# What is the general method to do analytic continuation and the obstruction to do analytic continuation?

Given an analytic function on some domain. What is the general method to do analytic continuation and the obstruction to do analytic continuation? For example, $g(z)=1+z+z^2+\cdots$ is analytic on $|z|<1$. $f(z)=\frac{1}{1-z}$ is an analytic continuation of $g$ since $f$ is analytic on $\{z \mid z\neq 0\}$ which contains $\{z \mid |z|<1 \}$. Is the general method to find the analytic continuation computing the sum of the series? Is there an analytic function which is analytic on $\mathbb{C}$ which is a analytic continuation of $g(z)$? Why analytic continuation is important?

-
I'm confused. First you say $f$ is an analytic continuation of $g$ (correct, except you mean $z\ne1$), then you ask if $g$ has an analytic continuation.... – anon Oct 1 '11 at 21:24
...and then there are functions that you just can't analytically continue, like $$\prod_{k=0}^\infty (1-q^{k+1})$$... – J. M. Oct 1 '11 at 23:44

For analytic continuation there is Borel summation for divergent series. No, there is not such analytic continuation of $g$ since the difference $f-g\;$ is equal to zero in the unit circle and therefore its continuation is zero in $\mathbb C$. Analytic continuation is important for many reasons as discussed in complex analysis. For example, the behavior of the analytic continuation of the series $\sum_{n=1}^\infty n^{-z}\;$ in the critical strip is closely related with the distribution of primes.