# Analytic continuation of “composite” zeta function

Let us define the Dirichlet series $$\mathcal C(s):=\sum_{n\text{ composite}}\frac{1}{n^s},\quad P(s):=\sum_{p\text{ prime}}\frac{1}{p^s}.$$ They are absolutely convergent in the half-plane $\sigma>1$, where we have also$$1+\mathcal C(s)+P(s)=\zeta(s),$$ $\zeta$ being the Riemann zeta function. Now, $\zeta(s)$ can be meromorphically continued on the whole complex plane $\mathbb C$, with a simple pole at $s=1$; the prime zeta function $P(s)$ can be analytically continued to the half plane $\sigma>0$, and the vertical strip $0<\sigma\leq 1$ is full of logarithmic singularities (at each point $s=\frac{\rho}{k}$ with $\rho$ equal to 1 or a non-trivial zero of $\zeta(s)$, and $k$ being a squarefree integer). The line $\sigma=0$ defines a natural boundary for the function $P(s)$.

What can be said about the maximal domain of analytic continuation of the composite zeta function $\mathcal C(s)$?

The constraint $1+\mathcal C+P=\zeta$ must be valid also under analytic continuation: can we deduce that the maximal domain of $\mathcal C(s)$ is the half-plane $\sigma>0$ with logarithmic singularities at the same points of $P(s)$, in such a way that in the sum only a simple pole at $s=1$ does not annihilate?

More in general, what can we say about analytic continuations of sums and sums of analytic continuations for Dirichlet series?

• It seems you've already answered your own question. – Christian Remling Aug 19 '16 at 0:06
• $P(s)$ is holomorphic for $Re(s) > 1$, the Riemann hypothesis is that $P(s) + \ln(s-1)$ is holomorphic on $Re(s) > 1/2$, and in every case it has a natural boundary on $Re(s) > 0$ (since $\zeta(s)$ has an infinity of zeros $1/2\pm i \gamma$ with $\gamma$ unbounded, then $P(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \zeta(ns)$ has a singularity at every $2/k + i \gamma /k$, that is dense in $Re(s) = 0$). $\zeta(s)$ being meromorphic, it is clear that the same is true for $\zeta(s)-P(s)$. – reuns Aug 19 '16 at 2:29
• a typo, I meant $P(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ (converging for every $Re(s) > 0$) – reuns Aug 19 '16 at 8:42