87

We have $$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} \quad \mbox{for} \ Re(s)>1.$$ Taking the derivative with respect to $s$, we get the following $$- \frac{\zeta'(s)}{\zeta(s)^2} = - \sum_{n=1}^{\infty} (\log n) \frac{\mu(n)}{n^s} \quad \mbox{for} \ Re(s)>1.$$ If we plug in $s=0$ to the second equation (which is not allowed, because ...


75

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 essential singularity at $z=0$, hence $\zeta(s)$ misses at most one value. If $\zeta(s)=w$ then $\zeta(\overline{s})=\overline{w}$, hence $\zeta(s)$ can only miss a ...


70

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 to positive solutions rather quickly, when they exist. (The Kadison-Singer problem is a recent exception which required radically new tools for a positive ...


68

$\zeta$ 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 $g(z)=(z-1)e^{az+b}$, by Hadamard's factorization theorem, so $$\zeta(z)=(z-1)^{-1}e^{-az-b}+c,$$ which is absurd. The formula I wrote is the general form of a non-...


55

One can show that $\sum_{n=1}^{\infty} \mu(n)/\sqrt{n}$ diverges. Suppose to the contrary that it converges, which as you note implies RH. Put $M_0(x)=\sum_{n\le x} \mu(n)/\sqrt{n}$, and our assumption is that $M_0(x)=C+o(1)$ as $x\to \infty$. Note that for any $s=\sigma+it$ with $\sigma>1/2$ we have $$ \int_0^{\infty} sM_0(e^x)e^{-sx} dx = \sum_{n=...


47

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 numerical evidence at the time indicating that $\pi(x)\le \mathrm{li}(x)$. Part of the point is that the first reversal only occurs at a rather large number. ...


45

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 (2015), 778--788.


42

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 $$\zeta(2)=\sum_{n=1}^\infty \frac{1}{n^2}.$$ This problem goes back to 1650, it was posed by Pietro Mengoli and John Wallis computed the sum to three decimal places. Ayoub ...


42

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 nontrivial zeros (after "unfolding" them to have average spacing 1, assuming they are on a vertical line to begin with). Odlyzko found striking numerical support ...


40

The Riemann hypothesis is true, if primes are random in certain ways.


37

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 $$ \frac{I_{n,m}}{I_{m+1,n-1}}=\frac{n}{m+1}. $$ All of your examples are special cases of this identity.


35

Just for your reference, the equation: $$ \sum_{j = 1}^\infty \delta(k - k_j) = - \frac{i}{\pi} \sum_{p} \sum_{m = 1}^\infty \frac{\ln p}{p^{m/2}} e^{i k \ln{p^m}} $$ seems to be a re-statement, in a distributional setting, of Riemann's explicit formula. It can be proven unconditionally, with appropriate test functions on each side. See for example Lemma 1 ...


35

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 Analytic Number Theory comes the closest to helping you see how you might have come up with the key ideas, but even Newman occasionally pulls things out of a hat (...


34

As long as a result remains unproved it can be pure speculation about whether it is really hard or just nobody has found the right simple idea, though in this case it seems plausible that the desired nonvanishing is going to be require a profound new idea. (For more on the task of deciding if a problem is hard before it has been solved, look at the answers ...


31

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 certain $L$-functions nonnegativity of the central value of certain $L$-functions subconvex bounds for certain $L$-functions strong lower and upper bounds for ...


30

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 that it is true (I do not know), only that it becomes more plausible when you take into account that its violation implies a sort of conspiracy between primes ...


28

So of course it does not converge. The behavior is interesting however. Below is the graph of the $\ln$ of the product going up to $5000.$ There can be runs very rich in square free integers with an even number of divisors which cause a dramatic shift. At $509$ the product is about $5\times 10^{-13}.$ Of the next $45$ square free integers ,$15$ provide a ...


28

Following the suggestion I made in a comment, the integral can be rewritten as the contour integral $$ I_{3,2} = \frac{1}{2\pi i} \oint \frac{\operatorname{tahn}^3 z}{z^2} \log(-z) \, dz , $$ where the clockwise contour tightly encircles the positive real axis, which coincides with the branch cut of the logarithm. The reason that this integral is ...


27

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 to a proof of the so-called “Riemann hypothesis” or even to a proof of the existence of infinitely many zeros of the zeta function on the critical line are not ...


23

As indicated in my comment, some of these integrals are essentially known, and involve the hyperbolic "Beukers-Kolk-Calabi" change of variables. In particular, in this paper, Z. Silagadze shows in (27) that \begin{equation*} \zeta(n) = \frac{2^n}{2^n-1}\int_0^1\cdots\int_0^1 \frac{dx_1\cdots dx_n}{1-x_1^2\cdots x_n^2}, \end{equation*} which clearly yields (...


22

Regarding (1), The American Institute of Mathematics survey of the Riemann Hypothesis refers to $\kappa=1$ as the "100% hypothesis", see http://www.aimath.org/WWN/rh/articles/html/35a/ "In contrast to most of the other conjectures in this section, the 100% Hypothesis is not motivated by applications to the prime numbers. Indeed, at present there are no ...


22

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 polynomial (with constant coefficients).


22

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 real function (simple up to a good approximation, approximation usually called the Riemann Siegel formula) - the Hardy function, $Z(t)$ - multiplied by a ...


21

The identity of distributions resembles the Weil-Guinand explicit formula, see here: Wikipedia, Explicit formulae (L-function). To my knowledge, you either have to include the trivial zeros of zeta or the distributions coming from $\Gamma$-factors and poles, so the formula as stated seems wrong from that perspective. On the history: Guinand was able to ...


21

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 be found here with five (commented and linked) references. On the other hand, this function is linked with many other transcendental special functions like ...


20

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, among others, motivated the question of linear independence. As Felipe commented, however, not much is known, apart from the following article: Rivoal, ...


20

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 subtracting $1$ from both sides and squaring, we have that $$(1-P(s))^{2}+1-\frac{2}{\zeta(s)}=1-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{...


20

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 a method to obtain variations thereof, such as as well as similar iterates in terms of the Dirichlet function $\lambda(s)=\sum_{n=0}^\infty(2n+1)^{-s}$,


20

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 limiting distribution of the zeta function at the critical line. The proof I found used an unexpectedly high amount of firepower; perhaps there is a more ...


20

$\zeta(s - z)$ 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 $$\prod_i \zeta(s - z_i) = \prod_p \prod_i \frac{1}{1 - p^{z_i - s}}.$$ We want to show that these monomials are linearly independent. Now here's an observation: it's quite hard for ...


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