The Prime Number Theorem
1.Introduction
The +ve integers other than 1 may be devided into two classes.Prime numbers, which do not
admit of resolution into smaller factors and composite numbers, which do.The prime numbers
derive their peculiar importance from the ’fundamental theorem of arithmetic’.Euclid around
200 B.C. announced his theorem which states that there are infinitely many primes.This theorem naturally suggests the question of determining the no.of primes not exceeding a given +ve
real number x. To study this question we write π(x) to denote the no.of primes ≤ x,for a real
no.x ≥ 1.Needless to say, an exact answer in the form of an expression involving x is not available.After that it is natural to ask for a formula that approximates π(x).The prime number theπ(x) log x
x
.In other words lim
= 1.This
orem answers our question which says that π(x) ∼
x→∞
log x
x
statement is called ‘the prime number theorem without error term’.
The prime number theorem was apparently first conjectured in the late 18th century,by Legendre and Gauss (independently).In particular Gauss conjectured an equivalent but more appealing Z
form of the prime number theorem in 1792 which states that π(x) ∼ li(x) where
x
dt
li(x) =
, which we call it as logarithmic integral.The prime number theorem proved
2 log t
much later by Hadamard and de la Vallee poussin in 1896,almost simultaneously but independently,following ideas introducing by Riemann in 1859.In summary the principle is to study the
function ζ(s),defined for complex numbers s with Re(s) > 1,by the relation
ζ(s) =
X 1
.
s
n
n≥1
(1)
Much before Riemann ,Euler had considered this function for real values of s.In effect,Euler
recast the fundamental theorem of arithmetic which states that every natural number is expressible in unique way as a product of prime numbers,in terms of the following remarkable
identity,called the Euler product f ormula.
Y
X 1
Y
1
1
1
=
1 + s + 2s + . . . =
(2)
ζ(s) =
1 .
s
n
p
p
1
−
s
p
p
p
n≥1
for all real s > 1.Formally,the identity (2) is easy to verify by distributivity and the fundamental
theorem of arithmetic and indeed (2) is equivalent to this theorem.We will recount this proof
together with a justification for this identity for all complex numbers s with Re(s) > 1 later in
this notes.
Euler as we have said ,studied the function ζ(s) for real values of s.Riemann,on the other
1
hand,showed that this function extends to a meromorphic function C whose only pole is a simple
pole at s = 1 with residue 1 and this extended function satisfies a simple functional equation.He
also observed that there is an explicit connection between the zeroes of ζ(s) extended as a
meromorphic function on C and the distribution of prime numbers on account of the identity
(2) viewed as a relation valid for all complex numbers s with Re(s) > 1.
Pursuing the direction shown by Riemann,Hadamard and de la Vallee poussin succeeded in
proving the prime number theorem by first proving that there is no zero of ζ(s) on the line
Re(s) = 1,which they then combined ,in quite different ways,with growth properties of ζ(s).
Following the work of Hadamard and de la Vallee Poussin and through the efforts of number of
mathematicians such as E.Landau,G.H.Hardey,J.E.Littlewood much simpler proof of the prime
number theorem were discovered,although these proofs still relied on growth properties of ζ(s)
together with the meromorphic continuation of ζ(s) to all points of the Re(s) ≥ 1 or a slightly
larger region and the non-vanishing of ζ(s) in this region.
It thus came as surprise when,in 1931,S.Ikehara using ideas of N.Weiner,succeeded in deducing
the prime number theorem without reference to the growth properties of ζ(s) and using only
that ζ(s) admits a meromorphic continuation to all points in the region Re(s) ≥ 1 and that ζ(s)
does not vanish in this region.Finally,in 1933 Bochner,Landau and Heilbornn produced,using
Ikehara’s work,what is arguably the easiest route to the prime number theorem which we shall
use in notes.
Z∞
dx
,we then have
If we define the function li(x) for x ≥ 2 to be the integral
log x
2
1
2
π(x) = li(x) + O x exp −c(log x)
.
This relation means that there are positive real numbers A and c such that
1
2
|π(x) − li(x)| ≤ Ax exp −c(log x) ,
(3)
(4)
for all x ≥ 2.The above equation is said to be ‘the prime number theorem with error term’.
In these notes we take up the proof of Weiner-Ikehara in section(5),the sections preceding which
describe the preliminaries required.
2.Equivalent forms of the Prime number theorem
In this section we consider a no.of asympotic relations that are equivalent to the prime number
x
theorem π(x) ∼
.
log x
We define the arithmetical function Λ(n) called von Mangold function ,
(
log p if n = pm for some prime p and m ∈ Z + ;
Λ(n) =
0
otherwise
Theorem: The following statements are equivalent.
x
(1).π(x) ∼
log x
X
(2).
log p ∼ x, where the summation on the left hand side is running over all those primes
p≤x
2
not exceeding x.
X
(3).
Λ(n) ∼ x. where the summation on the left hand side is running over all those +integers
n≤x
not exceeding x.
Proof: For our convenience we denote the sum in the (2) by θ(x) and the sum in (3) by ψ(x).
x
.
(1) ⇒ (2):Assume π(x) ∼
log x
Since dropping a finite number of primes doesn’t effect on the asympotic behaviour of π(x) we
can assume that prime numbers start from 3. Now,
!
X
X
log p ≤
3≤p≤x
1 log x
3≤p≤x
⇒ θ(x) ≤ π(x) log x.
Taking limits as x → ∞,
lim sup
x→∞
π(x) log x
θ(x)
≤ lim sup
.
x
x
x→∞
(5)
For 3 ≤ y ≤ x,
log y
X
X
1 ≤ log y
3≤P ≤y
3≤p≤x
X
1+
log p ≤ log yπ(x) + θ(x)
y≤p≤x
log yπ(x) ≤ log yπ(y) + θ(x).
Let y = xθ , θ ∈ (0, 1),
θ lim inf
x→∞
Since lim inf
x→∞
π(x) log x
log xπ(xθ )
θ(x)
≤ θ lim inf
+ lim inf
.
x→∞
x→∞
x
x
x
log xπ(xθ )
= 0, we will get
x
lim inf
x→∞
π(x) log x
θ(x)
≤ lim inf
x→∞
x
x
∴ by (5) & (6), we have θ(x) ∼ x.
Similarly we can prove (2) ⇒ (1).
(2) ⇔ (3) :
3
(6)
We have
ψ(x) =
X
Λ(n)
n≤x
=
X
log p
pm ≤x
log x
=
log p
log p
p≤x
X
X
log x
log p
=
log p
+
log p
1
1
X
x 2 ≤p≤x
1≤p≤x 2
1
2
1
=θ(x) − θ(x ) + O(log xπ(x 2 ))
1
=θ(x) + O(x 2 log x).
And hence
ψ(x)
θ(x)
=
+O
x
x
log x
.
1
x2
As x → ∞, we will get the desired result.In the last section we will prove the prime number
theorem of the form ψ(x) ∼ x.
3.Dirichlet series
A Dirichlet series is a series of the form
X an
n≥1
ns
, where an ’s are Real or Complex numbers.The
Dirichlet series associated to an arithmetical function f is the series
X f (n)
n≥1
ns
,which we denote
by ζ(f, s).
A word about notation:We denote a complex number by the letter s and the real and imaginary
parts of s by σ and τ respectively.We will sometimes also use t to denote the imaginary part of
s.The arithmetical function f which takes 1 at every natural number, we denote it by 1. The
X 1
Dirichlet series associated to the arithmetical function 1 is the series
. In place of writing
s
n
n≥1
ζ(1, s) for this series we simply write ζ(s).
X an
Note that If the series
is absolutely convergent for s = s0 = σ0 + iτ0 , it is then absolutely
s
n
n≥1
convergent for all s with σ ≥ σ0 , as is evident from the following inequalities
X an X |an | X |an | X an ≤
=
s.
s =
n
nσ
nσ0
n
n≥1
n≥1
n≥1
n≥1
It follows that if we define σa to be the infimum of all σ such that
X an
is absolutely cons
n
n≥1
vergent for s = σ, then the series is absolutely convergent for all complex numbers s in the
half plane determined
X an by Re(s) > σa . We call σa the abscissa of absolute convergence of the
Dirichlet series
and the half plane determined by Re(s) > σa the half plane of absolute
ns
n≥1
convergence.
4
We know that the Dirichlet series
X 1
is absolutely convergent for Re(s) > 1 and hence repns
n≥1
resent a holomorphic function in the half plane for which Re(s) > 1 by Weierstrass theorem.So
X logn
We can differentiate term by term.Then we will get ζ 0 (s) = −
. we can easily show that
s
n
n≥1
ζ(Λ, s) and ζ 0 (s) are absolutely convergent for Re(s) > 1 and hence they also represent holomorphic functions for Re(s) > 1.The holomorphic function ζ(s) which is defined for Re(s) > 1
is called the Riemann-zeta function.The proof the prime number theorem completely depend
on the non vanishing of ζ(s) on the line Re(s) = 1. We have a relation among these three
holomorphic functions.
proposition: −ζ 0 (s) = ζ(s)ζ(Λ, s) for Re(s) > 1.
Proof: Consider
X 1 X Λ(m)
ζ(s)ζ(Λ, s) =
ns m≥1 ms
n≥1
X X Λ(m)
=
(nm)s
n≥1 m≥1
XP
nm=k Λ(m)
=
ks
k≥1
P
X d/k Λ(d)
=
ks
k≥1
X log k
=
ks
k≥1
= − ζ 0 (s).
The rearrangement of terms in the proof being justified by the absolute convergent of ζ(s) and
ζ(Λ, s) for Re(s) > 1.
−ζ 0 (s)
This implies that
is a holomorphic in the half plane Re(s) > 1 and hence that ζ(s) 6= 0
ζ(s)
for all s in this half plane.
X Λ(n)
We can see that
is absolutely convergent in the half plane Re(s) > 1 and hence
s
(logn)n
n≥2
is a holomorphic function of s in this half plane.Let L(φ(s)) denote
φ0 (s)
for a holomorphic
φ(s)
function φ(s).For all Re(s) > 1 we then have
L exp
X
n≥2
Λ(n)
(log n)ns
!!
= −ζ(Λ, s).
On the other hand ,we have L(ζ(s)) = −ζ(Λ, s).Since half plane is simply connected and con!
X Λ(n)
nected,it follows that there is a complex number c 6= 0 such that ζ(s) = c exp
(log n)ns
n≥2
for all s in Re(s) > 1.On letting s → +∞ along the real line,it is easily seen that c = 1 and
hence that
!
X Λ(n)
(7)
ζ(s) = exp
s
(log
n)n
n≥2
5
for all s in Re(s) > 1.
Λ(n)
1
is
when n = pk for an integer k ≥ 1 and a
log n
k
prime p and is 0 for all other n.Therefore we have
X Λ(n)
XX 1
X
1
(8)
=
=−
log 1 − s
s
ks
(log
n)n
kp
p
p
p
n≥2
k≥1
From the definition of Λ(n) we see that
for all s in Re(s) > 1. On combining the above two equations we obtain
ζ(s) =
1
1 − p1s
Y
p
for all s in Re(s) > 1, which is the Euler product formula.
Now define for any n ∈ Z + and for any real no.x ≥ 1,
(
0 if x < n;
f (n, x) =
1 if x ≥ n
and consider
∞
X Λ(n)
n≥1
ns
X Z Λ(n)s
=
dx
xs+1
n≥1
n
∞
X Z f (n, x)Λ(n)
dx
=s
s+1
x
n≥1
1
=s
Z∞ X
1 n≥1
Z∞
f (n, x)Λ(n)
dx
xs+1
X Λ(n)
=s
1 n≤x
Z∞
xs+1
dx
ψ(x)
dx.
xs+1
=s
1
Z∞
∴ ζ(Λ, s) = s
ψ(x)
dx
xs+1
1
for Re(s) > 1.
4.ζ(s) 6= 0 for Re(s) = 1
To prove prime number theorem we must prove that ζ(s) 6= 0 forRe(s) ≥ 1. We already
proved in the last section that ζ(s) has no zeroes for Re(s) > 0.But it remains to show that
ζ(1 + it) 6= 0. In this section we will prove that the Riemann zeta function ζ(s) can be extended
as a meromorphic function to the half plane for which Re(s) > 0 and it has no zeroes on the
line Re(s) = 1.
6
Lemma: Let λ1 , λ2 , · · · be a real sequence which increases and has the limit infinity, and let
X
C(X) =
cn where {cn } may be real or complex and the notation indicates a summation
λn ≤X
over the finite set of positive integers n for which λn ≤ x. Then, if X ≥ λ1 and φ(x) has
continuous derivative,we have
X
ZX
cn φ(λn ) = −
λn ≤X
C(x)φ0 (x)dx + C(X)φ(X).
λ1
Proof: Consider
C(X)φ(X) −
X
X
cn φ(λn ) =
λn ≤X
cn (φ(X) − φ(λn ))
λn ≤(X)
X
X Z
=
λn ≤X λ
=
cn φ0 (x)dx
n
ZX X
cn φ0 (x)dx
λ1 λn ≤x
ZX
=
C(x)φ0 (x)dx
λ1
Hence the result follows.
X 1
defined for Re(s) > 1 admits analytic
Theorem(1): The Riemann zeta function ζ(s) =
s
n
n≥1
continuation over the half plane Re(s) > 0 having as its singularity in this half plane a simple
pole with residue 1 at s = 1.
ZX
X 1
[x]
[X]
−s
Proof: In the above theorem ,take λn = n,cn = 1,φ(x) = x then
=−
dx+ s
s
s+1
n
x
X
n≤X
if X ≥ 1.
Writing [x] = x − (x), so that 0 ≤ (x) < 1.So we obtain
1
X
Z
X 1
s
s
(x)
1
(X)
=
−
−
s
+
−
.
s
s−1
s+1 dx
s−1
s
n
s
−
1
(s
−
1)X
X
X
X
n≤x
1
Now letting X → ∞, we get
s
ζ(s) =
−s
s−1
Z∞
(x)
dx,
xs−1
1
if Re(s) > 1.
x 1
Since s+1 < σ+1 the last integral is uniformly convergent for σ > δ,where δ is any fixed
x
x
positive number and there fore represent a holomorphic function of s in Re(s) > 0.
7
This proves the theorem and the equation provides the continuation of ζ(s) over the half plane
re(s) > 0.
Theorem(2): ζ(s) has no zeroes on the line Re(s) = 1.
Proof: We prove this theorem based on the following inequality
3 + 4cosθ + cos2θ ≥ 0.
The above inequality holds for all real θ since the left hand side is nothing but 2(1 + cos2θ)2 .
And also we have the following for Re(s) > 1
log ζ(S) = −
X
log(1 − p−s ) =
p
X
p,m
1
mpms
By the equation in section(3). We have for Re(s) > 1,
log|ζ(σ + iτ )| = Re
∞
X
cn n
−σ
n=2
− iτ =
∞
X
cn n−σ cos(t log n)
n=2
1
if n is the m0 th power of prime,and 0 otherwise.Hence
m
X
log|ζ 3 (σ)ζ 4 (σ + iτ )ζ(σ + 2iτ )| =
cn n−σ 3 + 4cos(τ log n) + cos(2τ log n) ≥ 0
where cn is
n≥2
since cn ≥ 0. Thus
ζ(σ + iτ ) 4
|ζ(σ + 2iτ )| ≥ 1
|(σ − 1)ζ(σ)| σ−1 σ−1
3
holds for σ > 1.This shows that the point 1 + iτ (τ ≷ 0) cannot be zero of ζ(s).
For ,if it were then since ζ(s) is regular at the points 1 + iτ, 1 + 2iτ and has simple pole at the
point 1,the left hand side would tend to a finite limit and the right hand side tend to infinite
when σ → 1 + 0.
Hence we are done.
5.Taubarian theorem and proof of the PNT
In this section we will prove the prime number theorem in the form of ψ(t) ∼ t using the fact
that ζ(s) is distinct from 0 at all points on the line Re(s) = 1,which we proved in the last
section.The passage from this fact to the prime number theorem is effected by means of the
following theorem ,which is a basic form of the W einer − Ikehara T auberian theorem.
We define Fourier transform of a function in Cc∞ by
fˆ(y) =
Z∞
f (x)e−ixy dx.
−∞
Z
By this def.we can observe that
Z
f ĝ =
fˆg.
Let D(R) to denote the set of all Cc∞ functions on R and E(R) to denote the subset of D(R)
for which φ̂(t) ≥ 0.
8
We write ψ Zto denote a fixed element of R normalised
so that its support is contained in
t
ψ̂(t) = 1. Let ψλ (t) to denote ψ
(−1, 1) and
for each λ > 0.Then ψˆλ (t) = λψˆλ (t) and
λ
Z
ψˆλ (t)dt = 1 for all λ > 0.
For each l > 0 we define δλ (l) by the relation
Z−l
δλ (l) =
Z∞
ψˆλ (t)dt +
−∞
ψˆλ (t)dt
(9)
l
Since ψ̂(t) is integrable on R we see that for a fixed λ > 0,δλ (l) tends to 0 as l → +∞ and for
a fixed l > 0,δλ (l) tends to 0 as λ → +∞.
dt
Finally write d∗ t to denote ,the Haar measure on R∗ and write et+ to denote the function
t
which is et when t > 0 and 0 when t < 0.
Z∞
f (t)d∗ t has a finite
Theorem: Let f be a positive increasing function on R∗+ such that
1
abscissa of convergence a. Suppose that there is a real number ca such that
Z∞
f (t)t−s d∗ t −
ca
s−a
(10)
1
extends continuously to the closed strip Re(s) ≥ a and |Im(s)| ≤ T for some T > 0.
For all Λ in (0, T ), we then have that
Z∞
lim
x→∞
f (et )e−at ψˆλ (t − x)dt = ca .
(11)
0
Z∞
Proof: Let us write h(s) to denote
f (t)t−s d∗ t −
ca
and its continuous extension.Thus
s−a
1
h(s) is analytic for Re(s) > a,and is continuous on Re(s) ≥ a and |Im(s)| ≤ T.
−(σ−a)t
Further, when σ > a the functions f (et )e−σt
and e+
are integrable on R and we have
t
Z∞
−(σ−a)t
h(σ + iτ ) = (f (t)t−σ − ca t−(σ−a) )t−iτ d∗ t = F(f (et )e−σt
)(τ )
+ − ca e +
(12)
1
for every σ > a and all t in R. Thus for each σ > a and each φ in D(R) we deduce using that
Z∞
Z∞
t −σt
−(σ−a)t
(f (e )e − ca e
)φ̂(t)dt =
h(σ + iτ )φ(τ )dτ.
(13)
−∞
0
Let x be any real number ≥ 1. and let λ ∈ (0, T ). When ψλ (t) is in D(R) so is ψλ (t)eixt and
F(ψλ (t)eixt ) is ψˆλ (t − x). Applying (13) to φ(t)eixt in place of φ(t) and passing to the limit in
(13) as σ tends a with σ > a we obtain
Z∞
Z∞
t −(σ−a)t ˆ
lim (f (e )e
)ψλ (t − x)dt = lim
h(σ + iτ )ψλ (τ )eixτ dτ.
σ→a
σ→a
−∞
0
9
(14)
Since f and ψˆλ are positive functions and σ > a,the limit and integral signs on the left hand
side of (14) may be interchanged by the monotone convergence theorem.Since h(s) is continuous
on Re(s) ≥ a and |Im(s)| ≤ T and the support of ψλ is in (−T, T ),h(σ + iτ )φ(τ ) converges
uniformly in τ to h(a + iτ )φ(τ ) as σ tends to a with σ > a. Thus the limit and integral signs
on the right hand side of (14) may interchanged.Consequently,
Z∞
f (et )e−at ψˆλ (t − x)dt = ca
Z∞
ψˆλ (t − x)dt +
h(a + iτ )ψλ (τ )eixτ dτ.
(15)
−∞
0
0
Z∞
On making the change of variable t − x → t in the first integral on the right hand side of (15)
and recalling that ψˆλ (t)dt = 1,we obtain
Z∞
t
−at
f (e )e
ψˆλ t − xdt = ca + ca
Z−x
ψˆλ (t − x)dt +
h(a + iτ )ψλ (τ )eixτ dτ
(16)
−∞
−∞
0
Z∞
for all λ ∈ (0, T ).The function h(a + iτ )ψλ (τ )eixτ is continuous and of compact support and
thus integrable on R.Thus the third term on the right hand side of (16) tends to 0 as x tends
+∞,by the Riemann -Lebesgue lemma.Since ψˆλ (t) is integrable on R,the second term on the
right hand side of (16) also tends to 0 as x tends to +∞.Thus the theorem follows on passing
to the limit in (16) as x tends to +∞.
Proposition: Let f be a positive increasing function on (1, ∞) and a a real number.Suppose
that for some Λ > 0 and a real number ca we have
Z∞
lim
x→+∞
f (et )e−at ψˆλ (t − x)dt = ca .
(17)
0
Then the function f (ex )e−ax is bounded above on R∗+ . Moreover ,if K is an upper bound for
f (ex )e−ax on R∗+ ,then for all l > 0 we have
ca − Kδλ (l)
ca e2al
x −ax
x −ax
≤
lim
inf
f
(e
)e
≤
lim
sup
f
(e
)e
≤
.
x→∞
e2al
1 − δλ (l)
x→∞
Proof: Now, for any l ≤ xWe have
Z∞
t
−at
f (e )e
Zx+l
ψˆλ (t − x)dt ≥
f (et )e−at ψˆλ (t − x)dt
0
x−l
≥f (ex−l )e−a(x+l)
Zx+l
x−l
=f (ey )e−ay e−2al
Zl
ψˆλ (t)dt
−l
∴ ca ≥ lim sup f (ey )e−ay e−2al (1 − δλ (l))
y→∞
10
(18)
ca e2al
≥ lim sup f (ex )e−ax .
1 − δλ (l)
x→∞
And hence f (ex )e−ax bounded above by K say . Now
Z∞
Zx−l
Z∞
Zx−l
t −at ˆ
t −at ˆ
ψˆλ t − xdt)
f (e )e ψλ (t − x)dt ≤K( ψˆλ (t − x)dt +
f (e )e ψλ (t − x)dt +
0
−∞
x+l
Z−l
=K
x+l
ψˆλ (t)dt +
−∞
Z∞
ψˆλ (t)dt
l
=Kδλ (l)
Zx+l
∴ lim inf
f (et )e−at ψˆλ (t − x) ≥ ca − Kδλ (l)
x→∞
x−l
lim inf f (ex+l )e−a(x−l)
Zl
ψˆλ (t)dt ≥ ca − Kδλ (l)
x→∞
−l
⇒ f (ex+l )e−a(x−l) ≥ ca − Kδλ (l)
f (ex )e−ax ≥
ca − Kδλ (l)
e2al
Hence the theorem follows.
Proof of prime number theorem: Let us take ψ in place of f in the above argument.Since
Z∞
−ζ 0 (s)
−ζ 0 (s)
1
ζ(λ, s)
−(s+1)
=
has abscissa of convergence 1 and
−
=
ψ(x)x
dt =
s
sζ(s)
sζ(s)
s−1
1
−1 ζ 0 (s)
1
1
+
− extends continuously to the closed strip Re(s) ≥ 1 and Im(s) ≤ T
s
ζ(s)
s−1
s
for every T > 0 (since ζ(s) 6= 0 on the line Re(s) = 1.)
Then by the above theorem,for any λ ∈ (0, T ) we have
1 − Kδλ (l)
e2l
x −x
−x
≤
lim
inf
ψ(e
)e
≤
lim
sup
ψ(x)e
≤
x→∞
e2l
1 − δλ (l)
x→∞
The above equation holds for any l, for every λ ∈ (0, T ) and T can be arbitarly large so that
we can tend λ to ∞ . Then we have
1 ≤ lim inf ψ(ex )e−x ≤ lim sup ψ(ex )e−x ≤ 1
x→∞
x→∞
By making change of variable the prime number theorem follows.
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