Spence's function

In mathematics, Spence's function, or dilogarithm, denoted as Li2(z), is a particular case of the polylogarithm. Lobachevsky's function and Clausen's function are closely related functions. Two related special functions are referred to as Spence's function, the dilogarithm itself, and its reflection with the variable negated:


\operatorname{Li}_2(\pm z) = -\int_0^z{\ln|1\mp\zeta| \over \zeta}\, \mathrm{d}\zeta = \sum_{k=1}^\infty {(\pm z)^k \over k^2};

William Spence, after whom the function was named by early writers in the field, was a Scottish mathematician working in the early nineteenth century.[1] He was at school with John Galt,[2] who later wrote a biographical essay on Spence.

Contents

Identities

\operatorname{Li}_2(-z)=-\operatorname{Li}_2\left(\frac{z}{1%2Bz}\right)-\frac{\ln^2(1%2Bz)}{2}
\operatorname{Li}_2({\rm{i}}z) =\frac{\operatorname{Li}_2(-z^2)}{4}%2B{\rm{i}} \operatorname{Li}_2(z)
\operatorname{Li}_2(z)%2B\operatorname{Li}_2(-z)=\frac{1}{2}\operatorname{Li}_2(z^2)
\operatorname{Li}_2(1-z)%2B\operatorname{Li}_2\left(1-\frac{1}{z}\right)=-\frac{\ln^2z}{2}
\operatorname{Li}_2(z)%2B\operatorname{Li}_2(1-z)=\frac{{\pi}^2}{6}-\ln z \cdot\ln(1-z)
\operatorname{Li}_2(-z)-\operatorname{Li}_2(1-z)%2B\frac{1}{2}\operatorname{Li}_2(1-z^2)=-\frac  {{\pi}^2}{12}-\ln z
\operatorname{Li}_2\left(\frac{1}{3}\right)-\frac{1}{6}\operatorname{Li}_2\left(\frac{1}{9}\right)=\frac{{\pi}^2}{18}-\ln^23
\operatorname{Li}_2\left(-\frac{1}{2}\right)%2B\frac{1}{6}\operatorname{Li}_2\left(\frac{1}{9}\right)=-\frac{{\pi}^2}{18}-\ln2\cdot \ln3-\frac{\ln^22}{2}-\frac{\ln^23}{3}
\operatorname{Li}_2\left(\frac{1}{4}\right)%2B\frac{1}{3}\operatorname{Li}_2\left(\frac{1}{9}\right)=\frac{{\pi}^2}{18}%2B2\ln2\ln3-2\ln^22-\frac{2}{3}\ln^23
\operatorname{Li}_2\left(-\frac{1}{3}\right)-\frac{1}{3}\operatorname{Li}_2\left(\frac{1}{9}\right)=-\frac{{\pi}^2}{18}%2B\frac{1}{6}\ln^23
\operatorname{Li}_2\left(-\frac{1}{8}\right)%2B\operatorname{Li}_2\left(\frac{1}{9}\right)=-\frac{1}{2}\ln^2{\frac{9}{8}}
36\operatorname{Li}_2\left(\frac{1}{2}\right)-36\operatorname{Li}_2\left(\frac{1}{4}\right)-12\operatorname{Li}_2\left(\frac{1}{8}\right)%2B6\operatorname{Li}_2\left(\frac{1}{64}\right)={\pi}^2

Special values

\operatorname{Li}_2(-1)=-\frac{{\pi}^2}{12}  A072691
\operatorname{Li}_2(0)=0
\operatorname{Li}_2\left(\frac{1}{2}\right)=\frac{{\pi}^2}{12}-\frac{\ln^2 2}{2}  A076788
\operatorname{Li}_2(1)=\frac{{\pi}^2}{6}  A013661
\operatorname{Li}_2(2)=\frac{{\pi}^2}{4}  A091476
\operatorname{Li}_2\left(-\frac{\sqrt5-1}{2}\right)=-\frac{{\pi}^2}{10}-\ln^2 \frac{\sqrt5-1}{2}
=-\frac{{\pi}^2}{10}-\operatorname{arcsch}^2 2
\operatorname{Li}_2\left(-\frac{\sqrt5%2B1}{2}\right)=-\frac{{\pi}^2}{15}%2B\frac{1}{2}\ln^2 \frac{\sqrt5-1}{2}  A152115
=-\frac{{\pi}^2}{15}%2B\frac{1}{2}\operatorname{arcsch}^2 2
\operatorname{Li}_2\left(\frac{3%2B\sqrt5}{2}\right)=\frac{{\pi}^2}{15}-\frac{1}{2}\ln^2 \frac{\sqrt5-1}{2}
=\frac{{\pi}^2}{15}-\frac{1}{2}\operatorname{arcsch}^2 2
\operatorname{Li}_2\left(\frac{\sqrt5%2B1}{2}\right)=\frac{{\pi}^2}{10}-\ln^2 \frac{\sqrt5-1}{2}
=\frac{{\pi}^2}{10}-\operatorname{arcsch}^2 2

References

Notes