Dilogaritmen

Dilogaritmen är en speciell funktion som är ett specialfall av polylogaritmen. Den definieras som

${\mathrm{Li}}_2(z)=-\underset{0}{\overset{z}{\int }}\frac{\ln (1-u)}{u}\mathrm{d}u\text{, }z\in ℂ\setminus [1,\mathrm{\infty })$

För $|z|<1$ kan den definieras som den oändliga serien

${\mathrm{Li}}_2(z)=\sum _{k=1}^{\mathrm{\infty }}\frac{z^k}{k^2}.$

Identiteter

${\mathrm{Li}}_2(z)+{\mathrm{Li}}_2(-z)=\frac{1}{2}{\mathrm{Li}}_2(z^2)$

${\mathrm{Li}}_2(1-z)+{\mathrm{Li}}_2(1-\frac{1}{z})=-\frac{1}{2}{\ln }^{2}z$

${\mathrm{Li}}_2(z)+{\mathrm{Li}}_2(1-z)=\frac{1}{6}{\pi }^2-\ln z\ln (1-z)$

${\mathrm{Li}}_2(-z)+{\mathrm{Li}}_2\left(\frac{z}{1+z}\right)=-\frac{1}{2}{\ln }^2(1+z)$

${\mathrm{Li}}_2(-z)-{\mathrm{Li}}_2(1-z)+\frac{1}{2}{\mathrm{Li}}_2(1-z^2)=-\frac{1}{12}{\pi }^2-\ln z\ln (1+z)$

${\mathrm{Li}}_2(-z)+{\mathrm{Li}}_2(-\frac{1}{z})=-\frac{1}{6}{\pi }^2-\frac{1}{2}{\ln }^{2}z$

För $x>1$,

${\mathrm{Li}}_2(x)+{\mathrm{Li}}_2\left(\frac{1}{x}\right)=\frac{1}{3}{\pi }^2-\frac{1}{2}{\ln }^{2}x-\mathrm{i}\pi \ln x$

${\mathrm{Li}}_2(xy)={\mathrm{Li}}_2(x)+{\mathrm{Li}}_2(y)-{\mathrm{Li}}_2\left(\frac{x(1-y)}{1-xy}\right)-{\mathrm{Li}}_2\left(\frac{y(1-x)}{1-xy}\right)-\ln \left(\frac{1-x}{1-xy}\right)\ln \left(\frac{1-y}{1-xy}\right)$

Speciella värden

${\mathrm{Li}}_2(-1)=-\frac{{\pi }^2}{12}$

${\mathrm{Li}}_2(0)=0$

${\mathrm{Li}}_2\left(\frac{1}{2}\right)=\frac{{\pi }^2}{12}-\frac{{\ln }^{2}2}{2}$

${\mathrm{Li}}_2(1)=\frac{{\pi }^2}{6}$

${\mathrm{Li}}_2(2)=\frac{{\pi }^2}{4}-i\pi \ln 2$

${\mathrm{Li}}_2(-\frac{\sqrt{5}-1}{2})=-\frac{{\pi }^2}{15}+\frac{1}{2}{\ln }^2\frac{\sqrt{5}-1}{2}$

$=-\frac{{\pi }^2}{15}+\frac{1}{2}{\mathrm{arcsch}}^{2}2$

${\mathrm{Li}}_2(-\frac{\sqrt{5}+1}{2})=-\frac{{\pi }^2}{10}-{\ln }^2\frac{\sqrt{5}+1}{2}$

$=-\frac{{\pi }^2}{10}-{\mathrm{arcsch}}^{2}2$

${\mathrm{Li}}_2\left(\frac{3+\sqrt{5}}{2}\right)=\frac{{\pi }^2}{15}-\frac{1}{2}{\ln }^2\frac{\sqrt{5}-1}{2}$

$=\frac{{\pi }^2}{15}-\frac{1}{2}{\mathrm{arcsch}}^{2}2$

${\mathrm{Li}}_2\left(\frac{\sqrt{5}+1}{2}\right)=\frac{{\pi }^2}{10}-{\ln }^2\frac{\sqrt{5}-1}{2}$

$=\frac{{\pi }^2}{10}-{\mathrm{arcsch}}^{2}2$

${\mathrm{Li}}_2(\pm \mathrm{i})=-\frac{1}{48}{\pi }^2\pm \mathrm{i}G$

där G är Catalans konstant

Identiteter för speciella värden

${\mathrm{Li}}_2\left(\frac{1}{3}\right)-\frac{1}{6}{\mathrm{Li}}_2\left(\frac{1}{9}\right)=\frac{{\pi }^2}{18}-\frac{{\ln }^{2}3}{6}$

${\mathrm{Li}}_2(-\frac{1}{2})+\frac{1}{6}{\mathrm{Li}}_2\left(\frac{1}{9}\right)=-\frac{{\pi }^2}{18}-\ln 2\cdot \ln 3-\frac{{\ln }^{2}2}{2}-\frac{{\ln }^{2}3}{3}$

${\mathrm{Li}}_2\left(\frac{1}{4}\right)+\frac{1}{3}{\mathrm{Li}}_2\left(\frac{1}{9}\right)=\frac{{\pi }^2}{18}+2\ln 2\ln 3-2{\ln }^{2}2-\frac{2}{3}{\ln }^{2}3$

${\mathrm{Li}}_2(-\frac{1}{3})-\frac{1}{3}{\mathrm{Li}}_2\left(\frac{1}{9}\right)=-\frac{{\pi }^2}{18}+\frac{1}{6}{\ln }^{2}3$

${\mathrm{Li}}_2(-\frac{1}{8})+{\mathrm{Li}}_2\left(\frac{1}{9}\right)=-\frac{1}{2}{\ln }^2\frac{9}{8}$

$36{\mathrm{Li}}_2\left(\frac{1}{2}\right)-36{\mathrm{Li}}_2\left(\frac{1}{4}\right)-12{\mathrm{Li}}_2\left(\frac{1}{8}\right)+6{\mathrm{Li}}_2\left(\frac{1}{64}\right)={\pi }^2$