Quasicrystals and the Riemann Hypothesis

Question

Let $0 < k_1 < k_2 < k_3 < \cdots $ be all the zeros of the Riemann zeta function on the critical line:

$$ \zeta(\frac{1}{2} + i k_j) = 0 $$

Let $f$ be the Fourier transform of the sum of Dirac deltas supported at these points. In other words:

$$ f(x) = \sum_{j = 1}^\infty e^{ik_j x} $$

This is not a function, but it's a tempered distribution. Matt McIrvin made a graph of it:

(source)

McIrvin seems to get an infinite linear combination of Dirac deltas supported at logarithms of powers of prime numbers. But Freeman Dyson seems to claim that this is only known to be true assuming the Riemann Hypothesis. So, my question is:

1) What do people know about $f$ without assuming the Riemann Hypothesis?

2) What do people know about $f$ assuming the Riemann Hypothesis?

3) Can we prove that $f$ is a linear combination of Dirac deltas supported at prime powers, if we assume the Riemann Hypothesis?

4) Is there some property of $f$, like being a linear combination of Dirac deltas supported at prime powers, that's known to imply the Riemann Hypothesis?

Part of why I'm confused is that J. Main, V. A. Mandelshtam, G. Wunner and H. S. Taylor have a paper containing some equations (equations 8 and 9) that seem to imply

$$ \sum_{j = 1}^\infty \delta(k - k_j) = - \frac{1}{\pi} \sum_{p} \sum_{m = 1}^\infty \frac{\ln p}{p^{m/2}} e^{i k \ln{p^m}} $$

where the sum over $p$ is a sum over primes. This seems to be just what we need to show $f$ is an infinite linear combination of Dirac deltas supported at logarithms of prime powers!

I may be getting some signs and factors of $2 \pi$ wrong, but I don't think that's the main problem: I think the problem is whether a formula resembling the above one is known to be true, or whether it's only known given the Riemann Hypothesis.

Here's what Freeman Dyson said about this in 2009. Unfortunately he omits some details I'm dying to know:

The proof of the Riemann Hypothesis is a worthy goal, and it is not for us to ask whether we can reach it. I will give you some hints describing how it might be achieved. Here I will be giving voice to the mathematician that I was fifty years ago before I became a physicist. I will talk first about the Riemann Hypothesis and then about quasi-crystals.

There were until recently two supreme unsolved problems in the world of pure mathematics, the proof of Fermat's Last Theorem and the proof of the Riemann Hypothesis. Twelve years ago, my Princeton colleague Andrew Wiles polished off Fermat's Last Theorem, and only the Riemann Hypothesis remains. Wiles' proof of the Fermat Theorem was not just a technical stunt. It required the discovery and exploration of a new field of mathematical ideas, far wider and more consequential than the Fermat Theorem itself. It is likely that any proof of the Riemann Hypothesis will likewise lead to a deeper understanding of many diverse areas of mathematics and perhaps of physics too. Riemann's zeta-function, and other zeta-functions similar to it, appear ubiquitously in number theory, in the theory of dynamical systems, in geometry, in function theory, and in physics. The zeta-function stands at a junction where paths lead in many directions. A proof of the hypothesis will illuminate all the connections. Like every serious student of pure mathematics, when I was young I had dreams of proving the Riemann Hypothesis. I had some vague ideas that I thought might lead to a proof. In recent years, after the discovery of quasi-crystals, my ideas became a little less vague. I offer them here for the consideration of any young mathematician who has ambitions to win a Fields Medal.

Quasi-crystals can exist in spaces of one, two, or three dimensions. From the point of view of physics, the three-dimensional quasi-crystals are the most interesting, since they inhabit our three-dimensional world and can be studied experimentally. From the point of view of a mathematician, one-dimensional quasi-crystals are much more interesting than two-dimensional or three-dimensional quasi-crystals because they exist in far greater variety. The mathematical definition of a quasi-crystal is as follows. A quasi-crystal is a distribution of discrete point masses whose Fourier transform is a distribution of discrete point frequencies. Or to say it more briefly, a quasi-crystal is a pure point distribution that has a pure point spectrum. This definition includes as a special case the ordinary crystals, which are periodic distributions with periodic spectra.

Excluding the ordinary crystals, quasi-crystals in three dimensions come in very limited variety, all of them associated with the icosahedral group. The two-dimensional quasicrystals are more numerous, roughly one distinct type associated with each regular polygon in a plane. The two-dimensional quasi-crystal with pentagonal symmetry is the famous Penrose tiling of the plane. Finally, the one-dimensional quasi-crystals have a far richer structure since they are not tied to any rotational symmetries. So far as I know, no complete enumeration of one-dimensional quasi-crystals exists. It is known that a unique quasi-crystal exists corresponding to every Pisot–Vijayaraghavan number or PV number. A PV number is a real algebraic integer, a root of a polynomial equation with integer coefficients, such that all the other roots have absolute value less than one${}^1$. The set of all PV numbers is infinite and has a remarkable topological structure. The set of all one-dimensional quasi-crystals has a structure at least as rich as the set of all PV numbers and probably much richer. We do not know for sure, but it is likely that a huge universe of one-dimensional quasi-crystals not associated with PV numbers is waiting to be discovered.

Here comes the connection of the one-dimensional quasi-crystals with the Riemann hypothesis. If the Riemann hypothesis is true, then the zeros of the zeta-function form a one-dimensional quasi-crystal according to the definition. They constitute a distribution of point masses on a straight line, and their Fourier transform is likewise a distribution of point masses, one at each of the logarithms of ordinary prime numbers and prime-power numbers. My friend Andrew Odlyzko has published a beautiful computer calculation of the Fourier transform of the zeta-function zeros${}^2$. The calculation shows precisely the expected structure of the Fourier transform, with a sharp discontinuity at every logarithm of a prime or prime-power number and nowhere else.

My suggestion is the following. Let us pretend that we do not know that the Riemann Hypothesis is true. Let us tackle the problem from the other end. Let us try to obtain a complete enumeration and classification of one-dimensional quasicrystals. That is to say, we enumerate and classify all point distributions that have a discrete point spectrum...We shall then find the well-known quasi-crystals associated with PV numbers, and also a whole universe of other quasicrystals, known and unknown. Among the multitude of other quasi-crystals we search for one corresponding to the Riemann zeta-function and one corresponding to each of the other zeta-functions that resemble the Riemann zeta-function. Suppose that we find one of the quasi-crystals in our enumeration with properties that identify it with the zeros of the Riemann zeta-function. Then we have proved the Riemann Hypothesis and we can wait for the telephone call announcing the award of the Fields Medal.

These are of course idle dreams. The problem of classifying one-dimensional quasi-crystals is horrendously difficult, probably at least as difficult as the problems that Andrew Wiles took seven years to explore. But if we take a Baconian point of view, the history of mathematics is a history of horrendously difficult problems being solved by young people too ignorant to know that they were impossible. The classification of quasi-crystals is a worthy goal, and might even turn out to be achievable.

[1] M.J. Bertin et al., Pisot and Salem Numbers, Birkhäuser, Boston, 1992.

[2] A.M. Odlyzko, Primes, quantum chaos and computers, in Number Theory: Proceedings of a Symposium, 4 May 1989, Washington, DC, USA (National Research Council, 1990), pp. 35–46.

Answer 1

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 in http://www.math.sjsu.edu/~goldston/article38.pdf

EDIT: If you don't assume the Riemann Hypothesis, then of course some of the $k_j$ will have to be complex numbers.

Answer 2

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 prove this formula assuming RH, Weil removed the RH condition. Such a formula actually holds true for every function with a certain vertical growth restriction, an Euler product and a functional equation, e.g., the Selberg zeta function or every automorphic L function.

The issue whether or not RH hold might be equivalent to which kind of test functions the formula can be applied, eg., if the zeros do not lie on a line, the Fourier transform must extent to a neighborhood in a unique fasion (holomorphic).

What do we know about your distribution $f$ is essentially (Fourier) dual to what we know about the explicit formula which is in some sense (modulo Fourier uncertainity) equivalent to the knowledge about the zeros.

There is a certain positivity condition known to be equivalent to RH due to Weil(?), which can be shown easily via the explicit formula. I can't recall a good reference for it but there are plenty of survey article about RH (either an article of Conrey or Sarnak).

Edit: See the review of Weil's article for precise versions of my vague statements about equivalences for RH

http://www.ams.org/mathscinet-getitem?mr=0053152:

Quote: "The author further proves that a necessary and sufficient condition for the validity of the Riemann hypothesis for L(s) is that the right-hand side of (I) is ≥0 for all functions F(x) of a certain class. He also gives a necessary and sufficient condition for the validity of the Riemann hypothesis for all functions L(s) belonging to k and this in the form that a certain distribution on the group of idèle-classes should be of positive type."

Answer 3

Too long for a comment, so I will add an answer.

I heard often about this, but so far I have yet to find an actual proof that RH implies that the zeroes of the RZF form a 1-dimensional quasicrystal. Actually there is no formal definition of quasicrystal yet, the new definition (1992) of a crystal is intentionally vague and we will probably not have a better formal definition until we understand them better.

Also note that the zeroes of RZF have arbitrarily large gaps, which is not something the quasicrystal community accepts. And there is a reason for that: The diffraction of an infinite quasicrystal is formally defined as the "limit" of diffraction of larger samples of the material. In general this limit might not exist, which happens when large samples of the solid have completely different diffraction, but it can easily be proven that all those finite diffractions are inside some compact spaces, thus we always have cluster points and we can get a limit by simply going to subsequences/subnets (depending if we work in $\mathbb{R}^d$ or arbitrary lcag). This means that some solids have different diffractions depending on how we average.

While this looks clumsy, in reality is not. It only happens if different larger and larger samples of our solid have completely different properties, which means that different large pieces of the solid are basically different materials.... Imagine finding a new material and drilling some samples to diffract. Unknown to us, the rock we find has two halves, one material $A$ and one material $B$. What will our diffraction be? Well the answer is: depends where we drill the sample. We could get the diffraction of $A$, of $B$ or many different diffractions of mixtures of $A$ and $B$. Those are exactly the cluster points we get.

We typically ask that the solid is nice (typically an uniquely ergodicity assumption), which leads to many nice properties and unique diffraction, but this is most of the times not a needed assumption.

So back to zeroes of RZF. The fact that the zeroes of RZF have arbitrarily large holes imply that $0$ is a diffraction measure of this system (i.e. we could drill by chance all our samples just from those holes). The system definitely has multiple diffractions, which makes the problem harder: deciding if this is a quasicrystal depends on which one we pick.

Since we are in $\mathbb{R}$, there are some choices which are more natural, so lets ignore this first issue... And lets simplify things, lets forget the measure approach and work directly with distributions.

If the tempered distribution $ \sum_{j = 1}^\infty e^{ik_j x}$ is a translation bounded discrete measure, then I think Hof/Lagarias proved that the set has pure point diffraction, which is understood to imply quasicrystal. This seems to be the case here.

But there is also to consider the following Theorem by Cordoba([1]), which I only know from a paper of Lagarias (I didn't read the original yet):

Theorem: Let $S \subset \mathbb{R}^d$ be an uniformly discrete set. If the Fourier transform of $\delta_S= \sum_{x \in S} \delta_x$ is a tempered distribution which is a translation bounded discrete measure, then $S$ is a finite union of translates of full rank lattices in $\mathbb{R}^d $.

This shows that the zeroes of RZF cannot fit this description. The only possibility left is that Fourier transform of $\delta_S= \sum_{x \in S} \delta_x$ is a distribution which is a sum of Dirac deltas, but is not a translation bounded measure. But then, as far as I know, there is nothing done in the quasycrystal community in this direction, and I don't know how this fits within our Theory (Hof might have done something in this direction though).

Last but not least, classifying all one dimensional quasicrystals seems like a problem which is impossible to solve, at least in a reasonable way. The issue comes from the fact that at any point in $\chi \in \hat{\mathbb{R^d} } $ we lose in the diffraction the phase information, which is a complex number on the unit circle. There are $c^c$ potential values the phase could take, not all of the potential values would work, but it is easy to show that in general, for a given diffraction, there are at least $c$ values which can work and I suspect that there are more than $c$ good values.

Because of this, each diffraction corresponds to infinitely many solids, some which are related and some which are not. If somehow we manage to get the classification of 1-D quasicrystals, we still have to face two huge issues :

• The classification will be an uncountable list of classes, each with uncountable elements. And given the many cases of models with the same diffraction, the classification is highly likely to be not nice..
• If we have a list of all quasicrystals, how do we check if the zeroes of RZF are or are not in the list, without knowing already the zeroes?

To me, this approach seems similar to trying to classify the zeroes of all analytic functions, and then check which ones correspond to the RZF....

([1]) A. Cordoba, Dirac Combs, Letter Math Physics, 17 (1989), 191-196

Answer 4

Thank you for posting this. I have some doubts about the worth of my own answer but I think the plots look nice so I will post them anyways.

A lazy but true formula for the Riemann zeta function when $s>0$ is:

$$\zeta(s)=\lim\limits_{k\to \infty}\left(\sum _{n=1}^k \frac{1}{n^s}+\frac{k^{1-s}}{s-1}-\frac{k^{-s}}{2}\right), \tag{1}$$

according to Mathematica:

Clear[s, n, k];
Limit[Sum[1/n^(s), {n, 1, k}] + k^(1 - (s))/((s) - 1) - (k^(-(s)))/2, 
 k -> Infinity, Assumptions -> s > 0] 

A plot of the above formula on the critical line:

Clear[t, m, s, M, r, b, M, v, d, x, k, q, z, nn, H, T, TT, g, t];
(*b=N[Sum[Sum[(BernoulliB[2*r]/((2*r)!))*(-d k^(1-2 r) E^(I x) \
Abs[StirlingS1[2*r-1,m]] Gamma[1+m,s Log[d*k]] \
Log[d*k]^(-1-m)),{m,1,2*r-1}],{r,1,q-1}],30];*)
b = 4;
M = 7;
v = 2;
k = 20;
q = 8;
c = 1;
dd = 2;
nn = 1;

z[t_] = Sum[1/n^(1/2 + I*t), {n, 1, k}] + 
   k^(1 - (1/2 + I*t))/((1/2 + I*t) - 1) - (k^(-(1/2 + I*t)))/2;
z1 = ListLinePlot[Table[N[Re[z[t]]], {t, 0, 60, 1/10}], 
   DataRange -> {0, 60}, ImageSize -> Large, 
   PlotStyle -> {Thickness[0.004]}];
z2 = Table[
   Graphics[{Arrowheads[0.025], 
     Arrow[{{Im[ZetaZero[n]], -1/2}, {Im[ZetaZero[n]], 0}}]}], {n, 1, 
    13}];
z3 = Table[
   Graphics[
    Style[Rotate[Text[N[Im[ZetaZero[n]]], {Im[ZetaZero[n]], -1}], 
      90 Degree], Medium]], {n, 1, 13}];
Show[z1, z2, z3]

which is the familiar looking zeta function.

The Fourier transform of the real part of $(1)$ is as in the question:

$$l(x)=\int_{t_1}^{t_2} \Re\left(\zeta \left(\frac{1}{2}+i t\right)\right)\ e^{i t x}\,dt = \sum _{n=1}^k \left(\frac{i d^{\frac{1}{2}-i t} n^{-\frac{1}{2}-i t} e^{i t x}}{2 (\log (d)+\log (n)-x)}-\frac{i d^{\frac{1}{2}+i t} n^{-\frac{1}{2}+i t} e^{i t x}}{2 (\log (d)+\log (n)+x)}\right)+\frac{i d^{\frac{1}{2}+i t} k^{-\frac{1}{2}+i t} e^{i t N[x]}}{4 (\log (d)+\log (k)+x)}-\frac{i d^{\frac{1}{2}-i t} k^{-\frac{1}{2}-i t} e^{i t x}}{4 (\log (d)+\log (k)-x)}-\frac{1}{2} i e^{x/2} \text{Ei}\left(\frac{1}{2} i (i+2 t) (x-\log (d k))\right)+\frac{1}{2} i e^{-\frac{x}{2}} \text{Ei}\left(\frac{1}{2} (2 i t+1) (x+\log (d k))\right) ,\tag{2}$$ when $d=1$.
A plot of the Fourier transform $l(x)$ when setting $t_1=-100$ and $t_2=+100$:

h[t_] = Sum[(I d^(1/2 - I t) E^(I t x) n^(-(1/2) - I t))/(
     2 (-x + Log[d] + Log[n])) - (
     I d^(1/2 + I t) E^(I t x) n^(-(1/2) + I t))/(
     2 (x + Log[d] + Log[n])), {n, 1, k}] + -(1/2) I E^(x/2)
     ExpIntegralEi[1/2 I (I + 2 t) (x - Log[d k])] + 
   1/2 I E^(-x/2) ExpIntegralEi[1/2 (1 + 2 I t) (x + Log[d k])] - (
   I d^(1/2 - I t) E^(I t x) k^(-(1/2) - I t))/(
   4 (-x + Log[d] + Log[k])) + (
   I d^(1/2 + I t) E^(I t N[x]) k^(-(1/2) + I t))/(
   4 (x + Log[d] + Log[k]));

t1 = -100;
t2 = +100;
L = Table[N[(h[t2] - h[t1])], {x, 1/200, dd, 1/200}];
l1 = ListLinePlot[
   Re[Sum[Sum[L*MoebiusMu[d], {d, Divisors[z]}]/z^c, {z, 1, nn}]], 
   DataRange -> {1/200, dd}, PlotRange -> {-80, 80}, 
   ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
l2 = Table[
   Graphics[{Arrowheads[0.025], 
     Arrow[{{Log[n], -35}, {Log[n], 0}}]}], {n, 1, 7}];
l3 = Table[
   Graphics[
    Style[Rotate[Text[Log[n], {Log[n], -52}], 90 Degree], Large]], {n,
     1, 7}];
Show[l1, l2, l3]

Notice that no scaling/shrinking/stretching of the x-axis is needed because the factor $2\pi$ has been left out as in $f(x)$. The frequency spikes are exactly at logarithms of $n$.

Now consider the relation:

$$\boxed{\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}} \tag{3}$$

confirmed here.

and plot:

$$m(t)=\sum _{z=1}^{\infty} \zeta \left(\frac{1}{2}+i t\right)\left(\sum\limits_{d \mid n} \frac{\mu (d) }{z^c d^{i t+\frac{1}{2}-1}}\right) \tag{4}$$

nn = k;
m[t_] = Sum[
  Sum[(Sum[1/n^(1/2 + I*t), {n, 1, k}] + 
       k^(1 - (1/2 + I*t))/((1/2 + I*t) - 1) - (k^(-(1/2 + I*t)))/2)*
     MoebiusMu[d]/N[d]^(1/2 + I*t - 1), {d, Divisors[z]}]/z^c, {z, 1, 
   nn}]; m1 = 
 ListLinePlot[Table[N[Re[m[t]]], {t, 1/10, 60, 1/10}], 
  DataRange -> {0, 60}, PlotRange -> {-3, 7 + 1/2}, 
  ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
m2 = Table[
   Graphics[{Arrowheads[0.025], 
     Arrow[{{Im[ZetaZero[n]], -1}, {Im[ZetaZero[n]], 0}}]}], {n, 1, 
    13}];
m3 = Table[
   Graphics[
    Style[Rotate[Text[N[Im[ZetaZero[n]]], {Im[ZetaZero[n]], -2}], 
      90 Degree], Medium]], {n, 1, 13}];
Show[m1, m2, m3]

The formula $(3)$ and $(4)$, for the von Mangoldt function is integrable and it is the right hand side of $(2)$:

A plot of this formula is:

nn = k;
t1 = -100;
t2 = 100;
T = Table[N[(h[t2] - h[t1])], {x, 1/200, dd, 1/200}];
f1 = ListLinePlot[
   Re[Sum[Sum[T*MoebiusMu[d], {d, Divisors[z]}]/z^c, {z, 1, nn}]], 
   DataRange -> {1/200, dd}, PlotRange -> {-80, 80}, 
   ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
f2 = Table[
   Graphics[{Arrowheads[0.025], 
     Arrow[{{Log[n], -35}, {Log[n], 0}}]}], {n, 1, 7}];
f3 = Table[
   Graphics[
    Style[Rotate[Text[Log[n], {Log[n], -52}], 90 Degree], Large]], {n,
     1, 7}];
Show[f1, f2, f3]

We notice that the heights of the spikes at logarithms are proportional to the first few terms of the von Mangoldt function:

$${\infty ,\log (2),\log (3),\log (2),\log (5),0,\log (7)\,...}$$

A Fourier transform that is closer to the picture in the question can be computed by setting the upper and lower integration limits $t_1=-500$ and $t_2=+500$:

nn = k;
h[t_] = Sum[(I d^(1/2 - I t) E^(I t x) n^(-(1/2) - I t))/(
     2 (-x + Log[d] + Log[n])) - (
     I d^(1/2 + I t) E^(I t x) n^(-(1/2) + I t))/(
     2 (x + Log[d] + Log[n])), {n, 1, k}] + -(1/2) I E^(x/2)
     ExpIntegralEi[1/2 I (I + 2 t) (x - Log[d k])] + 
   1/2 I E^(-x/2) ExpIntegralEi[1/2 (1 + 2 I t) (x + Log[d k])] - (
   I d^(1/2 - I t) E^(I t x) k^(-(1/2) - I t))/(
   4 (-x + Log[d] + Log[k])) + (
   I d^(1/2 + I t) E^(I t N[x]) k^(-(1/2) + I t))/(
   4 (x + Log[d] + Log[k]));
t1 = -500;
t2 = 500;
T = Table[N[(h[t2] - h[t1])], {x, 1/1000, dd, 1/1000}];
f1 = ListLinePlot[
   Re[Sum[Sum[T*MoebiusMu[d], {d, Divisors[z]}]/z^c, {z, 1, nn}]], 
   DataRange -> {1/200, dd}, PlotRange -> {-500, 500}, 
   ImageSize -> Large, PlotStyle -> {Thickness[0.003]}];
f2 = Table[
   Graphics[{Arrowheads[0.025], 
     Arrow[{{Log[n], -250}, {Log[n], -100}}]}], {n, 1, 7}];
f3 = Table[
   Graphics[
    Style[Rotate[Text[Log[n], {Log[n], -400}], 90 Degree], 
     Large]], {n, 1, 7}];
Show[f1, f2, f3]

Answer:

The integrable/Fourier transformable function that gives the plot of $$f(x) = \sum_{j = 1}^\infty e^{ik_j x}$$ is: $$f(x)=\int_{t_1=-\infty}^{t_2=+\infty} \Re\left(\sum _{z=1}^{\infty} \zeta \left(\frac{1}{2}+i t\right)\left(\sum\limits_{d \mid z} \frac{\mu (d) }{z^c d^{i t+\frac{1}{2}-1}}\right)\right)\ e^{i t x}\,dt = \sum _{z=1}^{\infty}\sum\limits_{d \mid z} \frac{\mu (d) }{z^c}\left(\sum _{n=1}^k \left(\frac{i d^{\frac{1}{2}-i t} n^{-\frac{1}{2}-i t} e^{i t x}}{2 (\log (d)+\log (n)-x)}-\frac{i d^{\frac{1}{2}+i t} n^{-\frac{1}{2}+i t} e^{i t x}}{2 (\log (d)+\log (n)+x)}\right)+\frac{i d^{\frac{1}{2}+i t} k^{-\frac{1}{2}+i t} e^{i t N[x]}}{4 (\log (d)+\log (k)+x)}-\frac{i d^{\frac{1}{2}-i t} k^{-\frac{1}{2}-i t} e^{i t x}}{4 (\log (d)+\log (k)-x)}-\frac{1}{2} i e^{x/2} \text{Ei}\left(\frac{1}{2} i (i+2 t) (x-\log (d k))\right)+\frac{1}{2} i e^{-\frac{x}{2}} \text{Ei}\left(\frac{1}{2} (2 i t+1) (x+\log (d k))\right)\right) ,\tag{5}$$

Instead of depending on the zeta zeros this formulation of $f(x)$ depends on the Möbius/Mertens function.

The Fourier transform of the term with the Bernoulli numbers can be found here, but it requires integration by parts twice and then expanding the product in the expression inside the remaining integral from the second integration by parts. The convergence is slow for the Fourier transform when including the Bernoulli numbers in the Euler Maclaurin formula.

Disclaimer, if anything is wrong in the latex expressions, trust the Mathematica code. Also the blocks of code must be run/executed in the order they have been placed in the answer. The whole block of Mathematica code

Apparently the partial sums of the von Mangoldt function Fourier Dirac combs, with the Mathematica Accumulate[] command, gives an approximation of the Chebyshev psi function:

"Mathematica start"
Clear[t, m, s, M, r, b, M, v, d, x, k, q, z, nn, H, T, TT, g, t];
(*b=N[Sum[Sum[(BernoulliB[2*r]/((2*r)!))*(-d k^(1-2 r) E^(I x) \
Abs[StirlingS1[2*r-1,m]] Gamma[1+m,s Log[d*k]] \
Log[d*k]^(-1-m)),{m,1,2*r-1}],{r,1,q-1}],30];*)
b = 4;
M = 7;
v = 2;
k = 20;
q = 8;
c = 1;
dd = 2;
nn = 1;
nn = k;
t1 = -100;
t2 = 100;
h[t_] = Sum[(I d^(1/2 - I t) E^(I t x) n^(-(1/2) - I t))/(2 (-x + 
         Log[d] + 
         Log[n])) - (I d^(1/2 + I t) E^(I t x) n^(-(1/2) + 
          I t))/(2 (x + Log[d] + Log[n])), {n, 1, 
    k}] + -(1/2) I E^(x/2)
ExpIntegralEi[1/2 I (I + 2 t) (x - Log[d k])] + 
  1/2 I E^(-x/2) ExpIntegralEi[
    1/2 (1 + 2 I t) (x + 
       Log[d k])] - (I d^(1/2 - I t) E^(I t x) k^(-(1/2) - 
        I t))/(4 (-x + Log[d] + 
       Log[k])) + (I d^(1/2 + I t) E^(I t N[x]) k^(-(1/2) + 
        I t))/(4 (x + Log[d] + Log[k]));

T = Table[N[(h[t2] - h[t1])], {x, 1/200, dd, 1/200}];
f1 = ListLinePlot[-1 + 1/20 + 
    1/2000*Accumulate[
      Re[Sum[Sum[T*MoebiusMu[d], {d, Divisors[z]}]/z^c, {z, 1, nn}]]],
    DataRange -> {1/200, dd}, PlotRange -> {-1, 1}, 
   ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
f2 = Table[
   Graphics[{Arrowheads[0.025], 
     Arrow[{{Log[n], -0.3}, {Log[n], 0}}]}], {n, 1, 7}];
f3 = Table[
   Graphics[
    Style[Rotate[Text[Log[n], {Log[n], -0.6}], 90 Degree], 
     Large]], {n, 1, 7}];
Show[f1, f2, f3]
"Mathematica end"

"Mathematica start"
Clear[t, m, s, M, r, b, M, v, d, x, k, q, z, nn, H, T, TT, g, t];
(*b=N[Sum[Sum[(BernoulliB[2*r]/((2*r)!))*(-d k^(1-2 r) E^(I x) \
Abs[StirlingS1[2*r-1,m]] Gamma[1+m,s Log[d*k]] \
Log[d*k]^(-1-m)),{m,1,2*r-1}],{r,1,q-1}],30];*)
b = 4;
M = 7;
v = 2;
k = 20;
q = 8;
c = 1;
dd = 2;
nn = 1;
nn = k;
t1 = -500;
t2 = 500;
h[t_] = Sum[(I d^(1/2 - I t) E^(I t x) n^(-(1/2) - I t))/(2 (-x + 
         Log[d] + 
         Log[n])) - (I d^(1/2 + I t) E^(I t x) n^(-(1/2) + 
          I t))/(2 (x + Log[d] + Log[n])), {n, 1, 
    k}] + -(1/2) I E^(x/2)
ExpIntegralEi[1/2 I (I + 2 t) (x - Log[d k])] + 
  1/2 I E^(-x/2) ExpIntegralEi[
    1/2 (1 + 2 I t) (x + 
       Log[d k])] - (I d^(1/2 - I t) E^(I t x) k^(-(1/2) - 
        I t))/(4 (-x + Log[d] + 
       Log[k])) + (I d^(1/2 + I t) E^(I t N[x]) k^(-(1/2) + 
        I t))/(4 (x + Log[d] + Log[k]));

T = Table[N[(h[t2] - h[t1])], {x, 1/800, dd, 1/800}];
f1 = ListLinePlot[-1.05 + 
    1/7000*Accumulate[
      Re[Sum[Sum[T*MoebiusMu[d], {d, Divisors[z]}]/z^c, {z, 1, nn}]]],
    DataRange -> {1/800, dd}, PlotRange -> {-0.6, 1.3}, 
   ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
f2 = Table[
   Graphics[{Arrowheads[0.025], 
     Arrow[{{Log[n], -0.25}, {Log[n], 0}}]}], {n, 1, 7}];
f3 = Table[
   Graphics[
    Style[Rotate[Text[Log[n], {Log[n], -0.4}], 90 Degree], 
     Large]], {n, 1, 7}];
Show[f1, f2, f3]
"Mathematica end"

Answer 5

Alain Connes (here is the link http://arxiv.org/abs/math/9811068) proved a similar statement in his Selecta paper in 1999, but his method shows above presentation is true only with zeros on critical line, (actually he showed more, it is sum of powers of x^it(logx)^j, where i is square root of -1, j is discrete index, "i.t" is the imaginary part of zero on the critical line ) still to show there is no other zero off the critical line.

This letter of Sarnak to Bombieri might be of some help http://web.math.princeton.edu/sarnak/BombieriLtr2002.pdf

Answer 6

The classical result that comes closest to this is Landau's (1911) formula: for any $x > 1$,

$\frac{1}{T} \sum_{0 < \Im \rho \leq T} x^\rho = -\frac{1}{2\pi} \Lambda(x) + O(\frac{\log T}{T})$

where the sum is taken over all zeta zeros in the critical strip $0 < \Re \rho < 1$ with imaginary part between $0$ and $T$; $\Lambda$ is the von Mangoldt function, which is supported on the prime powers.

Of course that sum is over all zeros and doesn't itself imply anything about the sum over the zeros on the critical line.