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I'm looking for an explicit bound for $f(x) = \sum_{N\mathfrak p \leq x}\chi(\mathfrak p)\ln(N\mathfrak p)$, where $\chi$ is a Hecke character for a number field $K$ of degree $n$, on the ideals $I_\mathfrak m$, trivial on $P_\mathfrak m^+$, the sum is over prime ideals $\mathfrak p$ of bounded norm. All I could find is $$\sum_{N\mathfrak p \leq x}\chi(\mathfrak p)\ln(N\mathfrak p) = O(nx^{1/2}\ln(x)\ln(xd_KN\mathfrak m)),$$ for non-principal characters, and $$\sum_{N\mathfrak p \leq x}\chi(\mathfrak p)\ln(N\mathfrak p) = x + O(nx^{1/2}\ln(x)\ln(xd_KN\mathfrak m)),$$ for the principal character, assuming GRH, from Henryk Iwaniec and Emmanuel Kowalski, Analytic number theory, p. 114. I would need to know more about the implied constant factor. Conditional bounds (GRH, ERH) are welcome.

I actually plan to use such a result to derive a bound on $\sum_{N\mathfrak p \leq x}\chi(\mathfrak p)$, so any explicit bound on the later sum is also welcome.

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    $\begingroup$ Can you explain a bit more what you require? For example there clearly are effective bounds for the sum $\sum_{N\mathfrak{p} \leq x} \chi(\mathfrak{p})$, such as $x/(\log x)$, however I'm guessing that you want something stronger. How strong a bound do you require? Is there some bound you know that holds in the case where $K=\mathbb{Q}$, and you want to know whether it holds in the general case? The bound you have given looks like it is effective, as in your notation there is no implied constant and all the dependencies on the number field and character are there. What else do you require? $\endgroup$ – Daniel Loughran Apr 16 '14 at 7:30
  • $\begingroup$ What doesn't satisfy me with the bound I give is that a constant factor hides in the big O notation, and I don't know anything about it. We can derive from this a bound $\sum_{N\mathfrak p \leq x}\chi(\mathfrak p) = O(n x^{1/2}\ln(xd_KN\mathfrak m))$, but still this big O hides some factor. I would like an actual bound that can be computed. Your bound $x/\ln(x)$ is pretty tight for the trivial character but not so great otherwise (by this I mean the bound I've presented is in $x^{1/2}$, not in $x$). $\endgroup$ – Calodeon Apr 16 '14 at 7:55
  • $\begingroup$ I think I see, so you want some explicit upper bound which says $\sum_{N\mathfrak{p} \leq x} \chi(\mathfrak{p}) \leq \textrm{something}$ for all sufficiently large $x$? Can you explain what $n$ is in your big $O$? Also I ask again, do you know a bound of the shape you want in the case where $K = \mathbb{Q}$? $\endgroup$ – Daniel Loughran Apr 16 '14 at 9:28
  • $\begingroup$ Yes I need something like $\sum_{N\mathfrak p \leq x} \chi(\mathfrak p) \leq f(x)$, ideally for all $x > 0$, but I can live with it if interesting bounds happen only asymptotically. Here $n = [K:\mathbb Q]$. I don't know more about $K = \mathbb Q$ than the general case. But the shape I would expect is simply the one I gave in the big O, but with an explicit constant. $\endgroup$ – Calodeon Apr 16 '14 at 17:20
  • $\begingroup$ Well, I do know something for $K = \mathbb Q$ when $\chi$ is the trivial character, $\mathrm{li}(x) - x^{1/2}/\ln(x) < \sum_{N\mathfrak p \leq x}\chi(\mathfrak p) = \pi(x) < \mathrm{li}(x) + x^{1/2}/\ln(x)$. It is a great bound in the sense that is has a precision $\tilde{O}(x^{1/2})$ and is completely explicit, but this is a very particular case... $\endgroup$ – Calodeon Apr 16 '14 at 17:34
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It is certainly possible to prove such bounds, for example by re-reading Iwaniec and Kowalski and being more careful and precise than them. There are no tricks and no shortcuts.

Such a computation would be messy. Since you are asking for conditional bounds, perhaps no one has bothered to work out the details. It would be take a long time, but not be too terribly difficult -- indeed it would be a great way to master the proofs.

I think Lior Silberman wrote up something along these lines (see the bottom of his website), but I was unable to get the link on his website to work.

By the way, your use of the terminology `effective' is mistaken. The bounds you stated are effective --- "effective" means that one could work out the constant, not that anyone has actually done it. An example of an ineffective bound is the famous bound $h(-d) \gg d^{1/2 - \epsilon}$, where the implied constant depends on properties of a hypothetical Siegel zero -- so that this bound can't be used by itself to prove, e.g. that $h(-d) > 1$ for $d > 163$.

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  • $\begingroup$ Thanks, I have changed "effective" into "explicit". I have tried to redo the proof carrying explicit bounds, but it is indeed extremely messy. I couldn't get Lior Silberman's link to work either ; I've just sent him an email. $\endgroup$ – Calodeon Apr 17 '14 at 5:13

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