Timeline for If a Dirichlet series converges Conditionally, how can I apply Euler product?

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when toggle format	what		by	license	comment
Dec 11, 2016 at 21:53	history	edited	Noam D. Elkies	CC BY-SA 3.0	clarify and give a few more details
Dec 11, 2016 at 19:45	comment	added	Noam D. Elkies		I didn't claim, nor (I think) implied, that it's true for every Euler product.
Dec 11, 2016 at 18:46	comment	added	reuns		so that if $L(s,\chi)$ has no zeros for $Re(s) >\sigma$ then $\sum_{p^k < x} \frac{\chi(p^k)}{k} = \mathcal{O}(x^{\sigma+\epsilon})$ and $\log L(s,\chi) = s \int_1^\infty (\sum_{p^k < x} \frac{\chi(p^k)}{k}) x^{-s-1}dx$ converges for $Re(s) >\sigma$.
Dec 11, 2016 at 18:45	comment	added	reuns		Yes, but this is not true for every Euler product. The Dirichlet L-functions $L(s,\chi)$ are special because their coefficients are periodic, so they are entire and have a functional equation, from which we know the vertical density of zeros $\beta$ in the critical strip, allowing us to write $\frac{L'(s,\chi)}{L(s,\chi)} = -\sum_\beta \frac{1}{s-\beta}$, from which we get the explicit formula $\sum_{p^k < x} \frac{\chi(p^k)}{k} = \sum_\beta li(x^{\beta})+\mathcal{O}(1)$,
Dec 11, 2016 at 6:50	history	answered	Noam D. Elkies	CC BY-SA 3.0