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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.

4 votes

What is the behavior of the argument of Riemann’s Zeta function on other verticals inside th...

I'm not sure what sign changes are in this context, as there is no simple multiplier which forces the zeta function to be real on this strip. Instead the argument is going to be moving around in the c …
Will Sawin's user avatar
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7 votes
Accepted

On some analytic property of the Riemann zeta function

$$\int_x^\infty \{ u\} u^{-s-1} du = \int_x^\infty \left( \{ u\}-\frac{1}{2}\right) u^{-s-1} du + \frac{1}{2} \frac{x^{-s}}{s}$$ and by integration by parts, $$\int_x^\infty \left( \{ u\}-\frac{1}{2} …
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13 votes

Jensen Polynomials for the Riemann Zeta Function

The GUE random matrix model predicts that the zeroes should satisfy the local statistics of random matrices. It doesn't predict that the zeroes should satisfy the global statistics of random matrices, …
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20 votes

Heuristic argument for the Riemann Hypothesis

The function field model came up in GH from MO's answer, and then also in user54038's. I just want to add some detail to explain how good of an analogy the function field model is. The Riemann zeta …
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26 votes

Why is the Simple Zeros Conjecture said to be stronger than the Riemann Hypothesis?

As Peter Humphries points out, the precise claim is that "RH + Simple Zeroes" is stronger than "RH". Of course, this is formally trivial. So what's really meant is that "RH + Simple Zeroes" is a natur …
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7 votes

Size of $\zeta'(s)$ at its zeros

Bounds for the value of zeta take the form $$ |\zeta(\sigma +it) | \ll t^{ f(\sigma)+\epsilon} $$ where $|t|>2$ and $f$ is some function of $\sigma$. We can take $f(1)=0$ and $f(0)=1/2$ and we can al …
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