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How to show that all possible collection of analytic continuations of $\displaystyle f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2} $ has singular point at $z = 1$. I know that $f(z)$ converges for $|z| \le 1$. Also is there a theorem that relates the singularity of analytic continuation with circle of convergence?

Mula Ko Saag
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  • The radius of convergence is the distance to the nearest singularity. So, from the known radius in this case, there is at least one singularity on the circle $|z|=1$. For more information in this case, look for the "dilogarithm" http://mathworld.wolfram.com/Dilogarithm.html – GEdgar May 14 '13 at 19:05
  • The post seems to have been edited so I'm not sure if the wording fits the data, but as far as I can see now the given series converges, and rather beautifully, for $,z=1,$ ... – DonAntonio May 14 '13 at 22:16
  • Hi, Mula Ko Saag. Any feedback to my answer? – 23rd May 17 '13 at 17:14
  • @Landscape Hi .. I am not much experienced on analytic continuation. There is a small chapter in my book and there is only one example which is also already asked here. Perhaps you could give me some probles and links analytic continuation. – Mula Ko Saag May 17 '13 at 17:18
  • @MulaKoSaag: Analytic continuation could be a big topic leading to, say the theory of Riemann surfaces. I have no good recommendation for a short introduction, so I can only restrict my explanation to this question itself. Please see the comments under my answer. – 23rd May 17 '13 at 17:56

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Note that when $|z|<1$, $$f'(z)=\sum_{n=1}^\infty\frac{z^{n-1}}{n}=-\frac{\log(1-z)}{z}.\tag{1}$$ Since the right hand side of $(1)$ has a unique singularity at $z=1$, it implies that on the unit circle, the analytic continuation of $f$ has a unique singularity at $z=1$.

For the general situation, note that for one complex variable, any (nonempty) open set in $\mathbb{C}$ is a domain of holomorphy, which implies that for any closed subset $S$ of the unit circle, there exists a holomorphic function $f$ defined on the unit disk, such that the collection of sigularities of the analytic continuation of $f$ on the unit circle is precisely $S$.

Remark: Just in case, $(1)$ follows from integrating $$(zf'(z))'=\sum_{n=0}^\infty z^n=\frac{1}{1-z}.$$

23rd
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  • also according to wikipedia article it can be extended to |z| ≥ 1 by the process of analytic continuation. which is great source of my confusion. I am doubting the question on my book. – Mula Ko Saag May 17 '13 at 17:21
  • @MulaKoSaag: Let us only restrict our discussion to holomorphic functions defined on some open set in $\mathbb C$. function. To me, the easiest way to define analytic continuation is as follows. Let $G\subset \mathbb C$ be an open set and let $f:G\to \mathbb C$ be holomorphic. An analytic continuation of $f$ is some holomorphic function $F$, defined on a larger open set $H\supsetneq G$, such that $F=f$ on $H$. – 23rd May 17 '13 at 17:49
  • @MulaKoSaag: In your question, $G$ is the unit disk, and for $H=\mathbb C\setminus [1,+\infty)$, the RHS of $(1)$ is holomorphic on $H$. Since $H$ is simply connected, you can integrate RHS of $(1)$ on $H$ to get a holomorphic function $F$ defined on $H$ with $F(0)=0$. Then $F(0)=f(0)$ and $F'=f'$ on $G$, so $F=f$ on $G$, i.e. $F$ is an analytic continuation of $f$. This implies that for every $z\ne 1$ with $|z|=1$, $f$ has analytic continuation around $z$. – 23rd May 17 '13 at 17:49
  • @MulaKoSaag: However, for $z=1$, $f$ cannot have any analytic continuation for any open set $H\supset G\cup{1}$. It is because, if $F$ is such a continuation, then $F'=f'$ on $G$, so from $(1)$ we know that $F'$, and hence $F$ has a singularity at $z=1$, i.e. $F$ cannot be holomorphic on $H$. – 23rd May 17 '13 at 17:51
  • all right I'll try to digest it. But before that, could you explain why we have $H=\mathbb C\setminus [1,+\infty)$ since we have singularity only on $1$, not on other point of $\Bbb R$. – Mula Ko Saag May 17 '13 at 18:40
  • And could you help me with short question: There is a theorem that states that $F(z)$ defined by all possible analytic continuation of of a series $\sum a_n z^n $ has at least one point on circle of convergence. What is the name of this theorem? could you give me links? – Mula Ko Saag May 17 '13 at 18:46
  • @MulaKoSaag: Do you mean at least one singular point on circle of convergence? I don't know its name, and I suspect that it has no well known name, because it follows from definitions immediately: assume no such point exists, then by the compactness of the circle of convergence, $f$ has analytic continuation to a larger disk, which means that the radius of convergence is actually larger, contradicting to the definition of circle of convergence. – 23rd May 17 '13 at 18:52
  • thanks actually you proved it!! – Mula Ko Saag May 17 '13 at 18:53
  • @MulaKoSaag: You are welcome. – 23rd May 17 '13 at 18:55
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This is one of the polylogarithm, also $s=2$ case is called dilogarithm as GEdgar pointed out. The definition is $$ \operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s} = z + {z^2 \over 2^s} + {z^3 \over 3^s} + \cdots \, $$ Your case it is $s=2$, so $\operatorname{Li}_2(z)$.

According to http://en.wikipedia.org/wiki/Polylogarithm, we have a integral representation of $\operatorname{Li}_s(z)$:

$$ \operatorname{Li}_{s}(z) = {1 \over \Gamma(s)} \int_0^\infty {t^{s-1} \over e^t/z-1} \,\mathrm{d}t \,. $$

So in your case $$ \operatorname{Li}_{2}(z) = \int_0^\infty {zt \over e^t - z} \,\mathrm{d}t \,.$$

This allows an analytic continuation to $z\in \mathbb{C}-[1,\infty)$.

Sungjin Kim
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