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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.
8
votes
Accepted
Upper bound for real part of Riemann Zeta function zeros
There is no known non-trivial (less than 1) bound for real parts of Zeta zeros (I guess, it is even called "weak Riemann conjecture" to find such a bound). So, your result is very-very interesting, ma …
3
votes
Accepted
On the convergence of Dirichlet series over the Mobius Mu function
We can not hope to the estimate of $M(n):=\sum_{k\leq n} \mu(k)$ better than $M(n)=O(n^{1/2+o(1)})$. Take $s\in (0,1/2)$. Rewrite partial sum of our series as
$$\sum_{k=n}^m (M(k)-M(k-1))/k^s=-M(n-1) …
15
votes
Accepted
Does there exist a rational polynomial $P(x)\in{\mathbb Q}[x]{}$ such that $P(\zeta(s))=\zet...
Extending the argument by GH from MO, $\zeta(P(s))$ has a pole for any $s$ such that $P(s)=1$, while $P(\zeta(s))$ has unique pole for $s=1$. Therefore if $\zeta(P(s))=P(\zeta(s))$, then $P(s)=1$ has …
14
votes
Accepted
A multiple integral that seems related to the $\zeta$ function at even integers
We have $$I_{2n}=\frac{(2n)!!}{(2n+1)!!}\cdot \frac1{2n+2}\cdot \pi^{2n}.$$
To see this, we follow the suggestion by Terry Tao in the comments and apply the diagonalization of the integral operator wi …
12
votes
Accepted
Can one show that $(-1)^{n-1} {(1/\zeta)}^{(n)}(x) >0$ for all real $x>1$?
In other words, you ask whether the function $f(x):=1-1/\zeta(1+x)$ is completely monotonic on
$[0,+\infty)$. We have $f(x)=\sum_{n>1} -\mu(n)/n^{1+x}=\int e^{-xt}d\lambda(t)$, where $\lambda=\sum_{n> …
7
votes
What is known about sums of the form $\sum_{n=2}^{\infty}[\zeta(n)-1]^{p} $?
An integral representation:$$\sum (\zeta(n)^2-1)=\sum_{n\geqslant 2, (a,b)\ne (1,1)}\frac1{a^nb^n}=\sum_{(a,b)\ne (1,1)}\frac1{ab(ab-1)}\\=
\sum_{(a,b)\ne (1,1)}\int_0^1 x^{ab-2}(1-x)dx=\int_0^1 (1-x) …