The Proof and Application of a Summation Formula

Yongmin Wang

Huimin Branch of Shandong Hengtai Engineering Group Co., Ltd., Binzhou, China.

DOI: 10.4236/apm.2022.129042                

Abstract

In this paper, some conclusions related to the prime number theorem, such as the Mertens formula are improved by the improved Abelian summation formula, and some problems such as “Dirichlet” function and “W(n)” function are studied.

Keywords

Abel Summation Formula, Mertens a Prime Number Theorem, Dirichlet Function, Conclusion Improvement

1. The Main Conclusions to Be Used in This Paper

1.1. Theorem A (Be Summation Formula) [1]

$b\left(n\right)\left(n=1,2,\cdots \right)$ is a plural column, The and function $B\left(n\right)={\displaystyle \underset{n\le u}{\sum }b\left(n\right)}$, to set up $0\le u_1<u_2$, f(u) interval $\left[u_1,u_2\right]$ continuous differentiable function of, so there are

${\displaystyle \underset{u_1<n\le u_2}{\sum }b\left(n\right)f\left(n\right)}=B\left(u_2\right)f\left(u_2\right)-B\left(u_1\right)f\left(u_1\right)-{\displaystyle {\int }_{u_1}^{u_2}B\left(u\right)f^{\prime }\left(u\right)\text{d}u}$ (1)

special: if $u_1=1,u_2=u>1$, Have a type:

${\displaystyle \underset{1\le n\le u}{\sum }b}\left(n\right)f\left(n\right)=B\left(u\right)f\left(u\right)-{\displaystyle {\int }_1^{u}B}\left(t\right)f^{\prime }\left(t\right)\text{d}t$ (2)

1.2. Theorem B (Prime Number Theorem) [2]

1) A. Walfsz the results of the:

$\theta \left(x\right)=x+O\left(x\exp \left(-c{\left(\log x\right)}^{\frac{3}{5}}{\left(\log \log x\right)}^{-}{}^{\frac{1}{5}}\right)\right)$

$\pi \left(x\right)=Lix+O\left(x\exp \left(-c{\left(\log x\right)}^{\frac{3}{5}}{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

2) In “Riemann” The prime number theorem under the condition that the conjecture is true (Vonkock the results of the).

$\pi \left(x\right)=Lix+O\left(x^{\frac{1}{2}}\log x\right)$

$\theta \left(x\right)=x+O\left(x^{\frac{1}{2}}{\log }^{2}x\right)$

3) Theorem C (Siegel-Walfisz) [3]

Set l, k is suitable for $\left(l,k\right)=1$ and $3\le k\le {\left(\log x\right)}^{k_0}$. The natural Numbers, Among them, $k_0$ is any normal number, then:

$\pi \left(x,k,l\right)=\frac{1}{\phi \left(k\right)}lix+O\left(x{\text{e}}^{-a_0\sqrt{\log x}}\right)$,

Here $a_0>0$, And with the “O” the relevant constants depend only on $k_0$.

4) Theorem D (Mertens 1874)

If $L\left(1,x\right)\ne 0$, for any model q Non-principal features hold, so for any of these $\left(a,q\right)=1$ the a there are:

${\displaystyle \underset{\begin{array}{c}p\le x\\ p\equiv a\left(\mathrm{mod}q\right)\end{array}}{\sum }\frac{1}{p}}=\frac{\log \log x}{\phi \left(q\right)}+c\left(a,q\right)+O\left(\frac{1}{\log x}\right)\left(x\to \infty \right)$,

Among them $c\left(a,q\right)=\frac{1}{\phi \left(q\right)}\left\{\gamma -{\displaystyle \underset{p}{\sum }\left(\log \left(\frac{1}{1-\frac{1}{p}}\right)-\frac{1}{p}\right)}+{\displaystyle \underset{x\ne x_0}{\sum }\bar{x}\left(a\right){\displaystyle \underset{p}{\sum }\frac{x\left(p\right)}{p}}}\right\}$.

2. The Proof of Summation Formula

2.1. Theorem 1

Set $\alpha \left(n\right),f\left(n\right)$, for the real function, $A\left(x\right)={\displaystyle \underset{n\le x}{\sum }a\left(n\right)}=g\left(x\right)+O\left(g_0\left(x\right)\right)$,

Meet the conditions:

1) $g\left(x\right),g_0\left(x\right),f\left(x\right),f^{\prime }\left(x\right)$ is interval $t\in \left[y,\infty \right)$, continuous function of;

2) For any small positive number $\epsilon$, in $t\in \left[y,\infty \right)$ Satisfy condition on: $\left|g_0\left(t\right)f^{\prime }\left(t\right)\right|\ll {\left(t{\left(\log t\right)}^{1+\epsilon }\right)}^{-1}$.

2.1.1. When $1\le y\ll 1$ When

$\begin{array}{l}{\displaystyle \underset{y<n\le x}{\sum }a\left(n\right)f\left(n\right)}=g\left(x\right)f\left(x\right)-{\displaystyle \underset{y}{\overset{x}{\int }}g\left(t\right)f^{\prime }\left(t\right)\text{d}t}-A\left(y\right)f\left(y\right)\\ -{\displaystyle \underset{y}{\overset{\infty }{\int }}\left(A\left(t\right)-g\left(t\right)\right)f^{\prime }\left(t\right)\text{d}t}+O\left({\displaystyle \underset{x}{\overset{\infty }{\int }}\left|g_0\left(t\right)f^{\prime }\left(t\right)\right|\text{d}t}+\left|g_0\left(x\right)f\left(x\right)\right|\right)\end{array}$ (3)

Among them $A\left(y\right)f\left(y\right)+{\displaystyle \underset{y}{\overset{\infty }{\int }}\left(A\left(t\right)-g\left(t\right)\right)f^{\prime }\left(t\right)\text{d}t}$, for only with “y” related constants.

2.1.2. When y Sufficiently Large

$\begin{array}{l}{\displaystyle \underset{y<n\le x}{\sum }a\left(n\right)f\left(n\right)}=g\left(x\right)f\left(x\right)-{\displaystyle \underset{y}{\overset{x}{\int }}g\left(t\right)f^{\prime }\left(t\right)\text{d}t}-g\left(y\right)f\left(y\right)\\ +O\left({\displaystyle \underset{y}{\overset{x}{\int }}\left|g_0\left(t\right)f^{\prime }\left(t\right)\right|\text{d}t}+\left|g_0\left(x\right)f\left(x\right)\right|+\left|g_0\left(y\right)f\left(y\right)\right|\right)\end{array}$ (4)

Prove:

1) When $\text{1}<y\ll 1<x$ when, by “(1) type” available

$\begin{array}{l}{\displaystyle \underset{y<n\le x}{\sum }a\left(n\right)f\left(n\right)}=A\left(x\right)f\left(x\right)-A\left(y\right)f\left(y\right)-{\displaystyle {\int }_y^{x}A\left(t\right)f^{\prime }\left(t\right)\text{d}t}\\ =g\left(x\right)f\left(x\right)+O\left(\left|g_0\left(x\right)f\left(x\right)\right|\right)-A\left(y\right)f\left(y\right)\\ -{\displaystyle \underset{y}{\overset{x}{\int }}\left(A\left(t\right)-g\left(t\right)\right)f^{\prime }\left(t\right)\text{d}t}-{\displaystyle \underset{y}{\overset{x}{\int }}g\left(t\right)f^{\prime }\left(t\right)\text{d}t}\end{array}$

$\begin{array}{l}=g\left(x\right)f\left(x\right)-{\displaystyle \underset{y}{\overset{x}{\int }}g\left(t\right)f^{\prime }\left(t\right)\text{d}t}-A\left(y\right)f\left(y\right)-{\displaystyle \underset{y}{\overset{\infty }{\int }}\left(A\left(t\right)-g\left(t\right)\right)f^{\prime }\left(t\right)\text{d}t}\\ +{\displaystyle \underset{x}{\overset{\infty }{\int }}\left(A\left(t\right)-g\left(t\right)\right)f^{\prime }\left(t\right)\text{d}t}+O\left(\left|g_0\left(x\right)f\left(x\right)\right|\right)\\ =g\left(x\right)f\left(x\right)-{\displaystyle \underset{y}{\overset{x}{\int }}g\left(t\right)f^{\prime }\left(t\right)\text{d}t}-A\left(y\right)f\left(y\right)-{\displaystyle \underset{y}{\overset{\infty }{\int }}\left(A\left(t\right)-g\left(t\right)\right)f^{\prime }\left(t\right)\text{d}t}\\ +O\left({\displaystyle \underset{x}{\overset{\infty }{\int }}\left|g_0\left(t\right)f^{\prime }\left(t\right)\right|\text{d}t}+\left|g_0\left(x\right)f\left(x\right)\right|\right)\end{array}$

Due to the $1<y\ll 1$, due to the

$\begin{array}{l}A\left(y\right)f\left(y\right)+{\displaystyle \underset{y}{\overset{\infty }{\int }}\left(A\left(t\right)-g\left(t\right)\right)f^{\prime }\left(t\right)\text{d}t}\ll 1+{\displaystyle \underset{y}{\overset{\infty }{\int }}\left|\left(A\left(t\right)-g\left(t\right)\right)f^{\prime }\left(t\right)\right|\text{d}t}\\ \ll 1+{\displaystyle \underset{y}{\overset{\infty }{\int }}\left|g_0\left(t\right)f^{\prime }\left(t\right)\right|\text{d}t}\ll 1+{\displaystyle \underset{y}{\overset{\infty }{\int }}{\left(t{\left(\log t\right)}^{1+\epsilon }\right)}^{-1}\text{d}t}\ll 1+{\log }^{-\epsilon }y\ll 1\end{array}$

Namely: $A\left(y\right)f\left(y\right)+{\displaystyle \underset{y}{\overset{\infty }{\int }}\left(A\left(t\right)-g\left(t\right)\right)f^{\prime }\left(t\right)\text{d}t}$ For only with “y” Related constants.

2) When $1<y<x$, and y Sufficiently large, by “(1) type” known:

$\begin{array}{l}{\displaystyle \underset{1<y<n\le x}{\sum }a\left(n\right)f\left(n\right)}=A\left(x\right)f\left(x\right)-A\left(y\right)f\left(y\right)-{\displaystyle \underset{y}{\overset{x}{\int }}A\left(t\right)f^{\prime }\left(t\right)\text{d}t}\\ =g\left(x\right)f\left(x\right)-g\left(y\right)f\left(y\right)-{\displaystyle \underset{y}{\overset{x}{\int }}g\left(t\right)f^{\prime }\left(t\right)\text{d}t}-{\displaystyle \underset{y}{\overset{x}{\int }}\left(A\left(t\right)-g\left(t\right)\right)f^{\prime }\left(t\right)\text{d}t}\\ +O\left(\left|g_0\left(x\right)f\left(x\right)\right|+\left|g_0\left(y\right)f\left(y\right)\right|\right)\\ =g\left(x\right)f\left(x\right)-g\left(y\right)f\left(y\right)-{\displaystyle \underset{y}{\overset{x}{\int }}g\left(t\right)f^{\prime }\left(t\right)\text{d}t}\\ +O\left({\displaystyle \underset{y}{\overset{x}{\int }}\left|g_0\left(t\right)f^{\prime }\left(t\right)\right|\text{d}t}+\left|g_0\left(x\right)f\left(x\right)\right|+\left|g_0\left(y\right)f\left(y\right)\right|\right)\end{array}$

Notes: Set some strings attached, Such as when $x\ge t\ge T$ (Among them: $T>y$ ) when, $\left|g_0\left(t\right)f^{\prime }\left(t\right)\right|\ll {\left(t{\left(\log t\right)}^{1+\epsilon }\right)}^{-1}$ To set up, At this time “ $A\left(y\right)f\left(y\right)+{\displaystyle \underset{y}{\overset{\infty }{\int }}\left(A\left(t\right)-g\left(t\right)\right)f^{\prime }\left(t\right)\text{d}t}$ ” As with the “y” “T” Related constants.

For not meeting the conditions $\left|g_0\left(t\right)f^{\prime }\left(t\right)\right|\ll {\left(t{\left(\log t\right)}^{1+\epsilon }\right)}^{-1}$ can be summed as follows:

2.2. Theorem1'

Set $a\left(n\right)$ It’s an arithmetic function, $A\left(x\right)={\displaystyle \underset{n\le x}{\sum }a\left(n\right)}=g\left(x\right)+O\left(g_0\left(x\right)\right)$,

Meet the conditions: $g\left(x\right),g_0\left(x\right),{g^{\prime }}_0\left(x\right),f\left(x\right),f^{\prime }\left(x\right)$ interval $\left[y,\infty \right)$.

Is a continuous function of, then

$\begin{array}{l}{\displaystyle \underset{1<y<n\le x}{\sum }a\left(n\right)f\left(n\right)}\\ ={\displaystyle \underset{y}{\overset{x}{\int }}{g}^{\prime }\left(t\right)f\left(t\right)\text{d}t}+O\left({\displaystyle \underset{y}{\overset{x}{\int }}\left|g_0\left(t\right)f^{\prime }\left(t\right)\right|\text{d}t}+\left|g_0\left(x\right)f\left(x\right)\right|+\left|g_0\left(y\right)f\left(y\right)\right|\right)\end{array}$ (5)

Prove: by Abel the summation formula is easy to know:

${\displaystyle \underset{1<y<n\le x}{\sum }a\left(n\right)f\left(n\right)}={\displaystyle \underset{y}{\overset{x}{\int }}f\left(t\right)\text{d}A\left(t\right)}={\displaystyle \underset{y}{\overset{x}{\int }}f\left(t\right)\text{d}\left(g\left(t\right)+O\left(g_0\left(t\right)\right)\right)}$

$\Rightarrow$ conclusion.

3. The Proof of Several Common Conclusions about “Prime Number Theorem” [3]

Theorem 2 (Mertens Improvement of Prime Number Theorem) [4]

Verification: 1) ${\displaystyle \underset{p\le x}{\sum }\frac{\log p}{p}}=\log x+A_1+O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-}{}^{\frac{1}{5}}\right)\right)$,

2) ${\displaystyle \underset{p\le x}{\sum }\frac{1}{p}}=\log x+A_2+O\left(\exp \left(-c_2{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-}{}^{\frac{1}{5}}\right)\right)$,

3) ${\displaystyle \underset{p\le x}{\prod }\left(1-\frac{1}{p}\right)}=\frac{{\text{e}}^{-Y}}{\log x}+O\left(\exp \left(-c_3{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$,

Among them A₁, A₂. For constant, c_i > 0 (i = 1, 2, 3), Y for Euler constant.

Prove: From the prime number theorem:

$\theta \left(x\right)={\displaystyle \underset{p\le x}{\sum }\log p}=x+O\left(x\exp \left(-c{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-}{}^{\frac{1}{5}}\right)\right)$,

Make $A\left(x\right)=\theta \left(x\right)$, $g\left(x\right)=x$, $g_0\left(x\right)=x\exp \left(-c{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-}{}^{\frac{1}{5}}\right)$, $f\left(x\right)=\frac{1}{x}$, $f^{\prime }\left(x\right)=-\frac{1}{x^2}$, Easy card with it.

The foot “(3) type” all conditions of, by “(3) type” available:

$\begin{array}{l}{\displaystyle \underset{1<p\le x}{\sum }\frac{\log p}{p}}=1+{\displaystyle {\int }_1^x\frac{1}{t}\text{d}t}+{\displaystyle {\int }_1^{\infty }\frac{\theta \left(t\right)-t}{t^2}\text{d}t}\\ +O\left({\displaystyle {\int }_x^{\infty }\frac{\exp \left(-c{\log }^{\frac{3}{5}}t{\left(\log \log t\right)}^{-\frac{1}{5}}\right)}{t}\text{d}t}+\exp \left(-c{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)\end{array}$

(when $c>c_1>0$ when, Easy card: $\exp \left(-c{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right){\log }^{2}x\ll \exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)$ )

$\begin{array}{l}=1+\log x+{\displaystyle {\int }_1^{\infty }\frac{\theta \left(t\right)-t}{t^2}\text{d}t}\\ +O\left({\displaystyle {\int }_x^{\infty }\frac{\exp \left(-c_1{\log }^{\frac{3}{5}}t{\left(\log \log t\right)}^{-\frac{1}{5}}\right)}{t{\log }^{2}t}\text{d}t}+\exp \left(-c{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)\end{array}$

$\begin{array}{l}=1+\log x+{\displaystyle {\int }_1^{\infty }\frac{\theta \left(t\right)-t}{t^2}\text{d}t}\\ +O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right){\displaystyle {\int }_x^{\infty }\frac{\text{d}t}{t{\log }^{2}t}}+\exp \left(-c{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)\end{array}$

$=1+\log x+{\displaystyle {\int }_1^{\infty }\frac{\theta \left(t\right)-t}{t^2}\text{d}t}+O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

$=\log x+\left({\displaystyle {\int }_1^{\infty }\frac{\theta \left(t\right)-t}{t^2}\text{d}t}+1\right)+O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

If the:

$A_1={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{\theta \left(t\right)-t}{t^2}\text{d}t}+1$,

If the $A_1$ For constant.

2) $A\left(x\right)=\theta \left(x\right)$, $g\left(x\right)=x$, $g_0\left(x\right)=x\exp \left(-c{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)$, $f\left(x\right)=\frac{1}{x\log x}$, $f^{\prime }\left(x\right)=-\frac{\log x+1}{x^2{\log }^{2}x}$,

It is easy to prove that it satisfies “(3) type” all conditions of, by “(3) type” available:

${\displaystyle \underset{1<P\le x}{\sum }\frac{1}{p}}=\frac{1}{2}+{\displaystyle \underset{2<p\le x}{\sum }\frac{1}{p}}=\frac{1}{2}+\frac{1}{\log x}+{\displaystyle \underset{2}{\overset{x}{\int }}\frac{\log t+1}{t{\log }^{2}t}\text{d}t}-\frac{1}{2}+{\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{\left(\theta \left(t\right)-t\right)\left(\log t+1\right)}{t^2{\log }^{2}t}\text{d}t}$

$O\left({\displaystyle \underset{x}{\overset{\omega }{\int }}\frac{\exp \left(-c{\log }^{\frac{3}{5}}t{\left(\log \log t\right)}^{-\frac{1}{5}}\right)\left(\log t+1\right)}{t{\log }^{2}t}\text{d}t}+\exp \left(-c{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

$\begin{array}{l}=\log \log x+\left(\frac{1}{\log 2}-\log \log 2+{\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{\left(\theta \left(t\right)-t\right)\left(\log t+1\right)}{t^2{\log }^{2}t}\text{d}t}\right)\\ +O\left(\exp \left(-c_2{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)\end{array}$

Among them: $A_2=\frac{1}{\log 2}-\log \log 2+{\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{\left(\theta \left(t\right)-t\right)\left(\log t+1\right)}{t^2{\log }^{2}t}\text{d}t}$, for constant.

3) By “Mertens” Formula to know: ${\displaystyle \underset{P\le x}{\sum }\frac{1}{p}}=\log \log x+B_1+O\left(\frac{1}{\log x}\right)$, among them $B_1=Y+{\displaystyle \underset{P}{\sum }\left(\log \left(1-\frac{1}{p}\right)+\frac{1}{p}\right)}$.

Knowing from the evidence before:

${\displaystyle \underset{p\le x}{\sum }\frac{1}{p}}=\log \log x+A_2+O\left(\exp \left(-c_2{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

$\Rightarrow B_1-A_2=O\left(\frac{1}{\log x}\right)=0$ ( $x\to \infty$ when)

$\Rightarrow \text{constant}{A}_2=B_1=Y+{\displaystyle \underset{P}{\sum }\left(\log \left(1-\frac{1}{p}\right)+\frac{1}{p}\right)}$

$\begin{array}{l}\Rightarrow {\displaystyle \underset{p\le x}{\sum }\frac{1}{p}}=\log \log x+Y+{\displaystyle \underset{p}{\sum }\left(\log \left(1-\frac{1}{p}\right)+\frac{1}{p}\right)}\\ +O\left(\exp \left(-c_2{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)\end{array}$

${\displaystyle \underset{p\le x}{\prod }\left(1-\frac{1}{p}\right)}=\exp \left({\displaystyle \underset{p\le x}{\sum }\left(\log \left(1-\frac{1}{p}\right)+\frac{1}{p}\right)}-{\displaystyle \underset{p\le x}{\sum }\frac{1}{p}}\right)$

$\begin{array}{l}\Rightarrow {\displaystyle \underset{p\le x}{\prod }\left(1-\frac{1}{p}\right)}=\exp (-\log \log x-Y-{\displaystyle \underset{p>x}{\sum }\left(\log \left(1-\frac{1}{p}\right)+\frac{1}{p}\right)}\\ +O\left(\exp \left(-c_2{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right))\end{array}$

$=\frac{{\text{e}}^{-Y}}{\log x}\exp \left({\displaystyle \underset{p>x}{\sum }O}\left(\frac{1}{p^2}\right)+O\left(\exp \left(-c_2{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)\right)$

$=\frac{{\text{e}}^{-Y}}{\log x}\exp \left(O\left(\exp \left(-c_2{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)\right)$

$=\frac{{\text{e}}^{-Y}}{\log x}+O\left(\exp \left(-c_3{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

Inference 1:

Verification: 1) ${\displaystyle \underset{y<p\le x}{\sum }\frac{\log p}{p}}=\log \left(\frac{x}{y}\right)+O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}y{\left(\log \log y\right)}^{-\frac{1}{5}}\right)\right)$,

2) ${\displaystyle \underset{y<p\le x}{\sum }\frac{1}{p}}=\log \left(\frac{\log x}{\log y}\right)+O\left(\exp \left(-c_2{\log }^{\frac{3}{5}}y{\left(\log \log y\right)}^{-\frac{1}{5}}\right)\right)$,

3) ${\displaystyle \underset{y<p\le x}{\prod }\left(1-\frac{1}{p}\right)}=\frac{\log y}{\log x}\left(1+O\left(-c_3{\log }^{\frac{3}{5}}y{\left(\log \log y\right)}^{-\frac{1}{5}}\right)\right)$.

Prove: It can be derived directly from theorem two.

Inference 2: Two identity relations:

1) ${\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{\theta \left(t\right)-\pi \left(t\right)\left(\log t-1\right)}{t^2}\text{d}t}=0$,

2) ${\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{\theta \left(t\right)\left(\log t-1\right)-\pi \left(t\right){\log }^{2}t}{t^2{\log }^{2}t}\text{d}t}=0$.

Proof: if so

$\begin{array}{l}A\left(x\right)=\pi \left(x\right)=\frac{x}{\log x}+\frac{x}{{\log }^{2}x}+O\left(\frac{x}{{\log }^{3}x}\right),\\ g\left(x\right)=\frac{x}{\log x}+\frac{x}{{\log }^{2}x},g_0\left(x\right)=\frac{x}{{\log }^{3}x}\end{array}$

1) Take $f\left(x\right)=\frac{\log x}{x}$, using “(3) type” ${\displaystyle \underset{1\le p\le x}{\sum }\frac{\log p}{p}}$, available $A_1={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{\left(\log t-1\right)\pi \left(t\right)-t}{t^2}\text{d}t}+1$.

2) Take $f\left(x\right)=\frac{1}{x}$, using “(3) type” ${\displaystyle \underset{1\le p\le x}{\sum }\frac{1}{p}}$, available $A_2={\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{\pi \left(t\right)-\frac{t}{\log t}}{t^2}\text{d}t}-\log \log 2$.

Take the above A₁, A₂ is compared with the corresponding value in “Therorem 2”.

Inference 3: A₁, A₂ other forms of values:

1) Take $f\left(x\right)$,

$A_1={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{\theta \left(t\right)-t}{t^2}\text{d}t}+1={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{\left(\log t-1\right)\pi \left(t\right)-t}{t^2}\text{d}t}+1=-Y-{\displaystyle \underset{P}{\sum }\frac{\log p}{p\left(p-1\right)}}\approx -1.332$,

2) $\begin{array}{l}{A}_2={\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{\left(\theta \left(t\right)-t\right)\left(\log t+1\right)}{t^2{\log }^{2}t}\text{d}t}+\frac{1}{\log 2}-\log \log 2\\ ={\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{\pi \left(t\right)-\frac{t}{\log t}}{t^2}\text{d}t}-\log \log 2=Y+{\displaystyle \underset{p}{\sum }\left(\log \left(1-\frac{1}{p}\right)+\frac{1}{p}\right)}\approx 0.216\end{array}$

Proof: a little.

Note: For A₁, A₂, the various expressions of value are derived from different methods used to prove “Theorem two”, some of which we have proved, some of which we will prove later, but it is worth mentioning that many times we just need to know that it is a constant.

Inference 4: An improvement on theorem D,

Make $f\left(t\right)=\frac{1}{t},g\left(t\right)=\frac{lit}{\phi \left(q\right)},g_0\left(t\right)=t{\text{e}}^{-a_0\sqrt{\log t}}$,

Easy card: $\left|g_0\left(t\right)f^{\prime }\left(t\right)\right|\ll {\left(t{\left(\log t\right)}^{1+\epsilon }\right)}^{-1}$ ( $\epsilon >0$ when),

If the conditions of “Theorem 1” are met, the following can be obtained from “Theorem 1” and “Theorem C”:

${\displaystyle \underset{\begin{array}{c}p\le x\\ p\equiv a\left(\mathrm{mod}q\right)\end{array}}{\sum }\frac{1}{p}}={\displaystyle \underset{\begin{array}{c}2<p\le x\\ p\equiv a\left(\mathrm{mod}q\right)\end{array}}{\sum }\frac{1}{p}}+\frac{\pi \left(2,q,a\right)}{2}$

$\begin{array}{l}=\frac{lix}{\phi \left(q\right)}\cdot \frac{1}{x}-{\displaystyle \underset{2}{\overset{x}{\int }}\frac{lit}{\phi \left(q\right)}\cdot \left(-\frac{1}{t^2}\right)\text{d}t}-{\displaystyle \underset{2}{\overset{\infty }{\int }}\left(\pi \left(t,q,a\right)-\frac{lit}{\phi \left(q\right)}\right)\left(-\frac{1}{t^2}\right)\text{d}t}\\ +O\left({\displaystyle \underset{x}{\overset{\infty }{\int }}\left|t{\text{e}}^{-a_0\sqrt{\log t}}\cdot \frac{1}{t^2}\right|\text{d}t}+\left|x{\text{e}}^{-a_0\sqrt{\log t}}\cdot \frac{1}{x}\right|\right)\end{array}$

$=\frac{\log \log x}{\phi \left(q\right)}+\frac{li2}{2\phi \left(q\right)}-\frac{\log \log 2}{\phi \left(q\right)}+{\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{\pi \left(t,q,a\right)-\frac{lit}{\phi \left(q\right)}}{t^2}\text{d}t}+O\left({\text{e}}^{-{a^{\prime }}_0\sqrt{\log x}}\right)\left({a^{\prime }}_0>0\right)$

It is easy to know from the above formula and “Theorem D”:

$c\left(a,q\right)=\frac{li2}{2\phi \left(q\right)}-\frac{\log \log 2}{\phi \left(q\right)}+{\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{\pi \left(t,q,a\right)-\frac{lit}{\phi \left(q\right)}}{t^2}\text{d}t}$,

Namely:

${\displaystyle \underset{\begin{array}{c}p\le x\\ p\equiv a\left(\mathrm{mod}q\right)\end{array}}{\sum }\frac{1}{p}}=\frac{\log \log x}{\phi \left(q\right)}+c\left(a,q\right)+O\left({\text{e}}^{-{a^{\prime }}_0\sqrt{\log x}}\right)\left({a^{\prime }}_0>0\right)$.

Theorem 2': if “Riemann” If the guess is true, then:

1) ${\displaystyle \underset{p\le x}{\sum }\frac{\log p}{p}}=\log x+A_1+O\left(x^{-\frac{1}{2}}{\log }^{2}x\right)$,

2) ${\displaystyle \underset{p\le x}{\sum }\frac{1}{p}}=\log \log x+A_2+O\left(x^{-\frac{1}{2}}\log x\right)$,

3) ${\displaystyle \underset{p\le x}{\prod }\left(1-\frac{1}{p}\right)}=\frac{{\text{e}}^{-Y}}{\log x}+O\left(x^{-\frac{1}{2}}\right)$.

Prove: using “Riemann” If the guess is true, then “ $\theta \left(x\right)=x+O\left(x^{\frac{1}{2}}{\log }^{2}x\right)$ ”. This conclusion, the proof process is the same as “Theorem 2”, so it is omitted.

Theorem 3: 1) ${\displaystyle \underset{p\le x}{\sum }\frac{\wedge \left(n\right)}{n}}=\log x-Y+O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$,

2) ${\displaystyle \underset{p\le x}{\sum }\frac{\log p}{p-1}}=\log x-Y+O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$.

Prove: 1) The first $A_1=-Y-{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}$,

${\displaystyle \underset{1\le n\le x}{\sum }\log n}={\displaystyle \underset{1\le p\le x}{\sum }\left[\frac{x}{p}\right]}\log p+{\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\left(\left[\frac{x}{p^2}\right]+\left[\frac{x}{p^3}\right]+\cdots +\left[\frac{x}{p^{\left[{\log }_{p}x\right]}}\right]\right)\log p}$

$={\displaystyle \underset{1\le p\le x}{\sum }\left[\frac{x}{p}\right]}\log p+{\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\left(\frac{x}{p^2}+\frac{x}{p^3}+\cdots +\frac{x}{p^{\left[{\log }_{p}x\right]}}+O\left({\log }_{p}x\right)\right)\log p}$

$={\displaystyle \underset{1\le p\le x}{\sum }\left[\frac{x}{p}\right]}\log p+{\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\left(\frac{x}{p\left(p-1\right)}\log p+O\left(\log x\right)\right)}$

$={\displaystyle \underset{1\le p\le x}{\sum }\left[\frac{x}{p}\right]}\log p+{\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\frac{x\log p}{p\left(p-1\right)}}+O\left(\sqrt{x}\right)$ (6)

$={\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\left[\frac{x}{p}\right]}\log p+{\displaystyle \underset{1\le n\le \sqrt{x}}{\sum }\left(\theta \left(\frac{x}{n}\right)-\theta \left(\sqrt{x}\right)\right)}+{\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\frac{x\log p}{p\left(p-1\right)}}+O(\; x\; )$

$=x{\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\frac{\log p}{p}}+{\displaystyle \underset{1\le n\le \sqrt{x}}{\sum }\theta \left(\frac{x}{n}\right)}-\sqrt{x}\theta \left(\sqrt{x}\right)+x{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}+O(\; x\; )$

A prime number theorem:

$\theta \left(x\right)=x+O\left(x^{\theta }\right)=x+O\left(x\exp \left(-c{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

$\begin{array}{l}\Rightarrow {\displaystyle \underset{1\le n\le x}{\sum }\log x}=x{\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\frac{\log p}{p}}+x{\displaystyle \underset{1\le n\le \sqrt{x}}{\sum }\frac{1}{n}}-x+x{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}\\ +O\left(\sqrt{x}+x^{\frac{1+\theta }{2}}+{\displaystyle \underset{1\le n\le \sqrt{x}}{\sum }{\left(\frac{x}{n}\right)}^{\theta }}\right)\end{array}$

$\begin{array}{l}=x{\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\frac{\log p}{p}}+x\left(\log \sqrt{x}+Y+O\left(\frac{1}{\sqrt{x}}\right)\right)-x+x{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}\\ +O\left(x\exp \left(-{c^{\prime }}_4{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)\end{array}$

$\begin{array}{l}=x{\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\frac{\log p}{p}}+x\left(\frac{1}{2}\log x+Y-1\right)+x{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}\\ +O\left(x\exp \left(-{c^{\prime }}_4{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)\end{array}$

${\displaystyle \underset{1\le n\le x}{\sum }\log n}=x\log x-x+O\left(\log x\right)$

$\Rightarrow {\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\frac{\log p}{p}}=\frac{1}{2}\log x-Y-{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}+O\left(\exp \left(-{c^{\prime }}_4{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

$\Rightarrow {\displaystyle \underset{1\le p\le x}{\sum }\frac{\log p}{p}}=\log x-Y-{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}+O\left(\exp \left(-{c^{\prime }}_4{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

Theorem 2

$\Rightarrow A_1=-Y-{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}$

2) ${\displaystyle \underset{1\le p\le x}{\sum }\frac{\wedge \left(n\right)}{n}}={\displaystyle \underset{1\le p\le x}{\sum }\frac{\log p}{p}}+{\displaystyle \underset{1<p\le \sqrt{x}}{\sum }\left(\frac{1}{p^2}+\frac{1}{p^3}+\cdots +\frac{1}{p^{\left[{\log }_{p}x\right]}}\right)\log p}$

$={\displaystyle \underset{1\le p\le x}{\sum }\frac{\log p}{p}}+{\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\left(\frac{\log p}{p\left(p-1\right)}+O\left(\frac{\log p}{x}\right)\right)}$

$={\displaystyle \underset{1\le p\le x}{\sum }\frac{\log p}{p}}+{\displaystyle \underset{1\le p\le \sqrt{x}}{\sum }\frac{\log p}{p\left(p-1\right)}}+O(\; 1\; x\; )$

Theorem 2

$\begin{array}{l}\Rightarrow {\displaystyle \underset{1\le p\le x}{\sum }\frac{\wedge \left(n\right)}{n}}=\log x-Y-{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}+O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)\\ +{\displaystyle \underset{1<p\le \sqrt{x}}{\sum }\frac{\log p}{p\left(p-1\right)}}+O(\; 1\; x\; )\end{array}$

$=\log x-Y-{\displaystyle \underset{p>\sqrt{x}}{\sum }\frac{\log p}{p\left(p-1\right)}}+O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

$=\log x-Y+O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

3) ${\displaystyle \underset{1\le p\le x}{\sum }\frac{\log p}{p-1}}={\displaystyle \underset{1\le p\le x}{\sum }\frac{\log p}{p}}+{\displaystyle \underset{1\le p\le x}{\sum }\frac{\log p}{p\left(p-1\right)}}$

Theorem 2

$\begin{array}{l}\Rightarrow {\displaystyle \underset{1\le p\le x}{\sum }\frac{\log p}{p-1}}=\log x-Y-{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}+O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)\\ +{\displaystyle \underset{1<p\le x}{\sum }\frac{\log p}{p\left(p-1\right)}}\end{array}$

$=\log x-Y-{\displaystyle \underset{p>x}{\sum }\frac{\log p}{p\left(p-1\right)}}+O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

$=\log x-Y+O\left(\exp \left(-c_1{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)$

Note: The paper proves “ $A_1=-Y-{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}$ ”. In fact, it is another proof method of theorem 2.

Theorem 3': If “Riemann” conjecture is true, then

1) ${\displaystyle \underset{1\le n\le x}{\sum }\frac{\wedge \left(n\right)}{n}}=\log x-Y+O\left(x^{-\frac{1}{2}}{\log }^{2}x\right)$.

2) ${\displaystyle \underset{1\le n\le x}{\sum }\frac{\log p}{p-1}}=\log x-Y+O\left(x^{-\frac{1}{2}}{\log }^{2}x\right)$.

Proof: using the “(1)” process in “Theorem 2” is the same as the proof of “Theorem 3”.

Corollary 1: Let N be even,

$\omega \left(p\right)=\frac{P}{P-1},\left(P,N\right)=1$,

the

$\begin{array}{l}{\displaystyle \underset{\omega \le p<z}{\sum }\frac{\omega \left(p\right)\log p}{p}}\\ =\{\begin{array}{l}\log \left(\frac{z}{\omega }\right)+O\left(\log \log N\right);\text{when}\omega \le \frac{\log N}{\log \log N}\\ \log \left(\frac{z}{\omega }\right)+O\left(\frac{\log N}{\omega }\right);\text{when}\frac{\log N}{\log \log N}<\omega \le \log N\exp \left(c_1\left({\log }^{\frac{2}{5}}\omega {\left(\log \log \omega \right)}^{-\frac{1}{5}}\right)\right)\\ \log \left(\frac{z}{\omega }\right)+O\left(\exp \left(-c_1{\log }^{\frac{1}{5}}\omega {\left(\log \log \omega \right)}^{-\frac{1}{5}}\right)\right);\text{when}\log N<\omega \exp \left(-c_1{\log }^{\frac{3}{5}}\omega {\left(\log \log \omega \right)}^{-\frac{1}{5}}\right)\end{array}\end{array}$

Prove: Club meets the condition $1<\omega \le p<z$, $p|N$ In the “p” If the number of is k₀, then we know:

$N=p_1^{l_1}{p}_2^{l_2}\cdots p_{k_0}^{l_{k_0}}\Rightarrow k_0\le k={\displaystyle \underset{m=1}{\overset{k_0}{\sum }}{l}_m}\Rightarrow$

${\omega }^k\le N\Rightarrow k\le \frac{\log N}{\log \omega }$,

${\displaystyle \underset{\begin{array}{l}\omega \le p<Z\\ p|N\end{array}}{\sum }\frac{\log p}{p-1}}\ll {\displaystyle \underset{\begin{array}{l}\omega \le p<z\\ p|N\end{array}}{\sum }\frac{\log \omega }{\omega }}\le k\frac{\log \omega }{\omega }$,

$\Rightarrow {\displaystyle \underset{\begin{array}{l}\omega \le p<z\\ p|N\end{array}}{\sum }\frac{\log p}{p-1}}\ll \frac{\log N}{\omega }$.

Easy to prove:

${\displaystyle \underset{\begin{array}{l}\omega \le p<z\\ p|N\end{array}}{\sum }\frac{\log p}{p-1}}\ll \log \log N\Rightarrow {\displaystyle \underset{\begin{array}{l}\omega \le p<z\\ p|N\end{array}}{\sum }\frac{\log P}{p-1}}\ll \min \left(\frac{\log N}{\omega };\log \log N\right)$,

${\displaystyle \underset{\omega \le p<z}{\sum }\frac{\omega \left(p\right)\log p}{P}}={\displaystyle \underset{\omega \le P<z}{\sum }\frac{\log P}{P-1}}-{\displaystyle \underset{\begin{array}{l}\omega \le p<z\\ p|N\end{array}}{\sum }\frac{\log p}{p-1}}$.

Theorem 3

Conclusion.

Note: The result we have obtained is actually an improvement on an important result cited in the screening method to prove goldbach’s conjecture. Of course, there are several related formulas that can also be improved, which will not be described here.

4. The Application of Summation Formula in “Dirichlet” Function [5] [6]

Theorem 4: If the ${\displaystyle \underset{k\le x}{\sum }d\left(k\right)}=x\left(\log x+2y-1\right)+O\left(x^{\theta }\right)\left(\frac{1}{4}\le \theta <\frac{1}{3}\right)$, the

1) ${\displaystyle \underset{k\le x}{\sum }\frac{d\left(k\right)}{k}}=\frac{1}{2}{\log }^{2}x+2Y\log x+c_4+O\left(x^{\theta -1}\right)$ (c₄ For a constant),

2) ${\displaystyle \underset{k\le x}{\sum }\left\{\frac{x}{k}\right\}}=\left(1-Y\right)x+O\left(x^{\theta }\right)$.

Prove: 1) make $A\left(x\right)={\displaystyle \underset{n\le x}{\sum }d}\left(n\right)$, $g\left(x\right)=x\left(\log x+2Y-1\right)$, $g_0\left(x\right)=x^{\theta }$, $f\left(x\right)=\frac{1}{x}$, $f^{\prime }\left(x\right)=-\frac{1}{x^2}$, It is easy to prove that it satisfies “(3) type” All conditions of, by “(3) type” Have to:

$\begin{array}{l}{\displaystyle \underset{1\le k\le x}{\sum }\frac{d\left(k\right)}{k}}=1+{\displaystyle \underset{1<k\le x}{\sum }\frac{d\left(k\right)}{k}}\\ =1+\left(\log x+2Y-1\right)+{\displaystyle \underset{1}{\overset{x}{\int }}\frac{\log t+2Y-1}{t}\text{d}t}-1\\ +{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{{\displaystyle \underset{1\le k\le t}{\sum }d\left(k\right)}-t\left(\log t+2Y-1\right)}{t^2}\text{d}t}+O\left({\displaystyle \underset{x}{\overset{\infty }{\int }}{t}^{\theta -2}\text{d}t}+x^{\theta -1}\right)\end{array}$

$=\frac{1}{2}{\text{log}}^{\text{2}}x+2Y\log x-1+2Y+{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{{\displaystyle \underset{1\le k\le t}{\sum }d\left(k\right)}-t\left(\log t+2Y-1\right)}{t^2}\text{d}t}+O\left(x^{\theta -1}\right)$

$=\frac{1}{2}{\text{log}}^{2}x+2Y\log x+\left({\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{{\displaystyle \underset{1\le k\le t}{\sum }d\left(k\right)}-t\left(\log t+2Y-1\right)}{t^2}\text{d}t}+2Y-1\right)+O\left(x^{\theta -1}\right)$

The constant $c_4={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{{\displaystyle \underset{1\le k\le t}{\sum }d\left(k\right)}-t\left(\log t+2Y-1\right)}{t^2}\text{d}t}+2Y-1$.

2) ${\displaystyle \underset{k\le x}{\sum }\left\{\frac{x}{k}\right\}}=x{\displaystyle \underset{k\le x}{\sum }\frac{1}{k}}-{\displaystyle \underset{k\le x}{\sum }d(\; k\; )}$

$=x\left(\log x+Y+O\left(\frac{1}{x}\right)\right)-\left(x\left(\log x+2Y-1\right)+O\left(x^{\theta }\right)\right)$

$=\left(1-Y\right)x+O(\; x\; \theta \; )$

Theorem 5:

If the ${\displaystyle \underset{k\le x}{\sum }d\left(k\right)}=x\left(\log x+2Y-1\right)+\Delta \left(x\right)$, the:

1) $\begin{array}{l}{\displaystyle \underset{k=1}{\overset{n}{\sum }}\frac{d\left(k\right)}{k}}=\frac{1}{2}{\log }^{2}n+2Y\log n+\left(Y^2-2{\displaystyle \underset{1}{\overset{\infty }{\int }}\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}}t\right)\\ +{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1-2\left\{\frac{n}{k}\right\}}{n}}+O\left(\frac{\log n}{n}\right)\end{array}$,

The constant $c_4=Y^2-2{\displaystyle \underset{1}{\overset{\infty }{\int }}\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}$.

2) ${\displaystyle \underset{1}{\overset{x}{\int }}\frac{\Delta \left(t\right)}{t^2}\text{d}t}={\left(1-Y\right)}^2-2{\displaystyle \underset{1}{\overset{\infty }{\int }}\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}+O\left(\frac{\log x}{x}\right)$,

3) ${\displaystyle \underset{1}{\overset{x}{\int }}\Delta \left(t\right)\text{d}t}=O\left(x\log x\right)$,

Prove:

1) ${\displaystyle \underset{k=1}{\overset{n}{\sum }}\frac{d\left(k\right)}{k}}={\displaystyle \underset{k=1}{\overset{n}{\sum }}\frac{1}{k}}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{\left[\frac{n}{k}\right]}\right)$

$=2{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1}{k}}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{\left[\frac{n}{k}\right]}\right)-{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1}{k}}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{\left[\sqrt{n}\right]}\right)$

$=2{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1}{k}}\left(\text{log}\left[\frac{n}{k}\right]+Y+\frac{1}{2\left[\frac{n}{k}\right]}+O\left(\frac{k^2}{n^2}\right)\right)-{\left({\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1}{k}}\right)}^2$

$=2{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1}{k}}\left(\log \left(\frac{n}{k}\right)+Y-\frac{\left\{\frac{n}{k}\right\}}{\frac{n}{k}}+\frac{1}{2\left(n/k\right)}+O\left(\frac{k^2}{n^2}\right)\right)-{\left({\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1}{k}}\right)}^2$

$=\left(2\log n+2Y-{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1}{k}}\right)\cdot {\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1}{k}}-2{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{\log k}{k}}+{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1-2\left\{\frac{n}{k}\right\}}{n}}+O(\; 1\; n\; )$

$\begin{array}{l}=\left(\frac{3}{2}\log n+Y+\frac{\left\{\sqrt{n}\right\}-\frac{1}{2}}{\sqrt{n}}+O\left(\frac{1}{n}\right)\right)\left(\frac{1}{2}\log n-\frac{\left\{\sqrt{n}\right\}-\frac{1}{2}}{\sqrt{n}}+Y+O\left(\frac{1}{n}\right)\right)\\ -2{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{\log k}{k}}+{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1-2\left\{\frac{n}{k}\right\}}{n}}+O(\; 1\; n\; )\end{array}$

$\begin{array}{l}=\left(\frac{3}{2}\log n+Y+\frac{\left\{\sqrt{n}\right\}-\frac{1}{2}}{\sqrt{n}}\right)\cdot \left(\frac{1}{2}\log n+Y-\frac{\left\{\sqrt{n}\right\}-\frac{1}{2}}{\sqrt{n}}\right)\\ -2{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{\log k}{k}}+{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1-2\left\{\frac{n}{k}\right\}}{n}}+O\left(\frac{\log n}{n}\right)\end{array}$

$\begin{array}{l}=\frac{3}{4}{\log }^{2}n+2Y\log n+Y^2-\frac{\left(\left\{\sqrt{n}\right\}-\frac{1}{2}\right)\log n}{\sqrt{n}}+{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1-2\left\{\frac{n}{k}\right\}}{n}}\\ -2{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{\log k}{k}}+O\left(\frac{\log k}{n}\right)\end{array}$ (7)

while

${\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{\log k}{k}}={\displaystyle {\int }_1^{\left[\sqrt{n}\right]+1}\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}+{\displaystyle {\int }_1^{\left[\sqrt{n}\right]+1}\frac{\log t}{t}\text{d}t}$

$=\frac{1}{2}{\log }^2\left(\left[\sqrt{n}\right]+1\right)+{\displaystyle {\int }_1^{\left[\sqrt{n}\right]+1}\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}$

$=\frac{1}{2}{\log }^2\left(\left[\sqrt{n}\right]+1\right)+{\displaystyle {\int }_1^{\infty }\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}-{\displaystyle {\int }_{\left[\sqrt{n}\right]+1}^{\infty }\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}$ (8)

1) $\frac{1}{2}{\log }^2\left(\left[\sqrt{n}\right]+1\right)=\frac{1}{2}{\left(\log \sqrt{n}+\frac{1-\left\{\sqrt{n}\right\}}{\sqrt{n}}+O\left(\frac{1}{n}\right)\right)}^2$

$=\frac{1}{8}{\log }^{2}n+\frac{\left(1-\left\{\sqrt{n}\right\}\right)\log \sqrt{n}}{\sqrt{n}}+O\left(\frac{\log n}{n}\right)$

2) ${\displaystyle {\int }_1^{\infty }\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}\ll {\displaystyle {\int }_1^{\infty }\left(\frac{\log t}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}<{\displaystyle {\int }_1^{\infty }\frac{\log t}{\left[t\right]t}\text{d}t}\ll {\displaystyle {\int }_1^{\infty }\frac{\log t}{t^2}\text{d}t}\ll 1$

$\Rightarrow {\displaystyle {\int }_1^{\infty }\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}$

For constant

3) ${\displaystyle {\int }_{\left\{\sqrt{n}\right\}+1}^{\infty }\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}={\displaystyle {\int }_{\left\{\sqrt{n}\right\}+1}^{\infty }\left(\frac{\log t-\frac{\left\{t\right\}}{t}+O\left(\frac{1}{t^2}\right)}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}$

$={\displaystyle {\int }_{\left\{\sqrt{n}\right\}+1}^{\infty }\left(\frac{\left\{t\right\}\left(\log t-1\right)}{\left[t\right]t}+O\left(\frac{1}{t^3}\right)\right)\text{d}t}$

$={\displaystyle {\int }_{\sqrt{n}}^{\infty }\frac{\left\{t\right\}\left(\log t-1\right)}{\left[t\right]t}\text{d}t}+O\left(\frac{\log n}{n}\right)$

$={\displaystyle {\int }_{\sqrt{n}}^{\infty }\frac{\left\{t\right\}\left(\log t-1\right)}{t^2}\text{d}t}+O\left(\frac{\log n}{n}\right)$

$=\frac{1}{2}{\displaystyle {\int }_{\sqrt{n}}^{\infty }\frac{\log t-1}{t^2}\text{d}t}+O\left(\frac{\log n}{n}\right)$

$=\frac{\log \sqrt{n}}{2\sqrt{n}}+O\left(\frac{\log n}{n}\right)$

will (1), (2), (3) Can be substituted into Equation (8), available:

${\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{\log k}{k}}=\frac{1}{8}{\text{log}}^{2}n+{\displaystyle {\int }_1^{\infty }\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}+\frac{\left(\frac{1}{2}-\left\{\sqrt{n}\right\}\right)\log \sqrt{n}}{\sqrt{n}}+O\left(\frac{\log n}{n}\right)$

Then substitute the above formula into “Formula (7)” and sort it out

$\begin{array}{l}{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{d\left(k\right)}{k}}=\frac{1}{2}{\text{log}}^{2}n+2Y\log n+\left(Y^2-2{\displaystyle {\int }_1^{\infty }\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}\right)\\ +{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1-2\left\{\frac{n}{k}\right\}}{n}}+O\left(\frac{\log n}{n}\right)\end{array}$ (9)

2) By

${\displaystyle \underset{k=1}{\overset{n}{\sum }}d\left(k\right)}=n\left(\log n+2Y-1\right)+\sqrt{n}-2{\displaystyle \underset{1\le k\le \sqrt{n}}{\sum }\left\{\frac{x}{n}\right\}}+O(\; 1\; )$

$=n\left(\log n+2Y-1\right)+{\displaystyle \underset{1\le k\le \sqrt{n}}{\sum }\left(1-2\left\{\frac{x}{n}\right\}\right)}+O(\; 1\; )$

$={\displaystyle \underset{k=1}{\overset{n}{\sum }}\frac{d\left(k\right)}{n}}=\log n+2Y-1+{\displaystyle \underset{1\le k\le \sqrt{n}}{\sum }\frac{1-2\left\{\frac{n}{k}\right\}}{n}}+O\left(\frac{1}{n}\right)$ (10)

From “Formula (2)” in “Theorem A” $\Rightarrow {\displaystyle \underset{k=1}{\overset{n}{\sum }}\frac{d\left(k\right)}{k}}-{\displaystyle \underset{k=1}{\overset{n}{\sum }}\frac{d\left(k\right)}{n}}={\displaystyle {\int }_1^n\frac{{\displaystyle \underset{1\le k\le t}{\sum }d\left(k\right)}}{t^2}\text{d}t}$.

(9)

(10)

$\begin{array}{l}\Rightarrow {\displaystyle {\int }_1^n\frac{{\displaystyle \underset{1\le k\le t}{\sum }d\left(k\right)}}{t^2}}=(\frac{1}{2}{\text{log}}^{2}n+2Y\log n+Y^2-2{\displaystyle {\int }_1^{\infty }\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}\begin{array}{c}\stackrel{}{}\\ \\ \end{array}\\ +{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1-2\left\{\frac{n}{k}\right\}}{n}}+O\left(\frac{\log n}{n}\right))-\left(\log n+2Y-1+{\displaystyle \underset{k=1}{\overset{\sqrt{n}}{\sum }}\frac{1-2\left\{\frac{n}{k}\right\}}{n}}+O\left(\frac{1}{n}\right)\right)\end{array}$

$\begin{array}{l}\Rightarrow {\displaystyle {\int }_1^n\frac{{\displaystyle \underset{1\le k\le t}{\sum }d\left(k\right)}}{t^2}\text{d}t}=\frac{1}{2}{\text{log}}^{2}n+\left(2Y-1\right)\log n+{\left(1-Y\right)}^2\\ -2{\displaystyle {\int }_1^{\infty }\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}+O\left(\frac{\log n}{n}\right)\end{array}$

$\Rightarrow {\displaystyle {\int }_1^n\frac{\Delta \left(t\right)}{t^2}\text{d}t}={\displaystyle {\int }_1^n\frac{{\displaystyle \underset{1\le k\le t}{\sum }d\left(k\right)-t\left(\log t+2Y-1\right)}}{t^2}\text{d}t}$

$={\left(1-Y\right)}^2-2{\displaystyle {\int }_1^{\infty }\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}+O\left(\frac{\log n}{n}\right)$.

namely,

${\displaystyle {\int }_1^n\frac{\Delta \left(t\right)}{t^2}\text{d}t}={\left(1-Y\right)}^2-2{\displaystyle {\int }_1^{\infty }\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}+O\left(\frac{\log n}{n}\right)$ (11)

3) (1) by

${\displaystyle \underset{1\le t\le x}{\sum }\frac{\Delta \left(t\right)}{t^2}}-{\displaystyle {\int }_1^x\frac{\Delta \left(t\right)}{t^2}\text{d}t}={\displaystyle {\int }_1^{\infty }\left(\frac{\Delta \left(\left[t\right]\right)}{{\left[t\right]}^2}-\frac{\Delta \left(t\right)}{t^2}\right)\text{d}t}-{\displaystyle {\int }_x^{\infty }\left(\frac{\Delta \left(\left[t\right]\right)}{{\left[t\right]}^2}-\frac{\Delta \left(t\right)}{t^2}\right)\text{d}t}+O(\; 1\; x\; )$

$\begin{array}{l}={\displaystyle {\int }_1^{\infty }\left(\frac{\Delta \left(\left[t\right]\right)}{{\left[t\right]}^2}-\frac{\Delta \left(t\right)}{t^2}\right)\text{d}t}-{\displaystyle {\int }_x^{\infty }(\left(\frac{1}{{\left[t\right]}^2}-\frac{1}{t^2}\right)\cdot {\displaystyle \underset{k\le t}{\sum }d\left(k\right)}-(\frac{\left[t\right]\left(\log \left[t\right]+2Y-1\right)}{{\left[t\right]}^2}}\\ -\frac{t\left(\log t+2Y-1\right)}{t^2}))\text{d}t+O(\; 1\; x\; )\end{array}$

$={\displaystyle {\int }_1^{\infty }\left(\frac{\Delta \left(\left[t\right]\right)}{{\left[t\right]}^2}-\frac{\Delta \left(t\right)}{t^2}\right)\text{d}t}+O\left({\displaystyle {\int }_x^{\infty }\left(t\log t\cdot \frac{1}{t^3}+\frac{\log t}{t^2}\right)\text{d}t}\right)+O(\; 1\; x\; )$

$={\displaystyle {\int }_1^{\infty }\left(\frac{\Delta \left(\left[t\right]\right)}{{\left[t\right]}^2}-\frac{\Delta \left(t\right)}{t^2}\right)\text{d}t}+O\left(\frac{\log x}{x}\right)$

$\Rightarrow {\displaystyle \underset{1\le t\le x}{\sum }\frac{\Delta \left(t\right)}{t^2}}={\displaystyle {\int }_1^x\frac{\Delta \left(t\right)}{t^2}\text{d}t}+{\displaystyle {\int }_1^{\infty }\left(\frac{\Delta \left(\left[t\right]\right)}{{\left[t\right]}^2}-\frac{\Delta \left(t\right)}{t^2}\right)\text{d}t}+O\left(\frac{\log x}{x}\right)$ (12)

2) ${\displaystyle \underset{1\le t\le x}{\sum }\Delta \left(t\right)}-{\displaystyle {\int }_1^x\Delta }\left(t\right)\text{d}t={\displaystyle {\int }_1^x\left(\Delta \left(\left[t\right]\right)-\Delta \left(t\right)\right)\text{d}t}+O\left(\Delta (\; x\; )\right)$

$={\displaystyle {\int }_1^x\left(t\left(\log t+2Y-1\right)-\left[t\right]\left(\log \left[t\right]+2Y-1\right)\right)\text{d}t}+O\left(x\right)\ll x\log x$

$\Rightarrow {\displaystyle {\int }_1^x\Delta }\left(t\right)\text{d}t={\displaystyle \underset{1\le t\le x}{\sum }\Delta }\left(t\right)+O\left(x\log x\right)$

$={\displaystyle {\int }_1^x{t}^2\text{d}}\left({\displaystyle \underset{1\le m\le t}{\sum }\frac{\Delta \left(m\right)}{m^2}}\right)+O\left(x\log x\right)$.

(12)

$\begin{array}{l}\Rightarrow {\displaystyle {\int }_1^x\Delta \left(t\right)\text{d}t}={\displaystyle {\int }_1^x{t}^2\text{d}}\left({\displaystyle {\int }_1^t\frac{\Delta \left(m\right)}{m^2}\text{d}m}+{\displaystyle {\int }_1^{\infty }\left(\frac{\Delta \left(\left[m\right]\right)}{{\left[m\right]}^2}-\frac{\Delta \left(m\right)}{m^2}\right)\text{d}m}+O\left(\frac{\log t}{t}\right)\right)\\ +O\left(x\log x\right)\end{array}$

$={\displaystyle {\int }_1^x{t}^2\text{d}}\left({\displaystyle {\int }_1^t\frac{\Delta \left(m\right)}{m^2}\text{d}m}+O\left(\frac{\log t}{t}\right)\right)+O\left(x\log x\right)$.

(11)

$\Rightarrow {\displaystyle {\int }_1^x\Delta \left(t\right)\text{d}t}={\displaystyle {\int }_1^x{t}^2}\text{d}\left({\left(1-Y\right)}^2-2{\displaystyle {\int }_1^{\infty }\left(\frac{\log \left[t\right]}{\left[t\right]}-\frac{\log t}{t}\right)\text{d}t}+O\left(\frac{\log t}{t}\right)\right)+O\left(x\log x\right)$

$={\displaystyle {\int }_1^x{t}^2}\text{d}\left(O\left(\frac{\log t}{t}\right)\right)+O\left(x\log x\right)$.

$\ll x\log x$

Type ${\displaystyle {\int }_1^x\Delta \left(t\right)\text{d}t}=O\left(x\log x\right)$.

Note: 1) This paper proves that ${\displaystyle {\int }_1^x\Delta \left(t\right)}=O\left(x\log x\right)$, while ${\displaystyle {\int }_1^x{\Delta }^2\left(t\right)\text{d}t}=c^{\prime }{x}^{\frac{3}{2}}+O\left(x{\log }^{5}x\right)$ (Dong Guangchang, 1956) [7], if you combine these two equations, you can understand it better “ $\Delta \left(t\right)$ ”. Some variation rules of.

2) We can further prove this in other ways, ${\displaystyle \underset{1\le t\le x}{\sum }\Delta \left(t\right)}=\frac{1}{2}x\log x+O\left(x\right)$, and then get ${\displaystyle {\int }_1^x\Delta \left(t\right)\text{d}t}=O\left(x\right)$, I won’t go into details here for lack of space.

Theorem 6: The function w(n) (or Ω(n)) improvement of relevant conclusions [8] [9]:

1) ${\displaystyle \underset{n\le x}{\sum }w\left(n\right)}=x\log \log +A_{2}x-\left(1-Y\right)\frac{x}{\log x}+O\left(\frac{x}{{\log }^{2}x}\right)$.

2) ${\displaystyle \underset{n\le x}{\sum }{w}^2}\left(n\right)=x{\left(\log \log x\right)}^2+\left(1+2A_2\right)x\log \log x+O\left(x\right)$.

3) ${\displaystyle \underset{n\le x}{\sum }{\left(w\left(n\right)-\log \log x\right)}^2}=x\log \log x+O\left(x\right)$.

Prove: 1)

${\displaystyle \underset{n\le x}{\sum }w}\left(n\right)={\displaystyle \underset{p\le x}{\sum }\left[\frac{x}{p}\right]}=\left[\frac{x}{2}\right]+{\displaystyle \underset{2<p\le x}{\sum }\left[\frac{x}{p}\right]}$

$=\left[\frac{x}{2}\right]+{\displaystyle \underset{2}{\overset{x}{\int }}\frac{1}{\log t}\text{d}\left({\displaystyle \underset{p\le t}{\sum }\left[\frac{x}{p}\right]}\log p\right)}$

$=\left[\frac{x}{2}\right]+\frac{{\displaystyle \underset{p\le t}{\sum }\left[\frac{x}{p}\right]\log p}}{\log t}|{}_2^x+{\displaystyle \underset{2}{\overset{x}{\int }}\frac{{\displaystyle \underset{p\le t}{\sum }\left[\frac{x}{p}\right]\log p}}{t{\log }^{2}t}\text{d}t}$

$=\frac{{\displaystyle \underset{p\le x}{\sum }\left[\frac{x}{p}\right]\log p}}{\log x}+{\displaystyle \underset{2}{\overset{x}{\int }}\frac{{\displaystyle \underset{p\le t}{\sum }\frac{x}{p}\log p+O\left({\displaystyle \underset{p\le t}{\sum }\log p}\right)}}{t{\log }^{2}t}\text{d}t}$

$=x{\displaystyle \underset{2}{\overset{x}{\int }}\frac{\log t+A_1}{t{\log }^{2}t}\text{d}t}+x{\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{{\displaystyle \underset{p\le t}{\sum }\frac{\log p}{p}}-\left(\log t+A_1\right)}{t{\log }^{2}t}\text{d}t}$

$-x{\displaystyle \underset{x}{\overset{\infty }{\int }}\frac{{\displaystyle \underset{p\le t}{\sum }\frac{\log p}{p}-\left(\log t+A_1\right)}}{t{\log }^{2}t}\text{d}t}+\frac{{\displaystyle \underset{p\le x}{\sum }\left[\frac{x}{p}\right]\log p}}{\log x}+O\left(\frac{x}{{\log }^{2}x}\right)$ (13)

1) Theorem two is easy to prove:

${\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{{\displaystyle \underset{p\le t}{\sum }\frac{\log p}{p}}-\left(\log t+A_1\right)}{t{\log }^{2}t}\text{d}t}$,

For constant

${\displaystyle \underset{x}{\overset{\infty }{\int }}\frac{{\displaystyle \underset{p\le t}{\sum }\frac{\log p}{p}-\left(\log t+A_1\right)}}{t{\log }^{2}t}\text{d}t}=O\left(\frac{1}{{\log }^{2}x}\right)$.

2) By “(6) type” Easy card:

${\displaystyle \underset{p\le x}{\sum }\left[\frac{x}{p}\right]}\log p=x\log x-\left(1+{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}\right)x+O\left(\sqrt{x}\right)$.

3) Theorem two: $A_1=-Y-{\displaystyle \underset{p}{\sum }\frac{\log p}{p\left(p-1\right)}}$.

4) By

${\displaystyle \underset{p\le x}{\sum }\left[\frac{x}{p}\right]}=x{\displaystyle \underset{p\le x}{\sum }\frac{1}{p}}+O\left(\pi \left(x\right)\right)$.

Theorem 2

$\Rightarrow {\displaystyle \underset{p\le x}{\sum }w\left(n\right)}=x\log \log x+A_{2}X+O\left(\frac{x}{\log x}\right)$.

Will (1), (2), (3) (4) Can be substituted into equation (13), available:

Will

${\displaystyle \underset{n\le x}{\sum }w\left(n\right)}=x\log \log x+A_{2}x+\left(Y-1\right)\frac{x}{\log x}+O\left(\frac{x}{{\log }^{2}x}\right)$ (14)

2) $w\left(n\right)\left(w\left(n\right)-1\right)={\displaystyle \underset{pq|n}{\sum }1-{\displaystyle \underset{p^2|n}{\sum }1}}$

$\Rightarrow {\displaystyle \underset{n\le x}{\sum }{w}^2}\left(n\right)-{\displaystyle \underset{n\le x}{\sum }w\left(n\right)}={\displaystyle \underset{pq\le x}{\sum }\left[\frac{x}{pq}\right]}-{\displaystyle \underset{p^2\le x}{\sum }\left[\frac{x}{p^2}\right]}$

$={\displaystyle \underset{pq\le x}{\sum }\frac{x}{pq}}+O(\; x\; )$

$=2x{\displaystyle \underset{\begin{array}{l}pq\le x\\ p\le \sqrt{x}\end{array}}{\sum }\frac{1}{pq}}-x\left({\displaystyle \underset{\begin{array}{l}p\le \sqrt{x}\\ q\le \sqrt{x}\end{array}}{\sum }\frac{1}{p}}\right)+O(\; x\; )$

$=2x{\displaystyle \underset{p\le \sqrt{x}}{\sum }\frac{1}{p}}{\displaystyle \underset{q\le \frac{x}{p}}{\sum }\frac{1}{q}}-x{\left({\displaystyle \underset{p\le \sqrt{x}}{\sum }\frac{1}{p}}\right)}^2+O(\; x\; )$

Theorem 2

$\begin{array}{l}\Rightarrow {\displaystyle \underset{n\le x}{\sum }{w}^2}\left(n\right)-{\displaystyle \underset{n\le x}{\sum }w}\left(n\right)=2x{\displaystyle \underset{p\le \sqrt{x}}{\sum }\frac{\log \log \frac{x}{p}}{p}}-x{\left(\log \log x\right)}^2\\ +2\log 2\left(x\log \log x\right)+O(\; x\; )\end{array}$

$\underset{p\le \sqrt{x}}{\log \log \frac{x}{p}}=\log \log x+\log \left(1-\frac{\log p}{\log x}\right)=\log \log x+O\left(\frac{\log p}{\log x}\right)$

Theorem 2

(14)

$\Rightarrow {\displaystyle \underset{n\le x}{\sum }{w}^2\left(n\right)}=x{\left(\log \log x\right)}^2+\left(1+2A_1\right)x\log \log x+O(\; x\; )$

3) By (1), (2) The conclusions obtained are easy to prove:

${\displaystyle \underset{n\le x}{\sum }{\left(w\left(n\right)-\log \log x\right)}^2}=x\log \log x+O(\; x\; )$

Note: 1) The derivation process of “Ω(n)” function related conclusion is similar to “W(n)”.

2) In this paper, “ ${\displaystyle \underset{n\le x}{\sum }{\left(w\left(n\right)-\log \log x\right)}^2}=x\log \log x+O\left(x\right)$ ” slightly better than “Hardy-Ramanujan”, the results in.

3) With the o “ ${\displaystyle \underset{n\le x}{\sum }{w}^2}\left(n\right)$ ” The similarity method of values will be easy “Selber The formula” expressed in the following form:

$\begin{array}{l}\theta \left(x\right)\log x+{\displaystyle \underset{p\le x}{\sum }\theta \left(\frac{x}{p}\right)}\log p\\ =2x\log x+\left(2A_1-1\right)x+O\left(x\exp \left(-c{\log }^{\frac{3}{5}}x{\left(\log \log x\right)}^{-\frac{1}{5}}\right)\right)\end{array}$

Theorem of seven: if $\phi \left(n\right)$ Is euler function, then:

1) $\underset{n\to \infty }{\lim \inf }\phi \left(n\right)=\frac{n}{{\text{e}}^Y\log \log n}+O\left(n\exp \left(-c_5{\left(\log \log n\right)}^{\frac{3}{5}}{\left(\log \log \log n\right)}^{-\frac{1}{5}}\right)\right)$,

Among them “ $c_5$ ” For the normal number.

2) If “Riemnn” If the guess is true, then

$\underset{n\to \infty }{\lim \inf }\phi \left(n\right)=\frac{n}{{\text{e}}^Y\log \log n}+O\left(n{\log }^{-\frac{1}{2}}n\right)$.

Prove: This conclusion mainly uses “theorem 2 (Theorem 2’)”, the proof process is relatively simple, so skip [10].

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

References

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