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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).

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 $\Re(s)=1$?

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    $\begingroup$ I don't know any formal statements which say this, but the intuition definitely should be that analytic continuation beyond the obvious domain of convergence is rare, definitely not a norm. $\endgroup$
    – Wojowu
    Commented Jun 23, 2020 at 16:30
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    $\begingroup$ @Wojowu why does your intuition tell you this? $\endgroup$
    – Milo Moses
    Commented Jun 23, 2020 at 16:34
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    $\begingroup$ I expect the situation is the same as for power series. With poser series, analytic continuation beyond the radius of convergence "almost always" fails. ("Almost always" in the sense of category, and in the sense of measure.) $\endgroup$ Commented Jun 23, 2020 at 16:40
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    $\begingroup$ This title sounds promising... MR3457245 Bhowmik, Gautami(F-ULIL-LM); Matsumoto, Kohji(J-NAGO-GM) Analytic continuation of random Dirichlet series. Proc. Steklov Inst. Math. 282 (2013), suppl. 1, S67–S72. $\endgroup$ Commented Jun 23, 2020 at 16:58
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    $\begingroup$ I would actually expect continuation to $\Re(s)>1/2$ but no further (by looking at the variance) $\endgroup$
    – Will Sawin
    Commented Jun 23, 2020 at 16:59

2 Answers 2

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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 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.

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In addition to @KConrad's good answer: there were results already from the early 20th century (maybe partly due to Harald Bohr, but I can't find the ref) about natural boundaries of Dirichlet series with random coefficients from among $\{-1,0,1\}$ or similar.

More interesting to me is the Estermann phenomenon, from T. Estermann, "On certain functions represented by Dirichlet series", Proc. London Math. Soc. 27 (1928), 435-448. For example, he shows that $\sum d(n)^3/n^s$ has a natural boundary (where $d(n)$ is the number of divisors of $n$), while $\sum d(n)/n^s$ and $\sum d(n)^2/n^s$ have meromorphic continuations.

N. Kurokawa published considerable generalizations of this in 1986, in two papers in Proc. London Math. Soc. ("On the meromorphy of Euler products" (I) and (II)), showing, for example, that for three modular cuspforms with coefficients $a_n$, $b_n$, $c_n$, the naïve "triple product" $\sum a_nb_nc_n/n^s$ has a natural boundary. (In contrast to the naïve Rankin–Selberg $\sum a_nb_n/n^s$.)

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