Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $f$?

share|improve this question
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

1 Answer 1

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.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.