The Denseness of the Closure of Some Nyman–Beurling Linear Manifolds Implies the Absence of Zeroes of Certain Combinations of Riemann Zeta-Functions in the Critical Strip

Abstract

The famous Nyman–Beurling theorem states that the absence of zeroes in the Riemann zeta-function in the half-plane Res > 1/p, p > 1, is equivalent to the circumstance in which the closure of the linear manifold of the functions $f(x)={\displaystyle \sum _{k=1}^n{\alpha }_k\left\{\frac{{\vartheta }_k}{x}\right\}}$, where $0<{\vartheta }_k\le 1$, with the condition ${\displaystyle \sum _{k=1}^n{a}_k{\vartheta }_k}=0$, is dense in L_p(0,1). Here, we show that if the closure of linear manifolds of the same functions but with the conditions ${\displaystyle \sum _{k=1}^n{a}_k{\vartheta }_k^l}=0$ with l = 2, 3, 4 is dense in L_p(0,1), then certain combinations of Riemann zeta-functions are free from zeroes in the half-plane Res > 1/p, p > 1—like, e.g., the function $g_2(s)={\displaystyle \frac{2}{s-1}}\zeta (s-1)+\zeta (s)$ for l = 2. Similar results for larger integer l can be established. The connections between the Riemann zeta-function, including the question concerning the location of its zeroes, with different symmetry aspects of numerous physical systems are well established, and recently they were highlighted also for supersymmetric quantum mechanics.

1. Introduction

More than seventy years ago, Neumann and Beurling proposed the following now-famous criterion equivalent to the Riemann hypothesis [1,2]; see, e.g., [3,4] for the main properties of the Riemann zeta-function. Here and below, $\left\{y\right\}$ denotes the fractional part of non-negative number y, and ${\alpha }_i$ are arbitrary complex numbers.

Theorem 1. 

The Riemann zeta-function $\zeta (z)$ is free from zeroes in the half-plane Res > 1/p, p > 1, if and only if ${\overline{M}}_1$ is dense in the space $L_p(0,1)$. Here, M₁ is the linear manifold of the functions $f_1(x)={\displaystyle \sum _{k=1}^n{\alpha }_k\left\{\frac{{\vartheta }_k}{x}\right\}}$, where $0<{\vartheta }_k\le 1$, with the condition ${\displaystyle \sum _{k=1}^n{a}_k{\vartheta }_k}=0$, and ${\overline{M}}_1$ is the closure of these functions in the space $L_p(0,1)$.

Later on, this condition was much studied and modified, and nowadays the detailed list of the corresponding research contains at least a few tens of papers (in particular, see the already rather old, mostly bibliographical (but not only) reference [5]; see also [6] and the references in all other papers cited below). This prevents us to present it here in much detail. For completeness, we briefly indicate only the following aspects.

In our opinion, after the important contribution of Baez–Duarte [7], quite interesting strengthening and generalization of the Neumann–Beurling criterion was proposed by Nikolski in 1995 [8]. Very soon, in relation with this work, Vasyutin studied in detail the geometry of a set of locally constant functions in $L_2(0,1)$, which was used as a basis for the representation of a constant function [9]. Such an approach has evident connection with the Neumann–Beurling problem, and his work led to the introduction of the families of functions closely related with this problem (“cotangent sums”). Later on, these functions and sums in themselves became a subject of detailed studies; see, e.g., [10,11].

Another important direction of the research was certain specification of the parameters ${\vartheta }_k$. Baez-Duarte showed, after reformulation and slight modification of the problem to the space $L_2(0,\infty )$, that for the claim of the equivalence of the denseness of the corresponding appropriate manifold to the Riemann hypothesis, one can specify ${\vartheta }_k=1/k$ [12]. From the other side, the “randomization” of the set of ${\vartheta }_k$ also lead to the establishment of new characterizations and structures [13]. Hilbert space reformulation of the criterion in probability spaces also deserves to be mentioned [14].

These and other developments were certainly important. At the same time, we would like to underline that, as it stands today, the Neumann–Beurling manifold M₁ might produce an impression of a kind of unique and isolated object. This is not the case. The aim of the present note is to construct a whole class of somewhat similar linear manifolds. Namely, we introduce the manifolds M_l, l = 1, 2, 3… of the functions $f_l(x)={\displaystyle \sum _{k=1}^n{\alpha }_k\left\{\frac{{\vartheta }_k}{x}\right\}}$, where again $0<{\vartheta }_k\le 1$, but now with the conditions ${\displaystyle \sum _{k=1}^n{a}_k{\vartheta }_k^l}=0$. We show that the certain function, closely related with the Riemann zeta-function, is free from zeroes in the half-plane Res > 1/p, p > 1, if the closure of these manifolds is dense in the space $L_p(0,1)$.

The following theorems hold:

Theorem 2. 

Let M₂ be the linear manifold of the functions $f_2(x)={\displaystyle \sum _{k=1}^n{\alpha }_k\left\{\frac{{\vartheta }_k}{x}\right\}}$, where $0<{\vartheta }_k\le 1$, with the condition ${\displaystyle \sum _{k=1}^n{a}_k{\vartheta }_k^2}=0$, and ${\overline{M}}_2$ be the closure of these functions in the space $L_p(0,1)$.

The function

$$g_2(s):={\displaystyle \frac{2}{s-1}}\zeta (s-1)+\zeta (s)$$

is free from zeroes in the half-plane Res > 1/p, p > 1, if ${\overline{M}}_2$ is dense in the space $L_p(0,1)$.

Theorem 3. 

Let M₃ be the linear manifold of the functions $f_3(x)={\displaystyle \sum _{k=1}^n{\alpha }_k\left\{\frac{{\vartheta }_k}{x}\right\}}$, where $0<{\vartheta }_k\le 1$, with the condition ${\displaystyle \sum _{k=1}^n{a}_k{\vartheta }_k^3}=0$, and ${\overline{M}}_3$ be the closure of these functions in the space $L_p(0,1)$.

The function

$$g_3(s):={\displaystyle \frac{6}{(s-1)(s-2)}}\zeta (s-2)+{\displaystyle \frac{3}{(s-1)}}\zeta (s-1)+\zeta (s)$$

is free from zeroes in the half-plane Res > 1/p, p > 1, if ${\overline{M}}_3$ is dense in the space $L_p(0,1)$.

Theorem 4. 

Let M₄ be the linear manifold of the functions $f_4(x)={\displaystyle \sum _{k=1}^n{\alpha }_k\left\{\frac{{\vartheta }_k}{x}\right\}}$, where $0<{\vartheta }_k\le 1$, with the condition ${\displaystyle \sum _{k=1}^n{a}_k{\vartheta }_k^4}=0$, and ${\overline{M}}_4$ be the closure of these functions in the space $L_p(0,1)$.

The function

$$g_4(s):={\displaystyle \frac{24}{(s-1)(s-2)(s-3)}}\zeta (s-3)+{\displaystyle \frac{12}{(s-1)(s-2)}}\zeta (s-2)+{\displaystyle \frac{4}{s-1}}\zeta (s-1)+\zeta (s)$$

is free from zeroes in the half-plane Res > 1/p, p > 1, if ${\overline{M}}_4$ is dense in the space $L_p(0,1)$.

2. Proof of Sufficiency Condition of Neumann-Beurling Theorem

First, for completeness and as an introduction for further steps, we repeat the first part (sufficiency condition) of the proof of Beurling [2].

Proof of Theorem 1. 

Let us consider $I_1(a,b):={\displaystyle \underset{a}{\overset{b}{\int }}\left\{x\right\}{x}^{-s-1}dx}$. We have $\left\{x\right\}:=x-\left[x\right]$, so that for any s not equal to 0, 1:

$$\begin{array}{c}{I}_1(n,n+1)={\displaystyle \underset{n}{\overset{n+1}{\int }}{x}^{-s}dx}-n{\displaystyle \underset{n}{\overset{n+1}{\int }}{x}^{-s-1}dx}={\displaystyle \frac{1}{-s+1}}({\displaystyle \frac{1}{{(n+1)}^{s-1}}}-{\displaystyle \frac{1}{n^{s-1}}})-{\displaystyle \frac{n}{-s}}({\displaystyle \frac{1}{{(n+1)}^s}}-{\displaystyle \frac{1}{n^s}})=\\ {\displaystyle \frac{1}{-s+1}}({\displaystyle \frac{1}{{(n+1)}^{s-1}}}-{\displaystyle \frac{1}{n^{s-1}}})-{\displaystyle \frac{1}{-s}}({\displaystyle \frac{1}{{(n+1)}^{s-1}}}-{\displaystyle \frac{1}{n^{s-1}}}-{\displaystyle \frac{1}{{(n+1)}^s}}).\end{array}$$

Summing such integrals for all n = 1, 2, 3…, we obtain, for Res > 1, that

$${\displaystyle \sum _{n=1}^{\infty }{I}_1(n,n+1)}=-{\displaystyle \frac{1}{-s+1}}-{\displaystyle \frac{\zeta (s)}{s}},$$

Thus,

$$\zeta (s)={\displaystyle \frac{s}{s-1}}-s{\displaystyle \underset{1}{\overset{\infty }{\int }}\left\{x\right\}{x}^{-s-1}dx}.$$

(1)

By analytic continuation, this is valid also for $\mathrm{Re}s>0$, $s\ne 1$. Using the circumstance that for Res < 1, $I_1(0,1)={\displaystyle \frac{1}{-s+1}}$, for the critical strip $1>\mathrm{Re}s>0$, we have simply

$$\zeta (s)=-s{\displaystyle \underset{0}{\overset{\infty }{\int }}\left\{x\right\}{x}^{-s-1}dx}.$$

(2)

Now we introduce the functions $f_1(x)={\displaystyle \sum _{k=1}^n{\alpha }_k\left\{\frac{{\vartheta }_k}{x}\right\}}$, where $0<{\vartheta }_k\le 1$, with the condition ${\displaystyle \sum _{k=1}^n{a}_k{\vartheta }_k}=0$, and consider linear manifold of such functions M₁ and its closure ${\overline{M}}_1$ in the space $L_p(0,1)$. We have

$${\displaystyle \underset{1}{\overset{\infty }{\int }}\left\{\vartheta y\right\}{y}^{-s-1}dy}={\vartheta }^s{\displaystyle \underset{\vartheta }{\overset{\infty }{\int }}\left\{x\right\}{x}^{-s-1}dx}={\vartheta }^s{\displaystyle \underset{0}{\overset{\infty }{\int }}\left\{x\right\}{x}^{-s-1}dx}-{\vartheta }^s{\displaystyle \underset{0}{\overset{\vartheta }{\int }}\left\{x\right\}{x}^{-s-1}dx}=-{\displaystyle \frac{{\vartheta }^s\varsigma (s)}{s}}-{\displaystyle \frac{\vartheta }{-s+1}},$$

that is

$$s{\displaystyle \underset{1}{\overset{\infty }{\int }}\left\{\vartheta x\right\}{x}^{-s-1}dx}=-{\vartheta }^s\zeta (s)-\vartheta {\displaystyle \frac{s}{-s+1}},$$

(3)

and this is the central point.

Due to the condition ${\displaystyle \sum _{k=1}^n{a}_k{\vartheta }_k}=0$, we have, after summing over k,

$$-\zeta (s){\displaystyle \sum _{k=1}^n{\alpha }_k{\vartheta }_k^s}=s{\displaystyle \underset{1}{\overset{\infty }{\int }}{\displaystyle \sum _{k=1}^n{\alpha }_k}\left\{{\vartheta }_{k}y\right\}{y}^{-s-1}dy},$$

and, with the variable charge x = 1/y,

$$-\zeta (s){\displaystyle \sum _{k=1}^n{\alpha }_k{\vartheta }_k^s}=s{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \sum _{k=1}^n{\alpha }_k}\left\{\frac{{\vartheta }_k}{x}\right\}{x}^{s-1}dx},$$

so that

$$1-\zeta (s){\displaystyle \sum _{k=1}^n{\alpha }_k{\vartheta }_k^s}=s{\displaystyle \underset{0}{\overset{1}{\int }}[1+{\displaystyle \sum _{k=1}^n{\alpha }_k}\left\{\frac{{\vartheta }_k}{x}\right\}]x^{s-1}dx}.$$

(4)

Now take p > 1, $1/p+1/q=1$, and require $\mathrm{Re}(q(s-1))>-1$, that is Res > 1/p, and write Hölder’s inequality:

$$|{\displaystyle \underset{0}{\overset{1}{\int }}[1+{\displaystyle \sum _{k=1}^n{\alpha }_k}\left\{\frac{{\vartheta }_k}{x}\right\}]x^{s-1}dx}|\le ({{\displaystyle \underset{0}{\overset{1}{\int }}|1+{\displaystyle \sum _{k=1}^n{\alpha }_k}\left\{\frac{{\vartheta }_k}{x}\right\}{|}^{p}dx)}}^{1/p}\cdot ({\displaystyle \underset{0}{\overset{1}{\int }}\left|x^{q(s-1)}\right|dx}{)}^{1/q}.$$

(5)

This gives

$$|1-\zeta (s){\displaystyle \sum _{k=1}^n{\alpha }_k{\vartheta }_k^s}|\le {\displaystyle \frac{{\epsilon }^{1/p}}{|q(\mathrm{Re}s-1)+1{|}^{1/q}}}|s|,$$

(6)

where $\epsilon =\left|\right|1+f_1|{|}_p$ is the norm in the L_p space. Certainly, if $\epsilon$ can be made arbitrarily small, i.e., ${\overline{M}}_1$ is dense in the space $L_p(0,1)$, $\zeta (s)$ cannot be equal to zero because the l.h.s here is then simply one, and this finishes the proof of the sufficiency condition of the Nyman–Beurling theorem. □

Corollary 1. 

From (1) and Laurent expansion $\zeta (1+s)={\displaystyle \frac{1}{s}}+\gamma -{\gamma }_{1}s+O(s^2)$, where ${\gamma }_i$ are Stieltjes constants [3,4], we get the well-known

$${\displaystyle \underset{1}{\overset{\infty }{\int }}\left\{x\right\}{x}^{-2}dx}=1-\gamma .$$

(7)

We also have for larger s:

$$2{\displaystyle \underset{1}{\overset{\infty }{\int }}\left\{x\right\}{x}^{-3}dx}=2-{\pi }^2/6,$$

$$3{\displaystyle \underset{1}{\overset{\infty }{\int }}\left\{x\right\}{x}^{-4}dx}=3/2-\zeta (3),$$

etc. The differentiation over s of relations of type (1) and other examples given below, also might be interesting. For example, we have

$$-{\displaystyle \underset{1}{\overset{\infty }{\int }}\left\{x\right\}\ln x\cdot x^{-s-1}dx}=-{\displaystyle \frac{1}{{(s-1)}^2}}-{\displaystyle \frac{{\zeta }^{\prime }(s)}{s}}+{\displaystyle \frac{\zeta (s)}{s^2}},$$

which, in the limit of s tending to 1, gives

$$-{\displaystyle \underset{1}{\overset{\infty }{\int }}\left\{x\right\}\ln x\cdot x^{-2}dx}=\gamma +{\gamma }_1-{\displaystyle \frac{3}{2}},$$

and so on.

3. Proofs of Theorems 2–4

The proofs presented below are generalizations of the Neumann–Beurling proof of Theorem 1. We give the proofs of Theorems 2 and 3 in details, because they serve as an immediate model for the proofs for larger l. Only a few details are given for Theorem 4, and at the end of the section, we outline the approach for general l.

Proof of Theorem 2. 

Let us consider

$$I_2(a,b):={\displaystyle \underset{a}{\overset{b}{\int }}{\left\{x\right\}}^2{x}^{-s-1}dx}={\displaystyle \underset{a}{\overset{b}{\int }}(x^2-2x\left[x\right]+{\left[x\right]}^2)x^{-s-1}dx}.$$

We have, for Res < 2, $I_2(0,1)={\displaystyle \frac{1}{-s+2}}$. Then, further, for $s\ne 0, 1, 2$,

$${\displaystyle \underset{n}{\overset{n+1}{\int }}{x}^{-s+1}dx}={\displaystyle \frac{1}{-s+2}}({\displaystyle \frac{1}{{(n+1)}^{s-2}}}-{\displaystyle \frac{1}{n^{s-2}}}),$$

and also, using n = (n + 1) − 1:

$$-2n{\displaystyle \underset{n}{\overset{n+1}{\int }}{x}^{-s}dx}={\displaystyle \frac{-2n}{-s+1}}({\displaystyle \frac{1}{{(n+1)}^{s-1}}}-{\displaystyle \frac{1}{n^{s-1}}})={\displaystyle \frac{-2}{-s+1}}({\displaystyle \frac{1}{{(n+1)}^{s-2}}}-{\displaystyle \frac{1}{n^{s-2}}}-{\displaystyle \frac{1}{{(n+1)}^{s-1}}}),$$

while using $n^2={(n+1)}^2-2(n+1)+1$:

$$\begin{array}{l}{n}^2{\displaystyle \underset{n}{\overset{n+1}{\int }}{x}^{-s-1}dx}={\displaystyle \frac{n^2}{-s}}({\displaystyle \frac{1}{{(n+1)}^s}}-{\displaystyle \frac{1}{n^s}})={\displaystyle \frac{-1}{s}}[{\displaystyle \frac{{(n+1)}^2-2(n+1)+1}{{(n+1)}^s}}-{\displaystyle \frac{1}{n^{s-2}}}]=\\ {\displaystyle \frac{-1}{s}}({\displaystyle \frac{1}{{(n+1)}^{s-2}}}-{\displaystyle \frac{2}{{(n+1)}^{s-1}}}+{\displaystyle \frac{1}{{(n+1)}^s}}-{\displaystyle \frac{1}{n^{s-2}}})\end{array}.$$

Thus, for Res > 2,

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \underset{n}{\overset{n+1}{\int }}{x}^{-s+1}dx}=-{\displaystyle \frac{1}{-s+2}},$$

$${\displaystyle \sum _{n=1}^{\infty }}(-2n{\displaystyle \underset{n}{\overset{n+1}{\int }}{x}^{-s}dx})={\displaystyle \frac{-2}{-s+1}}(-1-(\zeta (s-1)-1))$$

and

$${\displaystyle \sum _{n=1}^{\infty }}{n}^2{\displaystyle \underset{n}{\overset{n+1}{\int }}{x}^{-s-1}dx}={\displaystyle \frac{-1}{s}}(-1-2(\zeta (s-1)-1)+\zeta (s)-1),$$

so that finally

$${\displaystyle \sum _{n=1}^{\infty }{I}_2(n,n+1)}={\displaystyle \frac{1}{s-2}}-{\displaystyle \frac{2}{s(s-1)}}\zeta (s-1)-{\displaystyle \frac{\zeta (s)}{s}}$$

and

$${\displaystyle \frac{2}{s-1}}\zeta (s-1)+\zeta (s)-{\displaystyle \frac{s}{s-2}}=-s{\displaystyle \underset{1}{\overset{\infty }{\int }}{\left\{x\right\}}^2{x}^{-s-1}dx}.$$

(8)

By analytic continuation, this is valid for Res > 0. If also Res < 2, then $I_2(0,1)={\displaystyle \frac{1}{-s+2}}$ and

$$I_2(0,\infty )={\displaystyle \frac{-2}{s(s-1)}}\zeta (s-1)-{\displaystyle \frac{\zeta (s)}{s}},$$

which proves

$${\displaystyle \frac{2}{s-1}}\zeta (s-1)+\zeta (s)=-s{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left\{x\right\}}^2{x}^{-s-1}dx}$$

(9)

for $0<\mathrm{Re}s<2$.

Next, we repeat the same steps as were performed above during the proof of the Nyman–Beurling criterion, but introducing the functions $f_2(x)={\displaystyle \sum _{k=1}^n{\alpha }_k\left\{\frac{{\vartheta }_k}{x}\right\}}$, where $0<{\vartheta }_k\le 1$, with the condition ${\displaystyle \sum _{k=1}^n{a}_k{\vartheta }_k^2}=0$, and considering the linear manifold of such functions M₂ and its closure ${\overline{M}}_2$ in the space $L_p(0,1)$.

First,

$${\displaystyle \underset{1}{\overset{\infty }{\int }}{\left\{\vartheta y\right\}}^2{y}^{-s-1}dy}={\vartheta }^s{\displaystyle \underset{\vartheta }{\overset{\infty }{\int }}{\left\{x\right\}}^2{x}^{-s-1}dx}={\vartheta }^s{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left\{x\right\}}^2{x}^{-s-1}dx}-{\displaystyle \frac{{\vartheta }^2}{-s+2}}=-{\displaystyle \frac{2}{s-1}}\zeta (s-1)-\zeta (s)-{\displaystyle \frac{{\vartheta }^2}{-s+2}},$$

and after summing, we obtain

$$1-[\zeta (s)+{\displaystyle \frac{2}{s-1}}\zeta (s-1)]{\displaystyle \sum _{k=1}^n{\alpha }_k{\vartheta }_k^s}=s{\displaystyle \underset{0}{\overset{1}{\int }}[1+{\displaystyle \sum _{k=1}^n{\alpha }_k}{\left\{\frac{{\vartheta }_k}{x}\right\}}^2]x^{s-1}dx}.$$

(10)

Using Hölder’s inequality, we obtain

$$|1-[\zeta (s)+{\displaystyle \frac{2}{s-1}}\zeta (s-1)]{\displaystyle \sum _{k=1}^n{\alpha }_k{\vartheta }_k^s}|<{\displaystyle \frac{{\epsilon }_1^{1/p}}{|q(\mathrm{Re}s-1)+1{|}^{1/q}}}|s|.$$

Here ${\epsilon }_1=\left|\right|1+f_2|{|}_p$, and we again require Res > 1/p. □

Corollary 2. 

The point s = 1 is of no peculiarities for Equation (9) so that, in particular, we have an easy proof of $\zeta (0)=-{\displaystyle \frac{1}{2}}$. We also establish, considering (9) for $s=1+\epsilon$ with $\epsilon \to 0$, that

$${\displaystyle \underset{0}{\overset{\infty }{\int }}{\left\{x\right\}}^2{x}^{-2}dx}=-2{\zeta }^{\prime }(0)-\gamma .$$

Thus, using [3,4],

$${\zeta }^{\prime }(0)=-{\displaystyle \frac{1}{2}}\ln (2\pi ).$$

we have

$${\displaystyle \underset{0}{\overset{\infty }{\int }}{\left\{x\right\}}^2{x}^{-2}dx}=\ln (2\pi )-\gamma \cong 1.2607.$$

(11)

We checked Equation (11) numerically.

From Equation (8) for $s=2+\epsilon$ with $\epsilon \to 0$, we get

$$2{\displaystyle \underset{1}{\overset{\infty }{\int }}{\left\{x\right\}}^2{x}^{-3}dx}=3-2\gamma -{\pi }^2/6.$$

(12)

Substituting s = 3 to Equation (8), we obtain

$$-3{\displaystyle \underset{1}{\overset{\infty }{\int }}{\left\{x\right\}}^2{x}^{-4}dx}={\pi }^2/6+\zeta (3)-3,$$

and so on. Numerous equalities of the type (11, 12), see also below, can be found in Furdui’s book [15].

Proof of Theorem 3. 

Let us consider

$$I_3(a,b):={\displaystyle \underset{a}{\overset{b}{\int }}{\left\{x\right\}}^3{x}^{-s-1}dx}={\displaystyle \underset{a}{\overset{b}{\int }}(x^3-3x^2\left[x\right]+3x{\left[x\right]}^2-{\left[x\right]}^3)x^{-s-1}dx}.$$

We have, for Res < 3, $I_3(0,1)={\displaystyle \frac{1}{-s+3}}$. Then, further, for $s\ne 0, 1, 2, 3$,

$$\begin{array}{l}{\displaystyle \underset{n}{\overset{n+1}{\int }}{x}^{-s+2}dx}={\displaystyle \frac{1}{-s+3}}({\displaystyle \frac{1}{{(n+1)}^{s-3}}}-{\displaystyle \frac{1}{n^{s-3}}}),\\ -3n{\displaystyle \underset{n}{\overset{n+1}{\int }}{x}^{-s+1}dx}={\displaystyle \frac{-3n}{-s+2}}({\displaystyle \frac{1}{{(n+1)}^{s-2}}}-{\displaystyle \frac{1}{n^{s-2}}})={\displaystyle \frac{-3}{-s+2}}({\displaystyle \frac{1}{{(n+1)}^{s-3}}}-{\displaystyle \frac{1}{n^{s-3}}}-{\displaystyle \frac{1}{{(n+1)}^{s-2}}}).\end{array}$$

Using $n^2={(n+1)}^2-2(n+1)+1$, we have

$$3n^2{\displaystyle \underset{n}{\overset{n+1}{\int }}{x}^{-s}dx}={\displaystyle \frac{3n^2}{-s+1}}({\displaystyle \frac{1}{{(n+1)}^{s-1}}}-{\displaystyle \frac{1}{n^{s-1}}})={\displaystyle \frac{3}{-s+1}}({\displaystyle \frac{1}{{(n+1)}^{s-3}}}-{\displaystyle \frac{1}{n^{s-3}}}-{\displaystyle \frac{2}{{(n+1)}^{s-2}}}+{\displaystyle \frac{1}{{(n+1)}^{s-1}}}).$$

And using $n^3={(n+1)}^3-3{(n+1)}^2+3(n+1)-1$, we have

$$-n^3{\displaystyle \underset{n}{\overset{n+1}{\int }}{x}^{-s-1}dx}={\displaystyle \frac{-n^3}{-s}}({\displaystyle \frac{1}{{(n+1)}^s}}-{\displaystyle \frac{1}{n^s}})={\displaystyle \frac{1}{s}}({\displaystyle \frac{1}{{(n+1)}^{s-3}}}-{\displaystyle \frac{1}{n^{s-3}}}-{\displaystyle \frac{3}{{(n+1)}^{s-2}}}+{\displaystyle \frac{3}{{(n+1)}^{s-1}}}-{\displaystyle \frac{1}{{(n+1)}^s}}).$$

Thus, finally, collecting everything together, we obtain

$$\begin{array}{l}{I}_3(n,n+1)={\displaystyle \frac{-3}{-s+2}}({\displaystyle \frac{1}{{(n+1)}^{s-3}}}-{\displaystyle \frac{1}{n^{s-3}}}-{\displaystyle \frac{1}{{(n+1)}^{s-2}}})\\ +{\displaystyle \frac{3}{-s+1}}({\displaystyle \frac{1}{{(n+1)}^{s-3}}}-{\displaystyle \frac{1}{n^{s-3}}}-{\displaystyle \frac{2}{{(n+1)}^{s-2}}}+{\displaystyle \frac{1}{{(n+1)}^{s-1}}})\\ +{\displaystyle \frac{1}{s}}({\displaystyle \frac{1}{{(n+1)}^{s-3}}}-{\displaystyle \frac{1}{n^{s-3}}}-{\displaystyle \frac{3}{{(n+1)}^{s-2}}}+{\displaystyle \frac{3}{{(n+1)}^{s-1}}}-{\displaystyle \frac{1}{{(n+1)}^s}}).\end{array}$$

Correspondingly,

$$\begin{array}{l}{\displaystyle \sum _{n=1}^{\infty }{I}_3(n,n+1)}=-{\displaystyle \frac{1}{-s+3}}+{\displaystyle \frac{3}{-s+2}}(1+\zeta (s-2)-1)\\ +{\displaystyle \frac{3}{-s+1}}(-1-2(\zeta (s-2)-1)+\zeta (s-1)-1)\\ +{\displaystyle \frac{1}{s}}(-1-3(\zeta (s-2)-1)+3(\zeta (s-1)-1)-\zeta (s)+1)=\\ -{\displaystyle \frac{1}{-s+3}}+{\displaystyle \frac{3}{-s+2}}\zeta (s-2)+{\displaystyle \frac{3}{-s+1}}(-2\zeta (s-2)+\zeta (s-1))\\ +{\displaystyle \frac{1}{s}}(-3\zeta (s-2)+3\zeta (s-1)-\zeta (s))={\displaystyle \frac{1}{s-3}}-{\displaystyle \frac{3}{s-2}}\zeta (s-2)-{\displaystyle \frac{3}{s-1}}(-2\zeta (s-2)+\zeta (s-1))\\ +{\displaystyle \frac{1}{s}}(-3\zeta (s-2)+3\zeta (s-1)-\zeta (s)).\end{array}$$

Thus, finally we get

$$s{\displaystyle \underset{1}{\overset{\infty }{\int }}{\left\{x\right\}}^3{x}^{-s-1}dx}={\displaystyle \frac{s}{s-3}}-{\displaystyle \frac{6}{(s-1)(s-2)}}\zeta (s-2)-{\displaystyle \frac{3}{s-1}}\zeta (s-1)-\zeta (s),$$

(13)

and prove, for $0<\mathrm{Re}s<3$,

$${\displaystyle \frac{6}{(s-1)(s-2)}}\varsigma (s-2)+{\displaystyle \frac{3}{s-1}}\varsigma (s-1)+\varsigma (s)=-s{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left\{x\right\}}^3{x}^{-s-1}dx}$$

(14)

The rest of the proof is quite similar to Theorems 1 and 2. □

Corollary 3. 

The point s = 1 is of no peculiarities for the r.h.s. of (14), which immediately gives $\zeta (-1)=-1/12$: this, of course, can be considered as one more proof of this equality. Further, using ${\zeta }^{\prime }(-1)={\displaystyle \frac{1}{12}}-\ln A$ [3,4], we get

$${\displaystyle \underset{0}{\overset{\infty }{\int }}{\left\{x\right\}}^3{x}^{-2}dx}=-6\ln A+{\displaystyle \frac{3}{2}}\ln (2\pi )-\gamma ,$$

(15)

where

$$\ln A={\lim }_{n\to \infty }({\displaystyle \sum _{k=1}^{n}k\ln k}-({\displaystyle \frac{n^2}{2}}+{\displaystyle \frac{n}{2}}+{\displaystyle \frac{1}{12}})\ln n+{\displaystyle \frac{n^2}{4}});$$

A = 1.282427… is the Glaisher–Kinkelin constant; see, e.g., [16]. We checked Equation (15) numerically.

In the vicinity of s = 2 here, we get

$${\displaystyle \underset{0}{\overset{\infty }{\int }}{\left\{x\right\}}^3{x}^{-3}dx}={\displaystyle \frac{3}{2}}\ln (2\pi )+{\displaystyle \frac{3}{2}}\gamma +{\displaystyle \frac{{\pi }^2}{12}}.$$

(16)

We checked Equation (16) numerically as well. By substitution of $s=3+\epsilon$ with $\epsilon \to 0$ into (13), we have

$$3{\displaystyle \underset{1}{\overset{\infty }{\int }}{\left\{x\right\}}^3{x}^{-4}dx}={\displaystyle \frac{11}{2}}-3\gamma -{\displaystyle \frac{{\pi }^2}{4}}-\zeta (3).$$

For s = 4, we have

$$4{\displaystyle \underset{1}{\overset{\infty }{\int }}{\left\{x\right\}}^3{x}^{-5}dx}=4-{\displaystyle \frac{{\pi }^2}{6}}-{\displaystyle \frac{3}{2}}\zeta (3)-{\displaystyle \frac{{\pi }^4}{90}},$$

etc.

Proof of Theorem 4. 

The proof of Theorem 4 is quite similar. We start from

$$I_4(a,b):={\displaystyle \underset{a}{\overset{b}{\int }}{\left\{x\right\}}^4{x}^{-s-1}dx}={\displaystyle \underset{a}{\overset{b}{\int }}(x^4-4x^3\left[x\right]+6x^2{\left[x\right]}^2-4x{\left[x\right]}^3+{\left[x\right]}^4)x^{-s-1}dx},$$

and then the rather lengthy repetition of the same steps, with the relation $n^4={(n+1)}^4-4{(n+1)}^3+6{(n+1)}^2-4(n+1)+1$ and analogous relations for $n^3,n^2$ used earlier, leads us to

$${\displaystyle \frac{24}{(s-1)(s-2)(s-3)}}\zeta (s-3)+{\displaystyle \frac{12}{(s-1)(s-2)}}\zeta (s-2)+{\displaystyle \frac{4}{s-1}}\zeta (s-1)+\zeta (s)=-s{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left\{x\right\}}^4{x}^{-s-1}dx},$$

(17)

valid for $0<\mathrm{Re}s<4$. Then, we prove the theorem statement analogous to Theorems 1–3. We omit these details here. □

Corollary 4. 

From (17), with s tending to 1, $\zeta (-2)=0$ immediately follows. Also, the integrals ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\left\{x\right\}}^4{x}^{-n}dx}$ for n = 2, 3, 4 can be estimated. We omit these estimations here.

To finish this section, we would like to underline how this line of research can be continued to larger integers n. For this, one needs to repeat the aforementioned procedure for $I_n(a,b):={\displaystyle \underset{a}{\overset{b}{\int }}{\left\{x\right\}}^n{x}^{-s-1}dx}$ with n = 5, 6, 7, etc., using the binomial formula

$$n^m={((n+1)-1)}^m={\displaystyle \sum _{k=0}^m{(-1)}^k{C}_m^k{(n+1)}^{m-k}}.$$

Here, $C_m^k={\displaystyle \frac{m!}{k!(m-k)!}}$ are binomial coefficients. In such a way, we find the closed expressions for the corresponding sums, ${\displaystyle \sum _{k=1}^{\infty }{I}_n(k,k+1)}$, and arrive to new equalities having the form

$$\begin{array}{l}{\displaystyle \frac{A_n}{(s-n+1)(s-n+2)\dots (s-1)}}\zeta (s-n+1)+{\displaystyle \frac{A_{n-1}}{(s-n+2)\dots (s-1)}}\zeta (s-n+2)+\dots {\displaystyle \frac{A_2}{s-1}}\zeta (s-1)+\zeta (s)=\\ -s{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left\{x\right\}}^n{x}^{-s-1}dx}\end{array}$$

Actually, the coefficients A_i here can be determined simply from the requirements that the points s = 1, 2… n are not the singular points of this relation using the known values of $\zeta (\pm n)$ [3,4] and $\zeta (1+\epsilon )={\displaystyle \frac{1}{\epsilon }}+\gamma +O(\epsilon )$.

4. Discussion and Conclusions

We spoke only about the sufficient conditions here and were unable to prove the necessity, i.e., the statement that the absence of zeroes of the corresponding functions in certain strips implies that the closures ${\overline{M}}_2$, ${\overline{M}}_3$, etc., are dense in L_p(0,1). In particular, the construction given by Beurling to prove the necessity condition of Theorem 1 in [2] apparently cannot be adapted to the cases of n > 1.

Next, we note that, in a similar fashion, a number of results like the following one can be proven.

Theorem 5. 

Let M_aux be the linear manifold of the functions $f_{aux}(x)={{\displaystyle \sum _{k=1}^n{\alpha }_k(\left\{\frac{{\vartheta }_k}{x}\right\}}}^2-\left\{{\displaystyle \frac{{\vartheta }_k}{x}}\right\})$, where $0<{\vartheta }_k\le 1$, with the condition ${\displaystyle \sum _{k=1}^n{\alpha }_k({\vartheta }_k^2-{\vartheta }_k)}=0$, and ${\overline{M}}_{aux}$ be the closure of these functions in the space $L_p(0,1)$.

Then, for any p > 1, the manifold ${\overline{M}}_{aux}$ is not dense in the space $L_p(0,1)$.

Proof. 

Proof is by contradiction. From $\zeta (s)=-s{\displaystyle \underset{0}{\overset{\infty }{\int }}\left\{x\right\}{x}^{-s-1}dx}$ and

$${\displaystyle \frac{2}{s-1}}\zeta (s-1)+\zeta (s)=-s{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left\{x\right\}}^2{x}^{-s-1}dx},$$

see Theorems 1 and 2, we immediately have

$${\displaystyle \frac{2}{s-1}}\zeta (s-1)=-s{\displaystyle \underset{0}{\overset{\infty }{\int }}({\left\{x\right\}}^2-\left\{x\right\})x^{-s-1}dx}$$

and thus, proceeding as above,

$$1-[{\displaystyle \frac{2}{s-1}}\zeta (s-1)]{\displaystyle \sum _{k=1}^n{\alpha }_k{\vartheta }_k^s}=s{\displaystyle \underset{0}{\overset{1}{\int }}[1+{\displaystyle \sum _{k=1}^n{\alpha }_k}({\left\{\frac{{\vartheta }_k}{x}\right\}}^2-\left\{\frac{{\vartheta }_k}{x}\right\})]x^{s-1}dx}.$$

Again, proceeding further exactly as above with the application of Hölder’s inequality, we see that if ${\overline{M}}_{aux}$ is dense in the space $L_p(0,1)$, the function ${\displaystyle \frac{2}{s-1}}\zeta (s-1)$ has no zeroes in the half-plane Res > 1/p, p > 1. But this function has infinite number of zeroes for Res = 3/2, that is, in the half-plane Res > 1/p for any p > 1. □

In relation with the obtained results, it seems interesting to look into the location of zeroes of certain functions closely related to the Riemann zeta-function. We put forward the following hypothesis, which is, of course, unproven.

Hypothesis 1. 

All zeroes of the function

$$g_2(s):={\displaystyle \frac{2}{s-1}}\zeta (s-1)+\zeta (s)$$

with $\mathrm{Im}s\ne 0$ are located on the critical line Res = 1/2.

All zeroes of the function

$$g_3(s):={\displaystyle \frac{6}{(s-1)(s-2)}}\zeta (s-2)+{\displaystyle \frac{3}{s-1}}\zeta (s-1)+\zeta (s)$$

with $\mathrm{Im}s\ne 0$ are located on the critical line Res = 1/2.

All zeroes of the function

$$g_4:={\displaystyle \frac{24}{(s-1)(s-2)(s-3)}}\zeta (s-3)+{\displaystyle \frac{12}{(s-1)(s-2)}}\zeta (s-2)+{\displaystyle \frac{4}{s-1}}\zeta (s-1)+\zeta (s)$$

with $\mathrm{Im}s\ne 0$ are located on the critical line Res = 1/2, and the list of similar hypothetical statements for other functions can be made much longer.

It seems that similar questions were not studied, contrary to the numerous investigations of the location of zeroes of the function of the type $\zeta ({\sigma }_0+s)\pm \zeta ({\sigma }_0-s)$; see, e.g., [17] and references therein. We believe that numerical research is truly desirable in the field.

Finally, we would like to note that the connections between the Riemann zeta-function, including the question concerning the location of its zeroes, and different symmetry aspects of numerous physical systems are well established [18], and recently they were also highlighted for supersymmetric quantum mechanics; see, e.g., [19].

The author sincerely hopes that the current results might be useful for research on the Riemann zeta-function. Certainly, at this moment, the concrete steps in this direction hardly can be predicted. However, it already happened more than once that the study of the whole family of certain objects, like the manifolds ${\overline{M}}_l$ introduced in this paper, might be, in a sense, easier and more straightforward than the study of one isolated object.