104
votes
Accepted
Is the Riemann zeta function surjective?
The Riemann zeta function is surjective. First, $\zeta(1/z)$ is holomorphic in the punctured disk $0<|z|<1$. Looking at $z=(1/2+it)^{-1}$ with $t\to\infty$ reveals that $\zeta(1/z)$ has an ...
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79
votes
Is the Riemann zeta function surjective?
$\zeta$ function has only one pole at $z=1$. It also has order $1$. If $\zeta$ omits $c\in C$ then $g:=1/(\zeta-c)$ is entire with one simple zero at $1$.
As it is of order $1$, it must be $g(z)=(z-1)...
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73
votes
"Long-standing conjectures in analysis ... often turn out to be false"
I don't know about analysis in general, but I think it's definitely fair to say "often" in functional analysis. My feeling is that we have a solid, thorough, elegant body of theory which usually leads ...
50
votes
Accepted
Why is so much work done on numerical verification of the Riemann Hypothesis?
People are interested in computing the zeros of $\zeta(s)$ and related functions not only as numerical support for RH. Going beyond RH, there are conjectures about the vertical distribution of the ...
- 838
48
votes
"Long-standing conjectures in analysis ... often turn out to be false"
If RH is "analysis", then surely Littlewood's 1914 theorem that $\pi(x)$ (the prime counting function) and $\mathrm{li}(x)$ (the logarithmic integral) alternate in size infinitely often... despite all ...
46
votes
Accepted
Could the Riemann zeta function be a solution for a known differential equation?
When posed properly, a long-standing open problem, but in the form you ask:
Robert A. Van Gorder, MR 3276353 Does the Riemann zeta function satisfy a differential equation?, J. Number Theory 147 (...
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43
votes
Accepted
Is there a closed form for $\int_0^\infty\frac{\tanh^3(x)}{x^2}dx$?
Following the suggestion I made in a comment, the integral can be rewritten as the contour integral
$$
I_{3,2} = \frac{1}{2\pi i} \oint \frac{\operatorname{tanh}^3 z}{z^2} \log(-z) \, dz ,
$$
where ...
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43
votes
Accepted
Why did Euler consider the zeta function?
This history is described in Euler and the Zeta Function by Raymond Ayoub (1974). In his early twenties, around 1730, Euler considered the celebrated problem to calculate the sum $$\zeta(2)=\sum_{n=1}^...
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42
votes
Heuristic argument for the Riemann Hypothesis
The Riemann hypothesis is true, if primes are random in certain ways.
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39
votes
Accepted
Why these surprising proportionalities of integrals involving odd zeta values?
For $n\geq 1$ and $m\geq 0$, an application of integration by parts ($u=\log^n(1-x)$, $dv=\log^m(x)\,dx/x$) followed by the substitution $x\mapsto 1-x$ shows that
$$
\frac{I_{n,m}}{I_{m+1,n-1}}=\frac{...
- 9,061
37
votes
Motivated account of the prime number theorem and related topics
To a certain extent, I think that analytic number theory really is magical, and there's a limit to how natural and motivated it can be. Of the accounts I have seen, the one in Donald Newman's book ...
- 82.6k
33
votes
Heuristic argument for the Riemann Hypothesis
There are many theoretical results that support the (generalized) Riemann hypothesis:
zero density estimates for the zeros of certain $L$-functions
infinitely many zeros on the critical line of ...
- 105k
30
votes
"Long-standing conjectures in analysis ... often turn out to be false"
The Riemann hypothesis is a conjecture in both analysis and number theory. Someone who tries to undermine it necessarily has to ignore the latter part or to declare it irrelevant. I am not suggesting ...
28
votes
Accepted
Riemann's attempts to prove RH
The short answer is no. If anyone were aware of such a record, it would surely have been Carl Siegel, who undertook a careful study of Riemann’s unpublished notes. However, Siegel wrote:
Approaches ...
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28
votes
$\psi(x)-x$ on average
In Theorem 1 of
Brent, Richard P.; Platt, David J.; Trudgian, Timothy S., The mean square of the error term in the prime number theorem, ZBL07569752.
it is shown that for sufficiently large $x$ one ...
- 114k
27
votes
Why is so much work done on numerical verification of the Riemann Hypothesis?
I would add a few more comments to the very pertinent ones above:
1: We are lucky to have two things that work in our favor - an excellent representation of $\zeta$ on the critical line by a simple ...
- 2,040
26
votes
Has Apéry's proof of the irrationality of $\zeta(3)$ ever been used to prove the irrationality of other constants?
Regarding your second question, Apéry's amazing formula
$$\zeta(3) = {5\over 2} \sum_{n\ge1} {(-1)^{n-1} \over n^3 {2n \choose n}}$$
has inspired the search for analogous formulas for other zeta ...
- 82.6k
26
votes
Accepted
Has Apéry's proof of the irrationality of $\zeta(3)$ ever been used to prove the irrationality of other constants?
The proof of irrationality of $\displaystyle\sum_{n=0}^{+\infty}\frac1{F_n}$ (where $F_n$ is the $n$-th Fibonacci number) by RIchard André-Jeannin is an adaptation of the original Apery's proof of the ...
- 3,996
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 + ...
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25
votes
Why is so much work done on numerical verification of the Riemann Hypothesis?
Part of the point is that such numerical checks can be demonstrations of the efficiency of this or that new algorithm. However, it is also the case that a finite check (that all the zeroes of $\zeta(s)...
- 545
24
votes
Accepted
Representations of $\zeta(3)$ as continued fractions involving cubic polynomials
See NOTE below.
This MO inquiry is over 3 yrs old now.
By the date the question about the $\zeta(3)$ CF with $k=8/7$ was made (Feb, 2019), it can be answered in the negative nowadays, since it was '(...
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23
votes
Could the Riemann zeta function be a solution for a known differential equation?
The fact that $\zeta$ satisfy no algebraic differential equation is due to its famous relation with the Gamma function which was proved by Hölder not to satisfy such an equation.
Detailed answer can ...
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22
votes
Could the Riemann zeta function be a solution for a known differential equation?
The Riemann zeta function is "hypertranscendental" in the sense shown HERE
It is not the solution $y(x)$ of a differential equation of the form
$$
F(x,y,y',y'',\dots,y^{(n)})=0
$$
where $F$ is a ...
- 41.1k
22
votes
A new way of approaching the pole of the Riemann zeta function - and a new conjectured formula
The paper entitled Euler constant as a renormalized value of Riemann zeta function at its pole by Andrei Vieru contains a derivation of the first formula in the OP (Benoȋt Cloitre's formula), and ...
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21
votes
Accepted
values of $\zeta$ function are linearly independent?
People have been exerting steady effort to prove/disprove irrationality of the Riemann zeta values in your list. Of course, $\zeta(3)$ is known to be irrational due to Roger Apery. Such investigations,...
- 43.1k
20
votes
Is there a closed form for $\int_0^\infty\frac{\tanh^3(x)}{x^2}dx$?
Rewrite the integrand and apply Taylor expansion to $\frac1{(1+e^{-2x})^3}$ so that
$$\frac{\tanh^3x}{x^2}=\sum_{j\geq0}(-1)^j\binom{j+2}2
\frac{(1-e^{-2x})^3}{x^2}\binom{j+2}2e^{-2jx}.$$
Integrate ...
- 43.1k
20
votes
Accepted
$P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-...}}}}$
This has been answered in the comments by Lucia. The identity $$P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-\cdots}}}}$$ is false. By ...
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 ...
- 148k
20
votes
Algebraic independence of shifts of the Riemann zeta function
Hmm, it was more difficult than I expected to leverage universality to establish the claim. But one can proceed by probabilistic reasoning instead, basically exploiting the phase transition in the ...
- 114k
20
votes
Accepted
Algebraic independence of shifts of the Riemann zeta function
$\zeta(s - z)$ has an Euler product $\prod_p \frac{1}{1 - p^{z-s}}$, and so a monomial $\prod_i \zeta(s - z_i)$ (with the $z_i$ not necessarily distinct) has an Euler product
$$\prod_i \zeta(s - z_i) =...
- 118k
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