"Li2" redirects here. For the molecule with formula Li
2 , see 
dilithium .
The dilogarithm along the real axis In mathematics , the dilogarithm  (or Spence's function ), denoted as Li2 (z ) , is a particular case of the polylogarithm . Two related special functions  are referred to as Spence's function, the dilogarithm itself: 
                                    Li                         2                                          (          z          )          =          −                     ∫                         0                                     z                                                                       ln                                (                1                −                u                )                            u                                d          u                     ,                     z          ∈                     C                              {\displaystyle \operatorname {Li} _{2}(z)=-\int _{0}^{z}{\ln(1-u) \over u}\,du{\text{, }}z\in \mathbb {C} }       and its reflection. For |z  , an infinite series  also applies (the integral definition constitutes its analytical extension to the complex plane ): 
                                    Li                         2                                          (          z          )          =                     ∑                         k              =              1                                     ∞                                                                       z                                 k                                                           k                                 2                                                              .                    {\displaystyle \operatorname {Li} _{2}(z)=\sum _{k=1}^{\infty }{z^{k} \over k^{2}}.}       Alternatively, the dilogarithm function is sometimes defined as  
                                    ∫                         1                                     v                                                                       ln                                t                                           1                −                t                                              d          t          =                     Li                         2                                          (          1          −          v          )          .                    {\displaystyle \int _{1}^{v}{\frac {\ln t}{1-t}}dt=\operatorname {Li} _{2}(1-v).}       In hyperbolic geometry  the dilogarithm can be used to compute the volume of an ideal simplex. Specifically, a simplex whose vertices have cross ratio  z  has hyperbolic volume  
                         D          (          z          )          =          Im                               Li                         2                                          (          z          )          +          arg                    (          1          −          z          )          log                               |                    z                     |                    .                    {\displaystyle D(z)=\operatorname {Im} \operatorname {Li} _{2}(z)+\arg(1-z)\log |z|.}       The function D (z )[ 1] Lobachevsky's function  and Clausen's function  are closely related functions. 
William Spence , after whom the function was named by early writers in the field, was a Scottish mathematician working in the early nineteenth century.[ 2] John Galt ,[ 3] 
 
Analytic structure  Using the former definition above, the dilogarithm function is analytic everywhere on the complex plane except at                         z          =          1                    {\displaystyle z=1}                               (          1          ,          ∞          )                    {\displaystyle (1,\infty )}                                          Li                         2                                          (          1          )          =                     π                         2                                           /                    6                    {\displaystyle \operatorname {Li} _{2}(1)=\pi ^{2}/6}       
 
Identities                                     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}).}       [ 4]                                    Li                         2                                          (          1          −          z          )          +                     Li                         2                                                     (                         1              −                                              1                  z                                                      )                    =          −                                                 (                ln                                z                                 )                                     2                                                              2                                .                    {\displaystyle \operatorname {Li} _{2}(1-z)+\operatorname {Li} _{2}\left(1-{\frac {1}{z}}\right)=-{\frac {(\ln z)^{2}}{2}}.}       [ 5]                                    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).}       [ 4] reflection formula .                                   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).}       [ 5]                                    Li                         2                                          (          z          )          +                     Li                         2                                                     (                                        1                z                                      )                    =          −                                                 π                                 2                                            6                                −                                                 (                ln                                (                −                z                )                                 )                                     2                                                              2                                .                    {\displaystyle \operatorname {Li} _{2}(z)+\operatorname {Li} _{2}\left({\frac {1}{z}}\right)=-{\frac {\pi ^{2}}{6}}-{\frac {(\ln(-z))^{2}}{2}}.}       [ 4]                         L                    (          x          )          +          L                    (          y          )          =          L                    (          x          y          )          +          L                               (                                                         x                  (                  1                  −                  y                  )                                                 1                  −                  x                  y                                                      )                    +          L                               (                                                         y                  (                  1                  −                  x                  )                                                 1                  −                  x                  y                                                      )                              {\displaystyle \operatorname {L} (x)+\operatorname {L} (y)=\operatorname {L} (xy)+\operatorname {L} \left({\frac {x(1-y)}{1-xy}}\right)+\operatorname {L} \left({\frac {y(1-x)}{1-xy}}\right)}       [ 6] [ 7]                         L                    (          z          )          =                                  π              6                                [                     Li                         2                                          (          z          )          +                                  1              2                                ln                    (          z          )          ln                    (          1          −          z          )          ]                    {\displaystyle \operatorname {L} (z)={\frac {\pi }{6}}[\operatorname {Li} _{2}(z)+{\frac {1}{2}}\ln(z)\ln(1-z)]}       Rogers  L-function  (an analogous relation is satisfied also by the quantum dilogarithm )
Particular value identities                                     Li                         2                                                     (                                        1                3                                      )                    −                                  1              6                                           Li                         2                                                     (                                        1                9                                      )                    =                                                                  π                                                 2                                            18                                −                                                 (                ln                                3                                 )                                     2                                                              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 3)^{2}}{6}}.}       [ 5]                                    Li                         2                                                     (                         −                                              1                  3                                                      )                    −                                  1              3                                           Li                         2                                                     (                                        1                9                                      )                    =          −                                                                  π                                                 2                                            18                                +                                                 (                ln                                3                                 )                                     2                                                              6                                .                    {\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 {(\ln 3)^{2}}{6}}.}       [ 5]                                    Li                         2                                                     (                         −                                              1                  2                                                      )                    +                                  1              6                                           Li                         2                                                     (                                        1                9                                      )                    =          −                                                                  π                                                 2                                            18                                +          ln                    2          ⋅          ln                    3          −                                                 (                ln                                2                                 )                                     2                                                              2                                −                                                 (                ln                                3                                 )                                     2                                                              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 3)^{2}}{3}}.}       [ 5]                                    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                    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\cdot \ln 3-2(\ln 2)^{2}-{\frac {2}{3}}(\ln 3)^{2}.}       [ 5]                                    Li                         2                                                     (                         −                                              1                  8                                                      )                    +                     Li                         2                                                     (                                        1                9                                      )                    =          −                                  1              2                                                        (                             ln                                                                    9                    8                                                              )                                     2                                .                    {\displaystyle \operatorname {Li} _{2}\left(-{\frac {1}{8}}\right)+\operatorname {Li} _{2}\left({\frac {1}{9}}\right)=-{\frac {1}{2}}\left(\ln {\frac {9}{8}}\right)^{2}.}       [ 5]                         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}.}       
Special values                                     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          )          =          ζ          (          2          )          =                                                                  π                                                 2                                            6                                ,                    {\displaystyle \operatorname {Li} _{2}(1)=\zeta (2)={\frac {{\pi }^{2}}{6}},}                               ζ          (          s          )                    {\displaystyle \zeta (s)}       Riemann zeta function .                                   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                                                                                                                                                                                                 5                                                                                            +                              1                                                        2                                                                                              )                                                             2                                                                                                                  =                  −                                                                                                          π                                                                         2                                                                    15                                                        +                                                          1                      2                                                                           arcsch                                         2                                                                          2.                                                                        {\displaystyle {\begin{aligned}\operatorname {Li} _{2}\left(-{\frac {{\sqrt {5}}-1}{2}}\right)&=-{\frac {{\pi }^{2}}{15}}+{\frac {1}{2}}\left(\ln {\frac {{\sqrt {5}}+1}{2}}\right)^{2}\\&=-{\frac {{\pi }^{2}}{15}}+{\frac {1}{2}}\operatorname {arcsch} ^{2}2.\end{aligned}}}                                                                                                          Li                                         2                                                                                             (                                         −                                                                                                                                                             5                                                                                      +                            1                                                    2                                                                                      )                                                                   =                  −                                                                                                          π                                                                         2                                                                    10                                                        −                                     ln                                         2                                                                                                                                                                                             5                                                                          +                        1                                            2                                                                                                                  =                  −                                                                                                          π                                                                         2                                                                    10                                                        −                                     arcsch                                         2                                                                          2.                                                                        {\displaystyle {\begin{aligned}\operatorname {Li} _{2}\left(-{\frac {{\sqrt {5}}+1}{2}}\right)&=-{\frac {{\pi }^{2}}{10}}-\ln ^{2}{\frac {{\sqrt {5}}+1}{2}}\\&=-{\frac {{\pi }^{2}}{10}}-\operatorname {arcsch} ^{2}2.\end{aligned}}}                                                                                                          Li                                         2                                                                                             (                                                                                         3                          −                                                                                  5                                                                                                      2                                                              )                                                                   =                                                                                                          π                                                                         2                                                                    15                                                        −                                     ln                                         2                                                                                                                                                                                             5                                                                          +                        1                                            2                                                                                                                  =                                                                                                          π                                                                         2                                                                    15                                                        −                                     arcsch                                         2                                                                          2.                                                                        {\displaystyle {\begin{aligned}\operatorname {Li} _{2}\left({\frac {3-{\sqrt {5}}}{2}}\right)&={\frac {{\pi }^{2}}{15}}-\ln ^{2}{\frac {{\sqrt {5}}+1}{2}}\\&={\frac {{\pi }^{2}}{15}}-\operatorname {arcsch} ^{2}2.\end{aligned}}}                                                                                                          Li                                         2                                                                                             (                                                                                                                                                 5                                                                                −                          1                                                2                                                              )                                                                   =                                                                                                          π                                                                         2                                                                    10                                                        −                                     ln                                         2                                                                                                                                                                                             5                                                                          +                        1                                            2                                                                                                                  =                                                                                                          π                                                                         2                                                                    10                                                        −                                     arcsch                                         2                                                                          2.                                                                        {\displaystyle {\begin{aligned}\operatorname {Li} _{2}\left({\frac {{\sqrt {5}}-1}{2}}\right)&={\frac {{\pi }^{2}}{10}}-\ln ^{2}{\frac {{\sqrt {5}}+1}{2}}\\&={\frac {{\pi }^{2}}{10}}-\operatorname {arcsch} ^{2}2.\end{aligned}}}       
In particle physics  Spence's Function is commonly encountered in particle physics  while calculating radiative corrections . In this context, the function is often defined with an absolute value  inside the logarithm: 
                                    Φ                              (          x          )          =          −                     ∫                         0                                     x                                                                       ln                                                 |                                1                −                u                                 |                                            u                                d          u          =                                  {                                                                                      Li                                             2                                                                                  (                    x                    )                    ,                                                       x                    ≤                    1                    ;                                                                                                                                                           π                                                     2                                                                          3                                                              −                                                                1                        2                                                              (                    ln                                        x                                         )                                             2                                                              −                                         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 x)^{2}-\operatorname {Li} _{2}({\frac {1}{x}}),&x>1.\end{cases}}}       
See also   
Notes   
References  Lewin, L. (1958). Dilogarithms and associated functions . Foreword by J. C. P. Miller. London: Macdonald. MR  0105524 . Morris, Robert (1979). "The dilogarithm function of a real argument" . Math. Comp . 33  (146): 778– 787. doi :10.1090/S0025-5718-1979-0521291-X MR  0521291 . Loxton, J. H. (1984). "Special values of the dilogarithm" . Acta Arith . 18  (2): 155– 166. doi :10.4064/aa-43-2-155-166 MR  0736728 . Kirillov, Anatol N. (1995). "Dilogarithm identities". Progress of Theoretical Physics Supplement . 118 : 61– 142. arXiv :hep-th/9408113 Bibcode :1995PThPS.118...61K . doi :10.1143/PTPS.118.61 . S2CID  119177149 . Osacar, Carlos; Palacian, Jesus; Palacios, Manuel (1995). "Numerical evaluation of the dilogarithm of complex argument". Celest. Mech. Dyn. Astron . 62  (1): 93– 98. Bibcode :1995CeMDA..62...93O . doi :10.1007/BF00692071 . S2CID  121304484 . Zagier, Don (2007). "The Dilogarithm Function". In Pierre Cartier; Pierre Moussa; Bernard Julia; Pierre Vanhove (eds.). Frontiers in Number Theory, Physics, and Geometry II (PDF) . pp. 3– 65. doi :10.1007/978-3-540-30308-4_1 . ISBN  978-3-540-30308-4 . 
Further reading   
External links