Sum of residue of two function

Question

Consider $P(s)$ to be the prime zeta function given by

$P(s)=\sum_{p}\frac{1}{p^{s}}=\sum_{n>0}\frac{\mu(n)Log\zeta(ns)}{n}$

Where $p$ is prime and $\mu$ is the mobius function which is equal to zero if $n$ has square factor . Now let consider the function $g(s)$ define by $g(s)=\zeta(s)-P(s)$ for $Re(s)>0$

We know that zeta have pole at $s=1$ but $P(s)$ have poles when $\zeta(ns)=0 $ or when $ns=1$

Now we have many cases and also when we comppute the residue at $s=1$ We find that $P(1)$ diverge to infinty !

My question is how we can determine the poles of $g(s)$ since it is the sum of two function? and how we can compute the residues of sum of two function have different poles? how we can write $g(s)$ as Euler product ?

Answer 1

$\log \zeta(s)$ thus $P(s)$ have a logarithmic singularity at $s=1$ not a pole, such a singularity doesn't have a residue

It works exactly the same at $\rho/n$ for $\rho$ a zero of $\zeta(s)$.

Try with $h_1(s)=1/(s-1)$ and $h_2(s)=\log (s-1)$ to see how $\int_{|s-1|=1/4} h_k(s)ds$ works.

If $h$ is meromorphic at $s=1$ and analytic on $0<|s-1|\le b$ then for all $0<a \le b$ $$\int_{|s-1|=a} g_k(s)ds=2i\pi Res(h(s),1)$$