On assumption of Riemann hypothesis when I write a complex zero (nontrivial zero) of zeta function as $\rho=\frac{1}{2}+it_\rho$, and I write $x^\rho$ as $\sqrt{x}e^{it_\rho \log x}$, then multiplying and dividing by the conjugate of $\rho$ in a well known formula (see [1]) involving the second Chebyshev function $\psi(x)$ we can write its real part $\Re\psi(x)=\psi(x)$ as

$$x-\sqrt{x}\cdot\Re\sum_\rho\frac{(\frac{1}{2}-it_\rho)\cdot(\cos(t_\rho\log x)+i\sin(t_\rho\log x))}{\frac{1}{4}+t_\rho^2}-\log 2\pi-\frac{1}{2}\log(1-\frac{1}{x^2}),$$ when (the genuine equation holds only if) $x$ is not a prime power.

Thus if I can compute $\Re\sum_\rho$ as $\sum_\rho\Re$, previous series (second summand) is computed as $$\sqrt{x}\cdot\sum_\rho\frac{\frac{1}{2}\cos(t_\rho\log x)+t_\rho\sin(t_\rho\log x)}{\frac{1}{4}+t_\rho^2}.$$

Then, from previous formula, Prime Number Theorem implies (I've used Dusart's theorem that states (see [2]), unconditionally $\psi(x)=x+o(x)$) $$\lim_{x\to\infty}\frac{1}{\sqrt{x}}\cdot\sum_\rho\frac{\frac{1}{2}\cos(t_\rho\log x)+t_\rho\sin(t_\rho\log x)}{\frac{1}{4}+t_\rho^2}=0.$$


Question. (On assumption of Riemann hypothesis) I don't know if I have convergence to claim $\Re\sum_\rho$ as $\sum_\rho\Re$ in previous computations. I don't know if the condition that $x$ isn't a prime power it affects the computation of previous limit. Truly I don't know currently what is the answer to this question: if $\rho$ is a such zero (we are assuming RH) then I can claim that $t_\rho\to\infty$ as $|\rho|\to\infty$? I know that there is an heuristic about the number of such zeros with $t_\rho\leq T$, $\frac{T}{2\pi}\log\frac{T}{2\pi e}$, but I don't know sure nothing conditionally/unconditionally about how grows $t_\rho$. Thanks in advance.

Remark: I would like ask more, but I believe that previous set of questions is more useful to me. If your want reference some in this site, please. I would like ask about if you can find an big oh asymptotic of previous series on assuption of RH. I don't know if it is an easy exercise. I've read, if there are no typo, that unconditionally $|x^\rho|=\sqrt{x}$ and $\sum_\rho x^\rho=O(\sqrt{x}\log x)$. Too I would like to ask if it is possible find a proof of previous limit unconditionally, this is you could to use PNT, but not RH. My doubts are since, in previous computations, I use the theory and formula simbolically. Then if you want say something about this please leave a comment or if is a good question, in the sense that it is possible that I to learn easy facts and computations from answer, I could encourage to me asking a new question in a new post. Thanks.


[1] Conrey, The Riemman hypothesis. The formula that I've cited is in first paragraph of page 343 of this free notes from AMS http://www.ams.org/notices/200303/fea-conrey-web.pdf

[2] Mathworld, Mangoldt function, statement (12) in http://mathworld.wolfram.com/MangoldtFunction.html

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    $\begingroup$ I'm in hurry so I will take a look later about your calculations, but I can answer now about your last question. Let $t_{k}$ the imaginary part of the $k$-th zero of the Zeta function, then $t_{k}\sim\frac{2\pi k}{\log\left(k\right)}.$ For the proof see math.stackexchange.com/questions/1217217/… $\endgroup$ – Marco Cantarini Nov 28 '15 at 8:28
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    $\begingroup$ I am trying to understand all statements in your post, when I have a concise question I will ask to you, very thanks much @MarcoCantarini you are Great! $\endgroup$ – user243301 Nov 28 '15 at 11:43
  • $\begingroup$ Now I've found your remark, concerning this part of my question in the section Counting zeros, of al book of Murty, about Analytic Number Theory. I will too understand your assertion in you referenced post, in I have doubts I write you, thanks @MarcoCantarini now I dont disturb more. $\endgroup$ – user243301 Nov 28 '15 at 13:11
  • $\begingroup$ I excuse to put a bounty [+50] going from myself reputation, since I would like to know if I can change the real part function with the sum, if I have convergence to claim it. If you want (optionally), since this is not explicit in my question, you can made a proof verification of previous claim involving the use of Dusart's theorem. Thanks in advance. $\endgroup$ – user243301 Nov 29 '15 at 11:16

So let answer your questions. The series over the non-trivial zeros is convergent, and you can see it simply because $\psi(x)$ is a number and you have an identity. If $x$ is a power of a prime there is no problem, in literature you will find this identity $$\psi_{0}\left(x\right)=x-\sum_{\rho}\frac{x^{\rho}}{\rho}-\log\left(2\pi\right)-\frac{1}{2}\log\left(1-x^{-2}\right)$$ where $\psi_{0}\left(x\right)=\psi\left(x\right)$ if $x$ is not a prime power, $\psi_{0}\left(x\right)=\psi\left(x\right)-\frac{1}{2}\Lambda(x)$ if $x$ is a prime power, so no problem about the convergence. This formula came from a (non trivial) application of the Perron's forumla, you can easily find it (but if you have problem I can give you some reference). Your calculation of the limit is right because, as you noticed, we know that holds $\psi(x)=x+o(x)$ and in the comment I wrote yesterday you can find a proof of the fact that $t_k\rightarrow\infty$ as $k\rightarrow\infty .$

  • $\begingroup$ Very thanks much @MarcoCantarini $\endgroup$ – user243301 Nov 29 '15 at 13:49

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