Prime Zeta function at 1

$$\zeta(s) = \prod_p \frac1{1-p^{-s}}, \qquad \Re(s) > 1$$ gives $$\log \zeta(s) =- \sum_p \log(1-p^{-s}) = \sum_{p^k} \frac{p^{-sk}}{k} = \sum_k \frac{P(sk)}{k}$$ so $$P(s) = \sum_k \frac{\mu(k)}{k} \log \zeta(sk)$$ at first for $\Re(s) > 1$ and by analytic continuation for $\Re(s) > 0$ so that $$ \lim_{s \to 1} P(s) - \log \zeta(s) = \sum_{k\ge2} \frac{\mu(k)}{k} \log \zeta(k)$$

At first for $\Re(s) > 1$ and by analytic continuation for $\Re(s) > 0$ $$\zeta(s)- \frac1{s-1} = \sum_n (n^{-s} -\int_n^{n+1} x^{-s}dx)$$ gives that $f(s)=(s-1)\zeta(s) $ is analytic at $s=1$ with $f(1)=1$ thus $F(s)=\log (s-1)\zeta(s)$ is analytic at $s=1$ with $F(1)=0$.

The $\gamma$ appearing in the Mertens constant $M$ comes from $\gamma = \sum_n (n^{-1} -\int_n^{n+1} x^{-1}dx)$