It's fairly classical that for $D>1$ and $(D,l)=1$ one has $$\sum_{\stackrel{p\leq x}{p\equiv l\; (mod \; D)}}\frac{\log p}{p} = \frac{\log x}{\phi(D)} + \textrm{O}(1)$$ where if I understand correctly the dependence on $D$ in the $\textrm{O(1)}$ is captured by something like $$\frac{1}{\phi(D)}\sum_{\textrm{non-principal} \; \chi}\frac{-L'(1,\chi)}{L(1,\chi)}$$ which is $\ll_{\epsilon}D^{\epsilon}$ for any $\epsilon>0$. If this is correct, suppose I am interested in the sum $$\sum_{\stackrel{p\leq x}{\left(\frac{-D}{p}\right)=1}}\frac{\log p}{p}$$ where $\left(\frac{-D}{p}\right)$ is the Legendre symbol. There are $\phi(D)/2$ residue classes mod $D$ that $p$ can lie in, and so then this is just my previous sum $\phi(D)/2$ times, and I get that it's $1/2 \log x +\textrm{O}(1)$, where the dependence on $D$ in $\textrm{O}(1)$ is now something like $\textrm{O}(D^{1+\epsilon})$. Is this the best one can do? I was hoping I could get it to be $\textrm{O}(D^{\epsilon})$ but if that's not even correct perhaps I should stop trying.
-
$\begingroup$ Just out of curiosity. Does the $O(1)$ term converge to some constant? Could you refer me to some literature on this question? : ) $\endgroup$– AfterMathAug 22, 2021 at 19:16
2 Answers
You're right that your proposed method for estimating the new sum, while correct, gives a worse error bound than we can obtain otherwise. Rather than quoting the classical result itself, I suggest going back to the proof of that classical result and modifying it to address the new sum.
Somewhat more precisely: the Legendre symbol $\big( \frac{-D}p \big)$, or rather its multiplicative generalization the Jacobi symbol $\big( \frac{-D}n \big)$, is a Dirichlet character (mod $D$)—call it $\chi_1(n)$. The sum you are interested in is just $$ \sum_{p\le x} \frac{\log p}p \bigg( \frac{1+\chi_1(p)}2 \bigg). $$ (Not exactly, since this expression counts primes dividing $D$ with weight $1/2$, but that's easily dealt with.) Therefore you're left with needing to understand $\sum_{p\le x} (\log p)/p$ (no problem) and $\sum_{p\le x} (\chi_1(p)\log p)/p$. The latter sum is going to be $O(1)$ in the same way that the error term in the classical problem is $O(1)$; it is going to be related to $-L'(1,\chi_1)/L(1,\chi_1)$. So in fact you don't have all the different $-L'(1,\chi)/L(1,\chi)$ error terms to deal with, only the single one where $\chi=\chi_1$.
-
1$\begingroup$ Thanks, now that I see that it seems obvious. I had tried to work it out that way (since it did seem unnecessary to go through all of the characters to get information basically just about one) but for some reason I was failing. $\endgroup$– ElenaMar 29, 2011 at 18:51
That's a good question, as it illustrates that one should try to find the "correct" harmonic analysis for a given problem (of this type...). What you can do is represent the characteristic function of your set of interest (the $p$ where $(-D/p)=1$) as a combination of Dirichlet characters; by quadratic reciprocity, you only need the principal character and a real character. So you can get your error term as the sum of two error terms coming from $$\sum_{p\leq x}{\chi(p)\log(p)/p}=\delta(\chi)\log(x)+(Error).$$ This should give what you want.
