Chebyshev’s theorem on the distribution of prime numbers
Nikolaos Stamatopoulos, Zhiang Wu 25 November 2021
1 The Chebyshev functions
Denote by π( x) the number of primes not exceeding x > 0. It is well known that there is infinitely many prime numbers, i.e., lim x→∞ π( x) → ∞. The famous prime number theorem tells us more, namely π( x) ∼ x/ log x.
Σ Q
In this paper, we are going to prove the Chebyshev’s theorem, which is an intermediate result of the prime number theory, and use similar methodology to derive a few other interesting results.
Theorem 1 (Euler) . The sum 1 /p and the product (1 − 1 /p) −1 are both divergent, as p runs through all the prime numbers.
Proof. We shall first show that the product diverges, and then deduce that the series also does. Let
P ( x) :=
pY⩽ x
1 −1
1 − p
and S(x) :=
1
Σ for x ∈ Z .⩾
p 2
p⩽ x
Note that for u ∈ (0, 1) and m ∈ Z⩾1, we have:
1
>
1 − u
1 um+1 m
−
1 − u = 1 + u + · · · + u
> 0 .
⩾
>
n
= log(x + 1),
y
Choose m ∈ Z⩾1, such that 2m ⩾ x. Note that the above inequality holds for all primes p ⩽ x with u = 1/p. By multiplying all of the resulting inequalities, we get
P (x) >
1 + p + · · · + pm
Y 1
p⩽x
1 (a) Σx 1
n=1
∫ x+1 dy
1
where (a) follows from that fact that, if n = pm1 pm2 ...pmk is the prime factorization of n ∈ {1 , 2 , ..., x},
1
2
k
then p1, p2, ...pk are all ⩽ x; and m1, m2, ...mk are all ⩽ m (since x ⩾ 2 m). Thus every term on the right hand side is a product of terms on the left hand side.
Hence the product Q(1 − 1 /p) −1 diverges.
To prove the divergence of the series, we consider the expansion:
∞
∞
<
n
= .
2 2(1 − u)
If u ∈ (0 , 1), we have:
log
1
− u
=
∞
Σ
n=1
un
n for u ∈ [−1 , 1) .
log
— u =
1
1 − u
Σ un
n=2
Σ un u2
n=2
Note that the above equation holds for all primes p ⩽ x with u = 1/p. By adding all of the resulting inequalities, we obtain:
log P (x) − S(x) =
log
— p
<
=
2(1 − 1/p) 2
Σ 1 1 ! Σ
p⩽x
1 − 1/p
p⩽x
p−2
1 Σ 1
p⩽x
1 Σ 1 1
n=2
It follows that S( x) > log P ( x) − 1 /2 > log log x − 1 /2, and hence the sum Σ 1 /p also diverges.
∞
<
p( p − 1) 2
= .
n( n − 1) 2
Definition. The Chebyshev functions ϑ and ψ are defined as follows:
p∈{q prime | q⩽x}
ϑ( x) := Σ log p = Σ
p⩽x
log p, x > 0; (1)
Σ
ψ(x) :=
pm⩽x
log p =
Σ
(p,m)∈{(q,l)∈N×N | q prime, l⩾1 and ql⩽x}
log p, x > 0. (2)
Σ
Alternatively, we can define ψ( x) := p⩽x Mp log p, where Mp is the highest exponent such that pMp ⩽ x holds, for example ψ(10) = 3 log 2 + 2 log 3 + log 5 + log 7 .
Σ, , ·
Note that Mp = ⌊log x/ log p⌋, and hence
ψ( x) = log x log p. (3)
log p
p⩽ x
Further, it follows from (1) and (2) that eϑ(x) equals the product of all primes p ⩽ x and, for x ⩾ 1 , eψ(x) is the least common multiple of all positive integers n ⩽ x. If pm ⩽ x, then p ⩽ x1/m, and conversely, hence (2) leads to the relation
Recall the von Mangoldt function
ψ( x) = ϑ( x) + ϑ x1/2 + ϑ x1/3 + · · · (4)
Λ( n) =
.
( log p if ∃ p prime, m ∈ Z ⩾1 : n = pm
0 otherwise
From (2) it is immediate that
Σ
ψ(x) = Λ(n) (5)
n⩽x
where the sum is finite because ∀ x < 2 : ϑ( x) = 0 .
We shall now establish a connection between the functions
Theorem 2. Let
π(x)
,
x/ log x
ϑ(x)
,
x
ψ(x)
.
x
l1 = lim inf
x→∞
π( x) x/ log x
ϑ( x)
, L1 = lim sup
x→∞
π( x)
,
x/ log x
ϑ( x)
l2 = lim inf
x→∞
l3 = lim inf
x→∞
x ψ(x) x
, L2 = lim sup
x→∞
, L3 = lim sup
x→∞
,
x
ψ(x)
.
x
Then l1 = l2 = l3, and L1 = L2 = L3.
Σ, , Σ Σlog x log x(3)
Proof. It follows from (3) that, for all x > 0,
ϑ(x) = Σ log p ⩽
ψ(x) = · log p ⩽ · log p = 1 · log x = π(x) log x
and hence
p⩽ x
p⩽ x
log p
p⩽ x
log p
p⩽ x
ϑ(x) ⩽ ψ(x) ⩽ π(x) log x.
For x → ∞, we get
ϑ(x)
x x
x→∞
ψ(x)
x
π(x)
lim sup
x→∞
⩽ lim sup
x
x→∞
x ⩽ lim sup x/ log x, i.e., L2 ⩽ L3 ⩽ L1. (6)
Now fix α ∈ (0, 1) and x > 1. Then
xα
⩽x
ϑ(x) = Σ log p ⩾ Σ
p⩽x
log p ⩾ Σ
α log x = α log x π(x) − π(xα) > α log x π(x) − xα ,
xα
⩽x
Dividing by x on both sides, we get
x
x
ϑ(x) > α · π(x) log x − αxα−1 log x.
Since α ∈ (0, 1), it follows that xα−1 log x → 0, as x → ∞, and thus
x→∞
ϑ(x) π(x)
lim sup
x→∞
x ⩾ α · lim sup x/ log x, i.e., L2 ⩾ αL1.
Because the above inequality holds for all α ∈ (0, 1), we deduce that L2 ⩾ L1. Combining this with (6), we get L1 = L2 = L3.
The proof of l1 = l2 = l3 follows analogously.
It follows from Theorem 2 that, if one of the three functions
π(x)
,
x/ log x
ϑ(x)
,
x
ψ(x) x
tends to a limit as x → ∞, then so do the others, and all three limits are the same. Thus in order to prove the prime number theorem, it is sufficient to show that lim x→∞ ψ( x) /x = 1.
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