Let $\zeta$ denote the Riemann zeta function and let $\rho$ denote one of its complex zeros. What is the best known upper bound for $\frac{1}{\zeta'(\rho)}$ assuming that all zeros are simple (SZC), but not assuming Riemann Hypothesis ?

Assuming both the RH and the SZC, one can mimick the proof of Theorem 15.6 of Montgomery-Vaughan's Multiplicative Number Theory and show that $$ \frac{1}{\zeta'(\rho)} \ll X,\label{1}\tag{1} $$ where $X$ is any real number $\geq |\rho| $ (actually the Montgomery-Vaughan argument seems to yield an upper bound of the form $o(X)$).
However, it looks like the bound \eqref{1} could hold assuming the SZC alone, as it appears that Montgomery-Vaughan only invoked the RH on bounding the $S(T)=\arg \zeta(\sigma + iT)$. On the RH, it is a classical fact that $$ S(T) \ll \frac{\log T}{\log \log T} $$ whilst $S(T) \ll \log T$ unconditionally. The unconditional bound seems sufficient for the purposes of showing that \eqref{1} comes from the Montgomery-Vaughan argument.

  • 6
    $\begingroup$ How does the argument of Montgomery-Vaughan Theorem 15.6 give (1)? If one has two extremely close simple zeroes $\rho_1,\rho_2$ it is extremely difficult to use contour integration to separate the contribution of $\frac{1}{\zeta'(\rho_1)}$ and $\frac{1}{\zeta'(\rho_2)}$ from each other. Indeed, given that $\frac{1}{\zeta'(\rho)}$ would become infinite if $\rho$ collided with another zero, it seems unreasonable to expect any upper bound on $\frac{1}{\zeta'(\rho)}$ whatsoever unless one assumed an explicit lower bound on the spacing between zeroes. $\endgroup$
    – Terry Tao
    Jun 2, 2021 at 16:20

1 Answer 1


We have an exact formula \begin{align*} \frac{1}{\zeta'(\rho)} &= \lim_{s \to \rho} \frac{s-\rho}{\zeta(s)} \\ &= \lim_{s \to \rho} \frac{(s-\rho) (s-1) \Gamma(1+s/2) \pi^{-s/2}}{\xi(s)} \\ &= (\rho-1) \Gamma(1+\rho/2) \pi^{-\rho/2} \lim_{s \to \rho} \frac{s-\rho}{\frac{1}{2} e^{Bs} \prod_{\rho'} (1-\frac{s}{\rho'}) e^{s/\rho'}} \\ &= -\frac{ 2 e^{1-B\rho} \rho(\rho-1) \Gamma(1+\rho/2) \pi^{-\rho/2}}{\prod_{\rho' \neq \rho} (1-\frac{\rho}{\rho'}) e^{\rho/\rho'}} \end{align*} where $B = -0.0230957\dots$ is the constant in Theorem 10.12 of Montgomery-Vaughan. All of the factors in the above formula are well understood except for the terms $1-\frac{\rho}{\rho'}$ for nearby zeroes $\rho' = \rho+O(1)$, which are proportional in magnitude to the distances from $\rho$ to the nearby zeroes $\rho'$. So the problem of upper bounding $1/\zeta'(\rho)$ is more or less equivalent to that of lower bounding the distance $|\rho-\rho'|$ to the nearest zero $\rho'$ (or more precisely the product of the distances to those zeroes $\rho'$ within $O(1)$ of $\rho$). An assumption of simple zeroes merely says that this distance is positive, but a more quantitative version of this hypothesis would be needed to get any quantitative upper bound.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.