# What is the best known upper bound for $\frac{1}{\zeta'(\rho)}$ assuming the SZC but not the RH?

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.

• 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. Jun 2, 2021 at 16:20

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.