3

For real $s>1$ the polylogarithm for $z=1$ reduces to the Riemann zeta function $\mathrm{Li}_s(1) = \zeta(s)$. For real $s<1$ Wolfram Alpha and the Wolfram function site give $\mathrm{Li}_s(1) = \infty, s<1$ whereas the mpmath package returns $\mathrm{Li}_s(1) = \zeta(s)$ for all $s\ne 1$. Is this just a convention or is it a kind of continuation for $s?$

Antonio Vargas
  • 24,993
gammatester
  • 18,827

1 Answers1

2

If you define the polylogarithm and zeta functions with the series :$$\operatorname{Li}_s(z):=\sum_{k=1}^\infty \frac {z^k}{k^s}\\\zeta(s):=\sum_{k=1}^\infty \frac 1{k^s}$$ then you'll be in the first case and have to suppose $\;\Re(s)>1\;$ for convergence but if you consider the analytic continuation of these two series as being the actual functions then things become much more interesting and considering every complex $s$ such that $\;s\ne 1\;$ should be the fine choice (see the discussions in the Wikipedia links or here concerning zeta).

Community
  • 1
Raymond Manzoni
  • 43,021
  • 5
  • 86
  • 140
  • Would $\mathrm{Li}_s(1) = \zeta(s)$ for $s \ne 1$ be the analytic continuation, because this relation holds for $\Re s >1$? – gammatester May 20 '15 at 14:02
  • Yes @gammatester. From the equality for $\Re s >1$ and since $\zeta$ admits a unique analytic continuation over the whole complex plane except $s=1$ we must have the equality for $s\neq 1$ (else $\operatorname{Li}_s(1)$ would not be analytic in $s$ !). – Raymond Manzoni May 20 '15 at 14:29