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Let $A>0$ and $q\leq (\log N)^A$. Then there exists a constant $c$ depending on $A$ such that $\displaystyle \sum_{n\equiv a \bmod q; n\leq N}\mu(n)\ll N\exp(-c\sqrt{\log N})$. I know this result because someone pointed me to Montgomery-Vaughan which mentions the result, but in an exercise. Could someone please suggest a source where this result is proved (so that I could cite it)? Thanks!

Edit: Someone = GH from MO.

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    $\begingroup$ The "someone" was me, see mathoverflow.net/questions/390774/… I think it is appropriate to quote Exercise 13 in Section 11.3 of Montgomery-Vaughan: Multiplicative number theory I., because it is broken down into 6 parts, and Exercises 6-12 prepare the scene for this exercise. That is, Montgomery-Vaughan gave a rather detailed sketch how to prove this result, and connecting the dots is really just an exercise. $\endgroup$
    – GH from MO
    Commented Sep 28, 2021 at 13:38
  • $\begingroup$ @GHfromMO, Sorry I wasn't sure about the etiquette here (about naming your user handle outright). I was hoping that if the proof is there in literature, I could save time (I am about to hand in my thesis and under severe time constraints) and just cite it, instead of trying to prove it. But thanks for your comment! $\endgroup$
    – user147650
    Commented Sep 28, 2021 at 13:44
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    $\begingroup$ You can cite the exercises in your thesis or in your papers. That was my point. It is a standard result even if the proof is not spelled out in full anywhere. $\endgroup$
    – GH from MO
    Commented Sep 28, 2021 at 16:37

2 Answers 2

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It is proved in Corollary 5.29 [p. 124] of the book Analytic Number Theory by Iwaniec and Kowalski, American mathematical Society, Colloquium Publications, Vol. 53, 2004.

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    $\begingroup$ Corollary 5.29 is a weaker result. It has $x(\log x)^{-A}$ instead of $x\exp(-c\sqrt{\log x})$. $\endgroup$
    – GH from MO
    Commented Sep 28, 2021 at 16:38
  • $\begingroup$ @GHfromMO, thanks. Shouldn't post that late at night. Just saw your related answer as well at mathoverflow.net/questions/117223/siegel-walfisz-theorem $\endgroup$
    – kodlu
    Commented Sep 28, 2021 at 21:51
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This is a year later after the question was asked but for the sake of completeness, the precise result is proved in the following paper of Davenport: On Some Infinite Series Involving Arithmetical Functions (II), The Quarterly Journal of Mathematics, Volume os-8, Issue 1, 1937, Pages 313–320. See Lemma 5 which proves it under the condition that $(a,q)=1$ and Lemma 6 which removes the said condition.

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