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## Estimates of the Bounds of ?(x) and ?((x + 1)2) ? ?(x2)

Author: Connor Paul Wilson (University of Michigan)

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### Estimates of the Bounds of ?(x) and ?((x + 1)2) ? ?(x2)

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Wilson, C., (2021) “Estimates of the Bounds of ?(x) and ?((x + 1)2) ? ?(x2)”, University of Michigan Undergraduate Research Journal 15. doi: https://doi.org/10.3998/umurj.1334

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##### Published on 21 Dec 2021

We show the following bounds on the prime counting function π(x) using principles from analytic number theory, giving an estimate

$\mathrm{log}2\le \underset{x\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\pi \left(x\right)}{x/\mathrm{log}x}\le \underset{x\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\pi \left(x\right)}{x/\mathrm{log}x}\le 2\mathrm{log}2$

We also conjecture about the bounding of π((x + 1)2) − π(x2), as is relevant to Legendre’s conjecture about the number of primes in the aforementioned interval such that:

$\begin{array}{l}⌊\frac{1}{2}\left(\frac{{\left(x+1\right)}^{2}}{\mathrm{log}\left(x+1\right)}-\frac{{x}^{2}}{\mathrm{log}x}\right)-\frac{{\left(\mathrm{log}x\right)}^{2}}{\mathrm{log}\left(\mathrm{log}x\right)}⌋\le \pi \left({\left(x+1\right)}^{2}\right)-\pi \left({x}^{2}\right)\le \\ ⌊\frac{1}{2}\left(\frac{{\left(x+1\right)}^{2}}{\mathrm{log}\left(x+1\right)}-\frac{{x}^{2}}{\mathrm{log}x}\right)-{\mathrm{log}}^{2}x\mathrm{log}\mathrm{log}x⌋\end{array}$

## 1. Introduction

Recall the definition:

$\pi \left(x\right):=\sum _{\begin{array}{c}p\le x\\ p \text{prime}\end{array}}1,$

and let us define the following:

$\Theta \left(x\right):=\sum _{\begin{array}{c}p\le x\\ p \text{prime}\end{array}}\mathrm{log}p,$

$\Psi \left(x\right):=\sum _{1\le n\le x}\Lambda \left(n\right)=\sum _{\begin{array}{c}{p}^{m}\le x\\ m\ge 1\\ p \text{prime}\end{array}}\mathrm{log}p,$

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

The following are simple statements from real analysis that are required for rigorousness’ sake: let {xn} be a sequence of real numbers and L be a real number with the following two properties: ∀ϵ > 0, ∃N such that xn < L + ϵ, ∀nN. ∀ϵ > 0 ^ N ≥ 1, ∃nN with xn > Lϵ. We thus define L as:

$\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}{x}_{n}=L$

Thus on the contrary we must have:

$\underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}{x}_{n}=-\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}-{x}_{n}$

## 2. Necessary Preliminary Results

Theorem 2.1. For all α ∈ (0, 1), and all x ≥ x0:

$\frac{\Theta \left(x\right)}{\mathrm{log}\left(x\right)}\le \frac{\Psi \left(x\right)}{\mathrm{log}\left(x\right)}\le \pi \left(x\right)\le \frac{\Theta \left(x\right)}{\alpha \mathrm{log}\left(x\right)}+{x}^{\alpha }$

Proof. Clearly Θ(x) ≤ Ψ(x), such that

$\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\Psi \left(x\right)}{x}\ge \underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\Theta \left(x\right)}{x}$

Also, if p is a prime and pmx < pm+1, then log p occurs in the sum for Ψ(x) exactly m times. 

$\Psi \left(x\right)=\sum _{\begin{array}{c}{p}^{m}\le x\\ p \text{prime}\\ m\ge 1\end{array}}\mathrm{log}p\phantom{\rule{0ex}{0ex}} =\sum _{\begin{array}{c}p\le x\\ p \text{prime}\end{array}}\left[\frac{\mathrm{log}x}{\mathrm{log}p}\right]\mathrm{log}p\phantom{\rule{0ex}{0ex}} \le \sum _{\begin{array}{c}p\le x\\ p \text{prime}\end{array}}\mathrm{log}x\phantom{\rule{0ex}{0ex}} =\pi \left(x\right)\mathrm{log}x$

$\underset{x\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\Psi \left(x\right)}{x}\le \underset{x\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\pi \left(x\right)}{x/\mathrm{log}x}$

Now fix α ∈ (0, 1). Given x > 1,

$\Theta \left(x\right)=\sum _{\begin{array}{c}p\le x\\ p \text{prime}\end{array}}\mathrm{log}p\ge \sum _{\begin{array}{c}{x}^{a}

It is clear that all p from the second sum satisfy: log p > α log x.

$\begin{array}{l}\Theta \left(x\right)>\alpha \mathrm{log}x\sum _{{x}^{\alpha }\alpha \mathrm{log}x\left(\pi \left(x\right)-{x}^{\alpha }\right)\end{array}$

϶

$\frac{\Theta \left(x\right)}{x}>\frac{\alpha \pi \left(x\right)}{x/\mathrm{log}x}-\frac{\alpha \mathrm{log}x}{{x}^{1-\alpha }}$

α ∈ (0, 1) we have:

$\underset{x\to \infty }{\mathrm{lim}}\frac{\alpha \mathrm{log}x}{{x}^{1-\alpha }}=0.$

Combining these we get:

$\underset{x\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\Theta \left(x\right)}{x}\ge \underset{x\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\alpha \pi \left(x\right)}{x/\mathrm{log}x}$

Once again, since our statement is true ∀α ∈ (0, 1),

$\underset{x\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\Theta \left(x\right)}{x}\ge \underset{x\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\pi \left(x\right)}{x/\mathrm{log}x}$

Similarly:

$\underset{x\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\Psi \left(x\right)}{x}\ge \underset{x\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\Theta \left(x\right)}{x}$

$\underset{x\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\pi \left(x\right)}{x/\mathrm{log}x}\ge \underset{x\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\Psi \left(x\right)}{x}$

Once again, we apply the same method:

$\underset{x\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\Theta \left(x\right)}{x}\ge \underset{x\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\pi \left(x\right)}{x/\mathrm{log}x},$

and have thus proven Theorem 2.1. □

## 3. Main Result

Theorem 3.1. We have:

$\mathrm{log}2\le \underset{x\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\pi \left(x\right)}{x/\mathrm{log}x}\le \underset{x\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\pi \left(x\right)}{x/\mathrm{log}x}\le 2\mathrm{log}2.$

Proof. First the lower bound. Take:

$S\left(x\right):=\sum _{1\le n\le x}\mathrm{log}n-2\sum _{1\le n\le x/2}\mathrm{log}n.$

Then,

$\sum _{\begin{array}{c}1\le d\le n\\ d|n\end{array}}\Lambda \left(d\right)=\sum _{\begin{array}{c}{p}^{r}|d\\ d|n\\ p \text{prime}\end{array}}\mathrm{log}p\phantom{\rule{0ex}{0ex}} =\sum _{i=1}^{l}\sum _{r=1}^{{e}_{i}}\mathrm{log}{p}_{i} \text{where} n={p}_{1}^{{e}_{1}}\cdots {p}_{l}^{{e}_{l}}\phantom{\rule{0ex}{0ex}} =\sum _{i=1}^{l}{e}_{i}\mathrm{log}{p}_{i}\phantom{\rule{0ex}{0ex}} =\mathrm{log}n$

$S\left(x\right)=\sum _{1\le n\le x}\sum _{d|n}\Lambda \left(d\right)-2\sum _{1\le n\le x/2}\sum _{d|n}\Lambda \left(d\right)$

Clearly {d, 2d, …, qd} is the set of n satisfying 1 ≤ n ≤ x and d | n (we can see this easily by writing x = r + qd with 0 ≤ r < d).

$\begin{array}{l}S\left(x\right)=\sum _{1\le d\le x}\Lambda \left(d\right)\left[\frac{x}{d}\right]-2\sum _{1\le d\le x/2}\Lambda \left(d\right)\left[\frac{x}{2d}\right]\\ =\sum _{1\le d\le x/2}\Lambda \left(d\right)\left(\left[\frac{x}{d}\right]-2\left[\frac{x}{2d}\right]\right)+\sum _{\left(x/2\right)

So,

$\frac{\Psi \left(x\right)}{x}\ge \frac{S\left(x\right)}{x}=\frac{1}{x}\sum _{1\le n\le x}\mathrm{log}n-\frac{2}{x}\sum _{1\le n\le x/2}\mathrm{log}n.$

log t is increasing,

${\int }_{1}^{x+1}\mathrm{log}t dt\ge \sum _{1\le n\le x}\mathrm{log}n,$

${\int }_{1}^{\left[x\right]}\mathrm{log}t dt\le \sum _{1\le n\le x}\mathrm{log}n.$

Actually, assuming$x\in {ℤ}^{+}$,

$\begin{array}{l}\frac{S\left(x\right)}{x}\ge \frac{1}{x}{\int }_{1}^{x}\mathrm{log}t dt-\frac{2}{x}{\int }_{1}^{\left(x/2\right)+1}\mathrm{log}t dt\\ =\frac{1}{x}\left(x\mathrm{log}x-x+1\right)-\frac{2}{x}\left(\frac{x+2}{2}\mathrm{log}\left(\frac{x+2}{2}\right)-\frac{x+2}{2}+1\right)\\ =\mathrm{log}x+\frac{1}{x}-\frac{x+2}{x}\mathrm{log}\left(x+2\right)+\frac{x+2}{x}\mathrm{log}2\\ >\mathrm{log}\left(\frac{x}{x+2}\right)-\frac{2}{x}\mathrm{log}\left(x+2\right)+\mathrm{log}2\end{array}$

Using Theorem 2.1, we get:

$\underset{x\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\pi \left(x\right)}{x/\mathrm{log}x}=\underset{x\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\Psi \left(x\right)}{x}\phantom{\rule{0ex}{0ex}} \ge \underset{x\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{S\left(x\right)}{x}\phantom{\rule{0ex}{0ex}} >\underset{x\to \infty }{\mathrm{lim}}\mathrm{log}\left(x/\left(x+2\right)\right)-\frac{2}{x}\mathrm{log}\left(x+2\right)+\mathrm{log}2\phantom{\rule{0ex}{0ex}} =\mathrm{log}2$

To complete the proof, we will need some auxiliary results taken from Murty’s Analytic Number Theory  in the form of three lemmas:

Lemma 3.2.

${\text{ord}}_{p}\left(m!\right)=\sum _{r\ge 1}\left[\frac{m}{{p}^{r}}\right],\forall m\in {ℤ}^{+}, prime p$

Proof. Fix an exponent r. The positive integers no larger than m that are multiples of pr are

${p}^{r},2{p}^{r},\dots ,\left[\frac{m}{{p}^{r}}\right]{p}^{r}$

and those that are multiples of pr+1 are

${p}^{r+1},2{p}^{r+1},\dots ,\left[\frac{m}{{p}^{r+1}}\right]{p}^{r+1}$

Thus there are precisely ${\left[m/p\right]}^{r}-{\left[m/p\right]}^{r+1}$ positive integers n ≤ m with ordp(n) = r.

${\text{ord}}_{p}\left(m!\right)=\sum _{n=1}^{m}{\text{ord}}_{p}\left(n\right)\phantom{\rule{0ex}{0ex}} =\sum _{r\ge 1}\sum _{\begin{array}{c}1\le n\le m\\ {\text{ord}}_{p}\left(n\right)=r\end{array}}r\phantom{\rule{0ex}{0ex}} =\sum _{r\ge 1}r\left(\left[m/{p}^{r}\right]-\left[m/{p}^{r+1}\right]\right)\phantom{\rule{0ex}{0ex}} =\sum _{r\ge 1}r\left[m/{p}^{r}\right]-\sum _{r\ge 1}r\left[m/{p}^{r+1}\right]\phantom{\rule{0ex}{0ex}} =\sum _{r\ge 1}r\left[m/{p}^{r}\right]-\sum _{r\ge 1}\left(r-1\right)\left[m/{p}^{r}\right]\phantom{\rule{0ex}{0ex}} =\sum _{r\ge 1}\left[\frac{m}{{p}^{r}}\right] \square$

Lemma 3.3.$\forall n\in {ℤ}^{+}$,

$\frac{{2}^{2n}}{2\sqrt{n}}<\left(\begin{array}{c}2n\\ 2\end{array}\right)<\frac{{2}^{2n}}{\sqrt{2n+1}}$

Proof.

${P}_{n}:=\prod _{i\le n}\frac{\left(2i-1\right)}{\left(2i\right)}\phantom{\rule{0ex}{0ex}} =\frac{\left(2n\right)!}{{2}^{2n}{\left(n!\right)}^{2}}\phantom{\rule{0ex}{0ex}} =\left(\begin{array}{c}2n\\ 2\end{array}\right)\frac{1}{{2}^{2n}}$

Since:

$\frac{\left(2i-1\right)\left(2i+1\right)}{{\left(2i\right)}^{2}}<1$

for all i ≥ 1.

$1>\left(2n+1\right){P}_{n}^{2},$

giving the upper bound. For the lower bound:

$1-\frac{1}{{\left(2i-1\right)}^{2}}<1$

i ≥ 1, such that

$1>\prod _{i=2}^{n}\left(1-\frac{1}{{\left(2i-1\right)}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\prod _{i=2}^{n}\frac{{\left(2i-1\right)}^{2}-1}{{\left(2i-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\prod _{i=2}^{n}\frac{\left(2i-2\right)\left(2i\right)}{{\left(2i-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{4n{P}_{n}^{2}}$

yielding our lemma. □

Lemma 3.4.$\forall n\in {ℤ}^{+}$,

$\Theta \left(n\right)<2n\mathrm{log}2$

Proof. By Lemma 3.3,

$\mathrm{log}\left(\left(\begin{array}{c}2n\\ n\end{array}\right)\frac{1}{2}\right)=\mathrm{log}\left(\left(\begin{array}{c}2n\\ n\end{array}\right)\right)-\mathrm{log}2\phantom{\rule{0ex}{0ex}} <2n\mathrm{log}2-\frac{1}{2}\mathrm{log}\left(2n+1\right)-\mathrm{log}2\phantom{\rule{0ex}{0ex}} =\left(2n-1\right)\mathrm{log}2-\frac{1}{2}\mathrm{log}\left(2n+1\right)$

since

$\left(\begin{array}{c}2n\\ n\end{array}\right)\frac{1}{2}=\frac{\left(2n\right)!}{{\left(n!\right)}^{2}}\frac{n}{2n}=\frac{\left(2n-1\right)!}{n!\left(n-1\right)!}=\left(\begin{array}{c}2n-1\\ n-1\end{array}\right).$

by Lemma 3.2:

$\mathrm{log}\left(\left(\begin{array}{c}2n\\ n\end{array}\right)\frac{1}{2}\right)=\mathrm{log}\left(\begin{array}{c}2n-1\\ n-1\end{array}\right)\phantom{\rule{0ex}{0ex}} =\sum _{p \text{prime}}{\text{ord}}_{p}\left(\left(2n-1\right)!\right)\mathrm{log}p-\sum _{p \text{prime}}{\text{ord}}_{p}\left(\left(n-1\right)!\right)\mathrm{log}p-\sum _{p \text{prime}}{\text{ord}}_{p}\left(n!\right)\mathrm{log}p\phantom{\rule{0ex}{0ex}} =\sum _{p \text{prime}}\mathrm{log}p\sum _{r\ge 1}\left[\left(2n-1\right)/{p}^{r}\right]-\left[n/{p}^{r}\right]\phantom{\rule{0ex}{0ex}} \ge \sum _{\begin{array}{c}p \text{prime}\\ n

Where

$\Theta \left(2n-1\right)-\Theta \left(n\right)<\left(2n-1\right)\mathrm{log}2-\frac{1}{2}\mathrm{log}\left(2n+1\right)$

We now proceed by induction. Proceeding from the trivialities, suppose m > 2 and the lemma is true for n < m,n,$m\in ℕ$. If m is odd, then m = 2n – 1 for some integer n ≥ 2 since m > 2. Thus by induction,

$\Theta \left(m\right)=\Theta \left(2n-1\right)<\Theta \left(n\right)+\left(2n-1\right)\mathrm{log}2-\frac{1}{2}\mathrm{log}\left(2n+1\right)\phantom{\rule{0ex}{0ex}} <2n\mathrm{log}2+\left(2n-1\right)\mathrm{log}2-\frac{1}{2}\mathrm{log}\left(2n\right)\phantom{\rule{0ex}{0ex}} =\left(4n-1\right)\mathrm{log}2-\frac{1}{2}\mathrm{log}\left(2n\right)\phantom{\rule{0ex}{0ex}} \le \left(4n-2\right)\mathrm{log}2 \left(since n\ge 2\right)\phantom{\rule{0ex}{0ex}} =2m\mathrm{log}2$

If m is even, then m = 2n for some integer n with m > n ≥ 2 and m is composite. Clearly Θ(m) = Θ(m − 1) and we know:

$\Theta \left(m\right)=\Theta \left(m-1\right)<2\left(m-1\right)\mathrm{log}2<2m\mathrm{log}2$

Lemma 3.4 gives

$2\mathrm{log}2\ge \mathrm{lim}\underset{x\to \infty }{\mathrm{sup}}\frac{\Theta \left(x\right)}{x}. \square$

The desired lower bound follows from Theorem 2.1

## 4. On Primes in the Gaps between Squares

The following is relatively aleatory compared to the previous workings, but it is worth mentioning considering the importance of the statement.

By Hassani , we have

$\begin{array}{l}⌊\frac{1}{2}\left(\frac{{\left(x+1\right)}^{2}}{\mathrm{log}\left(x+1\right)}-\frac{{x}^{2}}{\mathrm{log}x}\right)-\frac{{\mathrm{log}}^{2}x}{\mathrm{log}\mathrm{log}x}⌋\le \pi \left({\left(x+1\right)}^{2}\right)-\pi \left({x}^{2}\right)\\ \frac{1}{2}\left(\frac{{x}^{2}}{\mathrm{log}n}-\frac{{3}^{2}}{\mathrm{log}3}\right)-\sum _{j=3}^{x-1}\frac{{\mathrm{log}}^{2}x}{\mathrm{log}\mathrm{log}k}<\pi \left({x}^{2}\right)-\pi \left({3}^{2}\right)\end{array}$

And thus:

$\sum _{j=3}^{x-1}⌊\frac{1}{2}\left(\frac{{\left(j+1\right)}^{2}}{\mathrm{log}\left(j+1\right)}-\frac{{j}^{2}}{\mathrm{log}j}\right)-\frac{{\mathrm{log}}^{2}j}{\mathrm{log}\mathrm{log}j}⌋<\sum _{j=3}^{n-1}\pi \left({\left(j+1\right)}^{2}\right)-\pi \left({j}^{2}\right)$

$⌊\frac{1}{2}\left(\frac{{\left(x+1\right)}^{2}}{\mathrm{log}\left(x+1\right)}-\frac{{x}^{2}}{\mathrm{log}x}\right)-\frac{{\left(\mathrm{log}x\right)}^{2}}{\mathrm{log}\left(\mathrm{log}x\right)}⌋\le \pi \left({\left(x+1\right)}^{2}\right)-\pi \left({x}^{2}\right).$

And by the prime number theorem, which gives us the asymptotic estimate for some

$F\left(x\right):=\pi \left({\left(x+1\right)}^{2}\right)-\pi \left({x}^{2}\right)~\frac{1}{2}\left(\frac{{\left(x+1\right)}^{2}}{\mathrm{log}\left(x+1\right)}-\frac{{x}^{2}}{\mathrm{log}x}\right)$

We propose:

$\pi \left({\left(x+1\right)}^{2}\right)-\pi \left({x}^{2}\right)\le ⌊\frac{1}{2}\left(\frac{{\left(x+1\right)}^{2}}{\mathrm{log}\left(x+1\right)}-\frac{{x}^{2}}{\mathrm{log}x}\right)+{\mathrm{log}}^{2}x\mathrm{log}\mathrm{log}x⌋$

by the same method.

## References

Ram Murty, R., Problems in Analytic Number Theory, Second Edition — Graduate Texts in Mathematics, Springer, 2001.

Hassani, M., Counting primes in the interval (n2, (n + 1)2), AMS, 1997.