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The Riemann zeta function is the function of one complex variable $s$ defined by the series $\zeta(s) = \sum_{n \geq 1} \frac{1}{n^s}$ when $\operatorname{Re}(s)>1$. It admits a meromorphic continuation to $\mathbb{C}$ with only a simple pole at $1$. This function satisfies a functional equation relating the values at $s$ and $1-s$. This is the most simple example of an $L$-function and a central object of number theory.

12 votes

Sharpening of Lindelöf hypothesis

No. In fact it is known that for any $\varepsilon>0$ there are arbitrarily large values of $t$ with $$ |\zeta(1/2+it)| \ge \exp\left( (1-\varepsilon) \sqrt{\frac{\log t}{\log \log t}} \ \right).$$ Thi …
Micah Milinovich's user avatar
4 votes

On the critical line $ \Re \zeta'(s)/\zeta(s) =? 1/2 \log(\pi) - 1/2 \Re \psi(s/2)$ ?

Your formula follows from Hadamard's product formula for $\zeta(s)$ and the corresponding partial fraction decomposition of $\zeta'(s)/\zeta(s)$. See, for instance, section 2 of Soundararajan's paper: …
Micah Milinovich's user avatar
6 votes
Accepted

Riemann Z function, bounds on number of non-trivial zeros along horizontal lines, rather tha...

It $t$ is not an ordinate of a zero of $\zeta(s)$, define $$ S(t) = \frac{1}{\pi} \arg \zeta(1/2+it) = -\frac{1}{\pi} \Im \int_{1/2}^\infty \frac{\zeta'}{\zeta}(\sigma+it) d\sigma$$ and define $$ S(t) …
Micah Milinovich's user avatar
2 votes

Inequality for the modulus of Riemann zeta on horizontal lines and alleged partial result of...

GH has answered your question, but here is some additional remarks. The function you are calling $F(s)$ is often called $\chi(s)$ in the theory of the Riemann zeta-function (e.g. in Titchmarsh's book …
Micah Milinovich's user avatar
11 votes
Accepted

Axioms for Riemann $\zeta$ function

Perhaps you are looking for something like Hamburger's Theorem? It states, essentially, that the only Dirichlet series with a finite number of singularities satisfying the same functional equation a …
Micah Milinovich's user avatar
3 votes

Closed form for derivatives $\zeta^{(n)}(1/2)$

Edit: My original answer was incorrect. You can evaluate $\zeta'(\frac{1}{2})$ recursively in terms of $\zeta(\frac{1}{2})$ using the symmetric form of the functional equation: $$ \zeta(s)\Gamma(\tf …
Micah Milinovich's user avatar
5 votes

Possible locations for non trivial zeroes lying off the critical line

Let $\chi(s)=2^s \pi^{s-1}\sin(\pi s/2) \Gamma(1-s)$ so that $\zeta(s)=\chi(s)\zeta(1-s)$. You are asking about the curve $|\chi(s)|=1$. As you have observed, $|\chi(1/2+it)|=1$ for real $t$. There i …
Micah Milinovich's user avatar
7 votes
Accepted

Is this sum of reciprocals of zeta zeros correct?

To answer your modified question, according to Mathematica: $$ \lim_{s\to 0} \left(\frac{\hat{\zeta}'}{\hat{\zeta}}(s)+\frac{1}{s}\right) = -\frac{\gamma}{2} + \tfrac{1}{2}\log(4\pi).$$ This implies …
Micah Milinovich's user avatar
14 votes
Accepted

Upper bounds on the difference of consecutive zeta zeros

Littlewood was the first to prove that the gaps between the ordinates of successive zeros of $\zeta(s)$ tend to zero. This is proved, for instance, in Titchmarsh's book on the zeta-function (see Theor …
Micah Milinovich's user avatar
6 votes

Is this sum of reciprocals of zeta zeros correct?

Edit: I endorse Juan's answer to the original question. The sum $\displaystyle{\sum_{\rho} \tfrac{1}{|\rho|}}$, running over the non-trivial zeros $\rho$ of $\zeta(s)$, is known to diverge, so at best …
Micah Milinovich's user avatar
11 votes
Accepted

non-trivial zeros of partial zeta functions

The answer to question 1 is classical: Any Dirichlet series which has a finite abscissa of absolute convergence has a zero-free half-plane. Suppose the Dirichlet series $$ A(s)=\sum_{n=1}^\infty \fra …
Micah Milinovich's user avatar
11 votes

Explicit formula for Riemann zeros counting function

Assuming the Riemann Hypothesis, you can use a smooth approximation to the characteristic function of an interval in the Guinand-Weil explicit formula to approximately count the number of zeros of the …
Micah Milinovich's user avatar
11 votes
Accepted

Zeros of the derivative of Riemann's $\xi$-function

In exercise 1 on page 443 of their book "Multiplicative Number Theory," Montgomery & Vaughan outline a proof of the statement: "Assuming the Riemann Hypothesis, $\xi'(s)=0 \implies \mathrm{Re}(s)=1 …
Micah Milinovich's user avatar