Showing changes from revision #5 to #6:  
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    Define
Li  2 ( z ) ≔ ∑  n ≥ 1 z  n n  2 , | z | < 1 
  \operatorname{Li}_2(z) \coloneqq \sum_{n\geq 1} \frac{z^n}{n^2},\qquad |z|\lt 1
  
More generally, the polylogarithm m = 1 , 2 , … m=1,2,\dots 
Li  m ( z ) ≔ ∑  n ≥ 1 z  n n  m , | z | < 1 
  \operatorname{Li}_m(z) \coloneqq \sum_{n\geq 1} \frac{z^n}{n^m},\qquad |z|\lt 1
  
Note that
Li  1 ( z ) = − log ( 1 − z ) 
 \operatorname{Li}_1(z) = -\log(1-z)
  
and
d d z Li  m ( z ) = Li  m − 1 ( z ) 
  \frac{d}{d z} \operatorname{Li}_m(z) = \operatorname{Li}_{m-1}(z)
  
So we get an analytic continuation
Li  2 ( z ) = − ∫  0  z log ( 1 − u ) d u u 
  \operatorname{Li}_2(z) = -\int_0^z \log(1-u) \frac{d u}{u}
  
where the path from 0 0   to z z   is in ℂ ∖ [ 1 , ∞ ) \mathbb{C}\setminus [1,\infty) 
Functional equations:
Li  1 ( 1 − x y ) = Li  1 ( 1 − x ) + Li  1 ( 1 − y )   Li  2 = 5 terms (Spence 1809, Abel 1828, ...) 
\begin{gathered}
\operatorname{Li}_1(1-x y) = \operatorname{Li}_1(1-x) + \operatorname{Li}_1(1-y)\\
\operatorname{Li}_2 = \text{5 terms (Spence 1809, Abel 1828, ...)}
\end{gathered}
  
Monodromy (on Li  2 ( x ) , log ( x ) , 1 \operatorname{Li}_2(x),\log(x),1  )
γ  0 = ( 1   0   0   0   1   2 π i   0   0   1 ) , γ  1 = ( 1   − 2 π i   0   0   1   0   0   0   1 ) 
\gamma_0=\begin{pmatrix}1&0&0\\0&1&2\pi i\\0&0&1\end{pmatrix},
\gamma_1=\begin{pmatrix}1&-2\pi i&0\\0&1&0\\0&0&1\end{pmatrix}
  
generate a Heisenberg group
( 1   ℤ ( 1 )   ℤ ( 2 )   0   1   ℤ ( 1 )   0   0   1 ) 
\begin{pmatrix}1&\mathbb{Z}(1)&\mathbb{Z}(2)\\ 0&1&\mathbb{Z}(1)\\0&0&1\end{pmatrix}
  
Bloch-Wigner Dilogarithm 
D ( z ) ≔ Im Li  2 ( z ) + arg ( 1 − z ) log | z | 
D(z) \coloneqq \operatorname{Im} \operatorname{Li}_2(z) + \arg(1-z)\log|z|
  
is real-analytic in ℂ ∖ { 0 , 1 } \mathbb{C}\setminus\{0,1\}   and continuous in ℂ \mathbb{C}  .
D ( e  i θ ) = ∑  n ≥ 1 sin n θ n  2   D ( z ¯ ) = − D ( z ) 
\begin{gathered}
D\left(e^{i\theta}\right) = \sum_{n\geq 1} \frac{\sin n\theta}{n^2}\\
D(\overline{z}) = - D(z)
\end{gathered}
  
hence vanishes on ℝ \mathbb{R}  .
D ( z )   = D ( 1 − z  − 1 ) = D ( ( 1 − z )   − 1 )     − D ( z  − 1 ) = − D ( 1 − z ) = − D ( − z 1 − z ) 
\begin{split}
D(z)&= D\left(1-z^{-1}\right)= D\left({(1-z)}^{-1}\right)\\
& - D\left(z^{-1}\right) = - D(1-z) = -D\left(-\frac{z}{1-z}\right)
\end{split}
  
So we have a continuous real-vaued function on ℙ  1 ( ℂ ) \mathbb{P}^1(\mathbb{C})   with a maximum at z = ( 1 + − 3 ) / 2 z=(1+\sqrt{-3})/2  : D ( 1 + − 3 ) / 2 ) = 1.0149 … D(1+\sqrt{-3})/2)=1.0149\dots  .
Define recursively
z  n + 1 z  n − 1 = 1 − z  n 
  z_{n+1}z_{n-1} = 1-z_n
  
then z  n + 5 = z  n z_{n+5}=z_n  . If we call z  0 = x z_0=x  , z  1 = y z_1=y  , then we find
x , y , 1 − y x , x + y − 1 xy , 1 − x y 
x,y,\frac{1-y}{x},\frac{x+y-1}{xy},\frac{1-x}{y}
  
(Laurent phenomenon). (Cremona transformation of order 5 on ℙ  2 ( ℂ ) \mathbb{P}^2(\mathbb{C})   is ( x , y ) ↦ ( y , 1 − y x ) (x,y)\mapsto\left(y,\tfrac{1-y}{x}\right)  .)
The 5-term recursion relation is
∑  j = 0   4 D ( z  j ) = 0 
\sum_{j=0}^4 D(z_j)=0
  
This can be explained geometrically.
 
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In hyperbolic space, an ideal tetrahedron, with vertices at 0 , 1 , ∞ , z 0,1,\infty,z  , has volume D ( z ) D(z)  . (z = 1 + − 3 2  z=\tfrac{}{} z=\tfrac{1+\sqrt{-3}}{2}   is the regular tetrahedron, tetrahedron;  more generally, z z   is the cross ratio of the 4 vertices , which is invariant under PSL  2 ( ℂ ) = Isom ( ℍ ) PSL_2(\mathbb{C})=\operatorname{Isom}(\mathbb{H})  ) The 5-term recursion relation comes from taking 5 points in ℙ  1 ( ℂ ) \mathbb{P}^1(\mathbb{C})   and constructing five tetrahedra by taking the points 4 at a time
0 = ∑  j = 0   4 ( − 1 )   j Vol ( ( w  0 , … , w ^   j , … , w  4 ) ) 
   0 = \sum_{j=0}^4 {(-1)}^{j}\operatorname{Vol}((w_0,\dots,\hat{w}_j,\dots,w_4))
  
The cancellation is the 3-2 Pachner move.
 
 
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 Napier : Mirifici logorithorum canonis descriptio 
 Rule of circular parts
 Spherical trigonometry (navigation mathematics): 6 quantities (3 angles + 3 lengths or, equivalently, 6 angles). Fix one to be π / 2 \pi/2  .
 Parametrize parts wiith cross ratios of 4 points taken out of 5. The sides of the pentagon are A , B , b ′ , c , a ′ A,B,b',c,a'  .
 Any equation among the parts remains valid after a cyclic permutation around the pentagon. Denote the five triangles by ( a  i , B  i , c  i , A  i , b  i ) (a_i, B_i, c_i, A_i, b_i)  , i = 0 , 1 , … , 4 i=0,1,\dots,4  , with ( a  0 , B  0 , c  0 , A  0 , b  0 ) ≡ ( a , B , c , A , b ) (a_0, B_0, c_0, A_0, b_0)\equiv(a, B, c, A, b)  . Under a cyclic permutation
 ( a , B , c , A , A , b ) ↦ ( a  1 , B  1 , c  1 , A  1 , b  1 ) = ( A ′ , b ′ , a ′ , B , c ′ ) 
   (a,B,c,A,A,b)\mapsto (a_1,B_1,c_1,A_1,b_1) = (A',b',a',B,c')
 
 and
 ( a , B ′ , c ′ , A ′ , b ) ↦ ( A ′ , b , a , B ′ , c ′ ) 
 (a,B',c',A',b)\mapsto (A',b,a,B',c')
 
 Triangle can be solved if any two parts are known
 sin a = tan b tan B ′ = cos A ′ cos c ′ 
\sin a =\tan b \tan B' = \cos A' \cos c'
 
 The surface
 S : { 1 − z  1 = z  2 z  0   1 − z  2 = z  3 z  1   ⋮   1 − z  0 = z  1 z  4 Aut ( S ) ≃ S  5 
 S: \left\{\begin{gathered}1-z_1=z_2z_0\\ 1-z_2=z_3z_1\\ \vdots\\ 1-z_0 = z_1z_4 \end{gathered}\right.
\qquad
\operatorname{Aut}(S)\simeq S_5
 
 is a del Pezzo surface of degree 5.
 ∑  j = 0   4 z  j = 3 − s , − s = ∏  j = 0   4 z  j Schöne Gleichung   ( 1 − x ) ( 1 − y ) ( 1 − x − y ) − s xy = 0 
\begin{gathered}
\sum_{j=0}^4 z_j = 3-s,\quad -s=\prod_{j=0}^4 z_j\qquad\text{Schöne Gleichung}\\
(1-x)(1-y)(1-x-y) -s xy =0
\end{gathered}
 
 universal elliptic curve with a 5-torsion point x  1 ( 5 ) x_1(5)  .
 Coxeter: 5-cycle
 transmitted as mathematical gossip for a long time.
 Number Theory 
 Li  2 ( 1 ) = ζ ( 2 ) = π  2 6 (Euler 1768) 
  \operatorname{Li}_2(1)=\zeta(2) = \frac{\pi^2}{6}\qquad\text{(Euler 1768)}
 
 More generally,
 L ( χ , 2 ) = ∑  n ≥ 1 χ ( n ) n  2   χ : ( ℤ / N ℤ )   × → ℂ  × Dirichlet character 
\begin{gathered}
 L(\chi,2) = \sum_{n\geq 1} \frac{\chi(n)}{n^2}\\
\chi\colon {(\mathbb{Z}/N\mathbb{Z})}^\times \to \mathbb{C}^\times \qquad\text{Dirichlet character}
\end{gathered}