3 added 34 characters in body

This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/what-are-the-values-of-the-derivative-of-riemanns-zeta-function-at-the-known-non/80783#80783 (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 $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(\frac{1}{\log T}\right)\right)$$ \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}.$$If my reasoning is correct, M(1)=0 is equivalent to e^{i\phi(\rho)} being dense in \mathbb{T}, and M(n)=0, n\in\mathbb{N}, to them being uniformly distributed. Any advice, insights and corrections would be most welcome. EDIT 1: the arguments are 2\theta x- corrected. EDIT 2: O(T) inside the bracket- corrected. 2 added 68 characters in body This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/what-are-the-values-of-the-derivative-of-riemanns-zeta-function-at-the-known-non/80783#80783 (which is not a prerequisite for this version). As usual, let \rho=\beta+i\gamma denote the non-trivial zeros of \zeta(s). Essentially, the question is this: How are the numbers e^{i\phi(\rho)}=\lim_{s\rightarrow\rho}\zeta(s)/\overline{\zeta(s)} distributed in \mathbb{T}? Of course \phi\in\mathbb{R} and the limit could be replaced by quotients of derivatives if we knew the order of the zeros (in which case \phi is just twice their argument)argument there), but I don't think it is necessary to know that any way. Moreover, I think the answer is independent of RH. My reasoning is as follows. For x\in\mathbb{R}, one may consider the mean-value:$$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(\frac{1}{\log T}\right)\right)=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{-i\theta(\gamma)x}.$$\beta=1/2}e^{-i2\theta(\gamma)x}.$$

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.

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# What is the mean-value of a particular exponential sum related to the non-trivial zeros of Riemann's zeta function?

This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/what-are-the-values-of-the-derivative-of-riemanns-zeta-function-at-the-known-non/80783#80783 (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 their argument), 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 $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(\frac{1}{\log T}\right)\right)$$ $$=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{-i\theta(\gamma)x}.$$

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.