Theorem 1 (Euler)

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 u}m+1 _m

−
_{1 − u} = 1 + u + · · · + u


> 0.

⩾

>
n

= log(x + 1),
y
Choose m ∈ Z_⩾1, such that 2^m ⩾ 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 = p^m1 p^m2 ...p^mk is the prime factorization of n ∈ {1, 2, ..., x},

1

2

k
then p₁, p₂, ...p_k are all ⩽ x; and m₁, m₂, ...m_k 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

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) :=
p^m⩽x
log p =

Σ
(p,m)∈{(q,l)∈N×N | q prime, l⩾1 and q^l⩽x}
log p, x > 0. (2)

Σ
Alternatively, we can define ψ(x) := _p⩽x M_p log p, where M_p is the highest exponent such that p^Mp ⩽ x holds, for example ψ(10) = 3 log 2 + 2 log 3 + log 5 + log 7.

_Σ, , ·
Note that M_p = ⌊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 e^ϑ(x) equals the product of all primes p ⩽ x and, for x ⩾ 1, e^ψ(x) is the least common multiple of all positive integers n ⩽ x. If p^m ⩽ x, then p ⩽ x^1/m, and conversely, hence (2) leads to the relation

Recall the von Mangoldt function

ψ(x) = ϑ(x) + ϑ x^1/2 + ϑ x^1/3 + · · · (4)

Λ(n) =

.
⁽ log p if ∃p prime, m ∈ Z_⩾1 : n = p^m

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 Σ}log 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

x^α
⩽x
ϑ(x) = ^Σ log p ⩾ ^Σ

_p⩽x

log p ⩾ ^Σ
α log x = α log x π(x) − π(x^α) > α log x π(x) − x^α ,

x^α
⩽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 x^α−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.

2.