Recall the definition: π(x):=∑p≤xp prime1, and let us define the following: Θ(x):=∑p≤xp primelogp, Ψ(x):=∑1≤n≤xΛ(n)=∑pm≤xm≥1p primelogp, Λ(n)={logpif n=pk for some prime p and integer k≥10otherwise.

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

Thus on the contrary we must have: liminfn→∞xn=−limsupn→∞−xn

Theorem 2.1. For all α ∈ (0, 1), and all x ≥ x₀: Θ(x)log(x)≤Ψ(x)log(x)≤π(x)≤Θ(x)αlog(x)+xα

Proof. Clearly Θ(x) ≤ Ψ(x), such that limsupn→∞Ψ(x)x≥limsupn→∞Θ(x)x

Also, if p is a prime and p^m ≤ x < p^m+1, then log p occurs in the sum for Ψ(x) exactly m times. [1] Ψ(x)=∑pm≤xp primem≥1logp            =∑p≤xp prime[logxlogp]logp            ≤∑p≤xp primelogx            =π(x)logx limsupx→∞Ψ(x)x≤limsupx→∞π(x)x/logx

Now fix α ∈ (0, 1). Given x > 1, Θ(x)=∑p≤xp primelogp≥∑xa<p≤xp primelogp.

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

∴ Θ(x)>αlogx∑xα<p≤x1            =αlogx(π(x)−π(xα))            >αlogx(π(x)−xα)

϶ Θ(x)x>απ(x)x/logx−αlogxx1−α

∀α ∈ (0, 1) we have: limx→∞αlogxx1−α=0.

Combining these we get: limsupx→∞Θ(x)x≥limsupx→∞απ(x)x/logx

Once again, since our statement is true ∀α ∈ (0, 1), limsupx→∞Θ(x)x≥limsupx→∞π(x)x/logx

Similarly: liminfx→∞Ψ(x)x≥liminfx→∞Θ(x)x liminfx→∞π(x)x/logx≥liminfx→∞Ψ(x)x

Once again, we apply the same method: liminfx→∞Θ(x)x≥liminfx→∞π(x)x/logx, and have thus proven Theorem 2.1. □

Theorem 3.1. We have: log2≤liminfx→∞π(x)x/logx≤limsupx→∞π(x)x/logx≤2log2.

Proof. First the lower bound. Take: S(x):=∑1≤n≤xlogn−2∑1≤n≤x/2logn.

Then, ∑1≤d≤nd|nΛ(d)=∑pr|dd|np primelogp                      =∑i=1l∑r=1eilogpi   where n=p1e1⋯plel                      =∑i=1leilogpi                      =logn

∴ S(x)=∑1≤n≤x∑d|nΛ(d)−2∑1≤n≤x/2∑d|nΛ(d)

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).

∴ S(x)=∑1≤d≤xΛ(d)[xd]−2∑1≤d≤x/2Λ(d)[x2d]            =∑1≤d≤x/2Λ(d)([xd]−2[x2d])+∑(x/2)<d≤xΛ(d)[xd]           ≤∑1≤d≤x/2Λ(d)+∑(x/2)<d≤xΛ(d)            =Ψ(x).

So, Ψ(x)x≥S(x)x=1x∑1≤n≤xlogn−2x∑1≤n≤x/2logn. log t is increasing,

∴ ∫1x+1logt dt≥∑1≤n≤xlogn, ∫1[x]logt dt≤∑1≤n≤xlogn.

Actually, assumingx∈ℤ+, S(x)x≥1x∫1xlogt dt−2x∫1(x/2)+1logt dt            =1x(xlogx−x+1)−2x(x+22log(x+22)−x+22+1)            =logx+1x−x+2xlog(x+2)+x+2xlog2            >log(xx+2)−2xlog(x+2)+log2

Using Theorem 2.1, we get: liminfx→∞π(x)x/logx=liminfx→∞Ψ(x)x                       ≥liminfx→∞S(x)x                       >limx→∞log(x/(x+2))−2xlog(x+2)+log2                      =log2

To complete the proof, we will need some auxiliary results taken from Murty’s Analytic Number Theory [1] in the form of three lemmas:

Lemma 3.2.

Proof. Fix an exponent r. The positive integers no larger than m that are multiples of p^r are pr,2pr,…,[mpr]pr and those that are multiples of p^r+1 are pr+1,2pr+1,…,[mpr+1]pr+1

Thus there are precisely [m/p]r−[m/p]r+1 positive integers n ≤ m with ord_p(n) = r.

∴ ordp(m!)=∑n=1mordp(n)                     =∑r≥1∑1≤n≤mordp(n)=rr                     =∑r≥1r([m/pr]−[m/pr+1])                     =∑r≥1r[m/pr]−∑r≥1r[m/pr+1]                     =∑r≥1r[m/pr]−∑r≥1(r−1)[m/pr]                     =∑r≥1[mpr]        □

Lemma 3.3.∀n∈ℤ+, 22n2n<(2n2)<22n2n+1

Proof.

Since: (2i−1)(2i+1)(2i)2<1 for all i ≥ 1.

∴ 1>(2n+1)Pn2, giving the upper bound. For the lower bound: 1−1(2i−1)2<1 ∀i ≥ 1, such that 1>∏i=2n(1−1(2i−1)2)=∏i=2n(2i−1)2−1(2i−1)2=∏i=2n(2i−2)(2i)(2i−1)2=14nPn2 yielding our lemma. □

Lemma 3.4.∀n∈ℤ+, Θ(n)<2nlog2

Proof. By Lemma 3.3, log((2nn)12)=log((2nn))−log2                             <2nlog2−12log(2n+1)−log2                             =(2n−1)log2−12log(2n+1) since (2nn)12=(2n)!(n!)2n2n=(2n−1)!n!(n−1)!=(2n−1n−1). by Lemma 3.2: log((2nn)12)=log(2n−1n−1)                              =∑p primeordp((2n−1)!)logp−∑p primeordp((n−1)!)logp−∑p primeordp(n!)logp                              =∑p primelogp∑r≥1[(2n−1)/pr]−[n/pr]                              ≥∑p primen<p≤2n−1logp                              =Θ(2n−1)−Θ(n)

Where Θ(2n−1)−Θ(n)<(2n−1)log2−12log(2n+1)

We now proceed by induction. Proceeding from the trivialities, suppose m > 2 and the lemma is true for n < m,n,m∈ℕ. If m is odd, then m = 2n – 1 for some integer n ≥ 2 since m > 2. Thus by induction, Θ(m)=Θ(2n−1)<Θ(n)+(2n−1)log2−12log(2n+1)             <2nlog2+(2n−1)log2−12log(2n)             =(4n−1)log2−12log(2n)             ≤(4n−2)log2   (since n≥2)             =2mlog2

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: Θ(m)=Θ(m−1)<2(m−1)log2<2mlog2

Lemma 3.4 gives 2log2≥limsupx→∞Θ(x)x.        □

The desired lower bound follows from Theorem 2.1 □

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

By Hassani [2], we have ⌊12((x+1)2log(x+1)−x2logx)−log2xloglogx⌋≤π((x+1)2)−π(x2)  12(x2logn−32log3)−∑j=3x−1log2xloglogk<π(x2)−π(32)

And thus: ∑j=3x−1⌊12((j+1)2log(j+1)−j2logj)−log2jloglogj⌋<∑j=3n−1π((j+1)2)−π(j2)

∴ ⌊12((x+1)2log(x+1)−x2logx)−(logx)2log(logx)⌋≤π((x+1)2)−π(x2).

And by the prime number theorem, which gives us the asymptotic estimate for some F(x):=π((x+1)2)−π(x2)~12((x+1)2log(x+1)−x2logx)

We propose: π((x+1)2)−π(x2)≤⌊12((x+1)2log(x+1)−x2logx)+log2xloglogx⌋ by the same method.