9
$\begingroup$

Let $z$ be a complex number and $\omega(n)$ denote the number of distinct prime factors of the natural number $n$. I am considering the arithmetic functions $|\mu(n)|z^{\omega(n)}$ and their associated Dirichlet series $$F_z(s)=\sum_1^{\infty}\frac{|\mu(n)|z^{\omega(n)}}{n^s}=\prod_p \biggl(1+\frac{z}{p^s}\biggr).$$

Such functions arise in certain problems in probabilistic number theory, particularly questions about the distribution of $k$-almost primes and so on, though relatively little about them appears to be documented. I would like to understand more about the behaviour of the functions $F_z(s)$ beyond the region of convergence of the Dirichlet series, which is in general the half-plane $\sigma>1$.

Firstly, for each $z$, it can be seen that $F_z(s)$ looks a lot like $\zeta^z(s)$ throughout the half-plane $\sigma>1/2$ by observing that $$H_z(s)=\frac{F_z(s)}{\zeta^z(s)}$$ has an Euler product that is absolutely convergent in this region. This means they have the same asymptotics as $t\rightarrow\infty$, the same poles, zeros, branch-points and branch-cuts.

On the other hand, for $\sigma\leq 1/2$, the same is not true (take, for example, $z=1$). In this direction I am interested in finding a general formula for $F_z(s)$, say with $|z|<2$, by means of analytic continuation or a functional equation, valid in a larger region. Are such things known about?

It seems relevant to say that the absence of known closed forms for sums like $$\sum_1^{\infty}|\mu(n)|z^{\omega(n)}u^{n},$$ or even any obvious way to break such sums up into summable pieces, seems to be roadblock to proving analytic continuation/functional equations for $F_z(s)$ except in isolated cases, such as $z=\pm 1$. Perhaps this is a good reason to expect that a general method does not exist, especially in light of the vastly different character of these functions as $z$ varies.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.