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 $f$?
analytic continuation of a power series
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complex-analysis
power-series
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0It 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
1 Answers
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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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0@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