Math 680 Fall 2014
The Prime Number Theorem
Our goal is to prove the Prime Number Theorem, which is simply expressed as
π(x)
= 1.
x/ log x
lim
x→∞
(We’ll look at more precise statements later.) Recall that we previously proved that
lim sup
x→∞
Ψ(x)
π(x)
= lim sup
.
x
x→∞ x/ log x
Thus, for our purposes it will suffice to show that
Ψ(x)
= 1.
x
lim
x→∞
P
Towards that end, we have Ψ(x) = n≤x Λ(x). In other words, Ψ(x) is the “A(x)” for the Dirichlet
P
series n≥1 Λ(n)n−s . Via the Euler product, we’ve seen that this is the Dirichlet series for −ζ 0 (s)/ζ(s), and
has abscissa of convergence σc = 1. Now using Theorem 2 from the handout on Dirichlet series, we have
−
ζ 0 (s)
=s
ζ(s)
∞
Z
Ψ(u)u−(s+1) du
1
for all s = σ + it with σ > 1. Via the change of variables u = ex , we have
ζ 0 (s)
=
−
sζ(s)
(1)
∞
Z
Ψ(ex )e−sx dx
0
again for all s with σ > 1.
At the end of the Euler product handout we discussed the analytic properties of the function on the
left side of (1). Specifically, it has a simple pole (with residue 1) at s = 1. Any other poles must come from
zeros of the zeta function.
Theorem (Hademard and de la Vallée Poussin): We have ζ(1 + it) 6= 0 for all t 6= 0.
Corollary: The function
−
ζ 0 (s)
sζ(s)
is analytic on the set {s = σ + it : σ ≥ 1}, with the sole exception of a simple pole at s = 1 of residue 1.
We’ll prove the theorem of Hademard and de la Vallée Poussin shortly, but first see the connection to
the Prime Number Theorem.
Theorem (Wiener-Ikehara): Suppose A(x) ≥ 0 is non-decreasing on [0, ∞) and
Z
f (s) :=
∞
A(x)e−sx dx
0
1
is analytic on the set of s with <(s) ≥ 1 except for a simple pole of residue 1 at s = 1. Then
lim A(x)e−x = 1.
x→∞
Corollary(Prime Number Theorem): We have
π(x)
Ψ(x)
Ψ(ex )
= 1.
= lim
= lim
x→∞ x/ log x
x→∞
x→∞
x
ex
lim
We postpone the proof of this theorem for the time being, but note that ultimately it turns out that
the Prime Number Theorem follows (via the Euler product and more generic analysis) from the fact that
the zeta function is non-vanishing on the set of s with <(s) ≥ 1. In fact, the converse holds as well.
Proposition 1: Assuming that
lim
x→∞
π(x)
= 1,
x/ log x
we have ζ(σ + it) 6= 0 for all σ ≥ 1.
Proof: Let > 0. By the hypothesis, we have
Ψ(x)
= 1,
x→∞
x
lim
too, so that |Ψ(x) − x| < x for all x ≥ x() (a bound depending on ). Set
Φ(s) := −
1
ζ 0 (s)
−
.
sζ(s) s − 1
By (1) we have
Z
∞
Φ(s) =
1
Ψ(x) − x
dx
xs+1
for all s with <(s) > 1. Further, Φ is analytic on the half-plane {s = σ + it : σ ≥ 1} except for any zeros of
the zeta function that occur in this half-plane. As long as <(s) = σ > 1, we have
x()
|Ψ(x) − x|
|Φ(s)| <
dx +
x2
1
Z ∞
x−σ dx
< c()1 + Z
Z
∞
x()
dx
xσ
1
= c() + (σ − 1)−1 ,
where the c() depends only on . Hence |Φ(σ + it)|(σ − 1) ≤ c()(σ − 1) + for all > 0 and σ > 1, implying
that
lim |Φ(σ + it)|(σ − 1) = 0
σ→1+
for all t. In particular, Φ(s) has no poles on the line s = 1 + it except at s = 1, so that ζ(s) has no zeros on
this line.
2
We now prove the theorem of Hademard and de la Vallée Poussin. From the Euler product we get
!
Y
−s −1
log ζ(s) = log
(1 − p )
p 0
X
=−
log(1 − p−s )
p prime
=
X X
p prime m≥1
=
1
mp−ms
X an
ns
n≥2
for all s = σ + it with σ > 1, where
an =
1
k
if n = pk for some prime p,
0
otherwise.
Note that the coefficients of this Dirichlet series are non-negative. Using this series expansion, we have


X an

log |ζ(s)| = < log ζ(s) = < 
ns
n≥2
X an cos(t log n)
=
nσ
n≥2
for s = σ + it with σ > 1. In particular,
log |ζ 3 (σ)ζ 4 (σ + it)ζ(σ + 2it)| = 3 log |ζ(σ)| + 4 log |ζ(σ + it)| + log |ζ(σ + 2it)|
X an
=
(3 + 4 cos(t log n) + cos(2t log n)).
nσ
n≥2
Via the trigonometric identity
3 + 4 cos θ + cos 2θ = 2(1 + cos θ)2 ≥ 0,
we see that the terms of the series here are all non-negative, so that |ζ 3 (σ)ζ 4 (σ + it)ζ(σ + 2it)| ≥ 1 for all
σ > 1. Dividing by σ − 1 yields
(σ − 1)3 |ζ 3 (σ)|
(2)
|ζ 4 (σ + it)|
1
|ζ(σ + 2it)| ≥
(σ − 1)4
σ−1
for all σ > 1.
Now suppose ζ(1 + it0 ) = 0 for some t0 6= 0. We know that
(3)
lim (σ − 1)3 |ζ 3 (σ)| = 1.
σ→1+
Also, since ζ(s) is analytic on the set {s = σ + it : σ ≥ 1} \ {1},
(4)
lim |ζ(σ + 2it0 )| = |ζ(1 + 2it0 )|
σ→1+
3
and
lim+
(5)
σ→1
4
ζ(σ + it0 ) |ζ 4 (σ + it0 )| lim
=
σ→1+ (σ − 1) (σ − 1)4
= |ζ 0 (1 + 2it0 )|4 .
But taken together, (3)-(5) contradict (2), so that ζ(1 + it) 6= 0 for all t 6= 0.
Before proving the theorem of Wiener and Ikehara, we need a few auxiliary results.
Lemma 1: Let A(x) and f (s) be as in the statement of the theorem of Wiener and Ikehara. Set
B(x) = e−x A(x) and
g(s) =
f (s) −
0
1
s−1
if s 6= 1,
if s = 1.
Then g(s) is analytic on {s = σ + it : σ ≥ 1} and for all s with <(s) > 1,
Z ∞
B(x) − 1 e−(s−1)x dx.
g(s) =
0
This integral converges uniformly on any compact subset of {s = σ + it : σ > 1}.
Proof: The analytic property of g follows from the hypotheses on f . For all s with <(s) > 1,
Z ∞
Z ∞
B(x)e−(s−1)x dx =
A(x)e−sx dx
0
0
= f (s)
and
∞
−e−(s−1)x −e
dx =
s − 1 x=0
0
−1
=
.
s−1
Let C be a compact subset of {s = σ + it : σ > 1}. Then there is a δ > 1 such that <(s) ≥ δ for all
Z
∞
−(s−1)x
s ∈ C. Let > 0. By what we have already shown,
Z ∞
B(x)e−(δ−1)x dx = f (δ)
0
Z ∞
1
e−(δ−1)x dx =
,
δ
−
1
0
so that there is a c > 0 such that
Z
max
∞
−(δ−1)x
B(x)e
Z
∞
dx,
M
e
−(δ−1)x
dx < /2
M
for all M ≥ c. Now for all s ∈ C we have
Z ∞
Z ∞
−(s−1)x B(x) − 1 e
dx ≤
B(x) − 1 e−(s−1)x dx
M
ZM∞
B(x) − 1 e−(δ−1)x dx
≤
ZM∞
Z ∞
≤
B(x)e−(δ−1)x dx +
e−(δ−1)x dx
M
M
< .
4
Lemma 2: Suppose θ is a non-zero real number. Then for all b > 0
1
2
Z
2b
e
iθt
−2b
|t|
1−
2b
dt =
sin2 (bθ)
.
bθ2
Proof: Since eiθt = cos(θt) + i sin(θt), cosine is an even function and sine is an odd function,
1
2
2b
Z 2b
|t|
t
eiθt 1 −
dt =
cos(θt) 1 −
dt.
2b
2b
−2b
0
Z
We integrate by parts using u = 1 − t/2b and dv = cos(θt) dt, which yields
2b
Z
0
2b
Z 2b
t
1
t sin(θt) cos(θt) 1 −
+
sin(θt) dt
dt = 1 −
2b
2b
θ t=0 2bθ 0
2b
− cos(θt) =
2bθ2 t=0
1 − cos(2bθ)
=
2bθ2
2
sin (bθ)
=
.
bθ2
Lemma 3: Suppose h(t) is a (possibly) complex-valued continuous function on R and there is a C ≥ 0
such that h(t) = 0 for all t with |t| ≥ C. Then
∞
Z
h(t)eity dt = 0.
lim
y→±∞
−∞
Proof: Under the change of variables t = u + π/y we have du = dt and
Z
∞
h(t)e
ity
Z
∞
dt =
−∞
h(u + π/y)e
iy(u+π/y)
Z
∞
du = −
−∞
h(u + π/y)eiyu du,
−∞
so that
Z
∞
2
−∞
h(t)eity dt =
Z
∞
h(t) − h(t + π/y) eity dt.
−∞
Let > 0. Since h(t) is continuous, it is uniformly continuous on [−C − π, C + π]. Thus, there is a
bound c() > 0, depending only on , such that |h(t) − h(t + π/y)| < /(C + π) for all t ∈ [−C − π, C + π]
and |y| > c(). We may assume, without loss of generality, that c() ≥ 1. But then both h(t) and h(t + π/y)
equal 0 for all |t| ≥ C + π and y > c(). This shows that
Z
Z ∞
1 ∞
ity ity
h(t) − h(t + π/y) e dt
h(t)e dt = 2 −∞
−∞
Z
1 ∞
≤
|h(t) − h(t + π/y)| dt
2 −∞
Z
1 C+π
<
/(C + π) dt = 2 −C−π
for all |y| > c(). Since > 0 was arbitrary, this completes the proof.
5
Lemma 4: Suppose F (x, y) is a non-negative function on [0, ∞) × [0, b] for some positive b, F (x, a) →
R∞
F (x, 0) uniformly for all x in any compact subset and 0 F (x, a) dx converges uniformly for all a ∈ [0, b].
Then
∞
Z
lim
a→0+
Z
∞
F (x, a) dx =
F (x, 0) dx.
0
0
Proof: Let > 0. By hypothesis, there is a positive c such that
Z
∞
F (x, a) dx <
c
4
for all a ∈ [0, b]. Also, there is a δ > 0 such that for all positive a ≤ δ,
|F (x, 0) − F (x, a)| <
.
2c
Therefore, for all positive a ≤ δ we have
Z ∞
Z ∞
Z c
Z ∞
Z ∞
+
F
(x,
0)
dx
−
≤
F
(x,
0)
−
F
(x,
a)
dx
F
(x,
0)
dx
+
F
(x,
a)
dx
F (x, a) dx
0
0
c
0
c
Z c
<
|F (x, 0) − F (x, a)| dx +
2
Z0 c
dx +
<
2
0 2c
= .
Lemma 5: For all positive b
Z
lim
y→∞
0
∞
sin2 b(y − x)
dx = π.
b(y − x)2
Proof: With the change of variables u = b(y − x) we have
∞
Z
lim
y→∞
0
Z −∞
sin2 b(y − x)
sin2 u
dx
=
lim
−
du
2
y→∞
b(y − x)
u2
by
Z ∞
sin2 u
=
du
2
−∞ u
Z ∞
1 − cos(2u)
du.
=
u2
0
Now integrate by parts using v = 1 − cos(2u) and dw = u−2 du to get
Z
0
∞
1 − cos(2u)
1 − cos(2u)
1 − cos(2u)
du = lim+
− lim
+2
u→∞
u2
u
u
u→0
Z ∞
sin z
=2
dz,
z
0
Z
∞
0
where z = 2u. Next,
Z
lim 2
→0+
∞
sin z
dz = lim 2
z
→0+
6
Z
∞
Z
sin z
0
∞
e−rz dr dz.
sin(2u)
du
u
We want to change the order of integration above. Towards that end, suppose M ≥ . Then since
R∞
0
e−rz sin z dr converges uniformly for all z ∈ [, M ] we have
M
Z
e−rz sin z dr dz =
∞
Z
lim M →∞
Z
Z
e−rz sin z dz dr.
Z ∞
sin z sin z dr dz = lim dz M →∞
z
M
Z ∞
cos M
cos z = lim dz −
M →∞
M
z2
Z ∞ M
1
1
≤ lim
dz
+
2
M →∞ M
z
M
2
= lim
M →∞ M
∞
e
M
0
−rz
0
M
∞
Z
0
Since
∞
Z
=0
and
Z
lim M →∞ 0
∞
Z
∞
e
−rz
M
Z ∞ −rM
e
(r sin M + cos M ) sin z dz dr = lim dr
M →∞
1 + r2
Z 0∞
≤ lim
2e−rM dr
M →∞
0
2
= lim
M →∞ M
= 0,
we may change the order of integration to get
Z
2
∞
Z
∞
sin z
e−rz dr dz = 2
0
Z
∞
0
Z
∞
e−rz sin z dz dr
∞ −e−rz
r
sin
z
+
cos
z)
dr
2
1+r
z=
∞
∞
e−r (r sin + cos )
dr.
1 + r2
=2
0
Z
Z
=2
0
Note that the integrand here is non-negative and no greater than c/(1 + r2 ) for all ∈ [0, π/6], say, for some
positive c. Thus these integrals are uniformly convergent for such and we may invoke Lemma 4 to get
∞
Z
lim+ 2
→0
0
e−r (r sin + cos )
dr = 2
1 + r2
=
Z
∞
1
dr
1 + r2
0
∞
2 arctan r
r=0
= π.
Proposition 2: Let A(x) and f (s) be as in the theorem of Wiener and Ikehara. Set B(x) = e−x A(x)
and let b > 0. Then
Z
by
B(y − v/b)
lim
y→∞
−∞
7
sin2 v
dv = π.
v2
Proof: Let g(s) be as in the statement of Lemma 1 and let δ > 0. By Lemmas 1 and 2, for any real y
Z ∞
Z
Z
|t| iyt
|t| iyt
1 2b
1 2b
g(1 + δ + it) 1 −
1−
B(x) − 1 e−(δ+it)x dx dt
e dt =
e
2 −2b
2b
2 −2b
2b
0
!
Z ∞
Z 2b
−δx
1 i(y−x)t
|t|
(6)
=
B(x) − 1 e
e
1−
dt dx
2b
0
−2b 2
Z ∞
−δx sin2 b(y − x)
dx.
B(x) − 1 e
=
b(y − x)2
0
(The changing of the order of integration above is justified since the inner integral is uniformly convergent
for all t ∈ [−2b, 2b] by Lemma 1.)
We note that since limδ→0+ g(1 + δ + it) = g(1 + it) uniformly for all t ∈ [−2b, 2b], we may pass the
limit inside the integral on the left in (6):
Z 2b
Z 2b
|t| −iyt
|t| −iyt
g(1 + δ + it) 1 −
(7)
lim
e
e
dt =
g(1 + it) 1 −
dt.
2b
2b
δ→0+ −2b
−2b
We claim that we may also do this on the right-hand side of (6), i.e.,
Z ∞
Z ∞
2
sin2 b(y − x)
b(y − x)
−δx sin
dx =
dx
B(x)
(8)
lim
B(x)e
b(y − x)2
b(y − x)2
δ→0+ 0
0
and
Z
(9)
lim
δ→0+
∞
e
Z ∞
sin2 b(y − x)
b(y − x)
dx =
dx.
b(y − x)2
b(y − x)2
0
2
−δx sin
0
To see this, we first note that
0 ≤ e−δx
for all δ ≥ 0 and
R∞
0
1
b(y−x)2
sin2 b(y − x)
1
≤
b(y − x)2
b(y − x)2
dx converges. Thus the hypotheses of Lemma 4 are fulfilled to deduce (9).
Moreover, if we can prove the integral on the right-hand side of (8) converges, then a similar argument will
give (8). By what we have already proven,
Z
Z ∞
Z ∞
2
b(y − x)
sin2 b(y − x)
1 2b
|t| iyt
−δx sin
lim
dx =
g(1 + it) 1 −
e dt +
dx.
B(x)e
b(y − x)2
2 −2b
2b
b(y − x)2
δ→0+ 0
0
Denote this quantity by M . If the integral on the right-hand side of (8) diverges, there is a positive c such
that
Z
0
c
sin2 b(y − x)
dx ≥ 2|M |.
B(x)
b(y − x)2
But now for all positive δ < δ1 = (log 2)/c
Z ∞
Z c
2
2
b(y − x)
b(y − x)
−δx sin
−δx sin
M≥
B(x)e
dx ≥
B(x)e
dx
b(y − x)2
b(y − x)2
0
0
Z c
sin2 b(y − x)
>
B(x)e−δ1 c
dx
b(y − x)2
0
Z
sin2 b(y − x)
1 c
B(x)
dx
=
2 0
b(y − x)2
≥ |M |.
8
Thus the integral on the right-hand side of (8) converges, so that (8) holds as well as (9).
Combining (6)-(9) we have
1
2
(10)
2b
|t|
g(1 + it) 1 −
2b
−2b
Z
e
iyt
Z
∞
dt =
0
Z ∞
sin2 b(y − x)
sin2 b(y − x)
dx −
dx.
B(x)
b(y − x)2
b(y − x)2
0
This holds for all real y. Now set
(
h(t) =
1
2 g(1
+ it) 1 −
|t|
2b
0
if |t| ≤ 2b,
otherwise.
We note that h(t) satisfies the hypotheses of Lemma 3, with the C there equal to 2b here. Applying Lemma
3 yields
∞
Z
y→∞
1
y→∞ 2
h(t)eiyt dt = lim
lim
−∞
2b
|t| iyt
g(1 + it) 1 −
e dt = 0.
2b
−2b
Z
By (10) we get
∞
Z
lim
y→∞
0
Z ∞
sin2 b(y − x)
sin2 b(y − x)
B(x)
dx = lim
dx.
y→∞ 0
b(y − x)2
b(y − x)2
Now change variables to v = b(y − x). We have dx/b(y − x)2 = −dv/v 2 and v → −∞ as x → ∞ since
b > 0. Thus
lim
y→∞
sin2 b(y − x)
B(x)
dx
b(y − x)2
0
Z ∞
sin2 b(y − x)
= lim
dx
y→∞ 0
b(y − x)2
Z by
sin2 v
= lim
dv
y→∞ −∞ v 2
Z ∞
sin2 v
dv
=
2
−∞ v
by
sin2 v
B(y − v/b) 2 dv = lim
y→∞
v
−∞
Z
Z
∞
=π
by Lemma 5.
Proof of Wiener-Ikehara Theorem: Let B(x) = e−x A(x) be as above. Since both B(x) and
sin2 v
v2
non-negative, Proposition 2 implies that
Z
a
B(y − v/b)
lim sup
y→∞
−a
sin2 v
dv ≤ π
v2
for all positive a and b. Since ex B(x) = A(x) is non-decreasing, for all y > a/b and |v| ≤ a we have
ey−a/b B(y − a/b) ≤ ey−v/b B(y − v/b),
so that
B(y − v/b) ≥ B(y − a/b)e(v−a)/b ≥ B(y − a/b)e−2a/b .
9
are
Using this, we see that
Z
a
B(y − v/b)
π ≥ lim sup
y→∞
−a
a
Z
≥ lim sup
y→∞
sin2 v
dv
v2
B(y − a/b)e−2a/b
−a
Z
= lim sup B(y − a/b)e−2a/b
y→∞
= lim sup B(y)e−2a/b
Z
y→∞
= lim sup B(y)e−2/a
y→∞
Z
−a
2
a
−a
a
−a
a
sin2 v
dv
v2
sin2 v
dv
v2
sin v
dv
v2
sin2 v
dv
v2
(setting b = a2 ).
This holds for all positive a. Therefore, letting a tend to infinity and invoking Lemma 5 yields
1 ≥ lim sup B(y).
(11)
y→∞
One consequence of (11) is that B(x) ≤ c for some constant c ≥ 0 and all x ≥ 0. Fix a, b > 0 and
assume that y > a/b. We have
Z by
Z −a
Z ∞
Z a
sin2 v
sin2 v
sin2 v
sin2 v
(12)
B(y − v/b) 2 dv ≤ c
dv
+
c
dv
+
B(y − v/b) 2 dv.
2
2
v
v
v
v
−∞
−∞
a
−a
Using the fact that B(x)ex is non-decreasing once more, we see that for all |v| ≤ a
B(y − v/b)ey−v/b ≤ B(y + a/b)ey+a/b ,
whence
B(y − v/b) ≤ B(y + a/b)e(a+v)/b ≤ B(y + a/b)e2a/b .
Combining this with (12) and Proposition 2 yields
Z by
sin2 v
π = lim inf
B(y − v/b) 2 dv
y→∞
v
−∞
Z −a
Z ∞
Z a
2
sin v
sin2 v
sin2 v
2a/b
≤c
dv
+
c
dv
+
lim
inf
B(y
+
a/b)e
dv
2
y→∞
v2
v2
−∞
a
−a v
Z −a
Z ∞
Z a
1
1
sin2 v
2a/b
≤c
dv
+
c
dv
+
lim
inf
B(y)e
dv
2
2
y→∞
v2
−∞ v
a
−a v
Z a
2c
sin2 v
=
+ lim inf B(y)e2/a
dv
(setting b = a2 ).
2
y→∞
a
v
−a
This is true for all positive a, so letting a tend to infinity and invoking Lemma 5 gives
(13)
1 ≤ lim inf B(y).
y→∞
Finally, combining (11) and (13) gives the desired result:
1 = lim B(y) = lim A(x)e−x .
y→∞
x→∞
10