Chapter 4 The Prime Number Theorem
[7 lectures]
The main reference for this material is The Prime Number Theorem, G. J. O.
Jameson, LMS Student Texts, 53, 2003, though the actual choice of function
F (s) and the manner of moving the line of integration in Step 5 are seen in
Introduction to Analytic and Probabilistic Number Theory, G. Tenenbaum,
Cambridge Studies in Advanced Mathematics, 46, 1995.
Introduction. Plan of the proof of the Prime Number Theorem
(PNT):
Step 1. Analytic Properties of the Riemann zeta function
The function ζ (s) has been defined by a series valid only for Re s > 1. In
the proof we will need to define the zeta function for s = 1 + it, t 6= 0. So
in this section we find a function holomorphic for Re s > 0 except for a pole
at s = 1, that agrees with ζ (s) for Re s > 1. That is, we find an analytic
continuation of ζ (s) for Re s > 0.
[1 lectures]
Step 2. Relate ψ (x) to ζ ′ (s) /ζ (s) .
We will show that
Z x
Z c+i∞
1
ds
(ψ (t) − [t]) dt = −
F (s) xs+1
2πi c−i∞
s (s + 1)
1
where c > 1 and
ζ ′ (s)
+ ζ (s) .
(1)
F (s) =
ζ (s)
[1 lectures]
Step 3. ζ (1 + it) 6= 0.
We need that the F (s) in (1) is well-defined for Re s ≥ 1. Because of the zeta
function on the denominator we need to know that ζ (s) 6= 0 for Re s ≥ 1.
We already know that this is the case for Re s > 1, so all that needs to be
done is to show that ζ (s) 6= 0 for Re s = 1. Thus the results of Step 1 are
used here.
[2 lectures]
Step 4. Bounds on the Riemann zeta function.
The integral we study will be
Z c+i∞
F (s)
c−i∞
1
xs+1
ds,
s (s + 1)
(2)
and we need show that it converges absolutely for c ≥ 1. For this reason
we need bounds on F (s) on vertical lines in the complex plane of the form
|F (c + it)| ≤ |t|1−δ for some δ > 0, and any c ≥ 1, though we will do far
better than this. From (1) we see that
′
ζ (s) + |ζ (s)| .
|F (s)| ≤ ζ (s) This means we need upper bounds on |ζ (s)|, |ζ ′ (s)| and ζ −1 (s) or, equivalently, a lower bound on |ζ (s)|.
It is the necessity of absolute convergence of the integral (2) that requires
both s and s + 1 in the denominator, (so the integral (2) is
Z c+i∞
Z ∞
dt
xs+1
c+1
|F (c + it)|
,
ds ≤ x
F (s)
s (s + 1)
(1 + |t|)2
−∞
c−i∞
which converges). But it transpires
that having both s and s + 1 means that
Rx 2
back in Step 2 we related 1 ψ (y) dy, and not ψ (x), to the integral (2).
Rx
In fact, the integral (2) is the error in approximating 1 ψ 2 (y) dy by x2 /2.
[2 lectures]
Step 5 Moving the line of integration.
For any c > 1 we show that
Z c+i∞
Z 1+i∞
1
1
ds
ds
s+1
=
.
F (s) x
F (s) xs+1
2πi c−i∞
s (s + 1)
2πi 1−i∞
s (s + 1)
(3)
We could say that we have “moved the line of integration” from Re s =
c to Re = 1, and this is a very common technique in Analytic Number
Theory. The method of moving the line of integration is based on Cauchy’s
Theorem and the bounds on F (s) from Step 4 that show that the integrals
here converge very quickly.
Step 6 The Riemann-Lebesgue Lemma
In the integral on the Right Hand side of (3) in an integral along the line
Re s = 1, so we can write s = 1 + it when the xs+1 factor can be written as
x2+it = x2 eit log x . The Riemann-Lebesgue Lemma will tell us that
Z ∞
eit log x
F (1 + it)
dt → 0
(1 + it) (2 + it)
−∞
as x (and thus log x) →R ∞. Therefore the integral in (3) is o (x2 ), i.e. the
x 2
error in approximating
ψ (y) dy by x2 /2 is o (x2 ) .
[1 lecture]
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Step 7 Final deduction.
Finally
the Prime Number Theorem in the form ψ (x) ∼ x from
R x 2 we deduce
ψ (y) dy ∼ x2 /2.
[1 lecture]
Why does the proof work?
The first reason is that we have a weight function of the primes, namely
Λ (n), whose generating function ζ ′ (s) /ζ (s) has reasonable properties as a
complex-valued function.
In Step 5 we can talk about having “moved the line of integration” from
Re s = c to Re s = 1, but why is 1 important?
If we could have only moved the line to c0 > 1 the integrand in (3) would
then contain a factor x1+c0 in place of the x2 seen above, and the application
of the Riemann-Lebesgue Lemma in Step 6 would lead to the integral (2)
being o (x1+c0 ), which need not be smaller that the main term x2 /2.
What could go wrong on the line Re s = 1?
What could cause a problem is the pole of ζ (s) at s = 1 of residue 1, for this
becomes a pole of ζ ′ (s) /ζ (s) of residue −1. But if we add ζ ′ (s) /ζ (s) and
ζ (s) the poles cancel and this is the reason why we see F (s) in Step 2.
Another problem would be if ζ (s) were to have a zero on Re s = 1 (we already
know it doesn’t have a zero for Re s > 1), because a zero of ζ (s) becomes a
pole of ζ ′ (s) /ζ (s). So, as described before, an important part of the proof
is showing that ζ (s) has no zeros on the Re s = 1 line. This is where the
famous result, due to Mertens, 1898,
|ζ (σ)|3 |ζ (σ + it)|4 |ζ (σ + 2it)| ≥ 1,
t 6= 0, is used.
In fact it is generally asserted that the Prime Number Theorem is equivalent
to the fact that the Riemann zeta function has no zeros on the Re s = 1
line. Such an equivalence cannot be seen from the proof presented here, for
the proof we describe we need more than just the non-existence of zeros,
we also need bounds on F (s), as described in Step 4 above. To get upper
bounds on F (s) we require, because of the ζ (s) in the denominator of F (x) ,
lower bounds on ζ (s). Thus not only do we require that ζ (s) is non-zero on
Re s = 1 but we also have to prove that ζ is bounded away from 0.
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Does this proof give insight into the Prime Number Theorem?
Probably not. And this may due to the use of the Riemann-Lebesgue Lemma,
for what does this lemma mean? You can see a statement of it in background
Notes 0.10:
Definition
4.1 Let φ be a differentiable, complex-valued function on R such
R∞
that −∞ |φ (t)| dt is convergent. For such a function define
b (λ) =
φ
Z
∞
e−iλt φ (t) dt,
−∞
the Fourier Transform of φ.
Theorem 4.2 The Riemann-Lebesgue Lemma Let φ beR a complex∞
valued function on R with continuous derivative and such that −∞ |φ (t)| dt
b (λ) → 0 as λ → ∞.
is convergent. Then φ
This could be compared to the theory of Fourier Series. So if f is periodic,
period 1, and satisfies certain conditions then
f (x) =
∞
X
an e2πinx ,
n=−∞
and for the sum to be convergent you would expect an → 0 as n → ∞
or n → −∞ (though this has to be proved and requires conditions on f ,
especially if you want to know just how fast the an tend to 0).
Level 4
For level 4 students we will extend the results of Step 4 and prove bounds on
the Riemann zeta-function that are valid to the left of Re s = 1. In particular
we show that F (s) has no singularities just to the left of this vertical line so
in Step 6 we can move the line of integration further to the left. We will then
have no need to use the Riemann-Lebesgue, but we will, instead, be able to
show that the integral in (2) is ≪ x2 f (x) for some function that we can give
explicitly and for which f (x) → 0 as x → ∞. This eventually leads to the
Prime Number Theorem with an error term: there exists c > 0 such that
ψ (x) = x + O x exp −c log1/10 x .
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