I spend lots of time working with Dirichlet series with bounded coefficients, and I often need to find whether or not they have analytic continuations to the full complex plane. When proving that some mathematical object has some property, I like to know whether I'm working to prove that the object I'm looking has some strange property or if I'm working to prove that it's normal and the numbers aren't just conspiring against me.


For example, when trying to prove whether a number is irrational or not, I know that $100\%$ of numbers are irrational and so I'm trying to show that I didn't happen to pick one of those $0\%$ of numbers.


Sadly, I have no such intuition for analytic continuation. I think that my guess would be that either $100\%$ or $0\%$ of Dirichlet series have analytic continuations, but I could be wrong. To make my question more concrete,



> 
> If $\{a\_n\}$ is a sequence of complex numbers chosen uniformly randomly in the unit disk, and $F(s)=\sum\_{n=1}^{\infty}\frac{a\_n}{n^s}$ is its Dirichlet series, what is the probability that $F(s)$ has an analytic continuation past the the line $\Re(s)=1$ (not necessarily to the entire plane).
> 
> 
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Answers to variants of this question are also greatly appreciated, like if $a\_n$ is chosen uniformly randomly of $[0,1]$ or if we are looking for continuations to the entire complex plane.


EDIT: If it's too complicated to analyse analytic continuation, what about meromorphic ones? Should I expect functions to have meromorphic continuations to $\Re(s)=1$?



Should I expect functions to have analytic continuations?
I don't think it is reasonable to use "random" Dirichlet series as a guide if you are working with examples that are expected to have some actual structure to them (like most Dirichlet series that arise in practice in number theory). If you are working with Dirichlet series for reasons unrelated to number theory, then perhaps your question is reasonable. What are some reasons you are looking at Dirichlet series with bounded coefficients?


Let's describe a probabilistic model for Dirichlet $L$-functions and see what probability theory predicts about them.
The coefficients of a Dirichlet $L$-function are roots of unity (or $0$), which are on the unit circle, so should we consider a random
Dirichlet $L$-function to be $\sum z\_n/n^s$ where $\{z\_n\}$ is a sequence chosen independently and uniformly on
the unit circle? That doesn't reflect the multiplicativity of the coefficients of a Dirichlet $L$-function, so we will use a random Euler product, as follows: define a "random" Dirichlet $L$-function to be $L(s) = \prod\_p 1/(1 - z\_p/p^s)$ where $z\_p$ for each prime $p$ is
chosen from a uniform distribution on the unit circle.


For a random number $z = \cos \theta + i\sin \theta$ on the unit circle,
its real and imaginary parts have average value 0 ($\int\_0^{2\pi} \cos \theta \,d\theta/2\pi = 0$ and
$\int\_0^{2\pi} \sin \theta \,d\theta/2\pi = 0$) and variance 1/2 ($\int\_0^{2\pi} \cos^2 \theta \,d\theta/2\pi = 1/2$ and
$\int\_0^{2\pi} \sin^2 \theta \,d\theta/2\pi = 1/2$). Note we are not computing
the variance of $z^2$ on the unit circle, which would be 0: $z^2$ is not $\cos^2\theta + i\sin^2\theta$!
The product $L(s)$ converges absolutely and uniformly on compact
subsets of ${\rm Re}(s) > 1$. What happens if $0 < {\rm Re}(s) \leq 1$?


When ${\rm Re}(s) > 1$, a logarithm of $L(s)$ is
$$
\sum\_{p^k} \frac{z\_p^k}{kp^{ks}} = \sum\_{p} \frac{z\_p}{p^s} + \sum\_{\substack{p^k \\ k \geq 2}} \frac{z\_p^k}{kp^{ks}}, 
$$
where the sum involving $k \geq 2$ is absolutely convergent if ${\rm Re}(s) > 1/2$ since $|z\_p^k/kp^{ks}| = 1/kp^{k\sigma}$. The series over primes is an integral:
$$
\sum\_p \frac{z\_p}{p^s} = s\int\_1^\infty \frac{Z(x)}{x^{s+1}}\,dx, 
$$
where $Z(x) = \sum\_{p \leq x} z\_p = \sum\_{n \leq \pi(x)} z\_{p\_n}$. Applying the law of the iterated logarithm to the real and imaginary parts of $z\_p$ (or $2z\_p$ to make the variance 1), $|Z(x)| = O(\sqrt{\pi(x)\log\log \pi(x)})$ for almost all sequences $\{z\_p\}$, which makes the integral above absolutely convergent for ${\rm Re}(s) > 1/2$ since $\pi(x)\log\log \pi(x) \sim (x/\log x)\log \log x$.
Therefore for almost all sequences $\{z\_p\}$,
$\sum z\_p/p^s$ converges for ${\rm Re}(s) > 1/2$, so a random Dirichlet $L$-function as defined here is almost certain to have an analytic continuation from ${\rm Re}(s) > 1$ to ${\rm Re}(s) > 1/2$ (as the exponential of
the analytic continuation of its logarithm) and no zeros
with ${\rm Re}(s) > 1/2$.


There is something inconsistent between random Dirichlet $L$-functions as defined above and actual Dirichlet $L$-functions that go by this name in the "real world" of number theory: for the sequences with $z\_p \in \{\pm 1\}$ (a random quadratic Dirichlet $L$-function), with probability 1 the function $L(s)$ does not have an analytic continuation to ${\rm Re}(s) > 1/2 - \delta$ for $\delta > 0$ by Theorem 2 p. 550 of Queffélec's paper on random Euler products; see
"Propriétés presque sûres et quasi-sûres des séries de Dirichlet et des produits d'Euler" Canad. J. Math 32 (1980), 531-558. (I am not aware of a treatment of this issue for random Dirichlet $L$-functions with non-real $z\_p$.
The treatment of random Euler products $\prod\_p 1/(1 - z\_p/p^s)$ for $z\_p \in S^1$ in Kowalski's course notes [https://people.math.ethz.ch/~kowalski/probabilistic-number-theory.pdf](https://people.math.ethz.ch/%7Ekowalski/probabilistic-number-theory.pdf) focuses on ${\rm Re}(s) > 1/2$.) Actual
Dirichlet $L$-functions in number theory are not random objects, but highly structured ones, and being able to extend them beyond ${\rm Re}(s) > 1/2$ is a significant feature that you can't predict by only thinking of a "random" $L$-function in a probabilistic sense.


