}}={\frac {\mathrm {d} }{\mathrm {d} x}}{\biggl [}2\operatorname {Li} _{2}{\biggl (}{\frac {x}{{\sqrt {x^{2}+1}}+1}}{\biggr )}-{\frac {1}{2}}\operatorname {Li} _{2}{\biggl (}{\frac {{\sqrt {x^{2}+1}}-1}{{\sqrt {x^{2}+1}}+1}}{\biggr )}{\biggr ]}}$

${\displaystyle {\frac {\operatorname {arsinh} (x)}{x(x^{2}+1)}}={\frac {\mathrm {d} }{\mathrm {d} x}}{\biggl [}\operatorname {Li} _{2}{\biggl (}{\frac {x}{\sqrt {x^{2}+1}}}{\biggr )}-{\frac {1}{4}}\operatorname {Li} _{2}{\biggl (}{\frac {x^{2}}{x^{2}+1}}{\biggr )}{\biggr ]}={\frac {\mathrm {d} }{\mathrm {d} x}}{\biggl \{}\operatorname {Li} _{2}{\bigl [}1-{\bigl (}{\sqrt {x^{2}+1}}-x{\bigr )}^{2}{\bigr ]}-{\frac {1}{4}}\operatorname {Li} _{2}{\bigl [}1-{\bigl (}{\sqrt {x^{2}+1}}-x{\bigr )}^{4}{\bigr ]}{\biggr \}}}$