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

1. Introduction

The Chebyshev function ${\displaystyle \psi (x)=\sum _{n\le x}\Lambda (n)}$ 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 Erdős [4] in 1949, avoiding complex analysis, but the error term remained rather weak. In this paper, we will deduce the stronger error term that $\psi (x)-x=o(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ 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 the integral ${\displaystyle {\int }_1^{\infty }|\psi (x)-x|x^{-\frac{3}{2}-\epsilon }dx}$ for every $\epsilon >0.$

The proof mainly uses elementary estimates of the Chebyshev function, and Cauchy-Schwarz inequality, a dyadic decomposition, and Abel summation.

The paper is organized as follows.

Section 1 and Section 2 recall necessary notation and preliminary estimates. Section 3 proves an upper bound for ${\displaystyle U(N)={\displaystyle \sum _{k=1}^N}kA{(N/k)}^2,}$ 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 ${\displaystyle {\int }_1^{\infty }\frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}dx<\infty }$ 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, we provide a theoretical complement linking integral convergence, pointwise bounds and analyticity, where we conclude that the property of absolute convergence of the integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ for every $\epsilon >0$ is equivalent to the property of conditional convergence of the integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ for every $\epsilon >0,$ either of which is equivalent to the property of the o-bound: $|\psi (x)-x|=o(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ for every $\epsilon >0,$ and all of them imply the asymptotic relation of the O-bound: $\psi (x)-x=O(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ is also valid for every $\epsilon >0,$ which is the well-known equivalent form of the Riemann hypothesis, thus reconfirming the validity of the Riemann hypothesis.

To some extent, the third to ninth sections of this article are the core parts, which are the decisive and sufficient conditions for the validity of the entire proof, while the contents of other parts in the context are some necessary conditions for the core parts, which are considered as supplementary explanations or optional deductive results and do not play a decisive role in the entire proof.

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 $f(x)=O\left(g(x)\right).$ See, Hardy[8], pp.8-9, we have $logx=o(x^{\delta })$ for every positive $\delta ,$ and such that ${log}^{2}x=o(x^{2\delta })$ still holds for every positive $\delta .$

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

We write $f(x)=\Omega \left(g(x)\right)$ (as $x\to \infty$) if there exists a positive constant K independent of x such that $\left|f(x)\right|>Kg(x)$ for arbitrarily large values of $x,$ where $g(x)$ is positive for all sufficiently large x; thus ‘$\Omega$’ is the negation of the small oh ‘o’. We write $f(x)={\Omega }_{+}\left(g(x)\right)$ if $f(x)>Kg(x)$ for arbitrarily large $x,$ and $f(x)={\Omega }_{-}\left(g(x)\right)$ if $f(x)<-Kg(x)$ for arbitrarily large $x,$ where K is a positive number independent of $x,$ whereas $f(x)$ is a real function, and $g(x)$ is positive for all sufficiently large $x.$ Thus, ‘$\Omega$’ is equivalent to ‘either ${\Omega }_{+}$ or ${\Omega }_{-}$’, where we should choose one or both of them depending on positive or negative sign of the function $f(x)$ in the inequality $\left|f(x)\right|>Kg(x).$ We shall use the symbol ${\Omega }_{\pm }$ to denote ‘both ${\Omega }_{+}$ and ${\Omega }_{-}$’. It is to be observed that ${\Omega }_{\pm }$ implies, and is stronger than $\Omega .$

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

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

The empty set ∅ is a set with no elements. We say that a set I is an index set of the family of sets if for every $\alpha \in I,$ there is a set $S_{\alpha }$ in the family. Let S be a nonempty set and P be a collection of subsets ${\left\{S_{\alpha }\right\}}_{\alpha \in I}$ of S such that: (1) $S_{\alpha }\ne \varnothing$ for all $\alpha \in I;$ (2) $S={\bigcup }_{\alpha \in I}{S}_{\alpha };$ (3) if $S_{\alpha }\ne S_{\beta },$ then $S_{\alpha }\bigcap S_{\beta }=\varnothing .$ Then P is called a partition of the set $S.$

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 ${\displaystyle \psi (x)=\sum _{n\le x}\Lambda (n).}$

In fact, we observe that these two Chebyshev functions $\psi (x)$ and $\vartheta (x)$ are piecewise continuous.

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 Abscissa of Convergence and the Analyticity

Theorem 1.

Let c be any fixed positive number $>0,$ and δ is a fixed real number. Then the improper integral ${\displaystyle {\int }_c^{\infty }\frac{1}{x^{\delta }}dx}$ converges to a fixed positive number when $\delta >1;$ but this integral diverges to $+\infty$ when $\delta \le 1.$

Proof. 

We may write ${\left[f(x)\right]}_c^b=f(b)-f(c)$ when a function $f(x)$ of a real variable x is continuous on the closed interval $[c,b].$

Note that the real function $f(x)={\displaystyle \frac{1}{x^{\delta }}}$ is continuous on the open interval $(0,\infty ),$ but it is discontinuous at $x=0,$ where $\delta$ is a fixed real number.

When $\delta =1,$ we have ${\displaystyle \underset{b\to \infty }{lim}{\int }_c^b\frac{1}{x^{\delta }}dx=\underset{b\to \infty }{lim}{\int }_c^b\frac{1}{x}dx=\underset{b\to \infty }{lim}[logx{]}_c^b=+\infty .}$

When $\delta \ne 1,$ we have

$$\underset{b\to \infty }{lim}{\int }_c^b{\displaystyle \frac{1}{x^{\delta }}}dx=\underset{b\to \infty }{lim}[{\displaystyle \frac{x^{1-\delta }}{1-\delta }}{]}_c^b=\left\{\begin{array}{cc}+\infty ,\hfill & \mathrm{if}\delta <1;\hfill \\ \\ {\displaystyle \frac{c^{1-\delta }}{\delta -1}},\hfill & \mathrm{if}\delta >1.\hfill \end{array}\right.$$

Hence, the integral diverges to $+\infty$ when $\delta \le 1,$ but it converges to a fixed positive number ${\displaystyle \frac{c^{1-\delta }}{\delta -1}}$ when $\delta >1,$ where $c>0.$ Thus completing the proof. □

Theorem 2.

(cf. [7], pp.234-235.) (Weierstrass theorem): Let $\left\{f_n\right\}$ be a sequence of analytic functions on an open subset S of the complex plane, and assume that $\left\{f_n\right\}$ converges uniformly on every compact subset of S to a limit function $f,$ i.e., $lim{f}_n=f.$ Then f is analytic on S, and the sequence of derivatives $\left\{f_n^{\prime }\right\}$ converges uniformly on every compact subset of S to the derivative $f^{\prime }.$

Let $G(x)$ be a piecewise continuous function on the real-valued interval $\left[1,\infty \right).$ For a complex variable $s=\sigma +i\tau ,$ we write $Re(s)=\sigma .$ First note that if $\sigma \ge a$ we have $|x^{s+1}|\ge x^{a+1}$ for $x\ge 1,$ hence ${\displaystyle \left|\frac{G(x)}{x^{s+1}}\right|\le \frac{|G(x)|}{x^{a+1}}}$ for $x\ge 1.$ Therefore, if an integral ${\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$ converges absolutely for a complex number $s=a+ib,$ then by the comparison test it also converges absolutely for all s with $\sigma \ge a.$ This observation implies the following theorem.

Theorem 3.

Let a function $F(s)$ be represented by the integral form

$$F(s)={\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$$

where $G(x)$ is a piecewise continuous function on the real-valued interval $\left[1,\infty \right).$ Let’s write $Re(s)=\sigma .$

Suppose the integral

$${\int }_1^{\infty }\left|{\displaystyle \frac{G(x)}{x^{s+1}}}\right|dx$$

does not converge for all s or diverge for all $s.$ Then there exists a real number ${\sigma }_a,$ called the abscissa of absolute convergence, such that the integral

$${\int }_0^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$$

converges absolutely if $\sigma >{\sigma }_a$ but does not converge absolutely if $\sigma <{\sigma }_a.$

Proof. 

Let D be the set of all real numbers $\sigma$ such that ${\displaystyle {\int }_1^{\infty }\left|\frac{G(x)}{x^{s+1}}\right|dx}$ diverges. D is not empty because the integral does not converge for all $s,$ and D is bounded above because the integral does not diverge for all $s.$ Therefore D has a least upper bound which we call ${\sigma }_a.$ If $\sigma <{\sigma }_a$ then $\sigma \in D,$ otherwise $\sigma$ would be an upper bound for D smaller than the least upper bound. If $\sigma >{\sigma }_a$ then $\sigma \notin D$ since ${\sigma }_a$ is an upper bound for $D.$ This proves the theorem. □

Remark 1.

If the integral converges absolutely everywhere we define ${\sigma }_a=-\infty ,$ and if it converges absolutely nowhere we define ${\sigma }_a=+\infty .$

Theorem 4.

Let a function $F(s)$ be represented by the integral form

$$F(s)={\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx,$$

where $G(x)$ is a piecewise continuous function on the real-valued interval $\left[1,\infty \right).$ Let’s write $Re(s)=\sigma .$

If the integral form

$$F(s)={\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$$

converges for $s=s_0={\sigma }_0+i{\tau }_0.$ Then for each s with $\sigma >{\sigma }_0,$ we have

$$\left|{\int }_c^d{\displaystyle \frac{G(x)}{x^{s+1}}}dx\right|\le 2M{c}^{{\sigma }_0-\sigma }\left(1+{\displaystyle \frac{|s-s_0|}{\sigma -{\sigma }_0}}\right),$$

where $d\ge c\ge T_0\ge 1,$ and note that $\left(1+{\displaystyle \frac{|s-s_0|}{\sigma -{\sigma }_0}}\right)$ is independent of $c.$

Proof. 

Since the integral ${\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s_0+1}}}dx$ converges, then ${\int }_{T_0}^T{\displaystyle \frac{G(x)}{x^{s_0+1}}}dx$ also converges for any $T\ge T_0\ge 1,$ and we have $\left|{\int }_{T_0}^T{\displaystyle \frac{G(x)}{x^{s_0+1}}}dx\right|\le M,$ where M is a bound for ${\int }_{T_0}^T{\displaystyle \frac{G(x)}{x^{s_0+1}}}dx$ with $T\ge T_0\ge 1.$

For any $v\ge u\ge T_0\ge 1,$ putting $w(u):={\int }_1^u{\displaystyle \frac{G(x)}{x^{s_0+1}}}dx$ and $w(v)={\int }_1^v{\displaystyle \frac{G(x)}{x^{s_0+1}}}dx,$ then we have $\left|w(v)-w(u)\right|=\left|{\int }_u^T{\displaystyle \frac{G(x)}{x^{s_0+1}}}dx-{\int }_v^T{\displaystyle \frac{G(x)}{x^{s_0+1}}}dx\right|\le 2M,$ where M can be chosen to be a same bound between ${\int }_u^T{\displaystyle \frac{G(x)}{x^{s_0+1}}}dx$ and ${\int }_v^T{\displaystyle \frac{G(x)}{x^{s_0+1}}}dx.$

As a consequence, for $d\ge c\ge T_0\ge 1,$ we get

$$\begin{array}{cc}\hfill & \left|{\int }_c^d{\displaystyle \frac{G(x)}{x^{s+1}}}dx\right|=\left|{\int }_c^d{u}^{s_0-s}d\left(w(u)-w(c)\right)\right|\hfill \\ \hfill & =\left|d^{s_0-s}\left(w(d)-w(c)\right)+(s-s_0){\int }_c^d\left(w(u)-w(c)\right)u^{s_0-s-1}du\right|\hfill \\ \hfill & \le 2M{c}^{{\sigma }_0-\sigma }+\left|s-s_0\right|2M{\int }_c^d{u}^{{\sigma }_0-\sigma -1}du\hfill \\ \hfill & \le 2M{c}^{{\sigma }_0-\sigma }+\left|s-s_0\right|2M\left|{\displaystyle \frac{d^{{\sigma }_0-\sigma }-c^{{\sigma }_0-\sigma }}{{\sigma }_0-\sigma }}\right|\hfill \\ \hfill & \le 2M{c}^{{\sigma }_0-\sigma }\left(1+{\displaystyle \frac{|s-s_0|}{\sigma -{\sigma }_0}}\right),\hfill \end{array}$$

and note that $\left(1+{\displaystyle \frac{|s-s_0|}{\sigma -{\sigma }_0}}\right)$ is independent of c since s is independent of $c.$ Therefore, the theorem is proved. □

Theorem 5.

Let a function $F(s)$ be represented by the integral form

$$F(s)={\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx,$$

where $G(x)$ is a piecewise continuous function on the real-valued interval $\left[1,\infty \right).$ Let’s write $Re(s)=\sigma .$

(i) 

If the integral form

$$F(s)={\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$$

converges for $s=s_0={\sigma }_0+i{\tau }_0,$ then it also converges for all s with $\sigma >{\sigma }_0.$

Furthermore, if the integral form

$$F(s)={\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$$

diverges for $s=s_0={\sigma }_0+i{\tau }_0,$ then it diverges for all s with $\sigma <{\sigma }_0.$

(ii) 

If the integral form

$$F(s)={\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$$

converges absolutely for $s=s_0={\sigma }_0+i{\tau }_0,$ then it converges absolutely and uniformly for $\sigma \ge {\sigma }_0.$

(iii) 

If the function $F(s)={\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$ converges for $\sigma >{\sigma }_c,$ then the function $F(s)={\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$ converges uniformly on every compact subset lying interior to the half-plane of convergence $\sigma >{\sigma }_c.$ Moreover, the function $F(s)={\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$ is analytic on each of its convergent open domains.

Proof. 

In the assertion (i), the second statement follows from the first. To prove the first statement, choose any s with $\sigma >{\sigma }_0.$ Theorem 4 shows that

$$\left|{\int }_c^d{\displaystyle \frac{G(x)}{x^{s+1}}}dx\right|\le K{c}^{{\sigma }_0-\sigma }$$

for $c\ge 1,$ where K is independent of $c.$ Since $c^{{\sigma }_0-\sigma }\to 0$ as $c\to +\infty ,$ the Cauchy condition shows that

$${\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$$

converges for all s with $\sigma >{\sigma }_0.$ So the assertion (i) is proved.

The assertion (ii) is trivially verified by the estimate of the upper bound. Since

$${\int }_c^d\left|{\displaystyle \frac{G(x)}{x^{s+1}}}\right|dx\le {\int }_c^d\left|{\displaystyle \frac{G(x)}{x^{s_0+1}}}\right|dx$$

holds for $\sigma \ge {\sigma }_0,$ whereas ${\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$ converges absolutely for $\sigma \ge {\sigma }_0$ which implies the uniform convergence of partial sums ${\int }_c^d{\displaystyle \frac{G(x)}{x^{s+1}}}dx$ for $\sigma \ge {\sigma }_0$ with $d\ge c\ge T_0\ge 1,$ the Cauchy condition for uniform convergence is satisfied. So the assertion (ii) is proved.

To prove the assertion (iii), it suffices to show that ${\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$ converges uniformly on every compact rectangle $R=[\alpha ,\beta ]\times [a,b]$ with $\alpha >{\sigma }_c.$ To do this, we use the estimate obtained in Theorem 4, which is

$$\left|{\int }_c^d{\displaystyle \frac{G(x)}{x^{s+1}}}dx\right|\le 2M{c}^{{\sigma }_0-\sigma }\left(1+{\displaystyle \frac{|s-s_0|}{\sigma -{\sigma }_0}}\right),$$

(2)

where $s_0={\sigma }_0+i{\tau }_0$ is any point in the half-plane $\sigma >{\sigma }_c$ and s is any point with $\sigma >{\sigma }_0.$ We choose $s_0={\sigma }_0$ where ${\sigma }_c<{\sigma }_0<\alpha .$ Then if $s\in R$ we have $\sigma -{\sigma }_0\ge \alpha -{\sigma }_0$ and $|s-s_0|<C,$ where C is a constant depending on $s_0$ and R but not on $s.$ Then (2) implies $\left|{\int }_c^d{\displaystyle \frac{G(x)}{x^{s+1}}}dx\right|\le 2M{c}^{{\sigma }_0-\alpha }\left(1+{\displaystyle \frac{C}{\alpha -{\sigma }_0}}\right)=B{c}^{{\sigma }_0-\alpha },$ where B is independent of $s.$ Since $c^{{\sigma }_0-\alpha }\to 0$ as $c\to +\infty ,$ the Cauchy codition for uniform convergence is satisfied. Moreover, we apply the Weierstrass theorem to know that $F(s)={\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$ is analytic on each of its convergent open domains. So the assertion (iii) is proved. Therefore, the theorem is proved. □

Theorem 6.

Let a function $F(s)$ be represented by integrals of the form

$$F(s)={\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx,$$

where $G(x)$ is a piecewise continuous function on the real-valued interval $\left[1,\infty \right).$ Let’s write $Re(s)=\sigma .$

If the integral

$${\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$$

does not converge everywhere for all s or diverge everywhere for all $s,$ then there exists a real number ${\sigma }_c,$ called the abscissa of convergence, such that the integral

$${\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$$

converges for all s in the half-plane $\sigma >{\sigma }_c$ and diverges for all s in the half-plane $\sigma <{\sigma }_c.$

Proof. 

We argue as in the proof of Theorem 3, taking ${\sigma }_c$ to be the least upper bound of all $\sigma$ for which ${\displaystyle {\int }_1^{\infty }\frac{G(x)}{x^{s+1}}dx}$ diverges. This detail is repeated as follows.

Let D be the set of all real numbers $\sigma$ such that ${\int }_1^{\infty }{\displaystyle \frac{G(x)}{x^{s+1}}}dx$ diverges. D is not empty because the integral does not converge for all $s,$ and D is bounded above because the integral does not diverge for all $s.$ Therefore D has a least upper bound which we call ${\sigma }_c.$ If $\sigma <{\sigma }_c$ then $\sigma \in D,$ otherwise $\sigma$ would be an upper bound for D smaller than the least upper bound. If $\sigma >{\sigma }_c$ then $\sigma \notin D$ since ${\sigma }_c$ is an upper bound for $D.$ This proves the theorem. □

Remark 2.

If the integral converges everywhere we define ${\sigma }_c=-\infty ,$ and if it converges nowhere we define ${\sigma }_c=+\infty .$

2.4. 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 $\psi (x)-x.$

2.5. 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 ${\displaystyle Re(s)=\frac{1}{2}.}$

It is well-known that for $Re(s)>1$ the Riemann zeta function $\zeta (s)$ can be defined by the Euler product, which is a majorant of the series

$$\zeta (s)={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n^s}}=\prod _p{\left(1-p^{-s}\right)}^{-1},$$

and it represents an analytic function of s in the half-plane $Re(s)>1,$ where the product ${\prod }_p$ runs over all primes.

In fact, the only zeros of $\zeta (s)$ outside the “critical strip” $0\le Re(s)\le 1$ are at the negative even integers, $-2,$$-4,$$-6,$$-8,$$\dots .$

We have known that $\zeta (s)$ has no zeros on the real axis between 0 and $1,$ the non-trivial zeros of $\zeta (s)$ in the strip $0\le Re(s)\le 1$ are all complex, and $\zeta (s)$ has no zeros on $Re(s)\ge 1$ and no zeros on $Re(s)=0.$

The non-trivial zeros of $\zeta (s)$ in the strip $0\le Re(s)\le 1$ are symmetrically located about the critical line $Re(s)={\displaystyle \frac{1}{2}}$ and the real axis $Im(s)=0.$ The symmetry of those non-trivial zeros with respect to the line $Re(s)={\displaystyle \frac{1}{2}}$ in the strip $0\le Re(s)\le 1$ indicates that $\zeta (s)$ has no zeros for ${\displaystyle \frac{1}{2}}>Re(s)\ge 0$ if $\zeta (s)$ has no zeros for $Re(s)>{\displaystyle \frac{1}{2}}$ or $1\ge Re(s)>{\displaystyle \frac{1}{2}}.$

We have also known that $\zeta (s)$ has infinitely many zeros in the critical strip $0\le Re(s)\le 1,$ and that $\zeta (s)$ still has infinitely many zeros in the region $Re(s)\ge {\displaystyle \frac{1}{2}}$ due to the symmetry of the non-trivial zeros about the “critical line” $Re(s)={\displaystyle \frac{1}{2}}.$

It is known that the Riemann zeta function $\zeta (s)$ has an infinity of zeros on the critical line $Re(s)={\displaystyle \frac{1}{2}}.$ This was first proved by Hardy in 1914.

In addition, we have the following facts.

Theorem 7.

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

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

(3)

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

Theorem 8.

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

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

(4)

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 $Re(s)>1,$ which is a well-known formula and extends meromorphically to the half-plane $Re(s)>0$ with some poles at non-trivial zeros of $\zeta (s),$ but no other poles on the region $Re(s)>0.$

There is some information about the distribution of the imaginary parts of the complex zeros of the Riemann zeta function $\zeta (s),$ where $s=\sigma +i\tau$ is a complex variable. We denote a typical non-trivial zero of $\zeta (s)$ in the strip $0\le Re(s)\le 1,$ by $\rho =\beta +i\gamma .$ And we denote by $N(T),$ where $T>0,$ the number of zeros of $\zeta (s)$ in the rectangle $0\le \sigma \le 1,$$0\le \tau \le T,$ that is, the number of zeros $\rho =\beta +i\gamma$ of $\zeta (s),$ for which $0<\gamma \le T.$

Theorem 9.

(cf. [2], pp.70-71.) We have

$$\sum _{0<\gamma \le T}{\displaystyle \frac{1}{\left|\rho \right|}}=O\left({log}^{2}T\right),$$

(5)

and

$$\sum _{\gamma >T}{\displaystyle \frac{1}{{\gamma }^2}}=O\left({\displaystyle \frac{logT}{T}}\right)$$

(6)

as $T\to \infty .$

We shall apply the function ${\psi }_1(x)={\int }_0^x\psi (t)dt={\int }_1^x\psi (t)dt,$ where $\psi (t)$ is Chebyshev’s $\psi$-function defined earlier, and it is an increasing function. Indeed, we have the fundamental formula (cf. [2], pp.30-32):

$${\psi }_1(x)={\displaystyle \frac{1}{2\pi i}}{\int }_{c-i\infty }^{c+i\infty }{\displaystyle \frac{x^{s+1}}{s(s+1)}}\left(-{\displaystyle \frac{{\zeta }^{\prime }(s)}{\zeta (s)}}\right)ds(x>0,c>1),$$

(7)

where the path of integration is the straight line $\sigma =c,$ and we have the explicit formula for ${\psi }_1(x)$:

Theorem 10.

(cf. [2], pp.73-74) If $x\ge 1,$ then

$${\psi }_1(x)={\displaystyle \frac{x^2}{2}}-\sum _{\rho }{\displaystyle \frac{x^{\rho +1}}{\rho (\rho +1)}}-x{\displaystyle \frac{{\zeta }^{\prime }(0)}{\zeta (0)}}+{\displaystyle \frac{{\zeta }^{\prime }(1)}{\zeta (1)}}-{\displaystyle \sum _{r=1}^{\infty }}{\displaystyle \frac{x^{1-2r}}{2r(2r-1)}}.$$

(8)

In 1901, von Koch derived and obtained the estimate of the best error in the prime number theorem if the Riemann hypothesis is true, which is:

Theorem 11.

If the Riemann hypothesis is true, then $\psi (x)-x=O(x^{{\displaystyle \frac{1}{2}}}{log}^{2}x)$ is valid.

Proof. 

By using the explicit formula for ${\psi }_1(x),$ and since having the hypothesis that $\rho ={\displaystyle \frac{1}{2}}+i\gamma ,$ we can get

$$\begin{array}{cc}\hfill & \pm \left({\psi }_1(x\pm 1)-{\psi }_1(x)\right)\hfill \\ \hfill & =x\pm {\displaystyle \frac{1}{2}}-\sum _{\rho }{\displaystyle \frac{{(x\pm 1)}^{\rho +1}-x^{\rho +1}}{\rho (\rho +1)(\pm 1)}}-{\displaystyle \frac{{\zeta }^{\prime }(0)}{\zeta (0)}}+O\left({\displaystyle \frac{1}{x}}\right)\hfill \\ \hfill & =x+O\left(\left|\sum _{\rho }{\displaystyle \frac{{(x\pm 1)}^{\rho +1}-x^{\rho +1}}{\rho (\rho +1)}}\right|\right)+O(1)\hfill \\ \hfill & =x+O\left(\sum _{\left|\rho \right|\le x}{\displaystyle \frac{1}{\left|\rho \right|}}\left|{\int }_x^{x\pm 1}{u}^{\rho }du\right|\right)+O\left(\sum _{\left|\rho \right|>x}{\displaystyle \frac{{(x+1)}^{\frac{3}{2}}}{|\rho (\rho +1)|}}\right)\hfill \\ \hfill & =x+O\left(x^{{\displaystyle \frac{1}{2}}}·\sum _{\left|\gamma \right|\le x}{\displaystyle \frac{1}{\left|\rho \right|}}\right)+O\left(x^{{\displaystyle \frac{3}{2}}}·\sum _{\left|\gamma \right|>x}{\displaystyle \frac{1}{{\left|\rho \right|}^2}}\right)\hfill \\ \hfill & =x+O\left(x^{{\displaystyle \frac{1}{2}}}{log}^{2}x\right),\hfill \end{array}$$

where we need to apply the two relations (5) and (6) in Theorem 9, and note that the numerical value of the constant ${\displaystyle \frac{{\zeta }^{\prime }(0)}{\zeta (0)}}=log2\pi .$

Then, we get the two asymptotic relations

$$\left({\psi }_1(x)-{\psi }_1(x-1)\right)-x=O\left(x^{{\displaystyle \frac{1}{2}}}{log}^{2}x\right)$$

and

$$\left({\psi }_1(x+1)-{\psi }_1(x)\right)-x=O\left(x^{{\displaystyle \frac{1}{2}}}{log}^{2}x\right).$$

Also, we know $\psi (x)$ is the maximum of the function $\psi (t)$ in the closed interval $[x-1,x]$ and is the minimum of the function $\psi (t)$ in the closed interval $[x,x+1],$ where the length of the closed interval $[x-1,x]$ and the length of the closed interval $[x,x+1]$ are $1,$ and then

$${\psi }_1(x)-{\psi }_1(x-1)={\int }_{x-1}^x\psi (t)dt\le \psi (x)\le {\int }_x^{x+1}\psi (t)dt={\psi }_1(x+1)-{\psi }_1(x)$$

such that $\left|\psi (x)-x\right|\le \left|\left({\psi }_1(x)-{\psi }_1(x-1)\right)-x\right|+\left|\left({\psi }_1(x+1)-{\psi }_1(x)\right)-x\right|.$ So, if the Riemann hypothesis is true, then we have $\psi (x)-x=O(x^{{\displaystyle \frac{1}{2}}}{log}^{2}x).$ Therefore, the theorem is proved. □

Theorem 12.

The Riemann hypothesis is true if and only if the asymptotic form $\psi (x)-x=O\left(x^{{\displaystyle \frac{1}{2}}}{log}^{2}x\right)$ is valid, which implies that the asymptotic form $\psi (x)-x=O\left(x^{{\displaystyle \frac{1}{2}}}{log}^{2}x\right)$ is equivalent to the asymptotic form $\psi (x)-x=O\left(x^{{\displaystyle \frac{1}{2}}+ϵ}\right)$ for every $ϵ>0.$

Proof. 

The necessity immediately follows from Theorem 11.

For the sufficiency, the condition is closely connected with the formula

$$-{\displaystyle \frac{{\zeta }^{\prime }(s)}{\zeta (s)}}-{\displaystyle \frac{1}{s-1}}=1+s{\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{s+1}}}dx$$

(9)

which holds for $Re(s)>1$ from Theorem 8. If $\psi (x)-x=O\left(x^{{\displaystyle \frac{1}{2}}}{log}^{2}x\right)$ holds, we can obtain the relations $\left|\psi (x)-x\right|\le C{x}^{{\displaystyle \frac{1}{2}}}{log}^{2}x$ and $\left|\psi (x)-x\right|\le C{x}^{{\displaystyle \frac{1}{2}}+2\delta }$ for all sufficiently large x with any fixed positive number $\delta ,$ where C is a positive constant independent of $x,$ and we have $logx=o(x^{\delta })$ and ${log}^{2}x=o(x^{2\delta })$ for every positive $\delta ,$ then the integral on the right of the formula (9) converges absolutely and is analytic throughout the half-plane $Re(s)>{\displaystyle \frac{1}{2}}+2\delta$ for any $\delta >0,$ let $2\delta =ϵ$ for any $\delta >0,$ so that $ϵ$ is also any positive number, and note that $Re(s)>{\displaystyle \frac{1}{2}}+ϵ$ for any positive number $ϵ>0$ such that $Re(s)>{\displaystyle \frac{1}{2}}$ still holds based on the density of real numbers. By analytic continuation the right side (which is analytic by differentiation under the integral sign) must be equal to the left side throughout the half-plane $Re(s)>{\displaystyle \frac{1}{2}}+2\delta$ for any $\delta >0,$ which shows that $\zeta (s)$ could have no zeros in this half-plane $Re(s)>{\displaystyle \frac{1}{2}},$ then the function $\zeta (s)$ could have no zeros for $1\ge Re(s)>{\displaystyle \frac{1}{2}}$ and no zeros for ${\displaystyle \frac{1}{2}}>Re(s)\ge 0$ according to the symmetry of non-trivial zeros of the function $\zeta (s)$ about the line $Re(s)={\displaystyle \frac{1}{2}}$ in the strip $0\le Re(s)\le 1.$ Thus, all the non-trivial zeros of $\zeta (s)$ in the strip $0\le Re(s)\le 1$ must lie on the line $Re(s)={\displaystyle \frac{1}{2}},$ the Riemann hypothesis must be true. Therefore, the theorem is proved. □

Theorem 13.

(cf. [2], pp.90-92.) $\psi (x)-x={\Omega }_{\pm }\left(x^{{\displaystyle \frac{1}{2}}}\right)$ is valid.

Theorem 14.

(cf. [2], pp.100-103.) $\psi (x)-x={\Omega }_{\pm }\left(x^{{\displaystyle \frac{1}{2}}}logloglogx\right)$ is valid.

It is not hard to see that Theorem 14 actually implies Theorem 13.

Corollary 1.

The integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}}}dx}$ diverges.

Proof. 

The proof follows from the formula $\psi (x)-x={\Omega }_{\pm }\left(x^{{\displaystyle \frac{1}{2}}}\right),$ and is based on the fact that for $ϵ>0,$ the integral ${\int }_T^{\infty }{\displaystyle \frac{1}{x^{1+ϵ}}}dx$ converges and equals a finite positive number, where T can represent a fixed and sufficiently large number $\ge 1.$ In other words, these suffice to show that

$${\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}}}}>K{\displaystyle \frac{1}{x}}>K{\displaystyle \frac{1}{x^{1+ϵ}}}>0\mathrm{and}{\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}}}}<-M{\displaystyle \frac{1}{x}}<-M{\displaystyle \frac{1}{x^{1+ϵ}}}<0$$

for arbitrarily large x with $ϵ>0,$ where K and M are a positive constant and are independent of $x,$ the integral ${\int }_T^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}}}}dx$ changes sign infinitely often and diverges, but noting that the definite integral ${\int }_1^T{\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}}}}dx$ converges since it is a finite limit, then we apply the expression ${\int }_1^{\infty }={\int }_1^T+{\int }_T^{\infty }$ to conclude that ${\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}}}}dx$ diverges. Thus completing the proof. □

Remark 3.

According to Theorem 5 (iii) and the proof process of Theorem 12, we see that if the function $g(s):={\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{s+1}}}dx$ converges for $Re(s)>{\displaystyle \frac{1}{2}},$ then it is analytic for $Re(s)>{\displaystyle \frac{1}{2}},$ and then it implies the validity of the Riemann hypothesis, in which case we have $-{\displaystyle \frac{{\zeta }^{\prime }(s)}{\zeta (s)}}-{\displaystyle \frac{1}{s-1}}=1+s{\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{s+1}}}dx$ that holds for $Re(s)>1,$ and we can apply the analytic continuation and conclude that $-{\displaystyle \frac{{\zeta }^{\prime }(s)}{\zeta (s)}}-{\displaystyle \frac{1}{s-1}}$ will be extended to an analytic function in the half-plane $Re(s)>{\displaystyle \frac{1}{2}}.$ We note that $g(s)={\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{s+1}}}dx$ diverges for $Re(s)<{\displaystyle \frac{1}{2}}$ from Theorem 5 (i), since the integral ${\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}}}}dx$ diverges.

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

Remark 4.

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.

2.7. 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

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

(10)

Then

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

Proof. 

By (10), there exist constants $C_0>0$ and $N_0$ such that

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

(11)

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

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

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

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

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

hence $g(N)=O(h(N)).$ □

Corollary 2.

(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 ${\sum }_{k\in J_m}k\le {\displaystyle \frac{N}{m}}·\left|J_m\right|.$

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:

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

(12)

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 (12) holds. □

Remark 5.

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).

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

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

(13)

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 (13). □

2.8. 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:

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

(14)

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

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

(15)

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

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

Therefore, the lemma is proved. □

Corollary 3.

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

(16)

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,

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

(17)

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:

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

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

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

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

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

(18)

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 9 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 (18):

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

(19)

By (16), we have ${\displaystyle \frac{U(N)}{N}=O(NlogN),}$ and therefore ${\displaystyle \frac{U(N)}{N}=O(N{log}^{2}N).}$ Hence

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

Thus

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

(20)

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

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

(21)

By (20), 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 ${\sum }_{n=1}^{N}A{(n)}^2$:

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

(22)

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

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

(23)

8. Convergence of the Integral

Theorem 15.

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 (23) 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 the closed interval $[1,2],$ we obtain ${\displaystyle {\int }_1^{\infty }\frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}dx<\infty ,}$ this completes the proof. □

9. A Theorem on Integral Convergence, Pointwise Bound and Analyticity

Lemma 10.

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

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

(24)

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. □

Lemma 11.

If a Dirichlet series $D(s)={\sum }_{n=1}^{\infty }{a}_n{n}^{-s}$ converges at a point $s={\sigma }_0$ with ${\sigma }_0>0,$ then its coefficient partial sums satisfy ${\displaystyle A(N):=\sum _{n\le N}{a}_n=O(N^{{\sigma }_0}).}$

Proof. 

A short proof of $A(N)=O(N^{{\sigma }_0})$ using Abel summation is as follows. Let the series $H(N)={\sum }_{n=1}^N{a}_n{n}^{-{\sigma }_0}.$ Since the series $H(N)$ converges, $H(N)$ is bounded, say $|H(N)|\le M.$ Then by partial summation, we have

$$A(N)={\displaystyle \sum _{n=1}^N}{a}_n={\displaystyle \sum _{n=1}^N}{a}_n{n}^{-{\sigma }_0}·n^{{\sigma }_0}=H(N)N^{{\sigma }_0}-{\displaystyle \sum _{k=1}^{N-1}}H(k)\left({(k+1)}^{{\sigma }_0}-k^{{\sigma }_0}\right).$$

By the telescoping sum, we have ${\sum }_{k=1}^{N-1}\left({(k+1)}^{{\sigma }_0}-k^{{\sigma }_0}\right)=N^{{\sigma }_0}-1.$ Therefore,

$$\left|{\displaystyle \sum _{k=1}^{N-1}}H(k)\left({(k+1)}^{{\sigma }_0}-k^{{\sigma }_0}\right)\right|\le M(N^{{\sigma }_0}-1)=O(N^{{\sigma }_0}).$$

Thus, $A(N)=O(N^{{\sigma }_0}).$ □

Theorem 16.

Let $\psi (x)={\sum }_{n\le x}\Lambda (n)$ be the Chebyshev function. Define

$$a_n=\Lambda (n)-1,A(x)=\sum _{n\le x}{a}_n=\psi (x)-\lfloor x\rfloor .$$

For every $\epsilon >0$ define

$$I_{\mathrm{abs}}(\epsilon )={\int }_1^{\infty }{\displaystyle \frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}}dx,I_{\mathrm{cond}}(\epsilon )={\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}}dx.$$

Then the following statements are equivalent:

(1). Absolute convergence: $I_{\mathrm{abs}}(\epsilon )<\infty$ for all $\epsilon >0.$

(2). Conditional convergence: $I_{\mathrm{cond}}(\epsilon )$ converges for all $\epsilon >0.$

(3). Pointwise o-bound: $|\psi (x)-x|=o(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ for all $\epsilon >0$ (hence also $O(x^{{\displaystyle \frac{1}{2}}+\epsilon })$).

(4). Analyticity: The functions

$$f(s)={\int }_1^{\infty }{\displaystyle \frac{|\psi (x)-x|}{x^{s+1}}}dx,g(s)={\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{s+1}}}dx$$

are analytic in the half-plane $Re(s)>{\displaystyle \frac{1}{2}}.$

Moreover, the remaining statement (5) regarding the abscissa of convergence that the functions

$$f(s)={\int }_1^{\infty }{\displaystyle \frac{|\psi (x)-x|}{x^{s+1}}}dx\mathrm{and}g(s)={\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{s+1}}}dx$$

have the same the abscissa of convergence that ${\sigma }_c={\displaystyle \frac{1}{2}}.$ That is to say, the absolute convergence abscissa and the convergence abscissa of this function

$$g(s)={\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{s+1}}}dx$$

are consistent, i.e., ${\sigma }_a={\sigma }_c={\displaystyle \frac{1}{2}}.$ Conversely, if the absolute convergence abscissa and the convergence abscissa of the function $g(s)$ are consistent and equal to ${\displaystyle \frac{1}{2}},$ then the statements (1) and (2) are not only valid but also equivalent.

The details of the proof are presented below.

Proof. 

We prove the chain (1) ⇒ (3) ⇒ (1) and (2) ⇒ (3) ⇒ (2). Furthermore, (1) ⇒ (4), and (4) ⇒ (1). All of these form an equivalence chain.

Note that $A(x)=\psi (x)-\lfloor x\rfloor$ and $\psi (x)-x=A(x)-\left\{x\right\}$ where $\left\{x\right\}=x-\lfloor x\rfloor$ is the fractional part, bounded by $1.$ Hence the convergence of $I_{\mathrm{cond}}(\epsilon )$ is equivalent to the convergence of ${\int }_1^{\infty }A(x)x^{-{\displaystyle \frac{3}{2}}-\epsilon }dx$ (since the integral ${\int }_1^{\infty }\left\{x\right\}{x}^{-{\displaystyle \frac{3}{2}}-\epsilon }$ converges absolutely). Therefore we may work with $A(x)$ instead of $\psi (x)-x.$

Proof of (1) ⇒ (3). Assume $I_{\mathrm{abs}}(\epsilon )<\infty$ for every $\epsilon >0.$ Fix $\epsilon >0.$ If the o-bound were false, then there would exist $\delta >0$ and a sequence $x_n\to \infty$ such that

$$|\psi (x_n)-x_n|\ge \delta x_n^{1/2+\epsilon }.$$

(25)

Set $h_n=x_n^{1/2}.$ For sufficiently large n we have $h_n\ge 1$ and $h_n\le x_n;$ thus (24) applies. For any $t\in [x_n,x_n+h_n],$

$$|\psi (t)-\psi (x_n)|\le C{h}_{n}log{x}_n.$$

(26)

Then

$$|\psi (t)-t|\ge |\psi (x_n)-x_n|-|\psi (t)-\psi (x_n)|-|t-x_n|\ge \delta x_n^{1/2+\epsilon }-C{x}_n^{1/2}log{x}_n-x_n^{1/2}.$$

Since $x_n^{\epsilon }\to \infty ,$ for large n we have $\delta x_n^{\epsilon }\ge 2(Clog{x}_n+1),$ whence

$$|\psi (t)-t|\ge {\displaystyle \frac{\delta }{2}}{x}_n^{1/2+\epsilon }(t\in [x_n,x_n+h_n]).$$

(27)

Now estimate the integral over $[x_n,x_n+h_n].$ Since $t\le x_n+h_n\le 2x_n$ (as $h_n\le x_n$), and $t^{{\displaystyle \frac{3}{2}}+\epsilon }\le {(2x_n)}^{{\displaystyle \frac{3}{2}}+\epsilon }=2^{3/2+\epsilon }{x}_n^{3/2+\epsilon }.$ Using (27) we obtain ${\displaystyle \frac{|\psi (t)-t|}{t^{\frac{3}{2}+\epsilon }}\ge \frac{\delta }{2^{\frac{5}{2}+\epsilon }}{x}_n^{-1}.}$ Hence

$${\int }_{x_n}^{x_n+h_n}{\displaystyle \frac{|\psi (t)-t|}{t^{3/2+\epsilon }}}dt\ge {\displaystyle \frac{\delta }{2^{5/2+\epsilon }}}{x}_n^{-1}·h_n={\displaystyle \frac{\delta }{2^{5/2+\epsilon }}}{x}_n^{-1/2}.$$

(28)

Choose a subsequence (still denoted by $x_n$) such that the intervals $[x_n,x_n+h_n]$ are pairwise disjoint and satisfy $x_n\le n^2$ (this is always possible by thinning the sequence if necessary). Then ${\displaystyle \sum _n{x}_n^{-1/2}\ge \sum _{n}1/n=\infty .}$ Because the intervals are disjoint, (28) gives

$$I_{\mathrm{abs}}(\epsilon )\ge {\displaystyle \frac{\delta }{2^{5/2+\epsilon }}}\sum _n{x}_n^{-1/2}=\infty ,$$

contradicting the convergence of $I_{\mathrm{abs}}(\epsilon ).$ Hence $|\psi (x)-x|=o(x^{{\displaystyle \frac{1}{2}}+\epsilon }).$

Proof of (3) ⇒ (1). If $|\psi (x)-x|=o(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ for all $\epsilon >0,$ then for any fixed $\epsilon >0$ there exists $X_0$ such that for all $x\ge X_0,$ we have $|\psi (x)-x|\le x^{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{\epsilon }{2}}}.$ Thus ${\displaystyle \frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}\le x^{-1-\frac{\epsilon }{2}},}$ which is integrable on $[1,\infty ).$ Hence $I_{\mathrm{abs}}(\epsilon )<\infty .$

Proof of (2) ⇒ (3). Assume that $I_{\mathrm{cond}}(\epsilon )$ converges conditionally for every $\epsilon >0.$ Since $\psi (x)-x=A(x)-\left\{x\right\}$ and the integral ${\int }_1^{\infty }\left\{x\right\}{x}^{-{\displaystyle \frac{3}{2}}-\epsilon }$ converges absolutely, the conditional convergence of $I_{\mathrm{cond}}(\epsilon$) implies the convergence of ${\displaystyle {\int }_1^{\infty }\frac{A(x)}{x^{\frac{3}{2}+\epsilon }}dx.}$ Now write ${\displaystyle A(x)=\sum _{n\le x}{a}_n,}$ where $a_n=\Lambda (n)-1.$ For $Re(s)>1,$ integration by parts gives

$${\int }_1^{\infty }A(x)x^{-s-1}dx={\displaystyle \frac{1}{s}}{\displaystyle \sum _{n=1}^{\infty }}{a}_n{n}^{-s}.$$

(29)

The right-hand side is analytic for $Re(s)>1$ by standard results in Dirichlet series, which ${\sum }_{n=1}^{\infty }\Lambda (n)n^{-s}$ and ${\sum }_{n=1}^{\infty }{n}^{-s}$ are analytic for $Re(s)>1.$ According to Lemma 11, if the integral converges at a point $s={\sigma }_0$ (which is equivalent to the convergence of the Dirichlet series at $s={\sigma }_0$), then the partial sums of the coefficients satisfy

$$\sum _{n\le N}{a}_n=O(N^{{\sigma }_0}).$$

(30)

Here the convergence of the integral at $s={\displaystyle \frac{1}{2}}+\epsilon$ (for every $\epsilon >0$) implies that the Dirichlet series ${\sum }_{n=1}^{\infty }{a}_n{n}^{-s}$ converges at $s={\displaystyle \frac{1}{2}}+\epsilon .$ Therefore (30) holds with ${\sigma }_0={\displaystyle \frac{1}{2}}+\epsilon ,$ giving

$$\psi (N)-\lfloor N\rfloor =A(N)=O(N^{{\displaystyle \frac{1}{2}}+\epsilon }).$$

Extending the bound to real $x.$ Let $x\ge 2$ and set $N=\lfloor x\rfloor .$ Since $\psi$ is constant on $[N,N+1),$ we have $\psi (x)=\psi (N).$ Hence

$$|\psi (x)-x|=|\psi (N)-x|\le |\psi (N)-N|+|N-x|\le C{N}^{{\displaystyle \frac{1}{2}}+\epsilon }+1.$$

Because $N\le x$ and $1=O(x^{{\displaystyle \frac{1}{2}}+\epsilon }),$ we conclude $|\psi (x)-x|=O(x^{{\displaystyle \frac{1}{2}}+\epsilon }).$ Therefore $|\psi (x)-x|=O(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ for $\forall \epsilon >0,$ i.e., there exists an absolute constant $C>0$ such that for all sufficiently large $x,$ we have $|\psi (x)-x|\le C{x}^{{\displaystyle \frac{1}{2}}+\delta }$ for every $\delta >0.$ This actually implies the stronger o-bound: given any $\epsilon >0,$ choose $\delta =\epsilon /2;$ then $|\psi (x)-x|\le C{x}^{{\displaystyle \frac{1}{2}}+\delta }=o(x^{{\displaystyle \frac{1}{2}}+\epsilon }).$

Proof of (3) ⇒ (2). If $|\psi (x)-x|=o(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ for all $\epsilon >0,$ then for any fixed $\epsilon >0$ there exists $X_0$ such that for $x\ge X_0,$ we have $|\psi (x)-x|\le x^{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{\epsilon }{2}}}.$ Hence

$$\left|{\displaystyle \frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}}\right|\le x^{-1-{\displaystyle \frac{\epsilon }{2}}},$$

which is absolutely integrable. Therefore $I_{\mathrm{cond}}(\epsilon )$ converges absolutely (and hence conditionally).

Thus (1), (2), (3) are all equivalent. The equivalence with analyticity of $f(s)$ and $g(s)$ follows from standard properties of Dirichlet integrals: absolute convergence in a half-plane implies analyticity, and analyticity implies convergence on the boundary. That is to say, (4) has an analyticity equivalence with (1) and (2).

Proof of (1) ⇒ (4). Assume $I_{\mathrm{abs}}(\epsilon )<\infty$ for all $\epsilon >0.$ Let ${\sigma }_0>1/2$ be arbitrary and set $\epsilon =\left({\sigma }_0-{\displaystyle \frac{1}{2}}\right)/2.$ Then there exists a constant $C>0$ such that

$$|\psi (x)-x|\le C{x}^{{\displaystyle \frac{1}{2}}+\epsilon }$$

for all large $x.$ For $Re(s)\ge {\sigma }_0,$ we have

$$\left|{\displaystyle \frac{|\psi (x)-x|}{x^{s+1}}}\right|\le C{x}^{{\displaystyle \frac{1}{2}}+\epsilon -{\sigma }_0-1}=C{x}^{-1-\epsilon },$$

and ${\int }_1^{\infty }{x}^{-1-\epsilon }dx<\infty .$ Hence the integral defining $f(s)$ converges uniformly on any half-plane $Re(s)\ge {\sigma }_0.$ The integrand is analytic in $s,$ so by the Weierstrass theorem for parameter-dependent integrals, $f(s)$ is analytic in $Re(s)>{\displaystyle \frac{1}{2}}.$ The same estimate applies to $g(s)$ because $\left|g(s)\right|\le f\left(Re(s)\right).$ Thus (4) holds.

Proof of (4) ⇒ (1). If $f(s)$ is analytic for $Re(s)>1/2,$ then in particular for any $\epsilon >0,$ the point $s={\displaystyle \frac{1}{2}}+\epsilon$ lies in the domain of analyticity, so the integral $f({\displaystyle \frac{1}{2}}+\epsilon )={\int }_1^{\infty }|\psi (x)-x|x^{-{\displaystyle \frac{3}{2}}-\epsilon }dx$ converges, i.e., $I_{\mathrm{abs}}(\epsilon )<\infty .$ The same conclusion for $g(s)$ also gives $I_{\mathrm{cond}}(\epsilon )$ converges, but (1) follows directly. Hence (1) and (4) are equivalent. By the equivalences of (1), (2), and (3), so (4) is also equivalent to (2), and it is equivalent to (3). Thus (1), (2), (3), and (4) are all equivalent. And the remaining statement (5) regarding the abscissa of convergence ${\sigma }_c={\sigma }_a={\displaystyle \frac{1}{2}}$ is the results of the first four statements, since we have the fact that the function $g(s)={\int }_1^{\infty }{\displaystyle \frac{\psi (x)-x}{x^{s+1}}}dx$ is divergent for $Re(s)<{\displaystyle \frac{1}{2}}$ according to Corollary 1 and Remark 3. On the contrary, if the absolute convergence abscissa and the convergence abscissa of the function $g(s)$ are consistent and equal to ${\displaystyle \frac{1}{2}},$ i.e., ${\sigma }_a={\sigma }_c={\displaystyle \frac{1}{2}},$ then the statements (1) and (2) are not only valid but also equivalent. This completes the proof of Theorem 16. □

Remark 6.

For the Chebyshev function $\psi (x),$ the following are equivalent: Absolutely convergent integral ${\displaystyle {\int }_1^{\infty }|\psi (x)-x|x^{-\frac{3}{2}-\epsilon }dx}$ 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 $|\psi (x)-x|=o(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ 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 ψ and standard Dirichlet series theory. They are included as a theoretical complement to the 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 ${\int }_2^X{\left(\psi (t)-t\right)}^{2}dt=O(X^2{log}^{2}X),$ which is a suffficient condition for the Riemann hypothesis. All of them provide keen insight into the relationships between the properties for the Chebyshev function and the Riemann hypothesis.

Theorem 17.

The two asymptotic relations

$$\psi (x)-x=o(x^{{\displaystyle \frac{1}{2}}+\epsilon })\mathrm{and}\psi (x)-x=O(x^{{\displaystyle \frac{1}{2}}+\epsilon })\mathrm{are}\mathrm{valid}\mathrm{for}\mathrm{every}\epsilon >0.$$

In fact, the property of the asymptotic relation $\psi (x)-x=o(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ for every $\epsilon >0$ is equivalent to the property of the asymptotic relation $\psi (x)-x=O(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ for every $\epsilon >0.$

Proof. 

The proof follows from Theorem 15 and Theorem 16. The rest only needs to explain the equivalence between the two asymptotic relations.

If $\psi (x)-x=O(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ for $\forall \epsilon >0$ holds, i.e., there exists an absolute constant $C>0$ such that for all sufficiently large $x,$ we have $|\psi (x)-x|\le C{x}^{{\displaystyle \frac{1}{2}}+\delta }$ for every $\delta >0.$ This actually implies the stronger o-bound: given any $\epsilon >0,$ choose $\delta =\epsilon /2;$ then $|\psi (x)-x|\le C{x}^{{\displaystyle \frac{1}{2}}+\delta }=o(x^{{\displaystyle \frac{1}{2}}+\epsilon }).$

Conversely, the stronger o-bound definitely implies the weaker O-bound. That is to say, if $\psi (x)-x=o(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ for $\forall \epsilon >0$ holds, then it directly implies the property of the asymptotic relation $\psi (x)-x=O(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ for $\forall \epsilon >0.$

In general, the property of small o implies the property of big $O,$ the property of big O may not necessarily impliy the property of small o, but the two asymptotic relations are equivalent for every $\epsilon >0$ between them. □

Remark 7.

In a way, whether the (small o stronger bound) asymptotic relation $\psi (x)-x=o(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ holds for $\forall \epsilon >0$ constitutes an important criterion for testing the validity of the Riemann hypothesis. If this relation holds, it naturally implies the well-known (big O weaker bound) asymptotic relation $\psi (x)-x=O(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ for $\forall \epsilon >0$ equivalent to the Riemann hypothesis; if it fails to hold, the absolute convergence of the integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ for $\forall \epsilon >0$ in question will not hold, owing to the divergence of another well-documented integral ${\displaystyle {\int }_1^{\infty }\frac{1}{x}dx.}$ Fortunately, we are able to first prove that the integral in question is absolutely convergent, i.e., for $\forall \epsilon >0$ we have ${\displaystyle {\int }_1^{\infty }\frac{|\psi (x)-x|}{x^{\frac{3}{2}+\epsilon }}dx<\infty ,}$ which constitutes a sufficient condition for the truth of the Riemann hypothesis.

10. Conclusions

We have shown: Elementary estimates of the Chebyshev function lead to $U(N)=O(N^{2}logN),$ where ${\displaystyle U(N)={\displaystyle \sum _{k=1}^N}kA{(N/k)}^2}$ 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, the Cauchy-Schwarz inequality, and a dyadic decomposition, we obtain ${\displaystyle {\int }_2^X{\left(\psi (t)-t\right)}^{2}dt=O(X^2{log}^{2}X).}$ 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 $\epsilon >0.$ Since the integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ converges absolutely for every $\epsilon >0,$ which implies that ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ converges conditionally for every $\epsilon >0,$ whereas the integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ converges conditionally for every $\epsilon >0$$\iff \mathrm{RH},$ so then the Riemann hypothesis is true. In particular, we conclude that the property of absolute convergence of the integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ for every $\epsilon >0$ is equivalent to the property of conditional convergence of the integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ for every $\epsilon >0,$ either of which is equivalent to the property of the o-bound: $|\psi (x)-x|=o(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ for every $\epsilon >0,$ and all of them imply the O-bound: $\psi (x)-x=O(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ is also valid for every $\epsilon >0,$ which is the well-known equivalent form of the Riemann hypothesis, thus reconfirming the validity of the Riemann hypothesis.

Statements on the research background and significance of the work. This work was completed by the author through 13 years of independent research and endeavor. That is to say, the author first combined the Chebyshev function with the analytic convergence of an improper integral to study the proof of the Riemann hypothesis in 2013. Now, the new proof approach and new idea in this article is based on the author’s 13 years of research accumulation and persistent exploration from 2013 to the present (2026), which provides a clear and elementary treatment of the estimate of the mean-square integral ${\displaystyle {\int }_2^X{\left(\psi (t)-t\right)}^{2}dt=O(X^2{log}^{2}X)}$ involving the Chebyshev function, and presents some important properties of the improper integral ${\displaystyle {\int }_1^{\infty }\frac{\psi (x)-x}{x^{\frac{3}{2}+\epsilon }}dx}$ with absolute convergence (for every $\epsilon >0$) related to the estimate of the best error term in the prime number theorem involving the Chebyshev function, which can verify that $\psi (x)-x=o(x^{{\displaystyle \frac{1}{2}}+\epsilon })$ holds for every $\epsilon >0,$ and proves the validity of the Riemann hypothesis.