斯盆司函数

斯盆司函数（英文：Spence's function）也叫二重对数函数（英文：Dilogarithm），最早由欧拉提出，定义如下： Li 2 ⁡ ( z ) = − ∫ 0 z ln ⁡ ( 1 − u ) u d u ,  z ∈ C ∖ [ 1 , ∞ ) {\displaystyle \operatorname {Li} _{2}(z)=-\int _{0}^{z}{\ln(1-u) \over u}\,\mathrm {d} u{\text{, }}z\in \mathbb {C} \setminus [1,\infty )} Li 2 ⁡ ( z ) + Li 2 ⁡ ( − z ) = 1 2 Li 2 ⁡ ( z 2 ) {\displaystyle \operatorname {Li} _{2}(z)+\operatorname {Li} _{2}(-z)={\frac {1}{2}}\operatorname {Li} _{2}(z^{2})} Li 2 ⁡ ( 1 − z ) + Li 2 ⁡ ( 1 − 1 z ) = − ln 2 ⁡ z 2 {\displaystyle \operatorname {Li} _{2}(1-z)+\operatorname {Li} _{2}\left(1-{\frac {1}{z}}\right)=-{\frac {\ln ^{2}z}{2}}} Li 2 ⁡ ( z ) + Li 2 ⁡ ( 1 − z ) = π 2 6 − ln ⁡ z ⋅ ln ⁡ ( 1 − z ) {\displaystyle \operatorname {Li} _{2}(z)+\operatorname {Li} _{2}(1-z)={\frac {{\pi }^{2}}{6}}-\ln z\cdot \ln(1-z)} Li 2 ⁡ ( − z ) − Li 2 ⁡ ( 1 − z ) + 1 2 Li 2 ⁡ ( 1 − z 2 ) = − π 2 12 − ln ⁡ z ⋅ ln ⁡ ( 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)} Li 2 ⁡ ( z ) + Li 2 ⁡ ( 1 z ) = − π 2 6 − 1 2 ln 2 ⁡ ( − z ) {\displaystyle \operatorname {Li} _{2}(z)+\operatorname {Li} _{2}({\frac {1}{z}})=-{\frac {\pi ^{2}}{6}}-{\frac {1}{2}}\ln ^{2}(-z)} Li 2 ⁡ ( 1 3 ) − 1 6 Li 2 ⁡ ( 1 9 ) = π 2 18 − ln 2 ⁡ 3 6 {\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}}} Li 2 ⁡ ( − 1 2 ) + 1 6 Li 2 ⁡ ( 1 9 ) = − π 2 18 + ln ⁡ 2 ⋅ ln ⁡ 3 − ln 2 ⁡ 2 2 − ln 2 ⁡ 3 3 {\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}}} Li 2 ⁡ ( 1 4 ) + 1 3 Li 2 ⁡ ( 1 9 ) = π 2 18 + 2 ln ⁡ 2 ln ⁡ 3 − 2 ln 2 ⁡ 2 − 2 3 ln 2 ⁡ 3 {\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} Li 2 ⁡ ( − 1 3 ) − 1 3 Li 2 ⁡ ( 1 9 ) = − π 2 18 + 1 6 ln 2 ⁡ 3 {\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} Li 2 ⁡ ( − 1 8 ) + Li 2 ⁡ ( 1 9 ) = − 1 2 ln 2 ⁡ 9 8 {\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}}} 36 Li 2 ⁡ ( 1 2 ) − 36 Li 2 ⁡ ( 1 4 ) − 12 Li 2 ⁡ ( 1 8 ) + 6 Li 2 ⁡ ( 1 64 ) = π 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}} Li 2 ⁡ ( − 1 ) = − π 2 12 {\displaystyle \operatorname {Li} _{2}(-1)=-{\frac {{\pi }^{2}}{12}}} Li 2 ⁡ ( 0 ) = 0 {\displaystyle \operatorname {Li} _{2}(0)=0} Li 2 ⁡ ( 1 2 ) = π 2 12 − ln 2 ⁡ 2 2 {\displaystyle \operatorname {Li} _{2}\left({\frac {1}{2}}\right)={\frac {{\pi }^{2}}{12}}-{\frac {\ln ^{2}2}{2}}} Li 2 ⁡ ( 1 ) = π 2 6 {\displaystyle \operatorname {Li} _{2}(1)={\frac {{\pi }^{2}}{6}}} Li 2 ⁡ ( 2 ) = π 2 4 − i π ln ⁡ 2 {\displaystyle \operatorname {Li} _{2}(2)={\frac {{\pi }^{2}}{4}}-i\pi \ln 2} Li 2 ⁡ ( − 5 − 1 2 ) = − π 2 15 + 1 2 ln 2 ⁡ 5 − 1 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}}} = − π 2 15 + 1 2 arcsch 2 ⁡ 2 {\displaystyle =-{\frac {{\pi }^{2}}{15}}+{\frac {1}{2}}\operatorname {arcsch} ^{2}2} Li 2 ⁡ ( − 5 + 1 2 ) = − π 2 10 − ln 2 ⁡ 5 + 1 2 {\displaystyle \operatorname {Li} _{2}\left(-{\frac {{\sqrt {5}}+1}{2}}\right)=-{\frac {{\pi }^{2}}{10}}-\ln ^{2}{\frac {{\sqrt {5}}+1}{2}}} = − π 2 10 − arcsch 2 ⁡ 2 {\displaystyle =-{\frac {{\pi }^{2}}{10}}-\operatorname {arcsch} ^{2}2} Li 2 ⁡ ( 3 + 5 2 ) = π 2 15 − 1 2 ln 2 ⁡ 5 − 1 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}}} = π 2 15 − 1 2 arcsch 2 ⁡ 2 {\displaystyle ={\frac {{\pi }^{2}}{15}}-{\frac {1}{2}}\operatorname {arcsch} ^{2}2} Li 2 ⁡ ( 5 + 1 2 ) = π 2 10 − ln 2 ⁡ 5 − 1 2 {\displaystyle \operatorname {Li} _{2}\left({\frac {{\sqrt {5}}+1}{2}}\right)={\frac {{\pi }^{2}}{10}}-\ln ^{2}{\frac {{\sqrt {5}}-1}{2}}} = π 2 10 − arcsch 2 ⁡ 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。 Φ ⁡ ( x ) = − ∫ 0 x ln ⁡ | 1 − u | u d u = { Li 2 ⁡ ( x ) , x ≤ 1 ; π 2 3 − 1 2 ln 2 ⁡ ( x ) − Li 2 ⁡ ( 1 x ) , x > 1. {\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}}} Zagier Weisstein, Eric W. 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