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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... –  J. M. Oct 4 '11 at 5:37
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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? –  J. M. Oct 4 '11 at 5:46
    
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) –  Mariano Suárez-Alvarez Oct 4 '11 at 5:47
    
Oh, right. I was thinking of the principal value... –  J. M. Oct 4 '11 at 5:49
    
Mariano, Are you sure? The dilogarithm has an analytic continuation to the whole complex plane minus $(1,+\infty)$. –  Did Oct 4 '11 at 6:47
    
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 $[1,\infty)$. –  Robert Israel Oct 4 '11 at 7:05
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