Polar Density of a Set of Primes In Chapter 7 of Marcus' Number Fields, he defines the polar density of a set $A$ of primes of a number field $K$ as follows:

Definition: If some $n$th power of the function
  $$\zeta_{K,A}(s) = \prod_{P \in A} \left(1-\frac{1}{\|P\|^s} \right)^{-1}$$
  can be extended to a meromorphic function in a neighborhood of $s=1$, having a pole of order $m$ at $s=1$, then the polar density of $A$ is $\delta(A) = \dfrac{m}{n}$. 

Are there any relatively easily describable sets of primes such that this density would not be defined?
 A: We might as well restrict to $K=\mathbb Q$, so that $\zeta_A(s) = \prod_{p\in A} (1 - p^{-s})^{-1}$.
If the polar density of $A$ exists, then the Dirichlet density exists and is equal.  In particular, if the natural density and polar density of $A$ both exist, then they must be the same.
Since the definition of polar density is only capable of producing rational numbers, it suffices to name a set of primes $A$ having a natural density that is irrational.
I can think of several natural sets of primes which should have irrational density, but I haven't thought of one that is both aesthetically pleasing and of provably irrational density.
Some incomplete examples:


*

*The set of primes with $p-1$ squarefree has density equal to Artin's constant $\prod_p (1-\frac{1}{p(p-1)})$, see https://math.stackexchange.com/a/155117/30402.  Artin's constant is presumed but not known to be irrational.

*The set of primes for which $p-1$ has a prime factor greater than $\sqrt{p}$ is conjectured to be $\log 2$, which is the exact density of integers $n$ possessing a similar property.


Of course it is possible to describe less succinctly a set of primes having any density in $[0,1]$.  I still wonder if there is a simple variation that will give an obviously irrational singular series constant like $6/\pi^2$.
Another approach is to take a sparse set of primes for which $\prod_p(1-1/p)^{-1}$ still diverges slowly (for instance, if the counting function is $x/(\log x \log \log x)$).  This will have natural density $0$, but the singularity at $s=1$ prevents $\zeta_A(s)$ from being continued to a holomorphic function in a neighbourhood of $1$ (by a result of Landau).
The disjoint union of such a set with one already having polar density should also lack a polar density, even with a rational natural density.
