I am trying to understand whether some non trivial explicit estimate is known about the following $$\sum_{n \leq x} \mu(n) \chi(n)$$ as a function of $x$.
Here $\chi$ is a Dirichlet character modulo $q$ and $q > (\log x)^2$ (for example).
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3 Answers
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Estimating this sum is much the same as estimating the error term in the prime number theorem for arithmetic progressions. See Exercises 7-8 in Section 11.3 of Montgomery-Vaughan: Multiplicative number theory I (Cambridge University Press, 2006). (Your sum is denoted by $M(X,\chi)$ in this book, as introduced by (11.39) there.)
In particular, there is a constant $c>0$ such that for any $A>0$ we have $$ \sum_{n \leq X} \mu(n) \chi(n)\ll_A x\exp(-c\sqrt{\log x})$$ as long as $q\leq(\log x)^A$ .
answered Oct 14, 2016 at 15:29
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$\begingroup$ Yes, you're correct, I have forgotten to put that I need it in an explicit form. $\endgroup$– toshiCommented Oct 14, 2016 at 20:05
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2$\begingroup$ @toshi: Well, the constant $c>0$ can be calculated explicitly (and I guess it has been done, but I don't know for sure), while the implied constant is ineffective for any $A>2$ as it comes from Siegel's theorem on Siegel zeros. I don't think there is any effective result for any $A>2$. $\endgroup$ Commented Oct 14, 2016 at 21:15
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$\begingroup$ @GHfromMO do you have an idea why my answer is dis-approved ? $\endgroup$– reunsCommented Oct 26, 2016 at 20:41
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$\begingroup$ @user1952009: I don't know for sure. Usually, technical responses get fewer upvotes, and also users who have been around for longer get more upvotes and less downvotes. Your response is a bit technical, e.g. it does not reveal what can be said under the RH, or a weaker RH, or unconditionally (as in my response). Instead, it provides some intermediate information that could leave to certain bounds, but the OP wanted the bounds themselves. Finally, giving yourself a decent username (instead of "user" plus some number) would help in my opinion. Just my two cents. $\endgroup$ Commented Oct 26, 2016 at 21:08
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$\begingroup$ @GHfromMO I wanted to know if you disapproved too, tks (I have some difficulties proving how to make $\sum_\beta \frac{x^\beta}{\beta L'(\beta,\chi)}$ convergent) $\endgroup$– reunsCommented Oct 26, 2016 at 21:15
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Explicit estimates for the Moebius function are much harder than for the prime counting function. Ramare (From explicit estimates for the primes to explicit estimates for the Moebius function, Acta Arithmetica 157, (2013), 365-379) has shown how to translate estimates for primes to estimates for the Moebius function, as far as I know this approach is superior to a direct estimate using complex integration. It should be possible without too much work to change Sections 4, 5, 6 of that paper to the character case, as there is a lot of literature on various sums involving primes in arithmetic progressions (see the work of Rosser, Schoenfeld, McCurley, Rumely, Ramare). I would be worried about Lemma 9.1, but this is not necessary for your question.
answered Oct 15, 2016 at 10:15
Jan-Christoph Schlage-PuchtaJan-Christoph Schlage-Puchta
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$\begingroup$ Thanks for the reference, I was actually trying to adapt other results by Ramare, I will check whether it works. $\endgroup$– toshiCommented Oct 15, 2016 at 13:23
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There is an equivalent of the Riemann explicit formula for those :
If all the zeros of $L(s,\chi) = \sum_{n=1}^\infty \chi(n) n^{-s}$ are simple,
then $\displaystyle\frac{1}{L(s,\chi)} = \sum_{n=1}^\infty \mu(n)\chi(n) n^{-s}= s\int_1^\infty (\sum_{n < x}\mu(n) \chi(n))x^{-s-1}dx$ leads to $$\sum_{n < x} \mu(n) \chi(n) = \frac{1}{2i\pi}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{1}{L(s,\chi)}\frac{x^s}{s}ds = \frac{1}{L(0,\chi)} + \sum_\beta \frac{x^{\beta}}{\beta L'(\beta,\chi)}$$ by Mellin inverse transform and the residue theorem, where $\sigma > 1$,
$\beta$ are the zeros (trivial and non-trivial) of $L(s,\chi)$,
and $\sum_\beta$ is convergent only when grouping the terms correctly.
(note if some zeros are of order $k$, it is not so different except you'll get some additional terms $x^\beta (\ln x)^{m}, m < k$)
