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Dec 4, 2017 at 20:43 vote accept Johann Franke
Nov 20, 2017 at 16:51 answer added js21 timeline score: 5
Nov 20, 2017 at 14:31 comment added Johann Franke 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?
Nov 20, 2017 at 14:16 comment added Johann Franke Thank you very much so far. I am familiar with Delange's theorem but unfortunately the leading term does not suffice in my case.
Nov 19, 2017 at 23:22 comment added reuns 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 $x \to 0$
Nov 19, 2017 at 15:58 comment added Peter Humphries Delange's generalisation of Ikehara's Tauberian theorem gives you the leading order term, at the very least.
Nov 19, 2017 at 15:24 history edited Johann Franke CC BY-SA 3.0
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Nov 19, 2017 at 15:11 history asked Johann Franke CC BY-SA 3.0