ANALYTIC PROOF OF THE PRIME NUMBER THEOREM
RYAN SMITH, YUAN TIAN
Contents
1. Arithmetical Functions
2. Equivalent Forms of the Prime Number Theorem
3. The Relationship Between Two Asymptotic Relations
4. Dirichlet Series and Euler Products
5. Contour Integral Representation of ψ1 (x)/x2
6. ζ(s) near and on the Line σ = 1
7. Completion of the Proof of the Prime Number Theorem
References
1
3
6
7
9
12
16
18
Let π(n) be the prime counting function, that is, the function that gives the
number of primes less than or equal to n. The analytic proof of the prime number
theorem can then be summarized as follows:
1
π(n) log n
ψ1 (x) ∼ x2 as x → ∞ ⇒ ψ(x) ∼ x as x → ∞ ⇐⇒ lim
=1
n→∞
2
n
Z x
X
where ψ1 (x) =
ψ(t)dt, ψ(x) =
Λ(n) and
1
n≤x
(
Λ(n) =
log p if n = pm for some prime p and some m ≥ 1,
0
otherwise.
We first prove the last part, namely
(1)
ψ(x) ∼ x as x → ∞ ⇐⇒ lim
n→∞
π(n) log n
= 1.
n
1. Arithmetical Functions
We begin by defining arithmetical functions.
Definition. A real- or complex-valued function defined on the positive integers is
called an arithmetical function or a number-theoretic function.
The functions π(n) and Λ(n) referred to above are examples of arithmetical functions. We now formally define the functions Λ(n) and ψ(n).
Definition (Mangoldt function). For every integer n ≥ 1 we define
(
log p if n = pm for some prime p and some m ≥ 1,
Λ(n) =
0
otherwise.
1
2
SMITH & TIAN
Definition (Chebyshev’s ψ-function). For x > 0 we define Chebyshev’s ψ-function
by the formula
X
Λ(n).
ψ(x) =
n≤x
The following theorem will be useful later in our proof of the theorem.
Theorem 1 (Abel’s identity). For any arithmetical function a(n) let
X
a(n)
A(x) =
n≤x
where A(x) = 0 if x < 1. Assume f has a continuous derivative on the interval
[y, x], where 0 < y < x. Then
Z y
X
a(n)f (n) = A(x)f (x) − A(y)f (y) −
A(t)f 0 (t)dt.
(2)
x
y<n≤x
Proof. Let k = [x] and m = [y], so that A(x) = A(k) and A(y) = A(m). Then
X
a(n)f (n) =
k
X
n=m+1
y<n≤x
=
k
X
k−1
X
{A(n) − A(n − 1)} f (n)
n=m+1
A(n)f (n) −
k−1
X
A(n)f (n + 1)
n=m
n=m+1
=
k
X
a(n)f (n) =
A(n) {f (n) − f (n + 1)} + A(k)f (k) − A(m)f (m + 1)
n=m+1
=−
k−1
X
Z
=−
f 0 (t)dt + A(k)f (k) − A(m)f (m + 1)
n
n=m+1
k−1
X
n+1
A(n)
Z
n+1
A(t)f 0 (t)dt + A(k)f (k) − A(m)f (m + 1).
n
n=m+1
Now, observe that
A(k)f (k) = A(k)f (k) + A(k)f (x) − A(k)f (x) = A(x)f (x) − A(x) (f (x) − f (k))
Z x
Z x
0
= A(x)f (x) − A(x)
f (t)dt = A(x)f (x) −
A(t)f 0 (t)dt.
k
k
Similarly,
Z
m+1
A(t)f 0 (t)dt.
A(m)f (m + 1) = A(y)f (y) +
y
∴
X
y<n≤x
Z
k
a(n)f (n) = −
A(t)f 0 (t)dt + A(x)f (x) −
m+1
Z
x
A(t)f 0 (t)dt
k
Z
− A(y)f (y) −
m+1
A(t)f 0 (t)dt
y
Z
=A(x)f (x) − A(y)f (y) −
x
A(t)f 0 (t)dt.
y
Now we introduce another arithmetical function which we will use in the proof.
ANALYTIC PROOF OF THE PRIME NUMBER THEOREM
3
Definition (Chebyshev’s ϑ-function). If x > 0 we define Chebyshev’s ϑ-function
by the equation
X
ϑ(x) =
log p,
p≤x
where p runs over all primes less than or equal to x.
Also if x > 0 we let π(x) denote the number of primes not exceeding x. We will
now examine the relationship between ϑ(x) and π(x).
Theorem 2. For x ≥ 2 we have
Z
ϑ(x) = π(x) log x −
(3)
2
x
π(t)
dt
t
and
ϑ(x)
+
π(x) =
log x
(4)
Z
2
x
ϑ(t)
dt.
t log2 t
Proof. Let a(n) denote the characteristic function of the primes; that is
(
1 if n is prime,
a(n) =
0 otherwise.
Then we have
π(x) =
X
X
1=
p≤x
a(n),
1<n≤x
ϑ(x) =
X
log p =
n≤x
X
a(n) log n.
1<n≤x
Now we take f (x) = log x with y = 1 in Abel’s identity in Theorem 1 to get
Z x
X
X
π(t)
ϑ(x) =
log p =
a(n) log n = π(x) log x − π(1) log 1 −
dt,
t
1
n≤x
1<n≤x
which proves (3) since π(t) = 0 for t < 2.
Next, let b(n) = a(n) log n and write
π(x) =
X
3/2<n≤x
b(n)
1
,
log n
ϑ(x) =
X
b(n).
n≤x
Takeing f (x) = 1/ log x with y = 3/2 in Abel’s identity we obtain
Z x
ϑ(x)
ϑ(3/2)
ϑ(t)
π(x) =
−
+
2 dt,
log x log 3/2
t
log
t
3/2
which proves (4) since ϑ(t) = 0 if t < 2.
2. Equivalent Forms of the Prime Number Theorem
We will now show some equivalent forms of the prime number theorem. Before
actually showing these equivalencies, we will first introduce a new notation.
4
SMITH & TIAN
Definition (The big oh notation). If g(x) > 0 for all x ≥ a, we write
f (x) = O (g(x)) (read: “f (x) is big oh of g(x)”)
to mean that the quotient f (x)/g(x) is bounded for x ≥ a; that is, there exists a
constant M > 0 such that
|f (x)| ≤ M g(x) for all x ≥ a.
An equation of the form
f (x) = h(x) + O (g(x))
means
that f (x)
O (g(x)). We note that f (t) = O (g(t)) for t ≥ a implies
Z− xh(x) = Z x
g(t)dt for all x ≥ a.
f (t)dt = O
a
a
Theorem 3. The following relations are logically equivalent:
(5)
π(x) log x
= 1.
x→∞
x
(6)
ϑ(x)
= 1.
x→∞ x
lim
lim
(7)
lim
x→∞
ψ(x)
= 1.
x
Proof. We show this by first proving that (5) and (6) are equivalent and then that
(6) and (7) are equivalent. Recall equations (3) and (4) - from these we obtain,
respectively
Z
ϑ(x)
π(x) log x 1 x π(t)
=
−
dt
x
x
x 2
t
and
Z
π(x) log x
ϑ(x) log x x ϑ(t)
=
+
2 dt.
x
x
x
2 t log t
To show that (5) implies (6) we need only show that (5) implies
Z
1 x π(t)
dt = 0.
lim
x→∞ x 2
t
π(t)
1
But (5) implies
=O
for t ≥ 2 so
t
log t
Z x
Z
1 x π(t)
1
dt
dt = O
.
x 2
t
x 2 log t
Now
Z
2
so
x
√
Z x
Z x
dt
dt
dt
√
≤
+
√ log t
log
2
log
x
2
x
2
2
√
√
√
√
x− x
x−2 x− x
x
√ ≤
√
≤
+
+
log 2
log 2
log x
log x
dt
=
log t
Z
x
dt
+
log t
Z
x
Z
1 x dt
→ 0 as x → ∞.
x 2 log t
This shows that (5) implies (6).
ANALYTIC PROOF OF THE PRIME NUMBER THEOREM
5
To show that (6) implies (5) we need only show that (6) implies
Z
log x x ϑ(t)
lim
2 dt = 0.
x→∞ x
2 t log t
But (6) implies ϑ(t) = O(t) so
Z
Z
log x x ϑ(t)
log x x dt
.
2 dt = O
2
x
x
2 t log t
2 log t
Now
Z
2
x
√
dt
=
log2 t
Z
x
2
dt
+
log2 t
Z
x
√
x
√
√
x− x
dt
x
≤
+
√
log2 t
log2 2 log2 x
hence
log x
x
Z
x
2
dt
→ 0 as x → ∞.
log2 t
This proves that (6) implies (5), so (5) and (6) are equivalent. To show that (6)
and (7) are equivalent, we first notice that
X
(8)
ψ(x) =
ϑ(x1/m ).
m≤log2 x
Hence we have
X
ψ(x) − ϑ(x) =
ϑ(x1/m ) ≥ 0.
2≤m≤log2 x
But from the definition of ϑ(x) we have the trivial inequality
X
ϑ(x) ≤
log x ≤ x log x
p≤x
so
√
√
x1/m log(x1/m ) ≤ (log2 x) x log x
X
0 ≤ ψ(x) − ϑ(x) ≤
2≤m≤log2 x
=
√
√
x
x log2 x
log x
·
log x =
.
log 2 2
2 log 2
Now we divide all terms in the inequality by x to obtain
(9)
0≤
ψ(x) ϑ(x)
log2 x
−
≤ √
.
x
x
2 x log 2
But (9) implies that
lim
x→∞
ψ(x) ϑ(x)
−
x
x
which implies that (6) and (7) are equivalent.
=0
6
SMITH & TIAN
3. The Relationship Between Two Asymptotic Relations
Now we will show that the asymptotic relation
1
(10)
ψ1 (x) ∼ x2 as x → ∞
2
implies the asymptotic relation
ψ(x) ∼ x as x → ∞.
We will later show why (10) is true using properties of the Riemann zeta function.
We start with proving the following lemma.
Lemma 1. For any arithmetical function a(n) let
X
A(n) =
a(n),
n≤x
where A(x) = 0 if x < 1. Then
Z
X
(11)
(x − n)a(n) =
x
A(t)dt.
1
n≤x
Proof. We apply Abel’s identity (Theorem 1) with y = 1 to obtain
Z x
X
(12)
a(n)f (n) = A(x)f (x) −
A(t)f 0 (t)dt
1
n≤x
if f has a continuous derivative on [1, x]. Taking f (t) = t we have
X
a(n)f (n) =
n≤x
X
na(n)
X
and A(x)f (x) = x
n≤x
a(n)
n≤x
so (12) reduces to (11).
Z
X
Lemma 2. Let A(x) =
a(n) and let A1 (x) =
x
A(t)dt. Assume also that
1
n≤x
a(n) ≥ 0 for all n. If we have the asymptotic formula
(13)
A1 (x) ∼ Lxc as x → ∞
for some c > 0 and L > 0, then we also have
(14)
A(x) ∼ cLxc−1 as x → ∞.
In other words, formal differentiation of (13) gives a correct result.
Proof. The function A(x) is increasing since the a(n) are nonnegative. Choose any
β > 1 and consider the difference A1 (βx) − A1 (x). We have
Z βu
Z βu
A1 (βx) − A1 (x) =
A(u)du ≥
A(x)du = xA(x)(β − 1).
u
x
This gives us
1
{A1 (βx) − A1 (x)}
β−1
A(x)
1
A1 (βx) c A1 (x)
≤
β −
.
xc−1
β−1
(βx)c
xc
xA(x) ≤
or
ANALYTIC PROOF OF THE PRIME NUMBER THEOREM
7
Keep β fixed and let x → ∞ in this inequality. By (13) we find
lim sup
x→∞
A(x)
1
βc − 1
≤
(Lβ c − L) = L
.
c−1
x
β−1
β−1
Now let β → 1+. The quotient on the right is the difference quotient for the
derivative of xc at x = 1 and has the limit c. Therefore
(15)
lim sup
x→∞
A(x)
≤ cL.
xc−1
Now we consider any α with 0 < α < 1 and consider the difference A1 (x) − A1 (αx).
An argument similar to the above shows that
A(x)
1 − αc
≥
L
.
xc−1
1−α
lim inf
x→∞
As α → 1− the right member tends to cL. This, together with (15) shows that
A(x)/xc−1 tends to cL as x → ∞.
When a(n) = Λ(n), a(n) ≥ 0 holds and we have A(x) = ψ(x) and A1 (x) = ψ1 (x).
Therefore, we can apply Lemma 1 and Lemma 2 to obtain:
Theorem 4.
(16)
ψ1 (x) =
X
(x − n)Λ(n).
n≤x
Also, the asymptotic relation ψ1 (x) ∼ x2 /2 implies ψ(x) ∼ x as x → ∞.
4. Dirichlet Series and Euler Products
In this section, we will introduce some basic ideas in Dirichlet series and Euler
products which we will use later in our proof of the prime number theorem. A
Dirichlet series is a series of the form
∞
X
f (n)
ns
n=1
where f (n) is an arithmetical function. We call the f (n) the coefficients of the
corresponding Dirichlet series. We introduce the Riemann zeta function here as an
example of a dirichlet series.
Definition (Riemann zeta function). For any s ∈ C, we define
ζ(s) =
∞
X
1
.
ns
n=1
We will now prove a couple of lemmas.
Lemma 3. Let s0 = σ0 + it0 and assume that the Dirichlet series
has bounded partial sums, say
X
f
(n)
≤M
s
0
n≤x n X
f (n)n−s0
8
SMITH & TIAN
for all x ≥ 1. Then for each s with σ > σ0 we have
X f (n) ≤ 2M aσ0 −σ 1 + |s − s0 | .
(17)
s σ − σ0
a<n≤b n Proof. Let a(n) = f (n)n−s0 and let A(n) =
X
a(n). Then f (n)n−s0 = a(n)ns0 −s
n≤x
so we can apply Theorem 1 (with f (x) = xs0 −s ) to obtain
Z b
X f (n)
s0 −s
s0 −s
=
A(b)b
−
A(a)a
+
(s
−
s
)
A(t)ts0 −s−1 dt.
0
ns
a
a<n≤b
Since |A(x)| ≤ M this gives us
Z b
X f (n) ≤ M bσ0 −σ + M aσ0 −σ + |s − s0 |M
tσ0 −σ−1 dt
s n
a
a<n≤b
bσ0 −σ−aσ0 −σ σ0 −σ
≤ 2M a
+ |s − s0 |M σ0 − σ |s − s0 |
≤ 2M aσ0 −σ 1 +
.
δ − δ0
Lemma 4. Let {fn } be a sequence of functions analytic on an open subset S of the
complex plane, and assume that {fn } converges uniformly on every compact subset
of S to a limit function f . Then f is analytic on S and the sequence of derivatives
{fn0 } converges uniformly on every compact subset of S to the derivative f 0 .
Proof. Since fn is analytic on S we have Cauchy’s integral formula
Z
fn (z)
1
dz
fn (a) =
2πi ∂D z − a
where D is any compact disk in S, ∂D is its positively oriented boundary, and a is
any interior point of D. Because of uniform convergence we can pass to the limit
under the integral sign and obtain
Z
1
f (z)
f (a) =
dz
2πi ∂D z − a
which implies that f is analytic inside D. For the derivative we have
fn0 (a)
1
=
2πi
Z
∂D
fn (z)
dz
(z − a)2
1
and f (a) =
2πi
0
Z
∂D
f (z)
dz
(z − a)2
from which it follows easily that fn0 (a) → f 0 (a) uniformly on every compact
subset of S as n → ∞.
Now we are ready to prove the following theorems.
X
Theorem 5. A Dirichlet series
f (n)n−s converges uniformly on every compact
subset lying interior to the half-plane of convergence σ > σ0 .
ANALYTIC PROOF OF THE PRIME NUMBER THEOREM
9
X
Proof. It suffices to show that
f (n)n−s converges uniformly on every compact
rectangle R = [α, β] × [c, d] with α in the half-plane of convergence (the half-plane
of convergence is simply the half-plane in which the series converges). To do this
we use the estimate obtained in Lemma 3
X f (n) ≤ 2M aσ0 −σ 1 + |s − s0 |
(18)
s σ − σ0
a<n≤b n where s0 = σ0 + it0 is any point in the half-plane where of convergence. We choose
s0 = σ0 where σ0 < α. Then if s ∈ C we have σ − σ0 ≥ α − σ0 and |s0 − s| < C,
where C is a constant depending on s0 and R but not on s. Then (18) implies
X
f
(n)
≤ 2M aσ0 −σ 1 + C
= Baσ0 −σ
s α − α0
a<n≤b n where B is independent of s. Since aσ0 −α → 0 as a → +∞ the Cauchy condition
for uniform convergence is satisfied.
X
Theorem 6. The sum function F (s) =
f (n)n−s of a Dirichlet series is analytic
in its half-plane of convergence, and its derivative F 0 (s) is represented in this halfplane by the Dirichlet series
(19)
F 0 (s) = −
∞
X
f (n) log n
,
ns
n=1
obtained by differentiating term by term.
Proof. We apply Theorem 5 and Lemma 4 to the sequence of partial sum.
We now apply the previous theorem to ζ(s).
Differentiating ζ(s) term by term and summing over all n will give us
∞
X
log n
.
ζ (s) = −
ns
n=1
0
Applying Theorem 6 to ζ(s) and ζ 0 (s) will give us
(20)
−
∞
ζ 0 (s) X Λ(n)
=
.
ζ(s)
ns
n=1
5. Contour Integral Representation of ψ1 (x)/x2
Our next goal is to prove the asymptotic relation in (10) by representing ψ1 (x)/x2
as a contour integral. For the contour integral representation of ψ1 (x)/x2 , we need
some knowledge about the gamma function Γ(s) defined as
Z ∞
(21)
Γ(s) =
xs−1 e−x dx
0
for s = σ + it where σ > 0. A particularly useful property of Γ(s) is given by the
following functional equation:
(22)
Γ(s + 1) = sΓ(s).
10
SMITH & TIAN
Another useful property of the gamma function is that the gamma function has
simple poles at non-positive integers with residue (−1)n /n! at −n. With these
observations, we can now proceed to prove the following lemma.
Lemma 5. If c > 0 and u > 0, then for every integer k ≥ 1 we have

Z c+∞i
 1 (1 − u)k if 0 < u ≤ 1,
1
u−z
dz = k!
0
2πi c−∞i z(z + 1) · · · (z + k)
if u > 1,
the integral being absolutely convergent.
Proof. First we note that the integrand is equal to u−z Γ(z)/Γ(z + k + 1). This
follows by repeated use of the functional equation in (22). To prove the lemma we
apply Cauchy’s residue theorem to the integral
Z
u−z Γ(z)
1
dz,
2πi C(R) Γ(z + k + 1)
where C(R) is the contour shown in the following graph (a) if 0 < u ≤ 1 and (b)
if u > 1. The radius R of the circle is greater than 2k + c so that all the poles at
z = 0, −1, · · · , −k lie inside the circle.
_________________________
•
• • 
c
−2k −2 −1 









*
R





R










_________________________
c
J
(a) 0 < u ≤ 1
(b) u > 1
O
Now we show that the integral along each of the circular arcs tends to 0 as
R → ∞. If z = x + iy and |z| = R the integrand is dominated by
u−c
u−z
u−x
z(z + 1) · · · (z + k) = |z||z + 1| · · · |z + k| ≤ R|z + 1| · · · |z + k| .
The inequality u−x ≤ u−c follows from the fact that u−x is an increasing function
of x if 0 < u ≤ 1 and a decreasing function if u > 1. Now if 1 ≤ n ≤ k we have
R
|z + n| ≥ |z| − n = R − n ≥ R − k ≥
2
since R > 2k. Therefore the integral along each circular arc is dominated by
2πRu−c
= O R−k
k
R 21 R
and this tends to 0 as R → ∞ since k ≥ 1.
ANALYTIC PROOF OF THE PRIME NUMBER THEOREM
11
Z
(R) = 0. Letting R → ∞
If u > 1 the integrand is analytic inside C(R) hence
C
we find that the lemma is proved in this case.
If 0 < u ≤ 1 we evaluate the integral around C(R) by Cauchy’s residue theorem.
The integrand has poles at the integers n = 0, −1, · · · , −k, hence
1
2πi
Z
C(R)
k
X
u−z Γ(z)
u−z Γ(z)
dz =
Res
z=−n Γ(z + k + 1)
Γ(z + k + 1)
n=0
k
X
k
X
un
un (−1)n
Res Γ(z) =
Γ(k + 1 − n) z=−n
(k − n)!n!
n=0
n=0
k
1 X k
(1 − u)k
=
(−u)n =
.
k! n=0 n
k!
=
Letting R → ∞ we obtain the lemma.
Now we are ready to represent ψ1 (x)/x2 as a contour integral.
Theorem 7. If c > 1 and x ≥ 1 we have
0 Z c+∞i
1
xs−1
ζ (s)
ψ1 (x)
=
−
ds.
(23)
2
x
2πi c−∞i s(s + 1)
ζ(s)
X
Proof. From (16) we have ψ1 (x)/x =
(1 − n/x)Λ(n). Now use Lemma 5 with
n≤x
k = 1 and u = n/x. If n ≤ x we obtain
Z c+∞i
n
1
(x/n)s
1− =
ds.
x
2πi c−∞i s(s + 1)
Multiplying this relation by Λ(n) and summing over all n ≤ x we find
Z c+∞i
∞
X 1 Z c+∞i Λ(n)(x/n)s
X
1
ψ1 (x)
Λ(n)(x/n)s
=
ds =
ds
x
2πi c−∞i
s(s + 1)
2πi c−∞i
s(s + 1)
n=1
n≤x
since the integral vanishes if n > x. This can be written as
∞ Z
ψ1 (x) X c+∞i
1 Λ(n)(x/n)s
(24)
=
fn (s) ds, where fn (s) =
·
.
x
2πi
s(s + 1)
n=1 c−∞i
Next we wish to interchange the sum and integral in (24). For this it suffices to
prove that the series
∞ Z c+∞i
X
(25)
|fn (s)| ds
n=1
c−∞i
is convergent. The partial sums of this series satisfy the inequality
N Z
X
n=1
c+∞i
c−∞i
Z
N
∞
X
X
Λ(n) c+∞i
xc
Λ(n)
Λ(n)(x/n)c
ds =
ds
≤
A
,
c
|s||s + 1|
n
nc
c−∞i |s||s + 1|
n=1
n=1
12
SMITH & TIAN
where A is a constant, so (25) converges. Hence we can interchange the sum and
integral in (24) to obtain
Z c+∞i X
Z c+∞i
∞
∞
X
xs
ψ1 (x)
1
Λ(n)
ds
=
fn (s) ds =
x
2πi c−∞i s(s + 1) n=1 ns
c−∞i n=1
0 Z c+∞i
1
xs
ζ (s)
=
−
ds
2πi c−∞i s(s + 1)
ζ(s)
by (20). Now divide by x to obtain (23).
Theorem 8. If c > 1 and x ≥ 1 we have
2
Z c+∞i
1
1
ψ1 (x) 1
−
=
xs−1 h(s) ds,
1−
(26)
x2
2
x
2πi c−∞i
where
(27)
h(s) =
1
s(s + 1)
−
ζ 0 (s)
1
−
ζ(s)
s−1
.
Proof. This time we use Lemma 5 with k = 2 to get
2
Z c+∞i
xs
1
1
1
ds,
1−
=
2
x
2πi c−∞i s(s + 1)(s + 2)
where c > 0. Replace s by s − 1 in the integral (keeping c > 1) and subtract the
result from (23) to obtain Theorem 8.
Substituting s with c + it will yield
xs−1 = xc−1 xit = xc−1 eit log x
which converts (27) into
(28)
ψ1 (x) 1
−
x2
2
2
Z
1
xc−1 c+∞i
1−
=
h(c + it)eit log x dt.
x
2
c−∞i
Our next task is to show that the right member in (28) tends to 0 as x → ∞. We
want to first show that we can simply put c = 1 in (28). For this we need to study
ζ(s) in the neighborhood of the line σ = 1.
6. ζ(s) near and on the Line σ = 1
In this section, we will base our work on the following two facts. We will not
prove these facts, although the proofs can be easily found in books on analytic
number theory, complex analysis, or Fourier series (see references). First of all, for
all s = σ + it with σ > 0, we have
Z ∞
N
X
x − [x]
N 1−s
1
(29)
ζ(s) =
−
s
dx
+
.
ns
xs+1
s−1
N
n=1
Differentiating each member of (29) will give us the second fact that
Z ∞
Z ∞
N
X
log n
(x − [x]) log x
x − [x]
0
ζ (s) = −
+s
dx −
dx
s
s+1
n
x
xs+1
N
N
n=1
(30)
N 1−s log N
N 1−s
−
−
.
s−1
(s − 1)2
ANALYTIC PROOF OF THE PRIME NUMBER THEOREM
13
Now we are ready to obtain upper bounds for |ζ(s)| and |ζ 0 (s)|.
Theorem 9. For every A > 0 there exists a constant M (depending on A) such
that
|ζ(s)| ≤ M log t and |ζ 0 (s)| ≤ M log2 t
(31)
for all s with σ ≥ 1/2 satisfying
σ >1−
(32)
A
and t ≥ e.
log t
Proof. If σ ≥ 2 we have |ζ(s)| ≤ ζ(2) and |ζ 0 (s)| ≤ |ζ 0 (2)| and the inequalities in
(31) are trivially satisfied. Therefore we can assume σ < 2 and t ≥ e. We have
|s| ≤ σ + t ≤ 2 + t < 2t and |s − 1| ≥ t
so 1/|s − 1| ≤ 1/t. Estimating |ζ(s)| by using (29) we find
|ζ(s)| ≤
Z ∞
N
N
X
X
1
1
N 1−σ
2t
N 1−σ
1
+
2t
dx
+
=
+
+
.
σ
σ+1
σ
σ
n
t
n
σN
t
N x
n=1
n=1
Now we make N depend on t by taking N = [t]. Then N ≤ t < N + 1 and
log n ≤ log t if n ≤ N . The inequality (32) implies 1 − σ < A/ log t so
1
1
n1−σ
1 (1−σ) log n
1 A log n/ log t
1 A
=
=
e
<
e
≤
e
=
O
.
nσ
n
n
n
n
n
∴
N +1
N 1−σ
N 1
2t
≤
=
O(1)
and
=
≤O
σ
σN
N
t
t Nσ
N
X
1
∴ |ζ(s)| = O
n
n=1
1
N
= O(1),
!
+ O(1) = O(log N ) + O(1) = O(log t).
This proves the inequality for |ζ(s)| in (31). To obtain the inequality for |ζ 0 (s)| we
apply the same type of argument to (30). The only essential difference is that an
extra factor log N appears on the right. But log N = O(log t) so we get |ζ 0 (s)| =
O(log2 t) in the specified region.
Theorem 10. If σ > 1 we have
(33)
ζ 3 (σ)|ζ(σ + it)|4 |ζ(σ + 2it)| ≥ 1
Proof. Using the Euler product spanning over all primes and the Taylor series
expansion of log x, we get
(
)
(
)
(
)
∞
∞
X
YX
X
1
1
1
= exp log
= exp
log
ζ(s) = exp log
ns
pks
1 − 1/ps
p k=0
p
n=1
(
)
(
)
∞
X
XX
1
1
= exp −
log 1 − s
= exp
.
ms
p
mp
p
p m=1
14
SMITH & TIAN
(
∴ ζ(σ + it) = exp
∞
XX
)
1
(
∞
XX
1
)
= exp
mpmσ pimt
mpm(σ+it)
p m=1
(
)
(
)
−imt
∞
∞
XX
XX
p−imt
elog p
= exp
= exp
mpσ
mpmσ
p m=1
p m=1
(
)
)
(
∞
∞
XX
XX
cos(mt log p)
e−imt log p
= exp
.
= exp
mpmσ
mpmσ
p m=1
p m=1
(
)
∞
XX
3
∴ ζ 3 (σ) = exp
.
mpmσ
p m=1
)
(
∞
XX
4
cos(mt
log
p)
.
|ζ(σ + it)|4 = exp
mpmσ
p m=1
)
(
∞
XX
cos(2mt log p)
.
|ζ(σ + 2it)| = exp
mpmσ
p m=1
p m=1
Since mpmσ > 0, it suffices to show that
3 + 4 cos(mt log p) + cos(2mt log p) ≥ 0.
3 + 4 cos(mt log p) + cos(2mt log p) = 2 cos2 (mt log p) + 4 cos(mt log p) + 2
2
= 2 (cos(mt log p) + 1) ≥ 0.
Hence the theorem is proved.
Theorem 11. ∀t ∈ R, ζ(1 + it) 6= 0.
Proof. We only need to consider t 6= 0. Rewrite the previous theorem by dividing
both sides by σ − 1.
4
1
3 ζ(σ + it) (34)
{(σ − 1)ζ(σ)} |ζ(σ + 2it)| ≥
.
(σ − 1) σ−1
This is valid if σ > 1. Now let σ → 1+ in (34). The first factor approaches 1
since ζ(s) has residue 1 at the pole s = 1. The third factor tends to |ζ(1 + 2it)|. If
ζ(1 + it) were equal to 0, the middle factor could be written as
ζ(σ + it) − ζ(1 + it) 4
→ |ζ 0 (1 + it)|4 as σ → 1 + .
σ−1
Therefore, if for some t 6= 0 we had ζ(1 + it) = 0, the left member of (34) would
approach the limit
|ζ 0 (1 + it)|4 |ζ(1 + 2it)|
as σ → 1+. But the right member tends to ∞ as σ → 1+ and this gives a
contradiction.
Theorem 12. There is a constant M > 0 such that
1 7
and
ζ(s) < M log t
whenever σ ≥ 1 and t ≥ e.
0 ζ (s) 9
ζ(s) < M log t
ANALYTIC PROOF OF THE PRIME NUMBER THEOREM
15
Proof. For σ ≥ 2 we have
∞
∞
X
1 X
µ(n)
1
=
≤
≤ ζ(2)
ζ(s) s 2
n
n
n=1
n=1
and by (20) we have
0 X
∞
ζ (s) Λ(n)
=
,
ζ(s) n2
n=1
so the inequalities hold trivially if σ ≥ 2. Suppose, then, that 1 ≤ σ ≤ 2 and t ≥ e.
Rewrite Theorem 10 by dividing both sides by |ζ(σ + it)| and take the fourth root
of both sides.
1
∴
≤ ζ(σ)3/4 |ζ(σ + 2it)|1/4 .
|ζ(σ + it)|
Now (σ − 1)ζ(σ) is bounded by, say, M in the interval 1 ≤ σ ≤ 2, where M is an
absolute constant.
M
if 1 < σ ≤ 2.
∴ ζ(σ) ≤
σ−1
Also, ζ(σ + 2it) = O(log t) if 1 ≤ σ ≤ 2 by Theorem 9. So for 1 < σ ≤ 2 we have
A(log t)1/4
1
M 3/4 (log t)1/4
=
,
≤
3/4
|ζ(σ + it)|
(σ − 1)
(σ − 1)3/4
where A is an absolute constant. Therefore for some constant B > 0 we have
(35)
|ζ(σ + it)| >
B(σ − 1)3/4
, if 1 < σ ≤ 2 and t ≥ e.
(log t)1/4
This also holds trivially for σ = 1. Let α be any number satisfying 1 < α < 2.
Then if 1 ≤ σ ≤ α, t ≥ e, we may use Theorem 9 to write
Z α
|ζ(σ + it) − ζ(α + it)| ≤
|ζ 0 (u + it)|du ≤ (α − σ)M log2 t ≤ (α − 1)M log2 t
σ
Hence, by the triangle inequality,
|ζ(σ + it)| ≥ |ζ(α + it)| − |ζ(σ + it) − ζ(α + it)|
≥ |ζ(α + it)| − (α − 1)M log2 t ≥
B(α − 1)3/4
− (α − 1)M log2 t
(log t)1/4
This holds if 1 ≤ σ ≤ α, and by (35) it also holds for α ≤ σ ≤ 2 since (σ − 1)3/4 ≥
(α − 1)3/4 . In other words, if 1 ≤ σ ≤ 2 and t ≥ e we have the inequality
|ζ(σ + it)| ≥
B(α)3/4
− (α − 1)M log2 t
(log t)1/4
for any α ∈ (1, 2). Now we make α depend on t and choose α so that the first term
on the right is twice the second. This requires
4
1
B
α=1+
.
2M
(log t)9
Clearly α > 1 and also α < 2 if t ≥ t0 for some t0 . Thus, if t ≥ t0 and 1 ≤ σ ≤ 2
we have
C
.
|ζ(σ + it)| ≥ (α − 1)M log2 t =
(log t)7
16
SMITH & TIAN
This inequality also holds with (perhaps) a different C if e ≤ t ≤ t0 .
This proves that |ζ(s)| ≥ C log−7 t for all σ ≥ 1, t ≥ e, giving us a corresponding
upper bound for |1/ζ(s)|. To get the inequality for |ζ 0 (s)/ζ(s)| we apply Theorem 9
and obtain an extra factor of log2 t.
Now we are ready to finish the analytic proof of the prime number theorem,
which we do by proving that the asymptotic relation in (10) holds. We start by
proving the following lemma.
7. Completion of the Proof of the Prime Number Theorem
Lemma 6. If f (s) has a pole of order k at s = α then the quotient f 0 (s)/f (s) has
a first order pole at s = α with residue −k.
Proof. We have f (s) = g(s)/(s − α)k , where g is analytic at α and g(α) 6= 0. Hence
for all s in a neighborhood of α we have
g 0 (s)
kg(s)
g(s)
−k
g 0 (s)
0
f (s) =
−
=
+
.
(s − α)k
(s − α)k+1
(s − α)k s − α
g(s)
∴
−k
g 0 (s)
f 0 (s)
=
+
.
f (s)
s−α
g(s)
This proves the lemma since g 0 (s)/g(s) is analytic at α.
Theorem 13. The function
F (s) = −
ζ 0 (s)
1
−
ζ(s)
s−1
is analytic at s = 1.
Proof. By Lemma 6, −ζ 0 (s)/ζ(s) has a first order pole at 1 with residue 1, as does
1/(s − 1). Hence their difference is analytic at s = 1.
The final part of the proof makes use of the Riemann-Lebesgue
Z ∞ lemma from
Fourier analysis, which we do not prove here. It states that if
|f (x)| dx con−∞
Z ∞
Z ∞
verges, then lim
f (x) sin rx dx = 0 and lim
f (x) cos rx dx = 0 hold, and
r→∞ −∞
r→∞ −∞
Z ∞
therefore, so too does lim
f (x)eirx dx = 0.
r→∞
−∞
Theorem 14. For x ≥ 1 we have
2
Z ∞
1
1
ψ1 (x) 1
−
1
−
=
h(1 + it)eit log x dt,
x2
2
x
2π −∞
Z ∞
where the integral
|h(1+it)| dt converges. Therefore, by the Riemann-Lebesgue
−∞
lemma we have
(36)
ψ1 (x) ∼
x2
and hence ψ(x) ∼ x as x → ∞.
2
ANALYTIC PROOF OF THE PRIME NUMBER THEOREM
17
Proof. In Theorem 8 we proved that if c > 1 and x ≥ 1 we have
2
Z c+∞i
ψ1 (x) 1
1
1
−
=
xs−1 h(s) ds,
1
−
x2
2
x
2πi c−∞i
where
0
1
ζ (s)
1
h(s) =
−
−
.
s(s + 1)
ζ(s)
s−1
Our first task is to show that we can move the path of integration to the line σ = 1.
To do this we apply Cauchy’s theorem to the rectangle R shown below.
tO
TO
•
o
•
0
c
1
/σ
O
−T
•
/
•
The integral of xs−1 h(s) around R is 0 since the integrand is analytic inside and
on R. Now we show that the integrals along the horizontal segments tend to 0 as
T → ∞. Since the integrand has the same absolute value at conjugate points, it
suffices to consider only the upper segment, t = T . On this segment, we have the
estimates
1
1
≤ 1
≤ 1 ≤ 1 .
and s(s + 1) T 2
s(s + 1)(s − 1) T 3
T2
Also, there is a constant M such that |ζ 0 (s)/ζ(s)| ≤ M log9 t if σ ≥ 1 and t ≥ e.
Hence if T ≥ e we have
M log9 T
|h(s)| ≤
T2
so that
Z c
Z c
9
M log9 T
s−1
c−1 log T
xc−1
x h(s) ds ≤
dσ
=
M
x
(c − 1).
T2
T2
1
1
18
SMITH & TIAN
Therefore the integrals along the horizontal segments tend to 0 as T → ∞, and
hence we have
Z c+∞i
Z 1+∞i
xs−1 h(s) ds =
xs−1 h(s) ds.
c−∞i
1−∞i
On the line σ = 1 we write s = 1 + it to obtain
Z 1+∞i
Z ∞
1
1
s−1
x h(s) ds =
h(1 + it)eit log x dt.
2πi 1−∞i
2π −∞
Now we note that
Z
Z ∞
|h(1 + it)| dt =
e
Z
−e
−∞
Z ∞
∞
Z
−e
|h(1 + it)| dt.
|h(1 + it)| dt +
|h(1 + it)| dt +
e
−∞
|h(1 + it)| dt we have
For
e
M log9 t
|h(1 + it)| ≤
t
Z ∞
Z −e
so
|h(1 + it)| dt converges. Similarly,
|h(1 + it)| dt converges, so the entire
e
−∞
Z ∞
integral
|h(1+it)| dt converges. By the Riemann-Lebesgue lemma, this implies
−∞
that
Z
∞
h(1 + it)eit log x dt = 0
lim
x→∞
−∞
This gives us the asymptotic relation:
x2
⇒ ψ(x) ∼ x as x → ∞.
2
Hence the prime number theorem is proved.
ψ1 (x) ∼
References
[1] Apostol, T. M., Introduction to Analytic Number Theory, Springer-Verlag, New York, 1976
[2] Brown, J. W., Churchill, R. V., Fourier Series and Boundary Value Problems, McGraw-Hill,
Columbus, OH, 2000.
[3] Conway, J. B., Functions of One Complex Variable, Springer, New York, 1986.