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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
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3 Answers

up vote 13 down vote accepted

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
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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
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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 $\{1\}$.

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I asked for meromorphic continuation. It is meromorphic in all of C. –  Boris Bukh Nov 12 '09 at 23:59
    
I just want to note that though I did not find this answer helpful, I did not find it unhelpful either, and the downvote is not mine. –  Boris Bukh Nov 13 '09 at 0:43
    
I expect that Mariano was a bit quick in responding, and didn't read the question carefully enough. But the downvote seems an overreaction to me. (I might easily have fallen into the same trap, and so would probably others.) –  Harald Hanche-Olsen Nov 13 '09 at 1:25
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I'm not the down voter here, but I will say that I downvote any wrong answers that have nonnegative scores. This isn't always meant as criticism: I just think that the site will be more useful if the incorrect answers are sorteds to the bottom. –  David Speyer Nov 13 '09 at 1:40
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