A note on moments of \(\zeta'(1/2+i\gamma)\). (English) Zbl 1098.11042

Let \(\rho=\beta+i\gamma\) denote complex zeros of the Riemann zeta-funciton \(\zeta(s)\). The authors prove two results on the distribution of the \(\gamma\)’s, of which the second result is a generalization of the first one. Theorem 1 reads as follows: Let \(T\) be sufficiently large. Then, for \(0<k\leq1/2\) and \(T^{0.552}\leq H\leq T\), we have \[ \sum_{T\leq\gamma\leq T+H,\zeta'({1\over2}+i\gamma)\neq0} \left| \zeta'({\textstyle{1\over2}}+i\gamma)\right| ^{2k} \ll H(\log T)^{1+5k/2}. \] The exponent 0.552 comes from the second author’s work on simple zeros of \(\zeta(s)\) in short intervals [Acta Math. Hung. 96, 259–308 (2002; Zbl 1012.11080)]. If the Riemann hypothesis is true, then, for \(0<k<1\), \[ \sum_{0<\gamma\leq T,\zeta'({1\over2}+i\gamma)\neq0} \left| \zeta'({\textstyle{1\over2}}+i\gamma)\right| ^{2k}\gg T(\log T)^{5k-1}. \tag{(1)} \] It is remarkable that, under the Riemann hypothesis, the upper bound for \(k=1/2\) is of the order predicted by C. P. Hughes et al. [Proc. R. Soc. Lond. A 456, 2611–2627 (2000; Zbl 0996.11052)], who obtained an asymptotic formula for the sum on the left-hand side of (1), in case all the zeros are simple and the Riemann hypothesis holds.
Reviewer’s remark: Estimating trivially, by the Riemann–von Mangoldt formula, the sum \[ \sum_{T\leq\gamma\leq T+H,\zeta'({1\over2}+i\gamma)\neq0}1 \leq \sum_{T\leq\gamma\leq T+H,}1 \] as \(O(H\log T)\) in l.3 on p. 60, the range of validity in the upper bound in Theorem 1 can be increased to \(T^{1/2+\varepsilon}\leq H\leq T\).


11M06 \(\zeta (s)\) and \(L(s, \chi)\)
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