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Let $D(s) = \sum_{n=1}^\infty a_n n^{-s}$ be a Dirichlet series with $a_n ≥ 0$ and abscissa of convergence $\sigma_a = 1$. Further, we assume that $D(s)$ is holomorphic in each point $\Re(s) = 1$ except a pole of order $k > 0$ in $s = 1$ and that $D(s)$ possesses a meromorphic continuation on some half-plane $\Re(s) > 1 - \epsilon_0$ with some $\epsilon_0 > 0$.

My question is, whether under the above assumptions we can already say that

$$ \sum_{n ≤ x} a_n = xP(\log x) + o(x)$$

with some polynomial $P$ with degree $k-1$. Unfortunately I am not able to exclude other poles in the strip $1 > \Re(s) > 1 - \epsilon_0$.

In the simple case $k=1$ the question follows by the Wiener-Ikehara theorem, since then we have $\sum_{n ≤ x} a_n \sim Cx$ with $C = \mathrm{res}_{s=1}D(s)$.

Thank you!

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    $\begingroup$ Delange's generalisation of Ikehara's Tauberian theorem gives you the leading order term, at the very least. $\endgroup$
    – Peter Humphries
    Commented Nov 19, 2017 at 15:58
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    $\begingroup$ The $o(x)$ term depends on the density of poles and their residues. The first question is if $\Gamma(s) D(s)$ is fast decreasing so that $\sum_{n=1}^\infty a_n e^{-nx} = \sum_\beta x^{-\beta} P_\beta(\log x) +o(x^{-1+\epsilon})$ as $x \to 0$ $\endgroup$
    – reuns
    Commented Nov 19, 2017 at 23:22
  • $\begingroup$ Thank you very much so far. I am familiar with Delange's theorem but unfortunately the leading term does not suffice in my case. $\endgroup$
    – Johann Franke
    Commented Nov 20, 2017 at 14:16
  • $\begingroup$ This is interesting. In fact I know that the residues of all possible poles are bounded by a constant $K$ only dependent on the $\epsilon_0$ above (I also know that all poles of $D(s)$ have at most order $k$) such that $\Gamma(s)D(s)$ is kind of fast decreasing. Also I assume that there should be a pole free region which approaches the line $\Re(s) = 1$ when $t \to \infty$, but I haven't shown this yet. Does the inverse Mellin transform suffice to show the question? $\endgroup$
    – Johann Franke
    Commented Nov 20, 2017 at 14:31

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The answer is no, even if we assume there are no other poles than $1$ in $\sigma > 1- \epsilon_0$. I give an example below with $\epsilon_0=1$. This is a variant of an example given by Karamata in $1952$.

Let $b_n = 1 + \cos(\log^2(n)) \geq 0$ and consider $B(s) = \sum_{n \geq 1} b_n n^{-s}$. Let us write $$ \sum_{n \leq x} b_n = \int_{1}^x (1+\cos(\log^2(t))) dt + R(x) $$ where $R(x) = O(\log^2(x))$. Then for $\sigma >1$ we have $$ B(s) = \int_{1}^{+ \infty}(1+ \cos(\log^2(t)) )t^{-s}dt + s \int_{1}^{+ \infty} R(t) t^{-s-1} dt. $$ The last integral extends holomorphically to $\sigma >0$. The first integral is equal to $$ \frac{1}{s-1} + \frac{1}{2} I_+(s) + \frac{1}{2} I_-(s) $$ where $$ I_{\pm}(s) = \int_{0}^{+ \infty} \exp{((1-s)u \pm i u^2)} dt. $$ Using contour integration one can move the integration line to $u \mapsto e^{\pm \frac{i \pi}{4}}u$, and this yields $$ I_{\pm}(s) = e^{\pm \frac{i \pi}{4}} \int_{0}^{+ \infty} \exp{((1-s)u e^{\pm \frac{i \pi}{4}} - u^2)} dt. $$ Thus $I_+$ and $I_-$ are entire functions.

Let us now choose $a_n = b_n \log(n) \geq 0$ so that $D(s) = - B'(s)$ is meromorphic on $\sigma >0$, with a unique pole at $s=1$ of order $2$. One checks that $$ \sum_{n \leq x} a_n = x \log(x) + x \left( \frac{1}{2} \sin(\log^2(x))-1 \right) +O(\frac{x}{\log x}) $$ Remark : The error term in Delange's theorem can be improved provided growth assumptions (on a strip) are made on $D(s)$.

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  • $\begingroup$ A very fancy counter example, thank you...! $\endgroup$
    – Johann Franke
    Commented Dec 4, 2017 at 20:45

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