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This question arose from an earlier one and the MO user's useful answers there: What are the values of the derivative of Riemann's zeta function at the known non-trivial zeros? (which is not a prerequisite for this version). As usual, let $\rho=\beta+i\gamma$ denote the non-trivial zeros of $\zeta(s)$. Essentially, the question is this:

How are the numbers $e^{i\phi(\rho)}=\lim_{s\rightarrow\rho}\zeta(s)/\overline{\zeta(s)}$ distributed in $\mathbb{T}$?

Of course $\phi\in\mathbb{R}$ and the limit could be replaced by quotients of derivatives if we knew the order of the zeros (in which case $\phi$ is just twice their argument there), but I don't think it is necessary to know that any way. Moreover, I think the answer is independent of RH. My reasoning is as follows.

For $x\in\mathbb{R}$, one may consider the mean-value: $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$ Let $t>0$ and $\theta(t)$ denote the Riemann-Siegel theta-function. Firstly, by separating the sums over zeros on and off the line $\beta=1/2$ and, secondly, by the Bohr-Landau theorem (that the number of zeros with $\beta>1/2$ is $O(T)$), one gets $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(T\right)\right)$$ $$=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{-i2\theta(\gamma)x}.$$

If my reasoning is correct, $M(1)=0$ is equivalent to $e^{i\phi(\rho)}$ being dense in $\mathbb{T}$, and $M(n)=0$, $n\in\mathbb{N}$, to them being uniformly distributed. Any advice, insights and corrections would be most welcome.

EDIT 1: the arguments are $2\theta x$- corrected.

EDIT 2: $O(T)$ inside the bracket- corrected.

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It's not true that the number of zeroes with $\beta>1/2$ is $O(T)$; this would imply that almost all zeroes lie on the critical line which is not known. I think that the right estimate is $|\\{\rho:\beta>1/2+\epsilon,|\gamma|\le T\\}|\ll_\epsilon T$. – Dimitris Koukoulopoulos Nov 21 '11 at 2:04
Thanks Dimitri, indeed there should be a $\delta>0$ there. The interpretation being that all but a vanishing proportion lie within $\delta$ of the critical line. – Kevin Smith Nov 25 '11 at 17:18
It seems to me that the numbers $\phi(\rho)$ are not well-defined: if $\zeta(s)=g(s)(s-\rho)^m$, where $g(\rho)\neq0$, then we need to calculate the limit $\lim_{s\to\rho}\left(\frac{s-\rho}{\overline{s-\rho}}\right)^m$, which does not exist. – Dimitris Koukoulopoulos Nov 26 '11 at 1:17
Thank you for your comment, Dimitri, however I think it is generally incorrect (except in one particular circumstance, which I will suggest below): First note that $\zeta/\overline{\zeta}=e^{2i\arg(\zeta)+2\pi i n}$ whenever $\zeta\neq 0$. Admittedly, apart from when a zero is on the line $\sigma=1/2$, actually calculating the limit requires knowledge of the signs of the real and imaginary parts of the first non-vanishing derivative (otherwise you could get the sign wrong). – Kevin Smith Nov 26 '11 at 10:48
However, you would be correct if it happens that, at some $\rho$ not on the line $\sigma=1/2$, one has $\Im\zeta^{n}(\rho)= 0$, $n=0,1,...,m-1$; $\Im\zeta^{m}(\rho)\neq 0$ and $\Re\zeta^{m}(\rho)= 0$ $n=0,1,...,m$, because in this case the argument itself is singular. – Kevin Smith Nov 26 '11 at 10:57

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