The Prime Number Theorem
1
We study the distribution of primes via the function
π(x) = the number of primes ≤ x
6
5
4
3
2
1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
2
It’s easier to draw this way:
π(x) = the number of primes ≤ x
6
5
4
3
2
1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
3
Clearly, π(x) < x
(not every number is prime)
and lim π(x) = ∞
80
60
x→∞
40
(there are infinitely many primes)
20
100
200
300
400
500
How many primes are there among the first 1 000 000 000 integers?
By brute force, π(1 000 000 000) = 50 847 534.
Can we approximate π(x) using elementary functions?
4
Gauss’s Prime Number Conjecture (1792)
π(x) ≈
x
π(x)
x
for large x.
log(x)
x
log(x)
102
25
21.7
103
168
144.8
104
1229
1085.7
105
9592
8685.9
106 78 498 72 382.4
107 664 579 620 421.0
π(x) −
x
π(x) − x/ log(x)
log(x)
π(x)
3.3
23.2
143.3
906.1
6115.6
44 158.3
0.1320
0.1381
0.1166
0.0945
0.0779
0.0664
5
In terms of limits, we may write Gauss’s 1792 conjecture as
π(x) − x/ log(x)
= 0
x→∞
π(x)
lim
or
x/ log(x)
= 1.
x→∞
π(x)
lim
Gauss’s Conjecture of 1849
∫
π(x) ≈
For large x,
2
x
dt
log t
or
∫
lim
x→∞
x
dt
2 log t
= 1.
π(x)
def
∫
Aside: Li(x) =
µ
x
dt
.
log t
6
(
)
x
log(x)
lim ∫ x
= 1.
x→∞
dt
2 log t
Exercise:
(
Corollary:
If either lim
x→∞
(
lim
x→∞
x
log(x)
π(x)
)
=1
x
log(x)
π(x)
(Hint: l’Hôpital)
)
∫
x
dt
log(t)
or lim 2
exists, then
x→∞
π(x)
∫ x
dt
log(t)
is equivalent to
lim 2
= 1.
x→∞
π(x)
7
The Prime Number Theorem
40
π(x)
) = 1.
lim (
x→∞
x
log(x)
30
20
10
50
100
150
200
The convergence is very slow:
At x = 106, π(x)/(x/ log(x)) ≈ 1.08449
At x = 109, π(x)/(x/ log(x)) ≈ 1.05373
At x = 1012, π(x)/(x/ log(x)) ≈ 1.03915.
8
It’s also messy.
1.09040
1.09035
π(x)
)
y=(
x
log(x)
1.09030
1.09025
1.09020
1.09015
500 500
501 000
501 500
502 000
5.00002 ´ 108
5.00002 ´ 108
1.05582
1.07118
1.05582
1.05582
1.07117
1.05582
1.05582
1.07116
1.05582
1.00005 ´ 107
1.00010 ´ 107
1.00015 ´ 107
1.00020 ´ 107
5.00001 ´ 108
5.00001 ´ 108
9
The first proofs of the Prime Number Theorem
(1896) use the Riemann Zeta Function
∞
∑
1
1
1
1
ζ(s) =
=
+
+
+ ···
s
s
s
s
n
1
2
3
n=1
From calculus, we know
1 1 1
ζ(1) = + + + · · ·
1 2 3
1
1
1
ζ(2) = 2 + 2 + 2 + · · ·
1
2
3
1
1
1
ζ(4) = 4 + 4 + 4 + · · ·
1
2
3
Vallée-Poussin
diverges
π2
converges to
6
π4
converges to
90
Hadamard
10
The function
ζ(s) is defined for
2
all s = x + iy in C,
1
0.0
0
with a simple pole at s = 1.
0.5
20
1.0
0
1.5
What’s the connection with primes?
2.0
-20
80
60
40
20
100
200
300
400
500
Riemann
11
Fix a number M .
Let
Example: M = 8.
Then
PM = {p1, p2, . . . , pr :
pi ≤ M }
= set of primes ≤ M .
Let
P8 = {2, 3, 5, 7}
and
ΛM =
{
n ∈ Z+ : prime factors
of n are in PM } .
Note that ΛM is infinite and
Λ8 = {1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 12, 14, 15, 16, . . .}
= {2α1 3α2 5α3 7α4 :
each αi is a
non-negative integer}
contains {1, 2, 3, . . . , M }.
12
Recall that P8 = {2, 3, 5, 7}. Consider the product
(
) (
)
1
1
1
1
1
1
+
+
+
·
·
·
×
+
+
+ ···
20 21 22
30 31 32
) (
)
(
1
1
1
1
1
1
× 0 + 1 + 2 + ··· × 0 + 1 + 2 + ···
5
5
5
7
7
7
Each term in the expansion has the form
(
)
1
α1, α2, α3, α4 are
where
are non-negative integers
2α1 3α2 5α3 7α4
These denominators are exactly the elements of Λ8, and we get
)
∑ 1
∏ (1
1
1
+ + + ··· =
.
p0 p1 p2
n
p ∈P8
n∈Λ8
13
Furthermore, by the geometric series theorem, we know that
(
)
∞ ( )r
∑
1
1
1
1
1
+
+
+
·
·
·
=
=
1.
p0 p1 p2
p
1
−
p
r=0
So we get
)
∏ (1
∑ 1
1
1
=
+ + + ···
n
p0 p1 p2
p ∈P8
n∈Λ8
∏ 1
=
1
1
−
p
p ∈P8
(
)(
)(
)(
)
1
1
1
1
105
=
=
.
24
1 − 12
1 − 13
1 − 51
1 − 71
1+
1 1 1 1 1 1 1 1
1
1
1
1
1
1
1
1
+ + + + + + + +
+
+
+
+
+
+
+
+ ···
2 3 4 5 6 7 8 9 10 12 14 15 16 18 20 21
14
More generally, we have
Lemma:
Extension of Lemma:
∏
p ∈PM
Now
∑ 1
1
=
n
1 − p1
n∈Λ
M
ΛM ⊃ {1, 2, 3, . . . , M },
∞
∑
1
1
=
= ζ(s)
1
s
n
1 − ps
n=1
p
∏
all
so
M
∑
∑ 1
1
>
.
n
n
n=1
n∈ΛM
This gives us the
Corollary:
∏
p ∈PM
M
∑
1
1
>
n
1 − p1
n=1
15
As M → ∞,
∏
p ∈PM
1
→
1 − p1
∏
all
1
1
1
−
p
primes
and
M
∑
1
n=1
n
→ ∞
∏
so that
all
1
Corollary:
→
∞,
1
)
(
1
−
∏
p
1
primes
= 0
1
−
and we get
p
all p
That is,
1 2 4 6 10 12 16 18 22 28
× × × ×
×
×
×
×
×
× ···
2 3 5 7 11 13 17 19 23 29
can be made arbitrarily small.
16
A small digression:
Claim:
∏
all primes
(
)
1
1−
=0
p
implies
∑ 1
is divergent
p
p prime
Remark:
This gives another proof of the fact that there are infinitely
many primes. The proof of the claim itself follows from a
Lemma:
Let 1 > a1 > a2 > a3 >∑
· · · be a decreasing sequence of
positive numbers such that
an converges. Then
∞
∏
(1 − an) ̸= 0
n=1
.
17
Proof sketch
Because
∑
an converges, we can find N so that
∞
∑
n=N
S=
∞
∑
n=1
Write
1
an < .
2
an = a1 + a2 + · · · + aN −1 + |aN + aN{z+1 + · ·}·
|
{z
}
< 12
>S− 12
∞
N
−1
∞
∏
∏
∏
(1 − an).
(1 − an) =
(1 − an) ×
n=1
n=1
n=N
It will suffice to show that the second factor is non-zero.
18
Write
∞
∏
(1 − an) = (1 − aN )(1 − aN +1)(1 − aN +2) · · ·
n=N
= 1
−
∑
ai
i
+
∑
ai aj
i<j
−
∑
aiaj ak
i<j<k
+ ···
where all the indices start at N .
19
We can then show that
∑
1 ∑
1
ai < ,
ai aj < 2 ,
2
2
i<j
i
and
∑
ai >
i
∑
∑
ai aj ak <
i<j<k
ai aj >
i<j
∑
1
,
23
and so on,
aiaj ak > · · ·
i<j<k
That is, the right side of
∞
∏
n=N
(1 − an) = 1 −
∑
i
ai +
∑
i<j
ai aj −
∑
ai aj ak + · · ·
i<j<k
satisfies the hypotheses of the Alternating Series Test.
By the AST, the series (and therefore the product) converges to a number
∑
∑
1
ai. Since
ai < , we know that this number is
between 1 and 1 −
2
i
i
non-zero.
20
Old Stuff:
∑1
n
is divergent.
1
∑1
New Stuff:
p
is divergent.
1
1+
1
2
3
4
1 1 1 1 1 1
+ + + + + + ··· = ∞
2 3 4 5 6 7
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
1 1 1 1
1
1
+ + + +
+
+ ··· = ∞
2 3 5 7 11 13
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
The harmonic series diverges slowly:
The 1/p series diverges very slowly:
It takes about a million terms to get
to a sum of 12.
A million terms gets you only to
about 3.07.
But with either series, if you add up enough terms, you can make the sum as
large as you want.
21
?
Estimating π(x) Using Probability
?
?
M ??
Pick a random X with X > M and X “near” M .
What’s the probability that you get a prime?
(
)
1
1
Pr(X is not divisible by 2) = 2 = 1 − 2
(
)
Pr(X is not divisible by 3) = 32 = 1 − 13
(
)
4
1
Pr(X is not divisible by 5) = 5 = 1 − 5
...
Pr(X is not divisible by any prime in PM )
?
?
?
)
∏ (
1
1
=
1−
= ∑ .
1
p
p∈PM
n
n∈ΛM
22
Heuristic 1:
Example
Among the numbers between M and
M + ∆M , the proportion of primes
is about
1
∑ 1.
n
n∈ΛM
n∈Λ5 000 000
and
There are approximately
∆M
∑ 1
n∈ΛM
100
∑ 1 ≈ 3.6398
n
π(5 000 100) − π(5 000 000) = 4.
n
primes in the interval
[M, M +∆M ].
23
Heuristic 2:
We can approximate
M
∑ 1
∑
1
with
.
n
n
n=1
n∈ΛM
In fact,
M
∑
∑ 1
1
=
+
n
n
n=1
n∈ΛM
∑
n∈ΛM
n>M
1
.
n
What happens to the second summation on the right as M → ∞?
The number
M
∑
1
n=1
n
is called the M th harmonic number.
24
Heuristic 1 + Heuristic 2:
Example
There are approximately
∆M
M
∑
1
n=1
100
(2 000 000 ) ≈ 6.6287
∑ 1
primes
n
in the interval
[M, M + ∆M ].
n=1
n
and
π(2 000 100) − π(2 000 000) = 6.
25
Heuristic 3:
1.2
1.0
M
∑
1
n=1
n
0.8
0.6
≈ log(M + 1).
0.4
0.2
0
4
∑
1
1
∫
2
3
4
5
5
1
dx, which is log(5), plus the
n
x
1
n=1
dark red “steps,” whose total area is less than 1.
The picture shows that
is equal to
In fact, we have
]
[M
∑1
lim
− log(M + 1) → γ
M →∞
n
n=1
So for large M ,
M
∑
1
n=1
n
≈ log(M + 1) + γ.
where γ ≈ 0.5772.
26
Heuristic 1 + Heuristic 2 + Heuristic 3:
There are approximately
∆M
log(M + 1) + γ
primes in the interval
[M, M + ∆M ].
That is,
π(M + ∆M ) − π(M ) ≈
∆M
log(M + 1) + γ
or
π(M + ∆M ) − π(M )
≈
∆M
log(M
or
d
[π(M )] ≈
dM
log(M
1
+ 1) + γ
1
.
+ 1) + γ
27
Calculus?
In a Number Theory talk?
Now that we know (approximately) the derivative of π(x), we can use the
Fundamental Theorem of Calculus to find π(x) itself.
Using the fact that π(2) = 1, we get
40
∫
x
π(x) − π(2) ≈
2
so that
∫
π(x) ≈ 1 +
2
dt
log(t + 1) + γ
30
20
10
x
dt
.
log(t + 1) + γ
50
100
150
200
28
Comparing the approximations
π(x)
∫
1+
2
150
x
1
dt
log(t + 1) + γ
x
log(x)
100
50
200
400
600
800
1000
29
Our approximation
∫
π(x) ≈ 1 +
2
x
1
dt
log(t + 1) + γ
can be “simplified” to
e−γ [Ei(log(x + 1) + γ) − Ei(log(3) + γ)],
where Ei is the exponential integral function,
∫ ∞ −t
e
Ei(z) = −
dt.
t
−z
But it can’t be written in terms of elementary functions.
30
The good news is that you can use l’Hôpital’s rule to show that
(
)
∫ x
1
1+
dt
log(t
+
1)
+
γ
2
(
)
lim
= 1,
x→∞
x
log(x)
so by the real Prime Number Theorem,
our heuristic approximation also “works.” That is,
π(x)
) = 1.
x
1
dt
log(t
+
1)
+
γ
2
∫
lim (
x→∞
1+
31
Some numbers
x
x
log(x)
∫
1+
x
(log(x + 1) + γ)−1 dt
π(x)
2
106
72 382
75 182
78 498
109
48 254 942
49 394 762
50 847 532
1012
36 191 206 825
1015 28 952 965 460 217
36 807 618 212 37 607 912 018
29 338 656 062 654
???
32