Suppose that $f(s)$ is defined by some Euler product (absolutely convergent for $\sigma > 1$), and we happen to know that its logarithmic derivative $(f'/f)(s)$ is analytic for $\sigma > \delta$ with $\delta < 1$, say.
Then can we say anything about $f(s)$ for $\sigma > \delta$?
By hypothesis, $f(s)$ is of course analytic for $\sigma > 1$.
If $f(s)$ is analytic for $\sigma > \delta$ also, then does it necessarily follow that $(f'/f)(\sigma + it)$ grows as weak as, for instance, $(\zeta'/\zeta)(\sigma + it)$ as $t \to \infty$?
Just a problem I encountered.
Thanks in advance.
It's all going to come down to how best you can generate the coefficients that produce the Dirichlet series for $f(s)$. Restating what you have written: we have some multiplicative function $g$ such that $$f(s) = \sum_{n \geq 1} \frac{g(n)}{n^s} = \prod_{p\mathrm{\ prime}} \left(1-g(p) p^{-s}\right)^{-1}.$$ Now you have said that there is a "nice" definition for the logarithmic derivatives: $$\begin{align} L_f(s) & = \frac{f^{\prime}(s)}{f(s)} = \frac{d}{ds}\left[\log f(s)\right] \\ & = -\sum_p \frac{\log(p) \cdot g(p)}{p^s \cdot (1-g(p) p^{-s})} \\ & = -\sum_p \sum_{n \geq 0} \log(p) g(p)^{n+1} p^{-(n+1)s}. \end{align}$$ Now if $g(p)$ has nice behavior, for example if $g(p) \equiv 1$, then you might be able to recover some information about $L_f(s)$ (and hence, $f(s)$) from this series at this point. Otherwise, note that we have $$f(s) = \exp\left(\int \frac{L_f(s)}{s} ds\right),$$ which is a special case of when the Dirichlet series $f(s) := e^{G(s)}$ for some other Dirichlet series $G(s)$. In this case, we get that $$G(s) = \log g(1) + \sum_{n \geq 2} \frac{(g^{\prime} \ast g^{-1})(n)}{\log n \cdot n^s},$$ where $g^{\prime}(n) := \log n \cdot g(n)$. So proceeding by differentiation, we should get that $$L_f(s) = -s \times \sum_{n \geq 2} \frac{(g^{\prime} \ast g^{-1})(n)}{n^s}.$$ Does that help somewhat?