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Formulas for the analogous of the Z(t) function in the case of Dirichlet L-functions are known For example see: D. Davies, C. B Haselgrove, The evaluation of Dirichlet L-functions, Proc. of the Roya …

This is well known (Riemann could have writen it) $$\zeta(s)=\frac{1}{2}\frac{\pi^{s/2}}{(s-1)\Gamma(1+s/2)}\prod_{\Im\rho > 0}\Bigl\{ \Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{\overline{\rho}}\Bi …

The zeta function can be written $\zeta(1/2+it)=Z(t)e^{-i\vartheta(t)}$ where $Z(t)$ are real analytic and $\vartheta(t)$ is a simple function related to $\Gamma(s)$. The relation between $\zeta(1/2 …

This is not exactly an answer, but too long for a comment. Perhaps I have not been sufficiently clear in the comments. In the paper: T. Kotnik, Computational estimation of the order of $\zeta(1/2+i t …

I do not think this is so difficult as the Riemann hypothesis, I will only explain why this is so without giving complete proof. First on the line $s=1+it$ the functions are $$(\zeta(it)-\zeta(1+it)) …

It is known that the number of zeros with $T-1 < Im(\rho)<T+1$ is $O(\log(T))$. Therefore the multiplicity of a zero $\beta+\gamma i$ will be less than $C \log|\gamma|$ for some absolute constant $C …

I only get $$\sum_{p\le y}\frac{1}{p^{1-\delta}}\le \frac{y^{\delta}}{\log(y^\delta)} +e^2\log(1/\delta)+O\Bigl(\frac{y^\delta}{\delta^2(\log y)^2}+1\Bigr).$$ The first sum is $$\sum_{p\le e^{2/\delt …

Since we have $\Gamma(1/2+it)=\sqrt{\pi/\cosh(\pi t)}\exp[i(2 \vartheta(t)+t \log(2\pi)+\arctan(\tanh(\pi t/2)))]$ where $\vartheta(t)$ is the Riemann Siegel function. The zeros on the critical li …

The solutions of $\Re\zeta(1+it)=0$ are scarce. They limit small intervals where $\Re\zeta(1+it)<0$. The probability in the sense of the limit of the quotient of the measure of the set $\{0<t<T: \Re\ …

I will say that the natural normalisation of the zeros of zeta is $$\tilde\gamma=\frac{1}{\pi}\vartheta(\gamma)$$ where $$\vartheta(t)=\Im(\log\Gamma(\frac14+\frac{it}{2}))-\frac{t}{2}\log\pi=\frac{t} …

In fact for odd $n\ge3$ we have $$\Bigl.\frac{d^{n}}{ds^n}\log\zeta(s)\Bigr|_{s=\frac12}= \frac{(n-1)!}{2}\Bigl(2^n L(n,\chi)+(2^n-1)\zeta(n)\Bigr)$$ The proof (due to Voros) is the following: It …

This is Theorem 3 (p. 10) in the Report: J. van de Lune, Some inequalities involving Riemann's zeta-function, CWI Report ZW 50/75, CentruM Wiskunde & Informatica, Amsterdam,1975, in which the proo …

The figure of egg is not an elliptic curve. Since the curve pass through the point $(-2,0)$ and $(1,0)$ the equation will be of type $$y^2=(ax+b)(x+2)(x-1)$$ We find the numbers $y_0$ and $y_1$ wh …

In the paper J. Rosenblatt, Convergence of Series of Translations, Math. Ann. 230 (1977) 245-272 it is proved a general Theorem (Theorem 2.5) that in particular proves the convergence a. e. . Als …

I assume Riemann hypothesis on all this answer. I want a closed form for $$f(s,x):= \prod_{n=1}^\infty\Bigl(1-\frac{s}{\frac12+i x\gamma_n}\Bigr) \Bigl(1-\frac{s}{\frac12-i x\gamma_n}\Bigr).$$ Of co …