An Asymptotic Formula for the Chebyshev Theta Function

TL;DR

This paper derives an asymptotic formula for the Chebyshev theta function evaluated at prime indices, providing bounds and an approximation involving prime logarithms, advancing understanding of prime distribution asymptotics.

Contribution

It introduces a new asymptotic formula for the Chebyshev theta function at prime indices, refining previous bounds and approximations.

Findings

Derived bounds for $ heta(p_n)/n$ using $ ext{log } p_{n+1}$

Established an asymptotic expression for $ heta(p_n)/n$ involving $ ext{log } p_{n+1}$

Provided insights into the distribution of primes through asymptotic analysis.

Abstract

Let $\{p_n\}_{n\ge 1}$ be the sequence of primes and $\vartheta(x) = \sum_{p \leq x} \log p$, where $p$ runs over the primes not exceeding $x$, be the Chebyshev $\vartheta$-function. In this note we derive lower and upper bounds for $\vartheta(p_n)/n$ by comparing it with $\log p_{n+1}$ and deduce that $\vartheta(p_n)/n=\log p_{n+1}\left(1-\frac{1}{\log n}+\frac{\log\log n}{\log^2 n}\left(1+o(1)\right)\right).$

Equations

$$\frac{\vartheta (p_n)}{\log p_{n+1}}>n-\pi (n)(n\ge 2),$$

$$\frac{\vartheta(p_{n})}{\log p_{n+1}}>n-\pi(n)\Big{(}1-\frac{1}{\log n}\Big{)}\quad(n\geq 101).$$

$$1+\frac{1}{\log p_n}+\frac{2.7}{{\log }^2{p}_n}<\log p_n-\frac{\vartheta (p_n)}{n}<1+\frac{1}{\log p_n}+\frac{3.84}{{\log }^2{p}_n},$$

$$\displaystyle\frac{\vartheta(p_{n})}{n}=\log p_{n}-1-\frac{1}{\log p_{n}}+\Theta\Big{(}\frac{1}{\log^{2}p_{n}}\Big{)}.$$

$$n\Big{(}1-\frac{1}{\log n}+\frac{\log\log n}{4\log^{2}n}\Big{)}\leq\frac{\vartheta(p_{n})}{\log p_{n+1}}\leq n\Big{(}1-\frac{1}{\log n}+\frac{\log\log n}{\log^{2}n}\Big{)}.$$

$$n\Big{(}1-\frac{1}{\log n}+(1-\varepsilon)\frac{\log\log n}{\log^{2}n}\Big{)}\leq\frac{\vartheta(p_{n})}{\log p_{n+1}}\leq n\Big{(}1-\frac{1}{\log n}+\frac{\log\log n}{\log^{2}n}\Big{)}.$$

$$\frac{\vartheta (p_n)}{n}\ge G(n,2.1454),$$

$$\frac{\vartheta (p_n)}{n}\le G(n,2).$$

$$p_n\le n(\log n+\log \log n-0.8),$$

$$p_n\ge n(\log n+\log \log n-1).$$

$$p_n\le n(\log n+\log \log n-0.9385)$$

$$\mathcal{F}(n,0.25)\log p_{n+1}\le \vartheta (p_n)/n\le \mathcal{F}(n,1)\log p_{n+1}$$

$$\mathcal{F}\left(n,1-\epsilon \right)\log p_{n+1}\le \vartheta (p_n)/n\le \mathcal{F}(n,1)\log p_{n+1}.$$

$$\log p_{n+1}<\log n+\log \log n+\frac{\log \log n-0.8+0.018}{\log n}=U(n),$$

$$\log p_{n+1}>\log n+\log \log n+\frac{\log \log n-1}{\log n+0.5(\log \log n-1)}=V(n).$$

$$\frac{1}{x+0.4}>\log \left(1+\frac{1}{x}\right)>\frac{1}{x+0.5}.$$

$$\log p_{n+1}\le \log n+1+\log (\log n+1+\log \log n+1-0.8).$$

$$\log\log(n+1)<\log\log n+\log\Big{(}1+\frac{1}{(n+0.4)\log n}\Big{)}<\log\log n+\frac{1}{(n+0.4)\log n}.$$

$$\log p_{n+1}<\log n+\log \log n+\frac{\log \log n-0.8}{\log n}+\frac{1}{\log n}\cdot \frac{\log n+1+1/\log n}{n+0.4}.$$

$$\log p_{n+1}\ge \log (n+1)+\log (\log (n+1)+\log \log (n+1)-1).$$

$$\displaystyle\log\log(n+1)-\log\log n>\log\Big{(}1+\frac{1}{(n+0.5)\log n}\Big{)}>\frac{1}{(n+0.5)\log n+0.5}.$$

$$\log p_{n+1}$$

$$\displaystyle>\log n+\log\log n+\log\Big{(}1+\frac{\log\log n-1}{\log n}\Big{)}.$$

$$G(n,2.1454)\ge \mathcal{F}(n,0.25)\cdot U(n),$$

$$G(n,2)\le \mathcal{F}(n,1)\cdot V(n).$$

$$x+\log x-1+\frac{\log x-2.1454}{x}\geq\Big{(}1-\frac{1}{x}+\frac{\log x}{4x^{2}}\Big{)}\Big{(}x+\log x+\frac{\log x-0.8+0.018}{x}\Big{)},$$

$$\Big{(}\frac{3}{4}\log x+\frac{\log x}{x}\Big{)}+\Big{(}-2.1454-\frac{\log^{2}x}{4x}-\frac{\log^{2}x}{4x^{2}}\Big{)}+(0.8-0.018)\Big{(}1-\frac{1}{x}+\frac{\log x}{4x^{2}}\Big{)}\geq 0.$$

$$x^2+\log x(\log x-1)>\frac{x}{2}\log x(\log x-1)$$

$$\left(1-\frac{1}{x}+\frac{\log x}{x^2}\right)\frac{\log x-1}{x+0.5(\log x-1)}\ge \frac{\log x-2}{x}.$$

$$\frac{{\log }^{2}x}{x^2}+\left(1-\frac{1}{x}+\frac{\log x}{x^2}\right)\frac{\log x-1}{x+0.5(\log x-1)}\ge \frac{\log x-2}{x}$$

$$\vartheta (p_n)/n\ge G(n,2.1454)\ge \mathcal{F}(n,0.25)U(n)>\mathcal{F}(n,0.25)\log p_{n+1}.$$

Full text

1. Introduction

Let $\{p_{n}\}_{n\geq 1}$ be the sequence of the prime numbers and $\vartheta(x)=\sum_{p\leq x}\log p$, where $p$ runs over the primes not exceeding $x$, be the Chebyshev $\vartheta$-function. The type of bounds that we shall discuss here was introduced by Bonse [2], who showed that $\vartheta(p_{n})>2\log p_{n+1}$ holds for every $n\geq 4$ and $\vartheta(p_{n})>3\log p_{n+1}$ holds for every $n\geq 5$. Thereafter, Pósa [8] showed that, given any $k>1,$ there exists $n_{k}$ such that $\vartheta(p_{n})>k\log p_{n+1}$ holds for all $n\geq n_{k}.$ Panaitopol [7] showed that in Pósa’s result we can have $n_{k}=2k$ and also gave the bound

[TABLE]

where $\pi(n)$ is equal to the number of primes less or equal to $n$. Hassani [5] refined Panaitopol’s inequality to the following

[TABLE]

Recently, Axler [1, Propositions 4.1 and 4.5] showed that

[TABLE]

where the left-hand side inequality is valid for every integer $n\geq 218$ and the right-hand side inequality holds for every $n\geq 74004585$. This provides the following asymptotic formula

[TABLE]

For further terms, see Axler [1, Proposition 2.1].

In the present note, we show the following result, which is a refinement of (1.1).

Theorem 1**.**

For all $n\geq 6,$ we have

[TABLE]

The left-hand side inequality also holds for $2\leq n\leq 6.$

We also generalise the left-hand side of (1.2) to have the following result.

Theorem 2**.**

For every $0<\varepsilon<1,$ there exists $n_{\varepsilon}\in\mathbb{N}$ such that for every $n\geq n_{\varepsilon}$ it holds that

[TABLE]

Corollary 1**.**

We have $\displaystyle\frac{\vartheta(p_{n})}{n}=\log p_{n+1}\left(1-\frac{1}{\log n}+\frac{\log\log n}{\log^{2}n}\left(1+o(1)\right)\right).$

2. Preliminaries

Define $G(n,a)=\log n+\log\log n-1+\frac{\log\log n-a}{\log n}$. We shall use the following bounds for $\vartheta(p_{n})/n$.

Lemma 1**.**

For every $n\geq 3$, we have

[TABLE]

and for every $n\geq 198$, we have

[TABLE]

Proof.

The inequality (2.1) is due to Robin [9], and the inequality (2.2) was given by Massias and Robin [6]. ∎

Lemma 2**.**

For every $n\geq 227$, we have

[TABLE]

and for every $n\geq 2$,

[TABLE]

Proof.

For $n\geq 8602$, we have the following stronger bound

[TABLE]

given by Massias and Robin [6]. For $227\leq n\leq 8601$ we verify the inequality (2.3) by direct computation. The inequality (2.4) is due to Dusart [4]. ∎

For the sake of brevity, we shall define $\displaystyle\mathcal{F}(n,\lambda)=1-\frac{1}{\log n}+\lambda\frac{\log\log n}{\log^{2}n}$ and rewrite (1.2) as

[TABLE]

and rewrite (1.3) as

[TABLE]

3. Proof of Theorem 1

The proof of Theorem 1 is split into two lemmas. In the first lemma, we give lower and upper bounds for $\log p_{n+1}.$

Lemma 3**.**

For every $n\geq 140$, we have

[TABLE]

and for every $n\geq 2$, we have

[TABLE]

Proof.

First, we show that for every $x\geq 1$

[TABLE]

In order to prove this, we set $f_{a}(x)=\log(1+x)-\frac{x}{1+ax}$ and note that, $f_{a}^{\prime}(x)=\frac{x(a^{2}x+2a-1)}{(1+x)(1+ax)^{2}}.$ Hence, $f^{\prime}_{0.4}(x)<0$ for every $x\in(0,1.25)$ which yields $f_{0.4}(1/x)<f_{0.4}(0)=0$ for every $x\geq 1.$ On the other hand, $f^{\prime}_{0.5}(x)>0$ for all positive $x$, which gives $f_{0.5}(1/x)>f_{0.5}(0)=0$ for every $x\geq 1.$ This completes the proof of (3.3).

Next, we give a proof of (3.1). By (2.3), we have for $n\geq 227$,

[TABLE]

The left-hand side inequality of (3.3) implies $\displaystyle\log(n+1)<\log n+\frac{1}{n+0.4}.$ Using (3.3) once again, we get

[TABLE]

Applying this to (3.4), we obtain for $n\geq 227$,

[TABLE]

Now, $g(x)=\displaystyle\frac{\log x+1+1/\log x}{x+0.4}$ is a decreasing function for $x\geq 2$ with $g(e^{5.99})\leq 0.018$. Hence $g(x)\leq 0.018$ for every $x\geq 400>e^{5.99}$. Combined with (3.5), it shows out that $\log p_{n+1}<U(n)$ for every $n\geq 400$. For every $140\leq n\leq 399$ we check the inequality (3.1) with a computer. This completes the proof of (3.1).

To prove the inequality (3.2), first note that (2.4) gives for every $n\geq 1$,

[TABLE]

The right-side inequality of (3.3) gives $\displaystyle\log(n+1)>\log n+\frac{1}{n+0.5}.$ Using (3.3) once again, we get, for $n\geq 2,$

[TABLE]

Applying this to (3.6), we arrive at

[TABLE]

Applying (3.3) one more time, we get $\log p_{n+1}>V(n)$ for every $n\geq 2.$∎

Lemma 4**.**

For every $n\geq 396$, we have

[TABLE]

and for every $n\geq 2$, we have

[TABLE]

Here $U(n)$ and $V(n)$ are defined as in Lemma 3.

Proof.

We start with the proof of (3.7). Setting $x=\log n$, the inequality (3.7) can be rewritten as

[TABLE]

which is equivalent to

[TABLE]

The left-hand side is a sum of three increasing functions on the interval $[5.7,\infty)$ and at $x=5.99$ the left-hand side is positive. So the last inequality holds for every $x\geq 5.99$; i.e., for every $n\geq 400$. A direct computation shows that the inequality (3.7) also holds for every $n$ satisfying $396\leq n\leq 399$.

Next, we give a proof of (3.8). It is easy to see that

[TABLE]

for every $x>0$. Now, for $x\geq 1$, the last inequality is seen to be equivalent to

[TABLE]

Since $\frac{\log^{2}x}{x^{2}}\geq 0$ for every $x>0$, we get

[TABLE]

for every $x\geq 1$. Substituting $x=\log n$ in (3.9), we obtain the inequality (3.8) for every integer $n\geq 3$. We can directly check that (3.8) holds for $n=2$ as well.∎

Finally, we give a proof of Theorem 1.

Proof of Theorem 1..

We use (2.1), (3.7) and (3.1) respectively to see that for every $n\geq 396,$

[TABLE]

A direct computation shows that the left-hand side inequality of (2.6) also holds for every integer $n$ with $2\leq n\leq 395$.

In order to prove the right-hand side inequality of (2.6), we combine (2.2), (3.8) and (3.2) respectively to get

[TABLE]

for every $n\geq 198$. For smaller values of $n$, we use a computer. ∎

4. Proof of Theorem 2

The right-hand side of (2.7) has been established already. To show the left-hand side, we start with the following lemma.

Lemma 5**.**

For any $0<\varepsilon<1$, there exists $m_{\varepsilon}\in\mathbb{N}$ such that

[TABLE]

holds for every $n\geq m_{\varepsilon}.$ Here $U(n)$ is defined as in Lemma 3.

Proof.

Fix any $0<\varepsilon<1$. We denote $a=2.1454,b=0.8-0.018$ and set $x=\log n$ to transform the inequality (4.1) into

[TABLE]

This is equivalent to

[TABLE]

Now, the left-hand side is a sum of three functions, each of which is strictly increasing for all sufficiently large $x,$ and the limit of the left-hand side, as $x\to\infty$, is $+\infty.$ Therefore we conclude that the last inequality holds for all sufficiently large $x.$ ∎

Proof of Theorem 2..

For any $0<\varepsilon<1,$ we have $m_{\varepsilon}\in\mathbb{N}$ such that (4.1) holds for every $n\geq m_{\varepsilon}$. We combine this with (2.1) and (3.1) to obtain that for every $n\geq n_{\varepsilon}:=\max\{m_{\varepsilon},140\}$

[TABLE]

This completes the proof.∎

5. Remarks

1. (1)

For every $n\geq 599$, we have

[TABLE]

which was found by Dusart [3]. Using this and a computer, we get

[TABLE]

for every integer $n\geq 83$. Hence, (1.2) is an improvement of (1.1). 2. (2)

The bounds given in (1.2) are particularly useful for comparing $\vartheta(p_{n})/n$ with $\log p_{n+1}.$ To see a numerical example, we use a computer to find that for $n\geq 23$ the relative error in approximating $\vartheta(p_{n})/n$ with $\mathcal{F}(n,0.25)$ is less than $5\%$ and for $n\geq 114$ it is less than $2\%.$ An important feature of (1.2) is that it holds even for very small values of $n.$

Acknowledgements

I am thankful to Mridul Nandi (Indian Statistical Institute, Kolkata, India) and Mehdi Hassani (University of Zanjan, Iran) for their valuable suggestions.