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Consider the holomorphic function $f(z)=\sum\limits_{n=1}^{\infty}\frac{z^n}{n^2}$. How do I find the largest open set to which $f$ can be analytically continued? Is there a closed formula for $f$?

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    It can't be expressed elementarily. $-\int_0^z \frac{\log(1-t)}{t}\mathrm dt$ is what's termed as a *dilogarithm*...2011-10-04

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That function is called the polylogarithm $Li_2(z)$ or dilogarithm. It can be continued to the whole plane minus $0$ and $1$, which are then branch points; this is easy to see, using the integral representation that J.M. mentions in the comment above. The monodromy group is the Heisenberg group.

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    Wait, what? $0$ isn't a branch point... [right](http://www.wolframalpha.com/input/?i=plot%20PolyLog%5B2,x%5D%20for%20x%20from%20-1%20to%201)?2011-10-04
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    It is a branch point is some of the leaves (Morally, one of the branches of $\log(1-t)$ vanishes at zero, so there the pole of $1/t$ is killed; but on the other branches of $\log(1-t)$ the pole is quite there)2011-10-04
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    Oh, right. I was thinking of the principal value...2011-10-04
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    Mariano, Are you sure? The dilogarithm has an analytic continuation to the whole complex plane minus $(1,+\infty)$.2011-10-04
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    Also, $\infty$ is a branch point I think. So if you want a single-valued analytic continuation to a subset of the plane, that subset can be the complement of a curve starting at 1 and going to $\infty$. The "principal branch" has this cut on the real interval $[1,\infty)$.2011-10-04
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    @Didier, Robert: If you continue the function starting from the obvious branch around $\tfrac12$ along a curve turning positively around $1$, and then all the way to $-1$ and which finally makes several turns around 0, also in the positive direction, you find a logarithmic branching point there. This is discussed, for example, in http://links.jstor.org/sici?sici=1364-5021%2820031108%29459%3A2039%3C2807%3ATDFFCA%3E2.0.CO%3B2-F One does not see this on the obvious branch, which Robert mentions, but you do see it on the two branches to which you get when you cross Robert's cut.2011-10-04
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    Google turns out http://arxiv.org/abs/math.CA/0702243 which has a nice section doing the computation of the monodromy.2011-10-04
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    [Here](http://mathlab.snu.ac.kr/~top/articles/Dilogarithm%201.pdf) is a freely accessible version of the first article Mariano linked to.2011-10-04
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    Mariano, not sure I get your point. Are you saying that one can define a single-valued analytic function (which is what the OP asks for) on the complex plane minus $0$ and $1$? By the way, I think the correct answer to the OP is that there is no largest open set on which $f$ can be extended to a single-valued analytic function.2011-10-04
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    @Didier: I am saying that the if $\Sigma$ is the Riemann surface of the complete analytic function dilog and $p:\Sigma\to U$ is the map sending each germ in $\Sigma$ to its point (so that $U$ is an open set in $\mathbb C$, then $p$ is a branched covering space with $0$ and $1$ as branching points. I am using the vocabulary of Conway's *Functions of One Complex Variable*, Chapter IX.2011-10-04
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    Mariano: Right. Hence $\mathbb C\setminus\{0,1\}$ is NOT an open set on which $f$ can be extended to a single-valued analytic function.2011-10-04
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    @Didier: I never said that.2011-10-04
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    The trouble is that you wrote that *The function* (...) *can be continued to the whole plane minus $0$ and $1$*... To prevent misunderstandings, I mention being well aware of the rest of your post and knowing what it means. BUT the same might not apply to every reader here.2011-10-04
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    @MarianoSuárez-Alvarez I would have thought the largest open set was $\mathbb{C}-1$ since $0$ is a removable singularity of the integrand. I don't see how branch cuts enter the convo. Is that due to the log?2015-05-16