Editing Analytic continuation (section)

===Example I: A function with a natural boundary at zero (the prime zeta function)===
For <math>\Re(s) > 1</math> we define the so-called [[prime zeta function]], <math>P(s)</math>, to be

:<math>P(s) := \sum_{p\ \text{ prime}} p^{-s}.</math>

This function is analogous to the summatory form of the [[Riemann zeta function]] when <math>\Re(s) > 1</math> in so much as it is the same summatory function as <math>\zeta(s)</math>, except with indices restricted only to the [[prime numbers]] instead of taking the sum over all positive [[natural numbers]]. The prime zeta function has an analytic continuation to all complex ''s'' such that <math>0 < \Re(s) < 1</math>, a fact which follows from the expression of <math>P(s)</math> by the logarithms of the [[Riemann zeta function]] as

:<math>P(s) = \sum_{n \geq 1} \mu(n)\frac{\log\zeta(ns)}{n}.</math>

Since <math>\zeta(s)</math> has a simple, non-removable pole at <math>s := 1</math>, it can then be seen that <math>P(s)</math> has a simple pole at <math>s := \tfrac{1}{k}, \forall k \in \Z^{+}</math>. Since the set of points

:<math>\operatorname{Sing}_P := \left\{k^{-1} : k \in \Z^+\right\} = \left \{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},\ldots \right \}</math>

has accumulation point 0 (the limit of the sequence as <math>k\mapsto\infty</math>), we can see that zero forms a natural boundary for <math>P(s)</math>. This implies that <math>P(s)</math> has no analytic continuation for ''s'' left of (or at) zero, i.e., there is no continuation possible for <math>P(s)</math> when <math>0 \geq \Re(s)</math>. As a remark, this fact can be problematic if we are performing a complex contour integral over an interval whose real parts are symmetric about zero, say <math>I_F \subseteq \Complex \ \text{such that}\ \Re(s) \in (-C, C), \forall s \in I_F</math> for some <math>C > 0</math>, where the integrand is a function with denominator that depends on <math>P(s)</math> in an essential way.