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For $\sigma \in \mathbb{R}$, let each $\mathbb{C}_\sigma = \{s \in \mathbb{C} : \Re(s) > \sigma\}$. For a sequence $a_n \in \mathbb{C}$, consider the Dirichlet series $D(s) = \sum_{n\ge 0} a_n n^{-s}$. There exists some $\sigma_c(D), \sigma_a(D) \in \mathbb{R}\cup \{\pm \infty\}$ such that $D$ converges conditionally in $\mathbb{C}_{\sigma_c(D)}$ and absolutely in $\mathbb{C}_{\sigma_a(D)}$.

For a particular sequence $a_n$, we can ask whether the Dirichlet series $D$ has properties of an L-function: can it be written as an Euler product, then analytically continued to a meromorphic function on $\mathbb{C}$, and then shown to satisfy a functional equation.

I am interested when this can fail at the second step. Is there a sequence $a_n$ so that $D$ defines an analytic function on some half plane $\mathbb{C}_{\sigma}$ with the line $\Re(s) = \sigma$ as its natural boundary? Is there a sequence so that $D$ can be analytically continued to a meromorphic function on a non-trivial Riemann surface?

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    $\begingroup$ Maybe I am not understanding what you are asking, but what exactly do you mean by "Dirichlet Series that fail to be L-functions"? What is an L-function for you? $\endgroup$
    – M.G.
    Commented Sep 5 at 16:19
  • $\begingroup$ I tried to clarify in the statement. $\endgroup$ Commented Sep 5 at 16:46
  • $\begingroup$ If I understand your edit correctly, then the Prime Zeta Function is such an example. It extends analytically up to the natural boundary $\sigma = 0$. $\endgroup$
    – M.G.
    Commented Sep 5 at 16:56
  • $\begingroup$ Actually it does not quite, since it is not meromorphic up to the boundary. But it may be an example of the other question, since it has logarithmic singularities which lead to analytic continuation to a complicated Riemann surface. $\endgroup$ Commented Sep 5 at 18:12

2 Answers 2

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If $f(z)$ is a holomorphic function in a neighborhood of $z=0$ then $f( p^{-s})$ is a Dirichlet series for any prime (or really any natural number) $p$: we take $a_n=0$ if $n$ is not a power of $p$ and $a_{p^k}$ to be the coefficient of $z^k$ in $f(z)$.

Choosing $f$ to be holomorphic on a disc with a natural boundary on the disc gets you an analytic function on a half plane with a vertical line as its natural boundary.

Choosing $f$ to admit analytic continuation to some nontrivial Riemann surface produces a series $f(p^{-s})$ with analytic continuation to the fiber product of that Riemann surface with the Riemann surface for logarithm, which will be nontrivial (since the original surface is unbranched at zero and hence must be branched at some nonzero point, and the fiber product will be branched at the logarithms to base $p^{-1}$ of that point).

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Define the prime zeta function $P(s)$ as

$$ P(s)=\sum_pp^{-s}. $$

By Möbius inversion, one can verify that (configure $\log$ by specifying $\log\zeta(+\infty)=0$)

$$ P(s)=\sum_{n\ge1}{\mu(n)\over n}\log\zeta(ns), $$

Notice that as $n\to+\infty$, $\log\zeta(ns)\sim 2^{-ns}$, so that $P(s)$ extends to some analytic function regular at points in the right half plane apart from zeros of $\zeta(ns)$. Due to this irregularity $n$-factor in $\zeta(ns)$, it can be shown that $\Re(s)=0$ is a natural boundary for $P(s)$. Detailed derivation can be found in §9.5 of Titchmarsh's The theory of the Riemann zeta-function.

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