Search Results

In chapter 4, p. 15-18 of the book H. Iwaniec, Lectures on the Riemann Zeta Function, American Mathematical Society, University Lecture Series nº 62, 2014 there is an elementary proof of the prime …

This has almost nothing to do with the zeros of zeta. To show it you can substitute $\frac{\Im(\rho_n)}{2\pi}$ in your program with an adequate approximation. Since the number of zeros of zeta $$N( …

Read Theorem 8.12 in Titchmarsh: For $\frac12 \le \sigma <1$ take $0<\alpha <1-\sigma$. Then the inequality $|\zeta(\sigma+it)| > \exp(\log^\alpha t)$ is satisfied for indefinitely large values of …

Your question is a little imprecise, because you use $\arg(\zeta(1/2+it))$ without explaining what value to take. I assume that you refer to $\pi S(t)$ as defined in the book of Titchmarsh (section 9. …

You have also Guinand formula for $N(T)$, see, for example, in this answer https://mathoverflow.net/a/104570/7402

Equations of this type are known. You may see, for example, the classical book "Primzahlen" by Landau paragraph 67. "Continuation of the zeta function by partial integration" There it is proved the f …

For the first question about $2\log\zeta(\sigma_0)|$, I think the reasoning is this: For $\sigma>2$ we have $$|L(s,\chi)-1|\le \sum_{n=2}^\infty\frac{1}{n^\sigma}=\zeta(\sigma)-1<1.$$ Hence $\log L(s, …

The series $\sum_\rho \rho^{-1}$ over the non-trivial zeros is not absolutely convergent, this is proved in Davenport p. 80. But as Davenport says and proves in page 81-82 the series converges con …

Assuming the Riemann Hypothesis Montgomery consider the function $$F(\alpha)=F(\alpha,T)=\Bigl(\frac{T}{2\pi}\log T\Bigr)^{-1} \sum_{0<\gamma, \gamma'\le T} T^{i\alpha(\gamma-\gamma')}w(\gamma-\gamma' …

The function $$f(s):=\sin(\tfrac{\pi}{2}(s+1)) \zeta(s)+\sin(\tfrac{\pi s}{2}) L(\tfrac{s+1}{2})\tag{1}$$ where $L(s)$ is the second Dirichlet series $$L(s)=\sum\limits_{n=1}^{\infty } \frac{(-1)^{n-1 …

If $\varphi$ is in the class of Schwartz we have $$\int_{-\infty}^{+\infty}\varphi(t)\zeta(\frac12+it)\,dt= \sum_{n=0}^\infty\Bigl\{ \frac{1}{\sqrt{n+1}}\widehat{\varphi}\Bigl(\frac{1}{2\pi}\log (n+1) …

I think it is false: Consider a simple series with a zero at $s=2$. For example $$E(s)=1-\frac{1}{2^s}-\frac{12}{4^s}=P(2^{-s}),\quad \text{with} \quad P(x)=1-x-12x^2.$$ We have $E(2)=0$ and $E(1)=- …

In the paper: M. W. Coffey, "Asymptotic estimation of $\xi^{(2n)}(1/2)$: On a conjecture of Farmer and Rhoades", Mathematics of Computation, {\bf 78} (2009) 1147--1154 you may find the first terms o …

The formula is given by Guinand as said in his answer by Matthew Watkins but it can be found in page 111 of the paper and reads: assuming Riemann Hypothesis and $T>0$ $$\frac12(N(T+0)+N(T-0))=\frac{T} …

The asymptotic expansion of Cipolla starts $$p_n=n\log n+n\log\log n-n+n\frac{\log\log n}{\log n}+O(n(\log\log n/\log n)^2)$$ So the given approximations have errors $$p_n=n\frac{\Gamma'(n)}{\Gamma(n …