The Statement of the Prime Number Theorem
It will be shown in Chapter 5 that
π (x)
x
, i.e. lim
= 1.
x→∞ x/ log x
log x
We will not prove the Prime Number Theorem in this form but, just as
we deduced bounds on π (x) from those on ψ (x) , we will prove
π (x) ∼
ψ (x) ∼ x.
This is in fact equivalent to the Prime Number Theorem as we will now show.
P
Firstly, recall that θ (x) = p≤x log p, which thus differs from ψ (x) by
powers of primes. In fact, we can use Chebyshev’s Theorem to estimate the
contribution of the prime powers to give
Lemma 3.17 For x ≥ 2,
ψ (x) = θ (x) + O x1/2 ,
Proof From the definition of Λ (n) as log p if n = pk , 0 otherwise,
X
XX
XX
log p.
log p =
log p +
ψ (x) =
p≤x
p≤x k≥1
pk ≤x
(18)
(19)
p≤x k≥2
pk ≤x
From this we first get
ψ (x) ≥
X
log p = θ (x) .
(20)
p≤x
We next interchange the two sums in (19). to do so, we need to have an
upper bound on k, and we note that the largest k occurs when p is smallest
i.e. p = 2, when the condition is 2k ≤ x, i.e. k ≤ log x/ log 2. Let
K = log x/ log 2 ≪ log x.
Then on interchanging the double sum and rewriting pk ≤ x as p ≤ x1/k we
get
X
X X
XX
log p =
θ x1/k
log p =
p≤x k≥2
pk ≤x
2≤k≤K p≤x1/k
2≤k≤K
X
= θ x1/2 +
θ x1/k
3≤k≤K
20
Note what has been done here, the k = 2 term has been taken aside. For
the other terms we have
X
X
ψ x1/k
θ x1/k ≤
3≤k≤K
3≤k≤K
by (20). We now apply Theorem 3.11, in the form ψ (t) ≪ t, to get
X
X
ψ x1/k ≪
x1/k ≤ Kx1/3 ≪ x1/3 log x ≪ x1/2 .
3≤k≤K
3≤k≤K
Combining we have
ψ (x) = θ (x) + O x1/2 .
Check what would have happened if we had not taken the k = 2 term
aside in the above proof.
In the Question Sheet you will be asked to show that from this lemma
we can deduce that for all ε > 0 we have
(log 2 − ε) x < θ (x) < (2 log 2 + ε) x
for all x > x2 (ε). From this it is easy to show that
Corollary 3.18 Given c > 2 the interval [X, cX] contains a prime for all
X sufficiently large, depending on c.
Proof Left to the problem sheet.
We now wish toPgo between thePweighted sum θ (x) and the unweighted
π (x), i.e. between p≤x log p and p≤x 1.
Theorem 3.19 For x ≥ 2,
θ (x)
π (x) =
+O
log x
21
x
log2 x
.
(21)
Proof Start with partial summation, so
X
π (x) =
1=
p≤x
X
=
X log p
p≤x
log p
p≤x
log p
1
−
log x
1
1
−
log x log p
Z x
X
1 X
dt
=
log p
log p +
2
log x p≤x
p t log t
p≤x
θ (x)
+
=
log x
Z
x
θ (t)
2
θ (x)
=
+O
log x
Z
x
2
dt
t log2 t
dt
,
log2 t
since, by (18) and (7) , we have θ (t) ≪ t. The integral here
√ was bounded in
Question 5 on Problem Sheet 1. Splitting the integral at x we get
Z √x
Z x
Z x
dt
dt
dt
=
2
2 + √
2
log t
2
x log t
2 log t
√
x
x
x
≤
+
.
√ =O
log2 2 log2 x
log2 x
Here both integrals were estimated by
Z b
f (t) dt ≤ lubf (t) × (b − a) .
[a,b]
a
The following is now immediate,
Corollary 3.20
π (x) ∼
x
log x
if, and only if, ψ (x) ∼ x.
Proof From (18) and (21) we get
ψ (x) + O x1/2
x
x
ψ (x)
+O
+O
=
.
π (x) =
log x
log x
log2 x
log2 x
22
Thus
ψ (x)
π (x)
=
+O
x/ log x
x
1
log x
.
Therefore, on the assumption that the limits exist, they must satisfy
π (x)
ψ (x)
= lim
.
x→∞ x/ log x
x→∞
x
lim
23