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 $$T^{1/2+\varepsilon}\leq H\leq T$$.

MSC:

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

Citations:

Zbl 1012.11080; Zbl 0996.11052
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