Recall that given some prime $q$, the $q$-adic valuation of an integer $n\geqslant1$ is defined as : $$\nu_q(n)=\max\{\nu\geqslant0\;/\;q^\nu|n\}.$$
From the prime factorization, it follows that $\nu_q$ is a completely additive function. I know non-multiplicative functions usually do not behave well with Dirichlet series ($\Lambda$ being a counter-example to the statement though), but still, I asked myself the following :
What is the Dirichlet series of $\nu_q$ ?
What I mean is, similarly to $$\sum_{n=1}^{+\infty}\frac{\Lambda(n)}{n^s}=-\frac{\zeta'(s)}{\zeta(s)},$$
is there some nice formula to evaluate $$\sum_{n=1}^{+\infty}\frac{\nu_q(n)}{n^s}\;?$$
So far, I haven't been able to find anything on the web (including MSE), and my approach seemed like it lead nowhere ; what I tried was, using the fact that $\beta^{\nu_q}$ is completely multiplicative for any $\beta\in\mathbb{C}^\star$, we have the formal$^\dagger$ identity : $$\sum_{n=1}^{+\infty}\frac{\beta^{\nu_q(n)}}{n^s}=\prod_p\frac{1}{1-\beta^{\nu_q(p)}p^{-s}}=\prod_{p\neq q}\frac{1}{1-p^{-s}}\times\prod_{m=1}^{+\infty}\frac{1}{1-\beta^mq^{-s}},$$
where the products ranges over all primes, as usual. This can be re-written into : $$\sum_{n=1}^{+\infty}\frac{\beta^{\nu_q(n)}}{n^s}=(1-q^{-s})\zeta(s)\prod_{m=1}^{+\infty}\frac{1}{1-\beta^mq^{-s}}.$$
Moreover, similarly to the fact that $$(1-2^{-s})\zeta(s)=\sum_{m=0}^{+\infty}\frac{1}{(2m+1)^s},$$
I suspect that $$(1-q^{-s})\zeta(s)=\sum_{q\not{\,|}\,n}\frac{1}{n^s}.$$
From this point, I have no clue whether it is possible to manipulate the expression into a better thing. Moreover, I doubt it is relevant, as I find it unlikely this will allow to deduce the Dirichlet series for $\nu_q$. Is there a formula ? Am I missing a bit of theory for additive functions, as opposed to multiplicative ones ?
$\dagger$ : from $\nu_q(n)\in O(n)$, we see that the Dirichlet series for $\nu_q$ has $1$ as abscissa of absolute convergence. But $\beta^{\nu_q}$ needs not have finite abscissa, which is what I meant by a formal identity.
Best Answer
In your formal identity you made a mistake that makes things look harder than they are. If we define $$F(s,\beta) = \sum_{n = 1}^{\infty} \frac{\beta^{\nu_q(n)}}{n^s}$$ we get a series that converges absolutely for $\operatorname{Re} s > 1$ and $\lvert \beta\rvert < q$. Since $n \mapsto \beta^{\nu_q(n)}$ is (for every fixed $\beta$) completely multiplcative, the Euler product is $$F(s,\beta) = \prod_p \frac{1}{1 - \frac{\beta^{\nu_q(p)}}{p^s}} = \Biggl(\prod_{p \neq q} \frac{1}{1 - p^{-s}} \Biggr)\cdot \frac{1}{1 - \frac{\beta}{q^s}} = \frac{\zeta(s)\bigl(1 - q^{-s}\bigr)}{1 - \beta q^{-s}}\,.$$ Then $$\sum_{n = 1}^{\infty} \frac{\nu_q(n)}{n^s} = \frac{\partial F}{\partial \beta}(s,1)\,,$$ and since $$\frac{\partial F}{\partial \beta}(s,\beta) = \frac{\zeta(s)\bigl(1 - q^{-s}\bigr)}{\bigl(1 - \beta q^{-s}\bigr)^2}\cdot \frac{1}{q^s} = \frac{\zeta(s)\bigl(1 - q^{-s}\bigr)}{\bigl(q^s - \beta\bigr)\bigl(1 - \beta q^{-s}\bigr)}$$ we obtain $$\sum_{n = 1}^{\infty} \frac{\nu_q(n)}{n^s} = \frac{\zeta(s)}{q^s-1} = \zeta(s)\cdot \sum_{k = 1}^{\infty} \frac{1}{q^{ks}}\,.$$ Thus we know the analytic behaviour of the Dirichlet series of $\nu_q$ as well as we know the behaviour of $\zeta$.