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Say you are proving an explicit formula for $L(s,\chi)$ and/or the prime number theorem (in arithmetic progressions or not) in the usual way -- that is, shifting a line of integration from $\Re(s) = 1^+$ to the left. We need some kind of bound on $L'(s,\chi)/L(s,\chi)$ in the critical strip.

The usual method is a bit roundabout -- it goes through the Weierstrass-Hadamard product formula and results in a much stronger bound than what we need. It would seem to be enough to have $L'(s,\chi)/L(s,\chi) = o(|s|)$ away from zeroes -- in fact $o(|s|^2)$ is enough if you have smoothing.

Is there a way to obtain such "cheap" bounds? The case of a principal character $\chi$ may be hardest.

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    $\begingroup$ Well, any method would need to see the zeros as the logarithmic derivative has poles there. I am not sure what you mean by $L'(s,\chi)/L(s,\chi) = O(|s|)$ as this is false on the critical line. $\endgroup$
    – GH from MO
    Commented Aug 7, 2019 at 5:04
  • $\begingroup$ I meant "O(s) away from zeroes". $\endgroup$
    – H A Helfgott
    Commented Aug 7, 2019 at 11:26
  • $\begingroup$ How would that be false away from zeroes on the critical line? $\endgroup$
    – H A Helfgott
    Commented Aug 7, 2019 at 11:27
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    $\begingroup$ Not clear what it means "away from zeros" as there are infinitely many zeros. How far should we be away from a given zero for your bound to be true? Also, how can you prove this bound without talking about the zeros when the statement itself talks about the zeros? $\endgroup$
    – GH from MO
    Commented Aug 7, 2019 at 11:29
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    $\begingroup$ Let me be more precise. Take, say, Lemma 12.1 in Montgomery-Vaughan: $\zeta'(s)/\zeta(s) = -1/(s-1) + \sum_{\rho: |\Im \rho - \Im s|\leq 1} 1/(s-\rho) + O(\log \Im s + 4)$, where $\rho$ goes over the zeroes of $\zeta(s)$. Is it possible to give a proof of this or a weaker result of the same kind without going through the Weierstrass product formula? $\endgroup$
    – H A Helfgott
    Commented Aug 7, 2019 at 12:27

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