# Is the maximum domain to which a Dirichlet series can be continued always a halfplane?

Let $f(s)=\sum_n a_n n^{-s}$ be a Dirichlet series whose coefficients satisfy $\lvert a_n\rvert\leq n^{C}$. Then $f(s)$ converges absolutely in some halfplanes, and is conditionally convergent in (potentially different) halfplane. It might happen sometimes that $f(s)$ admits meromorphic continuation to a larger domain. Consider maximal such domain. Is this domain always a halfplane? For purposes of this question, I take it to mean that $\mathbb{C}$ is a halfplane.

The question is motivated by known results about analytic continuation of $L$-functions. It has vexed me: to what extent the analytic continuation is special to the algebraic world of $L$-functions, and how much its properties are common to the analytic world of Dirichlet series.

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This does not qualify as an answer, but: you could try "Basic analysis of regularized series and products" (Springer-Verlag LNM 1564) by Serge Lang and Jay Jorgenson. Their main theme is to extend analytic properties of L-functions to more general classes of Dirichlet series. –  Danny Calegari Nov 13 '09 at 0:02

It should be possible to make a Dirichlet series whose domain of meromorphicity is as screwy as you want. Notice that $\zeta(s - 1 - \alpha) = \sum n^{1+\alpha}/n^s$, so $\zeta(s - 1-\alpha)$ is a Dirichlet series, with pole at $\alpha$. Let $\gamma$ be a curve dividing the complex plane into two pieces, one of which contains all $z$ with $\Re(z)$ sufficiently large; and choose $\alpha_i$ a sequence of points of $\gamma$ which is dense in $\gamma$. Consider $$\sum c_i \zeta(s -1- \alpha_i)$$ where $c_i$ is some sequence which goes to zero very fast.

As long as the $c_i$ go to zero fast enough, this should converge absolutely on away from $\gamma$; and will be represented by a Dirichlet series on the half plane to the right of the rightmost point of $\gamma$. The dense set of poles will ensure that you can't continue past $\gamma$. Details are left to the reader. :)

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(There are some broken images, last I checked.) –  Theo Johnson-Freyd Nov 13 '09 at 3:25
Many thanks, David! Shame on me for not noticing this --- for a stupid reason, I tried taking products rather than sums. –  Boris Bukh Nov 13 '09 at 9:00
Alternately, for any function f holomorphic at 0, f(2^{-s}) is a Dirichlet series of the type you describe. –  moonface Nov 14 '09 at 3:42

The product $F(s) = \prod_{n=2}^{\infty}(1 - n^{-s})^{-1} = \sum_{n=1}^{\infty}c(n)n^{-s}$ is an explicit counterexample. The coefficient $c(n)$ is the number of ways of writing $n$ as a product of integers $\geq 2$ without regard to order, and the Dirichlet series has abscissa of convergence ${\sigma}_c = 1$. The function $F(s)$ has meromorphic continuation to $\sigma > 0$ minus the points $1,1/2,1/3,\ldots$ and has $\sigma = 0$ as a natural boundary. At the points $s = 1,1/2,1/3,\ldots$ it has essential singularities. One sees this by logarithmically differentiating the product. The essential singularities are obvious, but you have to do a little work to get the natural boundary.

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The Riemann zeta function is continuable to a set which is not a half plane, namely, the complement of $\{1\}$.