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Algorithms for polylogarithms

The polylogarithm is defined for $s,z\in \mathbb{C}$ with $|z|<1$ by

$${\mathrm{Li}}_s(z)=\sum _{k=1}^{\mathrm{\infty }}\frac{z^k}{k^s}$$

and for $|z|\ge 1$ by analytic continuation, except for the singular point $z=1$.

Computation for small z

The power sum converges rapidly when $|z|\ll 1$. To compute the series expansion with respect to $s$, we substitute $s\to s+x\in \mathbb{C}[[x]]$ and obtain

$${\mathrm{Li}}_{s+x}(z)=\sum _{d=0}^{\mathrm{\infty }}{x}^d\frac{(-1{)}^d}{d!}\sum _{k=1}^{\mathrm{\infty }}T(k)$$

where

$$T(k)=\frac{z^k{\log }^d(k)}{k^s}.$$

The remainder term $|\sum _{k=N}^{\mathrm{\infty }}T(k)|$ is bounded via the following strategy, implemented in mag_polylog_tail().

Denote the terms by $T(k)$. We pick a nonincreasing function $U(k)$ such that

$$\frac{T(k+1)}{T(k)}=z{\left(\frac{k}{k+1}\right)}^s{\left(\frac{\log (k+1)}{\log (k)}\right)}^d\le U(k).$$

Then, as soon as $U(N)<1$,

$$\sum _{k=N}^{\mathrm{\infty }}T(k)\le T(N)\sum _{k=0}^{\mathrm{\infty }}U(N{)}^k=\frac{T(N)}{1-U(N)}.$$

In particular, we take

$$U(k)=zB(k,max(0,-s))B(k\log (k),d)$$

where $B(m,n)=(1+1/m{)}^n$. This follows from the bounds

$$\begin{array}{r}{\left(\frac{k}{k+1}\right)}^s\le \{\begin{array}{ll}1& \text{if }s\ge 0\\ (1+1/k{)}^{-s}& \text{if }s<0.\end{array}\end{array}$$

and

$${\left(\frac{\log (k+1)}{\log (k)}\right)}^d\le {(1+\frac{1}{k\log (k)})}^d.$$

Expansion for general z

For general complex $s,z$, we write the polylogarithm as a sum of two Hurwitz zeta functions

$${\mathrm{Li}}_s(z)=\frac{\mathrm{\Gamma }(v)}{(2\pi {)}^v}[i^v\zeta (v,\frac{1}{2}+\frac{\log (-z)}{2\pi i})+i^{-v}\zeta (v,\frac{1}{2}-\frac{\log (-z)}{2\pi i})]$$

in which $s=1-v$. With the principal branch of $\log (-z)$, we obtain the conventional analytic continuation of the polylogarithm with a branch cut on $z\in (1,+\mathrm{\infty })$.

To compute the series expansion with respect to $v$, we substitute $v\to v+x\in \mathbb{C}[[x]]$ in this formula (at the end of the computation, we map $x\to -x$ to obtain the power series for ${\mathrm{Li}}_{s+x}(z)$). The right hand side becomes

$$\mathrm{\Gamma }(v+x)[E_1{Z}_1+E_2{Z}_2]$$

where $E_1=(i/(2\pi ){)}^{v+x}$, $Z_1=\zeta (v+x,\dots )$, $E_2=(1/(2\pi i){)}^{v+x}$, $Z_2=\zeta (v+x,\dots )$.

When $v=1$, the $Z_1$ and $Z_2$ terms become Laurent series with a leading $1/x$ term. In this case, we compute the deflated series ${\tilde{Z}}_1,{\tilde{Z}}_2=\zeta (x,\dots )-1/x$. Then

$$E_1{Z}_1+E_2{Z}_2=(E_1+E_2)/x+E_1{\tilde{Z}}_1+E_2{\tilde{Z}}_2.$$

Note that $(E_1+E_2)/x$ is a power series, since the constant term in $E_1+E_2$ is zero when $v=1$. So we simply compute one extra derivative of both $E_1$ and $E_2$, and shift them one step. When $v=0,-1,-2,\dots$, the $\mathrm{\Gamma }(v+x)$ prefactor has a pole. In this case, we proceed analogously and formally multiply $x\mathrm{\Gamma }(v+x)$ with $[E_1{Z}_1+E_2{Z}_2]/x$.

Note that the formal cancellation only works when the order $s$ (or $v$) is an exact integer: it is not currently possible to use this method when $s$ is a small ball containing any of $0,1,2,\dots$ (then the result becomes indeterminate).

The Hurwitz zeta method becomes inefficient when $|z|\to 0$ (it gives an indeterminate result when $z=0$). This is not a problem since we just use the defining series for the polylogarithm in that region. It also becomes inefficient when $|z|\to \mathrm{\infty }$, for which an asymptotic expansion would better.