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In wikipedia and also mathworld I find definitions, that the analytic continuation of the PolyLog() at $z=1$ is existent and equals that of the $\zeta()$, but Pari/GP as well as W|A seem to have implemented this only for $s \gt 1$ . W|A gives for the $\text{PolyLog}_s(1)$ at $s=-3$ the symbol $\infty$ but for $\zeta(-3)$ the well known finite value.

In the definition-parts in mathworld I did not see a restriction, but perhaps I've overlooked some thing (while on a second read in wikipedia I find the restriction of the equality to $s \gt 1$).

What is the definition for $\text{Polylog}_s(1) $ for $s\le 1$ ? Is there an analytic continuation?

Gottfried Helms
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2 Answers2

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According to the cited Wiki article the relation to the Zeta function is valid for $\Re(x) >1.$ For negative integers $n$ the polylog is a rational function (http://functions.wolfram.com/10.08.03.0033.01 , see also the series representations of the Lerch transcendent or http://mathworld.wolfram.com/Polylogarithm.html formulas (6ff)) ) $$\mathrm{Li}_{-n}(x) = \frac{1}{(1-x)^{n + 1}} \sum\limits_{m=1}^{n}\left( \sum\limits_{k=1}^{m} (-1)^{k+1}\binom{n+1}{k-1}(m-k+1)^n \right)x^m \quad (n>0) $$

Here $\mathrm{Li}_{-3}(x)=\frac{x^3+4x^2+x}{(1-x)^4}$, and this has no limit for $x\rightarrow 1.$

On the Mathworld site can additionally find the statement regarding the realation to $\zeta(s)$: Note, however, that the meaning of Li_s(z) for fixed complex s is not completely well-defined, since it depends on how s is approached in four-dimensional (s,z)

gammatester
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  • Hmm, so the mathworld-entry seems incomplete. Thanks so far - at the moment I'm still looking around for further references (Abramowitz& Stegun etc). – Gottfried Helms Aug 05 '17 at 08:45
  • Hmm, 1+polylog(0,1-h)-gamma(h) seems to approximate $\gamma_0 \approx 0.57721566$ for $h \to $ zero . But this can perhaps be seen by the series-representations... – Gottfried Helms Aug 05 '17 at 09:07
  • According to your Mathworld-finding: yes, I'd seen that but to work with two limit-processes at once (letting $x \to 0 $ and $s \to -n$ is over my head... so I gave up with that remark. – Gottfried Helms Aug 05 '17 at 09:11
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    Yes, confirmed with Maple limit(1+polylog(0,1-h)-GAMMA(h),h=0); gives $\gamma$ – gammatester Aug 05 '17 at 09:14
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    Playing further I get (1+polylog(-1,1-h)-gamma(h)/h)*h $\to \gamma-1$ and even (1+polylog(-2,1-h)-gamma(h)/h^2)*h^2 approximates $\to 2\gamma-3$ ... – Gottfried Helms Aug 05 '17 at 09:28
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    (1+polylog(-n,1-h)-n!gamma(h)/h^n)*h^n/n! seems to go to $\gamma - (n+1)/2$ for $n \in {1,2,3,...}$ – Gottfried Helms Aug 05 '17 at 09:39
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$$\text{Li}_s(z) = \sum_{n=1}^\infty z^n n^{-s}, \qquad \{|z| < 1\} \cup \{ |z| \le 1, \Re(s)> 1\} $$

$\lim_{z \to 1} \text{Li}_s(z)$ diverges for every $s, \Re(s) < 1$.

Now for $|z| < 1$, $\text{Li}_s(z)$ is entire in $s$, and summing by parts shows it is (locally uniformly) continuous in $z, |z| \le 1, z \ne 1$. Thus $\displaystyle\lim_{z \,\to\, -1} \text{Li}_s(z)$ is analytic in $s$, ie.

$$-(1-2^{1-s}) \zeta(s) = \lim_{z \,\to \,\color{red}{-1}} \text{Li}_s(z), \qquad s \in \mathbb{C}$$

reuns
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  • Hmm, for $z=-1$ it seems Euler-summation (of appropriate order) suffices to arrive at $\eta(s)$ for any $s in \Bbb C$. The problem I am facing is whether there is a defined analytic continuation at $z=1$ (this is part of study of a matrix-summation method which uses series transformation and whose properties I want to analyze for the divergent/non-Euler-summable cases) – Gottfried Helms Aug 05 '17 at 23:29
  • @GottfriedHelms As I said $\lim_{z \to 1} \text{Li}s(z)$ diverges for every $\Re(s) < 1$, while summation by parts shows $-(1-2^{1-s}) \zeta(s) = \lim{z ,\to ,\color{red}{-1}} \text{Li}_s(z)$. You should explain your concrete problem. – reuns Aug 05 '17 at 23:33
  • The question for me is, is there the same analytic continuation as for the $\zeta(s)$ which also "diverges for every $\mathfrak R(s)$" and still has meaningful analytic continuation. There is even the Ramanujan's summation of the $zeta(s)$ at $s=1$ with replacement by $\gamma$. So for $z=1$ the series-representation of the $\text{Li}_s(1)$ "looks" like the $\zeta(s)$ and still the software does not give the analytic-continued values for the divergent cases (which it does for the $\zeta(s)$ except for $s=1$). I want to understand the reason such that I can work with it reliably in all cases. – Gottfried Helms Aug 05 '17 at 23:46
  • If you are interested in this: I had a small essay on a matrix-summation-method and in rereading it with my improved knowledge I got aware, that my series-transformation is not compatible with the assumption of using zeta-series but is implicitely working as a transformation and limiting process of a polylog-series where the $z$-parameter has to be understood as limit going to $1$. I want now rewrite that essay (possibly adding more correct interpretations) It is chap 3 in http://go.helms-net.de/math/binomial_new/EulerianSumsV2.pdf – Gottfried Helms Aug 05 '17 at 23:52
  • @GottfriedHelms I answered to that in my answer : (at first) $\text{Li}_s(z)$ is a function of $s,z$ defined for $|z| < 1$ and $|z|\le1, \Re(s) > 1$. If you fix $z=1$ then you obtain a function of $s$ : $\zeta(s)$ having an analytic continuation. But this continuation is not compatible with what happens for $|z| < 1$. In contrary, at $z= -1$ (more generally $|z|=1,z \ne 1$) the continuation in $s$ and $z$ are compatible. – reuns Aug 05 '17 at 23:53
  • Your pdf is hard to follow, if you want some help, extract a clear motivating example. – reuns Aug 05 '17 at 23:59
  • reuns - I think the problem goes a bit over my head at the moment. Thank you for your effort anyway - I can come back to this if I got it clearer by myself and can get something more from your expositions. – Gottfried Helms Aug 06 '17 at 00:27
  • @GottfriedHelms The idea is just that if $f(s,z)$ is continuous at $(s_0,z_0)$ then $\lim_{s \to s_0} f(s,z_0)=\lim_{z \to z_0} f(s_0,z)$. So we need to find where $\text{Li}_s(z)$ is continuous in $s$ and $z$. If it is not continuous then we need to choose between different possible definitions of $f(s_0,z_0)$. – reuns Aug 06 '17 at 00:33
  • Yes, that aspect of being continuous at $(s_0,t_0)$ is what came up recently when I reread my study - and thus I got aware that I cannot work simply in the alms of the zeta-function but in that of polylogs. My matrix-transformation/-summation method transforms the terms $a_k=(1+k)^{-s}$ of a series $A(s)$ into $b_k=a_k/k!$ , defines then the function $g(x) = \sum b_k x^k $ and then calculates compositions of $g(x)$ at integer $x$ and their derivatives $d^c/dx^c g(x)$ as terms $d_{c}$ of a new series $D(s)$. For the divergent cases of $A(s)$ the series $D(s)$ ... (cntd...) – Gottfried Helms Aug 06 '17 at 01:11
  • (... cntd...) is again divergent but the structure of the terms $d_k$ is "surprising" and subject of my interest. That I became aware that I'm not simply working on an (impossible) zeta-regularization was the fact that I introduce $x$ and powers of $x$ as cofactors of the transformed terms $b_k$ and even derivatives $d/dx$ so I conclude, that in fact my method must reflect rather a "polylog-model" than a "zeta-model". Only because the definitions in WP and mathworld seemed to disagree with Pari/GP and W/A made me ask for more "authoritative" input for the cases $\text{Li}_s(z)$ for $z=1,s<=1$ – Gottfried Helms Aug 06 '17 at 01:19
  • @GottfriedHelms I can't follow what you write. Please find a clear motivating example and explain it clearly, in your question. $\sum_{n \ge 1} a_n n^{-s}$ and $\sum_{n \ge 1} a_n e^{-nx}$ are directly related by the Mellin transform. For $\sum_{n \ge 1} \frac{a_n}{n!} e^{-nx}$ it is more complicated. – reuns Aug 06 '17 at 01:25
  • Additional remark: that somehow complicated method is interesting because it has for the alternating series a similar power as the Borel-summation: it sums the geometric series for $1 \gt q \gt -\infty$ and even the Euler's alternating hypergeometric series apparently correct. – Gottfried Helms Aug 06 '17 at 01:27
  • Well, "matrix-summation" methods have chapters of text for their explanations, I think the chapter 13 of Konrad Knopp's monography on series, which explains summation-methods of divergent series, partly using matrix-based methods, is about 50 or 60 pages long, so I'll not be able to do that in short, I'm afraid... But good night so far, its 3:32 here and deep darkness outside the window and I'll go now to rest :) – Gottfried Helms Aug 06 '17 at 01:32
  • @GottfriedHelms I know many summation methods. My problem is that you didn't write a motivating example of what you'd like to show or investigate. For example choose a simple sequence $a_n$ and say "I would like to show $\displaystyle\overset{\mathfrak{A}}\sum a_n = \displaystyle\overset{\mathfrak{B}}\sum a_n$" where $\displaystyle\overset{\mathfrak{A}}\sum,\displaystyle\overset{\mathfrak{B}}\sum$ are two given summation methods. – reuns Aug 06 '17 at 01:43