斯盆司函数

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

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

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

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

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

${\displaystyle \operatorname {Li} _{2}\left({\frac {1}{3}}\right)-{\frac {1}{6}}\operatorname {Li} _{2}\left({\frac {1}{9}}\right)={\frac {{\pi }^{2}}{18}}-{\frac {\ln ^{2}3}{6}}}$ ^[2]

${\displaystyle \operatorname {Li} _{2}\left(-{\frac {1}{2}}\right)+{\frac {1}{6}}\operatorname {Li} _{2}\left({\frac {1}{9}}\right)=-{\frac {{\pi }^{2}}{18}}+\ln 2\cdot \ln 3-{\frac {\ln ^{2}2}{2}}-{\frac {\ln ^{2}3}{3}}}$ ^[2]

${\displaystyle \operatorname {Li} _{2}\left({\frac {1}{4}}\right)+{\frac {1}{3}}\operatorname {Li} _{2}\left({\frac {1}{9}}\right)={\frac {{\pi }^{2}}{18}}+2\ln 2\ln 3-2\ln ^{2}2-{\frac {2}{3}}\ln ^{2}3}$  ^[2]

${\displaystyle \operatorname {Li} _{2}\left(-{\frac {1}{3}}\right)-{\frac {1}{3}}\operatorname {Li} _{2}\left({\frac {1}{9}}\right)=-{\frac {{\pi }^{2}}{18}}+{\frac {1}{6}}\ln ^{2}3}$  ^[2]

${\displaystyle \operatorname {Li} _{2}\left(-{\frac {1}{8}}\right)+\operatorname {Li} _{2}\left({\frac {1}{9}}\right)=-{\frac {1}{2}}\ln ^{2}{\frac {9}{8}}}$ ^[2]

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

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

${\displaystyle \operatorname {Li} _{2}(0)=0}$ 

${\displaystyle \operatorname {Li} _{2}\left({\frac {1}{2}}\right)={\frac {{\pi }^{2}}{12}}-{\frac {\ln ^{2}2}{2}}}$ 

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

${\displaystyle \operatorname {Li} _{2}(2)={\frac {{\pi }^{2}}{4}}-i\pi \ln 2}$ 

${\displaystyle \operatorname {Li} _{2}\left(-{\frac {{\sqrt {5}}-1}{2}}\right)=-{\frac {{\pi }^{2}}{15}}+{\frac {1}{2}}\ln ^{2}{\frac {{\sqrt {5}}-1}{2}}}$ 

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

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

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

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

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

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

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

斯盆司函数在粒子物理领域中计算辐射校正时比较常见。此时该函数常用一个含绝对值的对数表达式定义。

${\displaystyle \operatorname {\Phi } (x)=-\int _{0}^{x}{\frac {\ln |1-u|}{u}}\,du={\begin{cases}\operatorname {Li} _{2}(x),&x\leq 1;\\{\frac {\pi ^{2}}{3}}-{\frac {1}{2}}\ln ^{2}(x)-\operatorname {Li} _{2}({\frac {1}{x}}),&x>1.\end{cases}}}$ 

1. ^ ^{1.0 1.1 1.2} Zagier
2. ^ ^{2.0 2.1 2.2 2.3 2.4 2.5 2.6} Weisstein, Eric W. (编). Dilogarithm. at MathWorld--A Wolfram Web Resource. Wolfram Research, Inc. （英语）. 

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