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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.
2
votes
Accepted
Error term in França-LeClair approximation of zeta zeros
There is no simple equation that given $n$ gives us $\gamma_n$, with sufficient approximation. Given a constant $c>0$, the error in both equations are at times greater than $c/\log n$ for any given c …
3
votes
Odlyzko's reformulation of Montgomery's pair correlation conjecture
I will say that the natural normalisation of the zeros of zeta is
$$\tilde\gamma=\frac{1}{\pi}\vartheta(\gamma)$$
where
$$\vartheta(t)=\Im(\log\Gamma(\frac14+\frac{it}{2}))-\frac{t}{2}\log\pi=\frac{t} …
1
vote
Zeroes of real and imaginary parts of $\zeta(1+it)$ separately (if any for $t>1$ say)
The solutions of $\Re\zeta(1+it)=0$ are scarce. They limit small intervals where
$\Re\zeta(1+it)<0$. The probability in the sense of the limit of the
quotient of the measure of the
set $\{0<t<T: \Re\ …
2
votes
Are Li's numbers $\lambda_n$ absolutely convergent for $n>1$?
Assuming the Riemann hypothesis $\rho=\frac12+i\gamma$, then
$$1-\frac{1}{\rho}=-\frac{\frac12-i\gamma}{\frac12+i\gamma}=e^{2i\theta},\qquad
\theta=\arctan\frac{1}{2\gamma}.$$
$$\sum_\rho\Bigl[1-\Big …
1
vote
Riemann–Von Mangoldt formula
You have also Guinand formula for $N(T)$, see, for example, in this answer
https://mathoverflow.net/a/104570/7402
3
votes
Accepted
A expression for the tangent function involving $\zeta(n),n=2,3,\ldots$
When I was writing my answer, Dan Romik answer appeared. Mine is the same but with more detail.
We have
$$\log\Gamma(1-x)=\gamma x+\sum_{n=2}^\infty\zeta(n)\frac{x^n}{n}$$
(this is known and is also a …
1
vote
Integrals involving $1/|\zeta(1+i t)|^2$: closed expressions?
This is not properly an answer,
after the comments of Tao it is difficult to give an answer. Only an explanation of my comment above. I still think that my series and the integral are equal.
We can w …
4
votes
Accepted
Derivatives of Riemann $\xi$ and traces of zeros
It is not a good idea to compute
$$\sum_{n=1}^\infty \gamma_n^{-k}$$ by computing a partial sum of several thousands of terms. The series converges but too slowly for this.
(see "Computation of the …
16
votes
Accepted
An integral involving the argument of the Gamma function and the Riemann Hypothesis
We prove that
$$I=-\frac{\pi}{4}(\gamma+\log 4).$$
$$I=\int_0^\infty\frac{t\arg\Gamma(\frac14+\frac{it}{2})}{(\frac14+t^2)^2}\,dt.$$
$I$ is the imaginary part of the complex integral
$$\int_0^\infty …
2
votes
Enquiry on an equality involving the Riemann zeta function
Your question is a little imprecise, because you use $\arg(\zeta(1/2+it))$ without explaining what value to take. I assume that you refer to $\pi S(t)$ as defined in the book of Titchmarsh (section 9. …
8
votes
Accepted
A recurrence relation for $\zeta(2n)$ - reference request
I have this recurrence in my collection of problems for my lessons in Analytic Number Theory. I have there the reference:
P. Ribenboim, Classical Theory of Algebraic Numbers, Springer, New York, 200 …
5
votes
Confusion about Montgomery's pair correlation conjecture
Assuming the Riemann Hypothesis Montgomery consider the function
$$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' …
12
votes
Accepted
Is this differential equation for zeta on the critical line? One can compute it from its der...
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 …
1
vote
$\sum_{d\leq x} (\mu(d)/d) \log x/d$: elementary estimates?
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 …
2
votes
Do the complex zeros of $\big(\zeta(s-1) -\zeta(s)\big) \pm \big(\zeta(1-s) - \zeta(2-s)\big...
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)) …