Proof of the Riemann Hypothesis via the Chebyshev Function and the Integral Convergence

1. Introduction

The Chebyshev function $\psi (x)={\sum }_{n\le x}\Lambda (n)$ plays a central role in the study of the distribution of prime numbers. The prime number theorem (PNT), proved independently by Hadamard and de la Vallée Poussin in 1896 using complex analysis, states that $\psi (x)\sim x$ , refer to Edwards [6].

In fact, the prime number theorem is equivalent to not only the asymptotic behavior $\psi (x)\sim x,$ but also to the statement that the error term $\psi (x)-x$ satisfies $\psi (x)-x=o(x)$ as $x\to \infty .$ A major breakthrough in the twentieth century was the discovery of elementary proofs of the prime number theorem by Selberg [3] and Erdos [4] in 1949, avoiding complex analysis, but the error term remained rather weak. In this paper, we will deduce the stronger error term that $\psi (x)-x=o(x^{\frac{1}{2}+\epsilon })$ holds for every $\epsilon >0.$

In this paper, we focus on the mean-square estimate of $\psi (x)-x.$ Our goal is to prove, by purely elementary means, that

$${\int }_2^X{\left(\psi (t)-t\right)}^{2}dt=O(X^2{log}^{2}X).$$

(1)

This estimate is unconditional and, as we shall see, already suffices to prove the convergence of ${\displaystyle {\int }_1^{\infty }|\psi (x)-x|x^{-\frac{3}{2}-\epsilon }dx}$ for every $\epsilon >0.$

The proof mainly uses elementary estimates of the Chebyshev function, and Cauchy-Schwarz inequality. In particular, we avoid any complex analysis or unproven hypotheses.

The paper is organized as follows.

Section 1 and Section 2 recall necessary notation and preliminary estimates. Section 3 proves an upper bound for $U(N)={\sum }_{k=1}^{N}kA{(N/k)}^2,$ where $A(x)=\psi (x)-\lfloor x\rfloor .$ Section 4 expands $U(N)/N.$ Section 5 establishes a pointwise lower bound for the weights in $U(N)/N.$ Section 6 using the lower bound for the weights and concluding the core estimate ${\displaystyle {\displaystyle \sum _{m=1}^N}\frac{A{(m)}^2}{m}=O(N{log}^{2}N).}$ Section 7 converts this discrete estimate into the continuous mean-square bound (1). Section 8 gives the proof of the convergence of ${\displaystyle {\int }_1^{\infty }|\psi (x)-x|x^{-\frac{3}{2}-\epsilon }dx}$ for every $\epsilon >0.$ Section 9 concludes with a summary, where we obtain ${\int }_1^{\infty }{\displaystyle \frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}}dx<\infty$ for every $\epsilon >0,$ thus the integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ converges absolutely for every $\epsilon >0,$ so that the integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ converges conditionally for every $\epsilon >0,$ whereas the integral converges conditionally $\iff \mathrm{RH},$ so then the Riemann hypothesis is true. In particular, an appendix provides a theoretical complement linking integral convergence, pointwise bounds and analyticity, where the absolute convergence of the integral is equivalent to the conditional convergence of the integral, either of which is equivalent to the o-bound: $|\psi (x)-x|=o(x^{\frac{1}{2}+\epsilon }),$ and all of them imply the O-bound: $\psi (x)-x=O(x^{\frac{1}{2}+\epsilon })$ is also valid for every $\epsilon >0,$ thus reconfirming the validity of the Riemann hypothesis.

2. Preliminary

2.1. Notation Conventions and Logical Symbols

We use the following notation throughout the paper:

$\mathbb{N}=\{1,2,3,\dots \}.$

p always denotes a prime.

$logx$ is the natural logarithm.

For a real number $x,$$\lfloor x\rfloor$ is the greatest integer $\le x,$ the fractional part is $\left\{x\right\}=x-\lfloor x\rfloor ,$ which satisfies $0\le \left\{x\right\}<1,$ and $\left\{x\right\}=0$ if and only if x is an integer.

$f(x)=O\left(g(x)\right)$ or $f(x)\ll g(x)$ means there exists a positive constant $C>0$ such that $|f(x)|\le Cg(x)$ for all sufficiently large $x.$ More importantly, if $g(x)$ is positive for all x in the domain, then there exists an absolute constant $C_0>0$ such that the asymptotic estimate holds for all sufficiently large x and all finite $x.$

$f(x)=o\left(g(x)\right)$ means ${\displaystyle \underset{x\to \infty }{lim}\left|f(x)/g(x)\right|=0.}$ It is to be observed that the asymptotic relation $f(x)=o\left(g(x)\right)$ implies, and is stronger than, the asymptotic relation $f(x)=O\left(g(x)\right).$

$f(x)\sim g(x)$ means ${\displaystyle \underset{x\to \infty }{lim}f(x)/g(x)=1.}$

Summations ${\sum }_{n\le x}$ are over integers n with $1\le n\le x.$

$A{(N/n)}^2$ means ${\left(A(N/n)\right)}^2.$

For the logical relations in theorems and lemmas we employ the standard symbols:

⇒: implies, if … then

⇐: is implied by

⇔: if and only if (iff)

The following terminology is used:

Sufficient condition: $a\Rightarrow b$ means that if a is true, then b is true; a guarantees $b.$

Necessary condition: $a\Leftarrow b$ means that if b is true, then a must be true; without a we cannot have $b,$ i.e., if a is not true, then b must be not true.

Necessary and sufficient condition (equivalence): $a\iff b$ means $a\Rightarrow b$ and $b\Rightarrow a$; a and b are either both true or both false.

2.2. Basic Definitions

The Chebyshev $\psi$-function is defined by

$$\psi (x)=\sum _{p^k\le x}logp,$$

where the sum runs over all prime powers $p^k$ (p prime, and integer $k\ge 1$).

The Chebyshev $\vartheta$-function is defined by ${\displaystyle \vartheta (x)=\sum _{p\le x}logp,}$ where the sum runs over primes $p\le x.$

The von Mangoldt function $\Lambda (n$) is defined by

$$\Lambda (n)=\left\{\begin{array}{cc}logp\hfill & \mathrm{if}n=p^k\mathrm{for}\mathrm{some}\mathrm{prime}p\mathrm{and}\mathrm{integer}k\ge 1,\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

A well-known identity connects it with the Chebyshev function ${\displaystyle \psi (x)=\sum _{n\le x}\Lambda (n).}$

Define ${\displaystyle a_n=\Lambda (n)-1\mathrm{and}A(x)={\sum }_{n\le x}{a}_n.}$ Then

$$A(x)=\sum _{n\le x}{a}_n=\sum _{n\le x}\Lambda (n)-\sum _{n\le x}1=\psi (x)-\lfloor x\rfloor =\psi (x)-x+\left\{x\right\}$$

with fractional part $0\le \left\{x\right\}<1.$ Thus, $\psi (x)=A(x)+\lfloor x\rfloor =A(x)+x-\left\{x\right\},$ and $\psi (x)-x=A(x)-\left\{x\right\}.$ For integer $n,$$\lfloor n\rfloor =n,$ so $A(n)=\psi (n)-n.$

2.3. The Prime Number Theorem

Let $\pi (x)$ denote the number of primes not exceeding $x(>0),$ which is the familiar prime-counting function.

The classical prime number theorem asserts that $\psi (x)\sim x$ as $x\to \infty ,$ or equivalently ${\displaystyle \pi (x)\sim \frac{x}{logx}.}$ While this result is not needed in our derivations (all estimates we employ are elementary or established results analogous to the PNT), it provides the historical context and motivation for studying the error term $\psi (x)-x.$

2.4. Non-trivial Zeros of Riemman’s Zeta Function and the Riemann Hypothesis

The Riemman zeta function is defined by ${\displaystyle \zeta (s)={\displaystyle \sum _{n=1}^{\infty }}\frac{1}{n^s},}$ where $\zeta (s)$ extends meromorphically to the whole complex plane with a simple pole at $s=1$ with residue $1.$

Non-trivial zeros of $\zeta (s)$ are the zeros of $\zeta (s)$ in the critical strip $0<Re(s)<1.$

As is well-known, the Riemann hypothesis (RH) states that all non-trivial zeros lie on the line ${\displaystyle Re(s)=\frac{1}{2}.}$

In addition, we have the following facts.

Theorem 1.

(cf.[2], p.17.) If $\Re e(s)>1,$ then

$$log\zeta (s)=-\sum _{p}log\left(1-p^{-s}\right)=\sum _{p,m}{\displaystyle \frac{p^{-ms}}{m}},$$

(2)

where p runs through all primes and m through all positive integers.

Theorem 2.

(cf.[2], pp.17-18.) For $\Re e(s)>1,$ then

$$-{\displaystyle \frac{{\zeta }^{\prime }(s)}{\zeta (s)}}=\sum _p{\displaystyle \frac{logp}{p^s-1}}=\sum _{p,m}(logp)p^{-ms}={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\Lambda (n)}{n^s}}=s{\int }_1^{\infty }{\displaystyle \frac{\psi (x)}{x^{s+1}}}dx$$

(3)

where p runs through all primes and m through all positive integers.

We see that ${\displaystyle -\frac{{\zeta }^{\prime }(s)}{\zeta (s)}-\frac{1}{s-1}=1+s{\int }_1^{\infty }\frac{\psi (x)-x}{x^{s+1}}dx}$ holds for $\Re e(s)>1,$ which is a well-known formula and extends meromorphically to the half-plane $\Re e(s)>0$ with some poles at non-trivial zeros of $\zeta (s),$ but no other poles on the region $\Re e(s)>0.$

2.5. Well-known Equivalent Forms of RH in Terms of the Chebyshev Function

There are well-known equivalent forms of the Riemann hypothesis in terms of the Chebyshev function $\psi (x)$:

(1). Standard arithmetical form: ${\displaystyle \psi (x)-x=O\left(x^{\frac{1}{2}}{log}^{2}x\right)\iff \mathrm{RH}.}$

(2). Another arithmetical form: For every $\epsilon >0,$

$${\displaystyle \psi (x)-x=O\left(x^{\frac{1}{2}+\epsilon }\right)\iff \mathrm{RH}.}$$

(3). Integral form (analyticity):

$${\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{s+1}}}dx\mathrm{is}\mathrm{analytic}\mathrm{for}{\displaystyle Re(s)>\frac{1}{2}}\iff \mathrm{RH}.$$

(4). Integral form (conditional convergence): For every $\epsilon >0,$

$${\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}}dx\mathrm{converges}\mathrm{conditionally}\iff \mathrm{RH},$$

since $f(s):={\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{s+1}}}dx$ is analytic for $Re(s)>\frac{1}{2}$ ⇔${\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}}dx$ converges conditionally.

Remark 1.

For instance, these celebrated equivalent formulations of the Riemann hypothesis can be found in the paper by von Koch [1] and on pages $83-85$ of the monograph by Ingham [2], which offer rich details and comprehensive perspectives.

2.6. Basic Analytical Tools

We shall employ the following standard results.

(i) Cauchy-Schwarz inequality.

For any real numbers $u_1,\dots ,u_N$ and $v_1,\dots ,v_N,$

$${\left({\displaystyle \sum _{n=1}^N}{u}_n{v}_n\right)}^2\le \left({\displaystyle \sum _{n=1}^N}{u}_n^2\right)\left({\displaystyle \sum _{n=1}^N}{v}_n^2\right).$$

We also use its integral form: for functions $f,g$ on an interval $I,$

$${\left({\int }_{I}f(x)g(x)dx\right)}^2\le \left({\int }_{I}f{(x)}^{2}dx\right)\left({\int }_{I}g{(x)}^{2}dx\right).$$

For a reference, please see, G. H. Hardy, J. E. Littlewood & G. Pólya [5], Theorems 7 and $181.$

(ii) A lemma on the O-notation.

Lemma 1.

(Bidirectional transfer). Let $f(N),g(N),h(N)$ be functions defined on positive integers and suppose

$$f(N)=g(N)+O\left(h(N)\right).$$

(4)

Then

$$g(N)=O\left(h(N)\right)\mathit{if}\mathit{and}\mathit{only}\mathit{if}f(N)=O\left(h(N)\right).$$

Proof. 

By (4), there exist constants $C_0>0$ and $N_0$ such that

$$|f(N)-g(N)|\le C_0|h(N)|\mathrm{for}\mathrm{all}N\ge N_0.$$

(5)

Sufficiency: Assume g(N) = O(h(N)). Then there exist constants $C_1>0$ and $N_1$ with $|g(N)|\le C_1|h(N)|$ for $N\ge N_1.$ For $N\ge max(N_0,N_1),$ we have

$$|f(N)|\le |f(N)-g(N)|+|g(N)|\le (C_0+C_1)|h(N)|,$$

hence $f(N)=O(h(N)).$

Necessity: Assume $f(N)=O(h(N)).$ Then there exist constants $C_2>0$ and $N_2$ with $|f(N)|\le C_2|h(N)|$ for $N\ge N_2.$ For $N\ge max(N_0,N_2),$ we have

$$|g(N)|\le |g(N)-f(N)|+|f(N)|\le (C_0+C_2)|h(N)|,$$

hence $g(N)=O(h(N)).$ □

Corollary 1.

(Absorption). If $h(N)=O(k(N)),$ then

$$f(N)=g(N)+O\left(h(N)\right)⟹f(N)=g(N)+O\left(k(N)\right).$$

Proof. 

By $h(N)=O(k(N)),$ there exist constants $C>0$ and $N_0$ such that $|h(N)|\le C|k(N)|$ for all $N\ge N_0.$ Hence any function that is $O(h(N))$ is also $O(k(N)).$ Therefore the error term $O(h(N))$ can be replaced by $O(k(N)).$ □

(iii) Two lemmas on the partition by floor values.

Lemma 2.

(Partition by Floor Values). For a fixed integer $N\ge 1,$ define for each integer $m\ge 1$ the set

$$J_m=\left\{k\in \{1,\dots ,N\}:⌊{\displaystyle \frac{N}{k}}⌋=m\right\}.$$

Then the collection $\{J_m:m=1,2,\dots ,N\}$ forms a partition of $\{1,2,\dots ,N\}.$ Furthermore, an equivalent description is

$$J_m=\left\{k\in \mathbb{N}:{\displaystyle \frac{N}{m+1}}<k\le {\displaystyle \frac{N}{m}}\right\}.$$

Proof. 

The proof of the lemma consists of several steps.

Step 1. The values of m range from 1 to $N.$ Since $1\le k\le N,$ we have $\lfloor N/k\rfloor$ is an integer between $\lfloor N/N\rfloor =1$ and $\lfloor N/1\rfloor =N.$ Hence $m=\lfloor N/k\rfloor$ takes values in $\{1,2,\dots ,N\}.$

Step 2. Disjointness. If $k\in J_m$ and also $k\in J_{m^{\prime }}$ with $m\ne m^{\prime },$ then $\lfloor N/k\rfloor$ would equal two different integers, which is impossible. So the sets $J_m$ are pairwise disjoint.

Step 3. Covering. For any $k\in \{1,\dots ,N\},$ let $m=\lfloor N/k\rfloor .$ Then $1\le m\le N$ and by definition $k\in J_m.$ Hence every element of $\{1,\dots ,N\}$ belongs to some $J_m.$ Thus ${\displaystyle {\left\{J_m\right\}}_{m=1}^N}$ is a partition of $\{1,\dots ,N\}.$

Step 4. Equivalent description. We show ${\displaystyle J_m=\left\{k\in \mathbb{N}:\frac{N}{m+1}<k\le \frac{N}{m}\right\}.}$ From $\lfloor N/k\rfloor =m$ we have ${\displaystyle m\le \frac{N}{k}<m+1.}$ Inverting (all terms positive) gives ${\displaystyle \frac{N}{m+1}<k\le \frac{N}{m}.}$ Conversely, if k satisfies the inequality ${\displaystyle \frac{N}{m+1}<k\le \frac{N}{m},}$ then ${\displaystyle m\le \frac{N}{k}<m+1,}$ so $\lfloor N/k\rfloor =m.$ The condition $k\in \mathbb{N}$ together with $k\le N/m\le N$ automatically implies $1\le k\le N.$ Hence the two descriptions coincide. Therefore, the lemma is proved. □

Lemma 3.

For $J_m=\left\{k\in \mathbb{N}:{\displaystyle \frac{N}{m+1}}<k\le {\displaystyle \frac{N}{m}}\right\},$ we have ${\displaystyle \sum _{k\in J_m}k=O\left(\frac{N^2}{m^2}\right).}$

Proof. 

The proof of the lemma consists of several steps.

Step 1. Every $k\in J_m$ satisfies $k\le N/m.$ Hence ${\sum }_{k\in J_m}k\le {\displaystyle \frac{N}{m}}·|J_m|.$

Step 2. The length of the interval is ${\displaystyle \frac{N}{m}}-{\displaystyle \frac{N}{m+1}}={\displaystyle \frac{N}{m(m+1)}}.$ Therefore the number of integers in it is at most $|J_m|\le {\displaystyle \frac{N}{m(m+1)}}+1.$

Step 3. Substituting, ${\sum }_{k\in J_m}k\le {\displaystyle \frac{N}{m}}\left({\displaystyle \frac{N}{m(m+1)}}+1\right)={\displaystyle \frac{N^2}{m^2(m+1)}}+{\displaystyle \frac{N}{m}}.$

Step 4. Bound each term: ${\displaystyle \frac{N^2}{m^2(m+1)}}\le {\displaystyle \frac{N^2}{m^3}},$ and ${\displaystyle \frac{N}{m}}={\displaystyle \frac{N^2}{m^2}}·{\displaystyle \frac{m}{N}}\le {\displaystyle \frac{N^2}{m^2}},$ where $m\le N.$

Step 5. Thus ${\sum }_{k\in J_m}k\le {\displaystyle \frac{N^2}{m^3}}+{\displaystyle \frac{N^2}{m^2}}\le {\displaystyle \frac{2N^2}{m^2}},$ which proves ${\sum }_{k\in J_m}k=O(N^2/m^2).$ Therefore, the lemma is proved. □

(iv) Dyadic decomposition. For sums over integers $n\ge 2,$ we can decompose the range using powers of two.

Lemma 4.

Let $N\ge 2$ be an integer and set $L=\lfloor {log}_{2}N\rfloor ,$ so that

$$2^L\le N<2^{L+1}.$$

Then the following identity holds:

$${\displaystyle \sum _{n=2}^N}f(n)={\displaystyle \sum _{k=1}^{\lfloor {log}_{2}N\rfloor }}\sum _{2^{k-1}<n\le min(N,2^k)}f(n).$$

(6)

Proof. 

For $k=1,2,\dots ,L-1,$ we have $2^k\le 2^{L-1}·2=2^L\le N,$ hence $min(N,2^k)=2^k.$ The inner sum becomes

$$\sum _{2^{k-1}<n\le 2^k}f(n),$$

which covers the integers $2^{k-1}+1,\dots ,2^k.$ For $k=L,$ since $N<2^{L+1},$ we have $min(N,2^L)=N,$ and the inner sum becomes ${\sum }_{2^{L-1}<n\le N}f(n),$ covering the remaining integers $2^{L-1}+1,\dots ,N.$ The intervals for $k=1,\dots ,L-1$ together with the last one partition the set $\{2,3,\dots ,N\}$ without overlap. Thus the equality (6) holds. □

Remark 2.

Note that when $N\to \infty ,$ we have $\lfloor {log}_{2}N\rfloor \to \infty ,$ so the number of dyadic intervals grows without bound, this property is essential for the convergence arguments in Section 8. If the sum includes $n=1,$ we treat it separately:

$${\displaystyle \sum _{n=1}^N}f(n)=f(1)+{\displaystyle \sum _{n=2}^N}f(n).$$

This decomposition is used in Section 8 to handle integrals via dyadic intervals.

(v) Abel summation (summation by parts), which will be applied in the appendix.

Lemma 5.

Let ${(a_n)}_{n\ge 1}$ and ${(b_n)}_{n\ge 1}$ be sequences of real numbers. Define $A_0=0$ and ${\displaystyle A_n={\displaystyle \sum _{k=1}^n}{a}_k}$ for $n\ge 1.$ Then for any $N\ge 1,$ we have

$${\displaystyle \sum _{n=1}^N}{a}_n{b}_n=A_N{b}_N+{\displaystyle \sum _{n=1}^{N-1}}{A}_n(b_n-b_{n+1})=A_N{b}_N-{\displaystyle \sum _{n=1}^{N-1}}{A}_n(b_{n+1}-b_n).$$

(7)

Proof. 

Write $a_n=A_n-A_{n-1}$ (with $A_0=0$). Then

$${\displaystyle \sum _{n=1}^N}{a}_n{b}_n={\displaystyle \sum _{n=1}^N}(A_n-A_{n-1})b_n={\displaystyle \sum _{n=1}^N}{A}_n{b}_n-{\displaystyle \sum _{n=1}^N}{A}_{n-1}{b}_n.$$

Shift the index in the second sum:

$${\displaystyle \sum _{n=1}^N}{A}_{n-1}{b}_n={\displaystyle \sum _{k=0}^{N-1}}{A}_k{b}_{k+1}={\displaystyle \sum _{k=1}^{N-1}}{A}_k{b}_{k+1}$$

(since $A_0=0$). Thus

$${\displaystyle \sum _{n=1}^N}{a}_n{b}_n={\displaystyle \sum _{n=1}^N}{A}_n{b}_n-{\displaystyle \sum _{n=1}^{N-1}}{A}_n{b}_{n+1}=A_N{b}_N+{\displaystyle \sum _{n=1}^{N-1}}{A}_n(b_n-b_{n+1}),$$

then we obtain (7). □

2.7. Elementary Estimates

We shall apply the following well-known elementary facts.

Lemma 6.

(Asymptotic expansion of harmonic numbers). For $n\to \infty ,$

$${\displaystyle \sum _{k=1}^n}{\displaystyle \frac{1}{k}}=logn+\gamma +O\left({\displaystyle \frac{1}{n}}\right),$$

where γ is the Euler-Mascheroni constant.

Proof. 

See Apostol [7], Chapter 3, Theorems 3.1 and 3.2. □

Lemma 7.

We have the elementary estimates for the bounds:

$$\vartheta (x)=O(x),\psi (x)=O(x),A(x)=O(x).$$

(8)

The two Chebyshev functions $\psi (x)$ and $\vartheta (x)$ satisfy $\psi (x)=\vartheta (x)+O(\sqrt{x}).$

The details of the proof are presented below, consisting of three steps.

Proof. 

Step 1. Proof of $\vartheta (x)=O(x)$ (Chebyshev upper bound)

Consider the binomial coefficient $\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right).$

Upper bound: $\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)\le {(1+1)}^{2n}=4^n.$

Lower bound via primes: every prime p with $n<p\le 2n$ divides $\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right).$ Hence

$$\prod _{n<p\le 2n}p\le \left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)\le 4^n.$$

Taking logarithms: $\vartheta (2n)-\vartheta (n)\le nlog4.$ Apply this repeatedly for $n=2^k$ and sum: $\vartheta (2^m)\le 2^{m}log4.$ For general $x,$ choose m such that $2^m\le x<2^{m+1}.$ Then

$$\vartheta (x)\le \vartheta (2^{m+1})\le 2^{m+1}log4\le 4xlog2=O(x).$$

Thus there exists $C>0$ such that $\vartheta (x)\le Cx$ for all $x\ge 1.$

Step 2. Proof of $\psi (x)=\vartheta (x)+O(\sqrt{x})$

Decompose $\psi (x)$ by prime power exponents:

$$\psi (x)=\sum _{k\ge 1}\sum _{p^k\le x}logp=\vartheta (x)+\vartheta (\sqrt{x})+\vartheta (\sqrt[3]{x})+\dots ,$$

where only terms with $x^{1/k}\ge 2$ (i.e., $k\le {log}_{2}x$) are non-zero.

Using $\vartheta (y)\le Cy$ from step 1,

$$\psi (x)-\vartheta (x)={\displaystyle \sum _{k=2}^{\lfloor {log}_{2}x\rfloor }}\vartheta (x^{1/k})\le C{\displaystyle \sum _{k=2}^{\lfloor {log}_{2}x\rfloor }}{x}^{1/k}.$$

Split the sum:

For $k=2$: term $C\sqrt{x}.$

For $k\ge 3:x^{1/k}\le x^{1/3},$ and there are at most ${log}_{2}x$ such terms. Hence

$${\displaystyle \sum _{k=3}^{\lfloor {log}_{2}x\rfloor }}{x}^{1/k}\le x^{1/3}·{log}_{2}x=o(\sqrt{x})(x\to \infty ).$$

This part is $O(x^{1/3}logx),$ which is certainly $O(\sqrt{x}).$ Therefore

$$\psi (x)-\vartheta (x)=O(\sqrt{x}).$$

(One often writes $O(\sqrt{x}logx),$ but the above shows $O(\sqrt{x})$ suffices because $x^{1/3}logx=o(\sqrt{x}).$)

Step 3. From $\vartheta (x)=O(x)$ and $\psi (x)=\vartheta (x)+O(\sqrt{x})$ we immediately obtain $\psi (x)=O(x).$ Hence $|A(x)|=|\psi (x)-\lfloor x\rfloor |\le \psi (x)+\lfloor x\rfloor =O(x),$ whereas $\lfloor x\rfloor \le x.$ Therefore, the lemma is proved. □

3. Estimating U(N)

Lemma 8.

Let ${\displaystyle U(N)={\displaystyle \sum _{k=1}^N}kA{(N/k)}^2.}$ Then

$$U(N)=O(N^{2}logN).$$

(9)

Proof. 

Chebyshev’s bound gives $|A(x)|=|\psi (x)-\lfloor x\rfloor |\le \psi (x)+\lfloor x\rfloor =O(x)$ for some constant $C>0$ and all $x\ge 1,$ where $\psi (x)=O(x)$ and $\lfloor x\rfloor \le x.$ Then

$$|U(N)|=\left|{\displaystyle \sum _{k=1}^N}kA{(N/k)}^2\right|\le {\displaystyle \sum _{k=1}^N}k·C^2{\displaystyle \frac{N^2}{k^2}}=C^2{N}^2{\displaystyle \sum _{k=1}^N}{\displaystyle \frac{1}{k}}=O(N^{2}logN).$$

Therefore, the lemma is proved. □

Corollary 2.

$${\displaystyle \frac{U(N)}{N}}=O(NlogN).$$

(10)

Proof. 

The proof follows immediately from Lemma 8. □

4. Expanding U(N)/N

For each $k=1,\dots ,N$ define $m=\lfloor N/k\rfloor .$ According to Lemma 2 (Partition by Floor Values), we know that the sets

$$J_m=\left\{k\in \{1,\dots ,N\}:⌊{\displaystyle \frac{N}{k}}⌋=m\right\}=\left\{k\in \mathbb{N}:{\displaystyle \frac{N}{m+1}}<k\le {\displaystyle \frac{N}{m}}\right\}$$

form a partition of $\{1,\dots ,N\}.$ Hence

$$U(N)={\displaystyle \sum _{k=1}^N}kA{(N/k)}^2={\displaystyle \sum _{m=1}^N}\sum _{k\in J_m}kA{(N/k)}^2.$$

For $k\in J_m$ we write $A(N/k)=A(m)+{\delta }_{m,k}.$ To bound ${\delta }_{m,k},$ note that

$${\delta }_{m,k}=A(N/k)-A(m)=\left(\psi (N/k)-\lfloor N/k\rfloor \right)-\left(\psi (m)-m\right)=\psi (N/k)-\psi (m),$$

because $\lfloor N/k\rfloor =m$ and $\lfloor m\rfloor =m.$ Since $m\le N/k<m+1,$ the interval $(m,N/k]$ has length ${\displaystyle \frac{N}{k}}-m<(m+1)-m=1,$ hence it contains at most one integer. Consequently, ${\displaystyle \psi (N/k)-\psi (m)=\sum _{m<n\le N/k}\Lambda (n)}$ is either 0 or $\Lambda (n_0)$ for a single integer $n_0$ (if it exists). In either case,

$$|\psi (N/k)-\psi (m)|\le logN,$$

(11)

because $\Lambda (n_0)\le log{n}_0\le logN$ when $n_0$ exists. Therefore $|{\delta }_{m,k}|\le logN.$

Expanding the square, we have

$$\begin{array}{cc}\hfill U(N)& ={\displaystyle \sum _{m=1}^N}\sum _{k\in J_m}kA{(N/k)}^2\hfill \\ & ={\displaystyle \sum _{m=1}^N}A{(m)}^2\sum _{k\in J_m}k+2{\displaystyle \sum _{m=1}^N}A(m)\sum _{k\in J_m}k{\delta }_{m,k}+{\displaystyle \sum _{m=1}^N}\sum _{k\in J_m}k{\delta }_{m,k}^2.\hfill \end{array}$$

For this $U(N),$ we can estimate the error terms in the cross-term and the quadratic term contribute at most $O(N^2{log}^{2}N).$

(1). Cross-term:

$$\left|{\displaystyle \sum _{m=1}^N}A(m)\sum _{k\in J_m}k{\delta }_{m,k}\right|\le logN{\displaystyle \sum _{m=1}^N}|A(m)|\sum _{k\in J_m}k.$$

Since ${\displaystyle \sum _{k\in J_m}k=O(N^2/m^2)}$ and $|A(m)|=O(m),$ we have

$${\displaystyle \sum _{m=1}^N}|A(m)|\sum _{k\in J_m}k\ll N^2{\displaystyle \sum _{m=1}^N}{\displaystyle \frac{1}{m}}=O(N^{2}logN).$$

Hence the cross-term is $O(N^2{log}^{2}N).$

(2). Quadratic term:

$$\left|{\displaystyle \sum _{m=1}^N}\sum _{k\in J_m}k{\delta }_{m,k}^2\right|\le {log}^{2}N{\displaystyle \sum _{m=1}^N}\sum _{k\in J_m}k={log}^{2}N{\displaystyle \sum _{k=1}^N}k=O(N^2{log}^{2}N).$$

Dividing by $N,$ we obtain

$${\displaystyle \frac{U(N)}{N}}={\displaystyle \sum _{m=1}^N}A{(m)}^2\sum _{k\in J_m}{\displaystyle \frac{k}{N}}+O(N{log}^{2}N).$$

(12)

5. A Pointwise Lower Bound for the Weights

Lemma 9.

For all $m\ge 1$ and $N\ge 2,$ we have ${\displaystyle \sum _{k\in J_m}\frac{k}{N}\ge \frac{1}{2m}.}$

Proof. 

From the explicit description $J_m=\{k:N/(m+1)<k\le N/m\},$ let $k_{min}$ be the smallest element of $J_m.$ Then $k_{min}>N/(m+1).$ Hence

$$\sum _{k\in J_m}{\displaystyle \frac{k}{N}}\ge {\displaystyle \frac{k_{min}}{N}}>{\displaystyle \frac{1}{m+1}}\ge {\displaystyle \frac{1}{2m}},$$

because $2m\ge m+1$ for all $m\ge 1.$ □

6. The Core Estimate

Since $A{(m)}^2\ge 0,$ multiplying the inequality ${\displaystyle \sum _{k\in J_m}\frac{k}{N}\ge \frac{1}{2m}}$ of Lemma 7.1 by $A{(m)}^2$ and summing over m yields ${\displaystyle {\displaystyle \sum _{m=1}^N}A{(m)}^2\sum _{k\in J_m}\frac{k}{N}\ge \frac{1}{2}{\displaystyle \sum _{m=1}^N}\frac{A{(m)}^2}{m}.}$ Insert this into (12):

$${\displaystyle \frac{U(N)}{N}}\ge \frac{1}{2}{\displaystyle \sum _{m=1}^N}{\displaystyle \frac{A{(m)}^2}{m}}+O(N{log}^{2}N).$$

(13)

By (10), we have ${\displaystyle \frac{U(N)}{N}=O(NlogN),}$ and therefore ${\displaystyle \frac{U(N)}{N}=O(N{log}^{2}N).}$ Hence

$$\frac{1}{2}{\displaystyle \sum _{m=1}^N}{\displaystyle \frac{A{(m)}^2}{m}}\le {\displaystyle \frac{U(N)}{N}}+O(N{log}^{2}N)=O(N{log}^{2}N).$$

Thus

$${\displaystyle \sum _{m=1}^N}{\displaystyle \frac{A{(m)}^2}{m}}=O(N{log}^{2}N).$$

(14)

This is the central estimate of the paper.

7. From the Core Estimate to an Upper Bound for the Mean-Square Integral

For $t\in [n,n+1),$ we have $\psi (t)=\psi (n),$ hence

$$\psi (t)-t=\psi (n)-t=\left(\psi (n)-n\right)+(n-t)=A(n)+(n-t).$$

Let $u=t-n\in [0,1).$ Then

$${\int }_n^{n+1}{\left(\psi (t)-t\right)}^{2}dt={\int }_0^1{\left(A(n)-u\right)}^{2}du=A{(n)}^2-A(n)+{\displaystyle \frac{1}{3}}.$$

Summing from $n=2$ to $\lfloor X\rfloor$ yields

$${\int }_2^X{\left(\psi (t)-t\right)}^{2}dt=\sum _{n\le X}A{(n)}^2-\sum _{n\le X}A(n)+O(X).$$

(15)

By (14), we have ${\displaystyle {\displaystyle \sum _{n=1}^N}\frac{A{(n)}^2}{n}=O(N{log}^{2}N),}$ and using the obvious inequality $1\le n\le N$, we can bound ${\sum }_{n=1}^{N}A{(n)}^2$:

$${\displaystyle \sum _{n=1}^N}A{(n)}^2={\displaystyle \sum _{n=1}^N}n·{\displaystyle \frac{A{(n)}^2}{n}}\le N{\displaystyle \sum _{n=1}^N}{\displaystyle \frac{A{(n)}^2}{n}}=O(N^2{log}^{2}N).$$

(16)

So, we have ${\displaystyle \sum _{n\le X}A{(n)}^2=O(X^2{log}^{2}X).}$ By Cauchy-Schwarz, we get

$$\left|\sum _{n\le X}A(n)\right|\le {\left(\sum _{n\le X}1\right)}^{1/2}{\left(\sum _{n\le X}A{(n)}^2\right)}^{1/2}=O(X^{1/2}·XlogX)=O(X^{3/2}logX),$$

which is negligible compared with $O(X^2{log}^{2}X).$ Therefore

$${\int }_2^X{\left(\psi (t)-t\right)}^{2}dt=O(X^2{log}^{2}X).$$

(17)

8. Convergence of the Integral

Theorem 3.

For every $\epsilon >0,$ the integral ${\displaystyle {\int }_1^{\infty }\frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}dx<\infty ,}$ i.e., the integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ converges absolutely.

Proof. 

For any $\epsilon >0,$ we have

$${\int }_1^{\infty }{\displaystyle \frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}}dx\le {\int }_1^2{\displaystyle \frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}}dx+{\displaystyle \sum _{n=1}^{\infty }}{\int }_{2^n}^{2^{n+1}}{\displaystyle \frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}}dx.$$

Let us examine the integral with weight $x^{-{\displaystyle \frac{3}{2}}-\epsilon }$ on dyadic interval $[2^n,2^{n+1}],$ using (17) we have

$${\int }_{2^n}^{2^{n+1}}{\left(\psi (x)-x\right)}^{2}dx=O\left(2^{2(n+1)}{log}^2(2^{n+1})\right)=O\left(2^{2n}{log}^2(2^{n+1})\right),$$

and by Cauchy-Schwarz, we get

$$\begin{array}{cc}\hfill {\int }_{2^n}^{2^{n+1}}{\displaystyle \frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}}dx& \le {(2^n)}^{-{\displaystyle \frac{3}{2}}-\epsilon }{\left({\int }_{2^n}^{2^{n+1}}{1}^{2}dx\right)}^{1/2}{\left({\int }_{2^n}^{2^{n+1}}{(\psi (x)-x)}^{2}dx\right)}^{1/2}\hfill \\ & \ll {(2^n)}^{-{\displaystyle \frac{3}{2}}-\epsilon }·{(2^n)}^{1/2}·\left(2^{n}log(2^{n+1})\right)=(n+1)2^{-n\epsilon }log2,\hfill \end{array}$$

the right-hand side is $\ll (n+1)2^{-n\epsilon }.$ Hence ${\displaystyle {\int }_{2^n}^{2^{n+1}}\frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}dx\ll (n+1)2^{-n\epsilon },}$ and the series ${\displaystyle {\displaystyle \sum _{n=1}^{\infty }}(n+1)2^{-n\epsilon }}$ converges. Adding the finite contribution from $[1,2],$ we obtain ${\int }_1^{\infty }{\displaystyle \frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}}dx<\infty ,$ this completes the proof. □

9. Conclusions

We have shown: Elementary estimates of the Chebyshev function lead to $U(N)=O(N^{2}logN),$ where $U(N)={\sum }_{k=1}^{N}kA{(N/k)}^2$ and $A(x)=\psi (x)-\lfloor x\rfloor .$ From this we derive the core estimate ${\displaystyle {\displaystyle \sum _{m=1}^N}\frac{A{(m)}^2}{m}=O(N{log}^{2}N).}$ Using only this core estimate and the Cauchy-Schwarz inequality, we obtain

$${\int }_2^X{\left(\psi (t)-t\right)}^{2}dt=O(X^2{log}^{2}X).$$

Consequently, ${\displaystyle {\int }_1^{\infty }\frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}dx<\infty }$ holds, thus the integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ converges absolutely for every $\epsilon >0.$ All arguments are purely elementary, avoiding complex analysis and unproven hypotheses. The estimate $O(X^2{log}^{2}X)$ is unconditional and best possible with elementary methods. Since we have proven the integral ${\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}}dx$ converges absolutely for every $\epsilon >0,$ which implies that ${\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}}dx$ converges conditionally for every $\epsilon >0,$ whereas the integral ${\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}}dx$ converges conditionally for every $\epsilon >0$$\iff \mathrm{RH},$ so then the Riemann hypothesis is true.

Appendix A.1. Local Bounded Variation Estimate

Lemma A1.

There exists an absolute constant $C>0$ such that for all $x\ge 2$ and all h with $1\le h\le x,$

$$|\psi (x+h)-\psi (x)|\le Chlogx.$$

(A1)

Proof. 

The interval $(x,x+h]$ contains at most $\lfloor h\rfloor +1\le h+1$ integers. For any integer n in this interval, if $n=p^k$ is a prime power, then

$$\Lambda (n)=logp\le logn\le log(x+h);$$

otherwise $\Lambda (n)=0,$ where we have

$$\psi (x+h)-\psi (x)=\sum _{x<n\le x+h}\Lambda (n).$$

Hence

$$|\psi (x+h)-\psi (x)|\le (h+1)log(x+h).$$

Because $h\le x,$ we have $x+h\le 2x$ and $log(x+h)\le log(2x)=log2+logx.$ For $h\ge 1$ we also have $h+1\le 2h.$ Thus ${\displaystyle |\psi (x+h)-\psi (x)|\le 2h(log2+logx).}$ Choosing $C=2(log2+1)$ gives the desired inequality.

Appendix B.1. Remarks

For the Chebyshev function $\psi (x),$ the following are equivalent: Absolutely convergent integral ${\int }_1^{\infty }|\psi (x)-x|x^{-{\displaystyle \frac{3}{2}}-\epsilon }dx$ for all $\epsilon >0;$ Conditionally convergent integral ${\int }_1^{\infty }\left(\psi (x)-x\right)x^{-{\displaystyle \frac{3}{2}}-\epsilon }dx$ for all $\epsilon >0;$ The pointwise estimate $|\psi (x)-x|=o(x^{\frac{1}{2}+\epsilon })$ for all $\epsilon >0;$ The analyticity of the Dirichlet-type integrals $f(s)$ and $g(s)$ in $Re(s)>1/2.$ These equivalences rely on the local bounded variation of $\psi$ and standard Dirichlet series theory. They are included as a theoretical complement to the main paper, where we have directly proved the absolute convergence of the integral ${\int }_1^{\infty }\left(\psi (x)-x\right)x^{-{\displaystyle \frac{3}{2}}-\epsilon }dx$ via the mean-square estimate ${\int }_2^X{\left(\psi (t)-t\right)}^{2}dt=O(X^2{log}^{2}X)$ and a dyadic decomposition. In a word, the main proof does not rely on this appendix; the appendix is only a theoretical complement that shows the equivalence of the integral convergence, the pointwise o-bound, and the analyticity of $f(s),$ as well as the consequence of the analyticity of $g(s).$ It provides additional insight into the relationships between these properties for the Chebyshev function.