California State University, Northridge Prime Number

Home , Prime Number Theorem

CALIFORNIA STATE UNIVERSITY, NORTHRIDGE

PRIME NUMBER THEOREM

A thesis submitted in partial fulﬁllment of the requirements for the degree of Master of Science in Mathematics

Delﬁno Nolasco

Dec 2020 The thesis of Delﬁno Nolasco is approved:

Werner Horn, Ph.D Date

Alberto Candel, Ph.D Date

Xiaolong Han, Ph.D, Chair Date

California State University, Northridge

ii Table of Contents

Signature page ii

Abstract iv

1 Introduction and Preliminaries 1 1.1 The Riemann Zeta Function & Prime Number Theorem ...... 1 1.2 History of the Prime Number Theorem ...... 5

2 Basic properties of the Zeta function 8 2.1 Gamma Function ...... 8 2.2 Zeta function and its analytic continuation ...... 10 2.3 Zeros of the Zeta function ...... 14

3 Prime Number Theorem 20 3.1 Tchebychev Functions ...... 20 3.2 Estimates of the Zeta functions ...... 25 3.3 Proof of the Prime Number Theorem ...... 33

Bibliography 39

iii ABSTRACT

PRIME NUMBER THEOREM

Delﬁno Nolasco

Master of Science in Mathematics

The prime number theorem is broadly considered one of the greatest theorems proved in all of number theory, taking more than a hundred years between being conjectured and being proved. This thesis examines the analytic proof of the theorem. An auxiliary function to Tchebychev’s ψ function is used in conjunction with a theorem showing the asymptotic behavior of the function is equivalent to the asymptotic behavior in the prime number

2 theorem. With these tools, it is then sufﬁcient to show that ψ1 is asymptotic to x /2. Ac- complishing this is surprisingly involved and requires producing an integral that connects the ψ function to the ζ function, integrating along the real line in the complex plane whose real part is greater than 1, acquiring estimates of the ζ function and its derivative, and ﬁ- nally ensuring ζ has no zeroes on the real line in order to shift the contour to the real line s=1.

iv Chapter 1

Introduction and Preliminaries

This thesis explores the Prime Number Theorem. What follows are some preliminary deﬁ- nitions, history, and motivation to contextualize the precise focus of the research.

1.1 The Riemann Zeta Function & Prime Number Theorem

Deﬁnition 1.1. The Riemann zeta function is initially deﬁned for real s > 1 by the con- vergent series ∞ X 1 ζ(s) = . ns n=1 Remark.

• In 1735 Euler [9] showed that

1 1 π2 1 1 π4 ζ(2) = 1 + + + ··· = and ζ(4) = 1 + + + ··· = . 22 32 6 24 34 90

1 1 • Letting s = 1 gives ζ(1) = 1 + 2 + 3 + ··· the harmonic series which diverges.

Proposition 1.1. The series deﬁning ζ(s) converges for <(s) > 1, and the function ζ is holomorphic in this half-plane.

Proof. Let s = σ + it for σ and t are real, then

−s −s log(n) −σ log(n) −σ n = e = e = n .

As a result, the series deﬁning ζ(s) converges if σ > 1 + δ > 1 since it is uniformly ∞ ∞ P 1+δ P 1 bounded by 1/n . Therefore the given series ns converges uniformly on every n=1 n=1 half-plne <(s) > 1 + δ > 1. Hence, deﬁning ζ as a holomorphic function <(s) > 1.

1 Theorem 1.2 (The Euler Product Formula).

∞ ! X 1 Y 1 = , <(s) > 1. ns 1 − 1 n=1 p, prime ps

Proof. Note that

∞ ! ∞ ∞ ∞ X 1 1 X 1 X 1 X 1 1 − = − = . ns 2s ns (2n)s ns n=1 n=1 n=1 n=1,2-n

∞ ! X 1 Y 1 X 1 1 − = = 1. ns ps ns n=1 p,prime n=1,p-n

Remark.

• By using the Euler product formula one can see that ζ(s) has no zeros on <(s) > 1.

Although Euler product formula was proved in 1737 [5] , Euler only considered s to be s > 1. It was then in 1859 [1] as stated in his paper ” On the Number of Prime less then a given Magnitude”, Bernard Riemann was able to extend the deﬁnition of ζ(s) to C. The Riemann zeta function enjoys a multitude of properties such as:

• The zeta function has a meromorphic continuation into the entire complex plane, with a simple pole at s = 1 with residue 1.

• The zeta function has a functional equation

1 1−s s− 2 π Γ( 2 )ζ(1 − s) ζ(s) = s Γ( 2 )

where Γ(s) is the Gamma function.

2 • The only zeros of ζ(s) outside the strip 0 ≤ <(s) ≤ 1 are the negative integers, −2, −4, −6, ··· which are sometimes called the trivial zeros of the zeta function.

We will prove these properties in detail in chapter 2.

The region that remains to be studied, 0 ≤ <(s) ≤ 1, is called the critical strip, which is now known as the Riemann hypothesis.

The original motivation for Riemann to consider the zeta function was to study the distribution of primes among integers.

Deﬁnition 1.2 (Prime counting function). We will deﬁne the prime counting function π(x) to be π(x) = #{ p is a prime : p ≤ x}.

Theorem 1.3 (The Prime Number Theorem). The Prime Number Theorem asserts

x π(x) ∼ as x → ∞, log(x) that is, π(x) lim = 1. x→∞ x/ log(x)

This theorem was proven independlty by Hadamard and dela Valle Pousin in 1896 [5], using the ideas introduced by Bernard Riemann. The most important fact of this proof is that ζ has no zeros on the line <(s) = 1.

As we already know ζ(s) has a pole at s = 1, so we can say that ζ(s) has no zeros in a neighborhood of this point. The question becomes how to prove it has no zeros on the line <(s) = 1.

In the great work of Riemann on the Prime Number Theorem, he introduced the use of complex analysis into the study of the real function ζ(s) and proved its functional equation;

3 he also wrote down explicit formula for determining the distribution of prime. Although Riemann was unable to prove his assertion, he did introduce many important new ideas into the subject. He also made one of the most famous conjectures which is now know as the Riemann hypothesis.

Conjecture 1.4. The Riemann hypothesis asserts that the zeros of ζ(s) in the critical strip lie on the line <(s) = 1/2

Riemann said about this [8]: ’It would certainly be desirable to have a rigorous demon- stration of this proposition; nevertheless, I have for the moment set this aside, after several quick but unsuccessful attempts, because it seemed unneeded for the immediate goal of my study.’ Although most mathematician assert that the Riemann hypothesis is indeed true, a proof or counter example remains to be discovered. The Riemann hypothesis has been ex- amined for over a century and remains one of the great unsolved problems in mathematics.

The proof of the Prime Number Theorem has to do with the behaviour of the zeta function near the line <(s) = 1; the basic object involved is the zeta logarithmic derivative ζ0(s)/ζ(s) which we know is analytic on <(s) > 1. So, we need to know about the growth of ζ0(s) and 1/ζ(s) in addition to ζ(s) 6= 0 on <(s) = 1.

In 1899 [1] Charles Jean de la Vallee Poussin published the following error term in the Prime Number Theorem

p π(x) = Li(x) + O(x exp (−C log(x))) for some C > 0 where x Z 1 Li(x) = dx log(x) 2 If the Riemann hypothesis is true, then it gives us an even better error term. In 1901 [6],

4 Helge von Koch showed that if the Riemann hypothesis is true, then

√ π(x) = Li(x) + O( x log(x))

We will proof the Prime Number Theorem in Chapter 3.

1.2 History of the Prime Number Theorem

Prime numbers play an important role in the study of Number theory. With its unusual properties such as the distribution of prime numbers have long been the subject of intensive study by many mathematicians. In this section we are going to give a brief history of the ideas which led up to the prime number theorem and to its proof.

The ﬁrst important thing that we know about prime numbers is that there is inﬁnitely many of them as demonstrated by Eulid around 300 B.C [3].

In the mid-18th century, Leonahard Euler introduced methods from mathematical analysis in his study of the distribution of primes. He introduced the zeta function and the Euler product formula.

Later, in 1798 [5] Adrien Marie Legendre stated the ﬁrst conjecture of the prime number theorem. Friedrich Gauss presented a conjecture different but equivalent to the prime number theorem perhaps prior to the 1800 [5]. Both Legendre and Gauss investigated the distribution of prime numbers up to millions and made the conjecture that

x π(x) ∼ log(x)

In 1851 [1] Pafnuty Lvoich Tchevychev made an important approach in the direction of the prime number theorem. He introduced an auxiliary function ψ(x) which is more convenient for studying the distribution of prime numbers than π(x).

5 Deﬁnition 1.3. We will deﬁne the Tchevychev’s θ(x) function to be

X θ(x) = log(p). p≤x

Deﬁnition 1.4. We will deﬁne the Tchevychev’s ψ(x) function to be

X ψ(x) = log(p) pm≤x where p runs over primes and m over positive integers.

Tchevychev proved that the Prime Number Theorem is equivalent to either of the two statements.

1. lim θ(x) = 1 , x→∞ x

2. lim ψ(x) = 1. x→∞ x

One of the easiest way to prove the Prime Number Theorem is to prove that the above are equivalent. In addition, Tchevychev proved that if the lim θ(x) exists, then its value must x→∞ x be 1. Unfortunately, Tchevychev couldn’t prove that the limit is 1. Tchevychev [5] also prove that

π(x) π(x) .9219 ≤ lim inf ≤ 1 ≤ lim sup ≤ 1.010555 x→∞ x/ log(x) x→∞ x/ log(x)

Over the years there were various improvements on Tchevychev’s work. In 1859, Berna- hard Riemann introduced an idea of how to prove the prime number theorem by using complex analysis. He was able to connect the zeros of the zeta function to the distribution of the prime number. Although Riemann never succeeded in proving the prime number theorem, but he conjectured that it could be done by showing that ζ(s) has no zeros when <(s) = 1.

6 The ﬁrst complete proof of the prime number theorem was established independently by Hadamard and de la Valle Pussin in 1896 using the method introduced by Riemann. Both mathematicians proved that ζ(s) has no zeros on the line <(s) = 1. This fact is essential in proving the prime number theorem.

After Hadamrd and de la Valle Pussin success in proving the prime number theorem using the methods of complex analysis, many mathematicians doubt that an elementary proof was even possible. However in 1948 [4] Atle Selberg and Paul Erdos independently gave an elementary proof of the prime number theorem using the simplest properties of the logarithm function. Both Atle Selberg [7] and Paul Erdos [2] published their work in 1949.

7 Chapter 2

Basic properties of the Zeta function

In this chapter we introduce the gamma function and we talk about some of its properties.

2.1 Gamma Function

Deﬁnition 2.1. For s > 0, the gamma function is deﬁned by

∞ Z Γ(s) = e−tts−1 dt.

Let s > 0. Then the integrand in the gamma function can be described as follows:

  s−1 t if 0 ≤ t ≤ 1, e−tts−1 ≤  − t e 2 if t 1.

s−1 − t In addition of this, t is integrable on [0, 1] while e 2 is integrable on [1, ∞). As a result, the gamma function Γ(s) is well deﬁned for s > 0. In fact, a similar argument implies that it is also deﬁned for all complex numbers s such that <(s) > 0.

Proposition 2.1. The gamma function Γ(s) extends to an analytic function in the half plane <(s) > 0, and is still given there by the integral formula.

We can prove that there exists a meromorphic function deﬁned on all of C that equals Γ in the half plane <(s) > 0. We say that this function is the analytic continuation of Γ, and we therefore continue to denote it by Γ.

In order to prove the analytic extension to a meromorphic function, there needs to be a lemma, which incidentally exhibits an important property of Γ.

8 Lemma 2.2. If <(s) > 0, then Γ(s + 1) = sΓ(s).

Remark.

∞ m R −t R −t −t m • Letting s = 1 gives Γ(1) = e = lim e dt = lim −e |0 = 1 1 m→∞ 1 m→∞

• By using Lemma 2.2

Γ(n + 1) = nΓ(n) = ··· = n(n − 1)(n − 2) ··· 2Γ(1) = n! for all n ∈ N.

Lemma 2.2 is one of the main factors which we can show that Γ(s) can be extended to a meromorphic function on C.

Theorem 2.3. The function Γ(s) initially deﬁned for <(s) > 0 has an analytic continuation to a meromorphic function on C whose only singularities are simple poles at the negative integers s = 0, −1, ··· . The residue of Γ at s = −n is (−1)n/n! .

Theorem 2.4. For all s ∈ C

π Γ(s)Γ(1 − s) = . sin(πs)

Remark.

• Notice that Γ(s) has simple poles at s = 0, −1, −2, −3,... while Γ(1−s) has simple

1 poles at s = 1, 2, 3, ··· . They agree with simple poles at sin(πs) at all integers.

1 1 √ • Letting s = 2 gives Γ( 2 ) = π.

• Note that π Γ(s) = , sin(πs)Γ(1 − s)

in which Γ(1 − s) has simple poles at positive integers and sin(πs) has simple zeros

1 at all integers. Therefore, Γ(s) has no zeros so Γ(s) is an entire function.

9 On the other hand, 1 sin(πs)Γ(1 − s) = Γ(s) π

has simple zeros at non-positive integers 0, −1, −2, −3 ··· .

2.2 Zeta function and its analytic continuation

Deﬁnition 2.2. The Riemann zeta function is initially deﬁned for real s > 1 by the con- vergent series ∞ X 1 ζ(s) = . ns n=1

Similarly to the gamma function, ζ can be continued into the complex plane. In this section we are going to show that it depends on the functional equation of ζ. We will ﬁrst provided

a simple extension of ζ to a half-plane in C.

Proposition 2.5. The series deﬁning ζ(s) converges for <(s) > 1, and the function ζ is holomorphic in this half-plane.

The analytic continuation of ζ to a meromorphic function in C is more challenging than in the case of the gamma function. The proof we present here relates ζ to Γ and another important function.

Definition 2.3 (Theta function). We will define the theta function which is defined for real t > 0 by ∞ X 2 v(t) = e−πn t. n=−∞

An application of the Poisson summation formula gives the functional equation satisﬁed by v, namely v(t) = t−1/2v(1/t).

10 The growth and decay of v we shall need are

v(t) ≤ Ct−1/2, as t → 0, and |v(t) − 1| ≤ Ce−πt for some C > 0, and all t ≥ 1.

The inequality for t tending to zero follows from the functional equation, while the behavior as t tends to inﬁnity follows from the fact that

X 2 X e−πn t ≤ e−πnt ≤ Ce−πt for t ≥ 1. n≥1 n≥1

We are now in a position to prove an important relation among ζ, Γ, and v.

Theorem 2.6. If <(s) > 1, then

∞ 1 Z π−s/2Γ(s/2)ζ(s) = u(s/2)−1[v(u) − 1] du. 2 0

Proof. Observe

∞ Z 2 e−πn uu(s/2)−1 du = π−s/2Γ(s/2)n−s if n ≥ 1.

t This statement is easily proved by using the substitution u = πn2

∞ ∞ Z (s/2)−1 Z −πn2( t ) t 1 −t (s/2)−1 2 −s/2 −s/2 −s e πn2 dt = e t dt(πn ) = π Γ(s/2)n πn2 πn2 0 0

Next notice that ∞ v(u) − 1 X 2 = e−πn u. 2 n=1

11 This follows immediately

∞ −1 ∞ ∞ X 2 X 2 X 2 X 2 v(u) = e−πn u = e−πn u + 1 + e−πn u = 2 e−πn u + 1. n=−∞ n=−∞ n=1 n=1

The estimates for v given before the statement of the theorem justify an interchange of the inﬁnite sum with the integral, and thus

∞ ∞ Z Z ∞ 1 X 2 u(s/2)−1[v(u) − 1] du = u(s/2)−1 e−πn u du 2 0 0 n=1 ∞ Z ∞ X 2 = u(s/2)−1e−πn u du

0 n=1 ∞ ∞ Z X 2 = u(s/2)−1e−πn u du

n=1 0 ∞ X = π−s/2Γ(s/2)n−s n=1 = π−s/2Γ(s/2)ζ(s).

In view of this, the modiﬁcation of the ζ function called the xi function, appear more symmetric.

Definition 2.4. We will define the xi function ξ(s) which is defined for <(s) > 1 by

ξ(s) = π−s/2Γ(s/2)ζ(s).

Theorem 2.7. The function ξ is holomorphic for <(s) > 1 and has an analytic continuation

to all of C as a meromorphic function with simple poles at s = 0 and s = 1. Moreover,

ξ(s) = ξ(1 − s) for all s ∈ C.

12 v(u)−1 Proof. Let ψ(u) = 2 . The functional equation for the theta function, namely v(u) = u−1/2v(1/u), implies 1 1 ψ(u) = u−1/2ψ(1/u) + − . 2u1/2 2

This follows immediately

− 1 −1/2 −1/2 −1/2 u 2 v(1/u) − 1 u v(1/u) − u u 1 ψ(u) = = + − . 2 2 2 2

Now, by Theorem 2.6 for <(s) > 1, we have

∞ Z π−s/2Γ(s/2)ζ(s) = u(s/2)−1ψ(u) du

0 1 ∞ Z Z = u(s/2)−1ψ(u) du + u(s/2)−1ψ(u) du

0 1 1 ∞ Z 1 1 Z = u(s/2)−1 u−1/2ψ(1/u) + − du + u(s/2)−1ψ(u) du 2u1/2 2 0 1 ∞ 1 1 Z = − + (u(−s/2)−1/2 + u(s/2)−1)ψ(u) du s − 1 s 1

whenever <(s) > 1. Therefore

∞ 1 1 Z ξ(s) = − + (u(−s/2)−1/2 + u(s/2)−1)ψ(u) du. s − 1 s 1

Since the function ψ has exponential decay at inﬁnity, the integral above deﬁnes an entire function, and we conclude that ψ has an analytic continuation to all of C with simple poles at s = 0 and s = 1. Moreover, it is immediate that the integral remains unchanged if we replace s by 1 − s. We conclude that ξ(s) = ξ(1 − s) as was to be shown.

The identity proved for ξ, thus we obtain the desired result for the zeta function: its analytic continuation and its functional equation.

13 Theorem 2.8. The zeta function has a meromorphic continuation into the entire complex plane whose only singularity is a simple pole at s = 1.

Proof. By using ξ(s) = π−s/2Γ(s/2)ζ(s) provides the meromorphic continuation of ζ, namely ξ(s) ζ(s) = πs/2 . Γ(s/2)

Recall that 1/Γ(s/2) is entire with simple zeros at 0, -2, -4, ··· so the simple pole of ξ(s) at the origin is cancelled by the corresponding zero of 1/Γ(s/2). As a consequence, the only singularity of ζ is a simple pole at s = 1.

Next, we will discuss the zeros of the zeta function.

2.3 Zeros of the Zeta function

Theorem 2.9 (The Euler Product Formula).

∞ ! X 1 Y 1 = , <(s) > 1. ns 1 − 1 n=1 p prime ps

Theorem 2.10. The only zeros of ζ outside the strip 0 ≤ <(s) ≤ 1 are the negative integers , −2, −4, −6, ··· .

Proof. This follows immediately from using the Euler Product Formula and Theorem 2.7. By the Euler Product Formula

∞ X 1 Y 1 ζ(s) = = , <(s) > 1. ns 1 − 1 n=1 p prime ps

This shows that ζ(s) has no zero when <(s) > 1. Now let’s show that ζ(s) has zeros when <(s) < 0.

14 By using Theorem 2.7 ξ(s) = ξ(1 − s)

−s/2 s − (1−s) 1 − s π Γ( )ζ(s) = π 2 Γ ζ(1 − s) 2 2 (1−s) 1−s − 2 π Γ( 2 )ζ(1 − s) ζ(s) = s s . − 2 π Γ( 2 )

Let <(s) < 0, then we will have the following:

1. ζ(1 − s) has no zeros when <(1 − s) > 1. Notice <(1 − s) > 1 if and only if <(s) < 0.

1−s 2. Γ( 2 ) has no zeros by the remark on Theorem 2.4.

1 3. By the remark on Theorem 2.4, s only has zeros when s = −2, −4, −6, ··· . Γ( 2 )

This allows us to see that ζ(s) has no zero when <(s) > 1 and the only zeros of ζ(s) outside of 0 ≤ <(s) ≤ 1 are when s = −2, −4, −6, ··· .

Speciﬁcally, it is for this reason that the zeros of ζ located outside the critical strip are sometimes called trivial zeros of the zeta function.

The region that remains to be studied is called the critical strip, 0 ≤ <(s) ≤ 1. A key fact in the proof of the Prime Number Theorem is that ζ has no zeros on the line <(s) = 1. As a simple consequence of this fact and the functional equation, it follows that ζ has no zeros on the line <(s) = 0.

In the rest of this section we shall limit ourselves to proving the following theorem, together with related estimates on ζ, which we shall use in the proof of the Prime Number Theorem. We already have shown ζ(s) has a pole at s = 1. So we can say that ζ(s) has no zeros in a neighborhood of this point. Following is the stronger property of what we are trying to prove.

ζ(1 + it) 6= 0 for all t ∈ R.

15 Theorem 2.11. ζ(s) has no zeros on the line <(s) = 1.

The following sequence of lemmas are going to be helpful in proving Theorem 2.11.

Lemma 2.12. If <(s) > 1, then

−ms ∞ X p X cn log(ζ(s)) = = for some c ≥ 0. m ns n p,m n=1

Proof. This proof follows immediately from the Euler Product Formula and by using the power series expansion for the logarithm. Lets ﬁrst assume s > 1.

Recall

∞ X −xm log(1 − x) = for 0 ≤ x < 1. m m=1

Now, by applying log to the Euler Product Formula and by using the power series expansion

1 for log( 1−x )

  ∞ Y 1 X 1 X X p−sm log(ζ(s)) = log = log = .  1 − p−s  1 − p−s m p prime p prime p m=1

Since the double sum converges absolutely, we need not specify the order of summation. By analytic continuation we can say that this formula then holds for all <(s) > 1. Since ζ(s) has no zeros on the simply connected half-plane <(s) > 1 we can say that log(ζ(s)) is well deﬁned on <(s) > 1. Finally,

 −ms ∞ c = 1 if n = pm, X p X cn  n m = , where m n−s p,m n=1  cn = 0 otherwise.

16 In ordered to prove Theorem 2.11 we will need the following equality.

Lemma 2.13. If θ ∈ R, then 3 + 4 cos(θ) + cos(2θ) ≥ 0.

Proof.

2(1 + cos(θ))2 = 2(1 + 2 cos(θ) + cos2(θ)) 1 + cos(2θ) = 2 1 + 2 cos(θ) + 2 = 3 + 4 cos(θ) + cos(2θ).

Corollary 2.14. If σ > 1 and t is real, then log |ζ3(σ)ζ4(σ + it)ζ(σ + 2it)| ≥ 0.

Proof. Let s = σ + it where σ > 1,

log |ζ3(σ)ζ4(σ + it)ζ(σ + i2t)|

= 3 log |ζ(σ)| + 4 log |ζ(σ + it)| + log |ζ(σ + i2t)|

= 3<(log(ζ(σ)) + 4<(log(ζ(σ + it))) + <(log(ζ(σ + i2t)).

By using Lemma 2.13.

∞ ∞ X −s X −s <(log(ζ(s))) = <( cnn ) = cn<(n ). n=1 n=1

Note that <(n−s) = <(e−s log(n)) = e−σ log(n) cos(t log(n)).

So we see ∞ X −σ log(n) <(log(ζ(s)) = cne cos(t log(n)). n=1

17 Similarly, we can show

∞ X −σ log(n) <(log(ζ(σ))) = cne cos(0 log(n)) n=1 ∞ X −σ log(n) <(log(ζ(σ + i2t))) = cne cos(2t log(n)). n=1

This allows us to see that

log |ζ3(σ)ζ4(σ + it)ζ(σ + i2t)| ∞ ∞ ∞ X −σ log(n) X −σ log(n) X −σ log(n) = 3 cne + 4 cne cos(t log(n)) + e cos(2t log(n)) n=1 n=1 n=1 ∞ X −σ log(n) = e (3cn + 4cn cos(t log(n)) + cos(2t log(n))). n=1

Now, by letting θn = t log(n) for all n ∈ N

∞ 3 4 X −σ log(n) log |ζ (σ)ζ (σ + it)ζ(σ + i2t)| = e (3cn + 4cn cos(θn) + cos(2θn)). n=1

By using Lemma 2.13, 3 + 4 cos(θ) + cos(2θ) ≥ 0 and cn ≥ 0 for all n ∈ N , we see

log |ζ3(σ)ζ4(σ + it)ζ(σ + 2it)| ≥ 0

We can now ﬁnish the proof of Theorem 2.11.

Proof. For the sake of contradiction assume ζ(1 + it0) = 0 for some t0 6= 0.

Since ζ(s) is holomorphic at 1 + it0, it must vanish at least to order 1 at this point, hence

|ζ(σ + it0)| ≤ C(σ − 1) as σ → 1, for some C > 0.

18 So we see

4 4 4 |ζ (σ + it0)| ≤ C (σ − 1) as σ → 1.

Also, we know that s = 1 is a simple pole for ζ(s), so that

|ζ(σ)| ≤ D(σ − 1)−1 as σ → 1, for some D > 0.

So we see |ζ3(σ)| ≤ D−3(σ − 1)−3 as σ → 1.

Lastly, since ζ(s) is holomorphic at the points σ + i2t0 , the quantity |ζ(σ + i2t0)| remains bounded as σ → 1. By combining these facts

3 4 |ζ (σ)ζ (σ + it0)ζ(σ + i2t0)| → 0 as σ → 1 but by Corollary 2.14

3 4 log |ζ (σ)ζ (σ + it0)ζ(σ + i2t0)| ≥ 0 for σ > 1.

So we see

3 4 1 ≤ |ζ (σ)ζ (σ + it0)ζ(σ + i2t0)| for σ > 1 and by taking the limit as σ → 1 we get 1 ≤ 0. This is clearly a contradiction. So ζ(s) has no zeros on the line <(s) = 1.

19 Chapter 3

Prime Number Theorem

The Prime Number Theorem is proved through the estimate of Tchebychev auxiliary functions. An important identity (Prop 3.6) provides the integral representation of this function involving the zeta functions on the complex plane. Then using the estimate of the zeta function along the line <(s) = 1, we apply the contour integral to prove the asymptotic of the Tchebychev function in Theorem 3.9. The ﬁrst section that follows introduces the key function that features in this equivalent statement, the equivalent statements themselves in Proposition 3.1, and the proof of their equivalence. Section 3.2 goes on to build tools and introduce other propositions and lemmas needed in the proof, and the ﬁnal section of this chapter puts it all together to accomplish the proof.

3.1 Tchebychev Functions

The key function to the central proof of this chapter was created by Tchebychev. This function’s behavior is to a large extent equivalent to the asymptotic distribution of primes, but is easier to manipulate than π(x).

Deﬁnition 3.1. Tchebychev’s ψ function is deﬁned to be

X ψ(x) = log p. pm≤x

The sum is taken over those integers of the form pm that are less than or equal to x. Here p is a prime number and m is a positive integer. Two other formulations of ψ are necessary

20 for our purposes. First, if we deﬁne

  m log p if n = p for some prime p and some m ≥ 1, Λ(n) =  0 otherwise, then it’s clear that X ψ(x) = Λ(n). 1≤n≤x Also, it is immediate that X log(x) ψ(x) = log(p) log(p) p≤x where [u] denotes the greatest integer ≤ u, and the sum is taken over the primes less than

m log(x) x. This formula follows from the fact that if p ≤ x then m ≤ log(p) .

Deﬁnition 3.2. We deﬁne Tchebychev’s ψ1 function to be

x Z ψ1(x) = ψ(u) du. 1

Proposition 3.1. The following statements are equivalent as x → ∞.

1. π(x) ∼ x/ log(x).

2. ψ(x) ∼ x.

2 3. ψ1(x) ∼ x /2.

Proof. Let us ﬁrst show that statement 1 and statement 2 are equivalent. First assume that statement 2 is true and show statement 1.

In order to show π(x) ∼ x/ log(x) as x → ∞ its sufﬁces to show

π(x) log(x) π(x) log(x) 1 ≤ lim inf and lim sup ≤ 1. x→∞ x x→∞ x

21 Then get an upper bound for ψ(x),

X log(x) X log(x) ψ(x) = log(p) ≤ log(p) = log(x)π(x). log(p) log(p) p≤x p≤x

Now, by dividing by x and taking the lim inf on both sides and using ψ(x) ∼ x as x → ∞

ψ(x) log(x)π(x) 1 = lim inf ≤ lim inf . x→∞ x x→∞ x

log(x)π(x) Next, we need to show lim sup x ≤ 1. We start by ﬁnding a lower bound for ψ(x). x→∞ Let 0 < α < 1,

X X X ψ(x) ≥ log(p) ≥ log(p) ≥ log(xα) = log(xα)(π(x) − π(xα)). p≤x xα

This allows us to conclude,

ψ(x) + log(xα)π(xα) ≥ log(xα)π(x)

ψ(x) + α log(x)π(xα) ≥ log(xα)π(x).

Clearly, π(xα) ≤ xα ψ(x) + αxα log(x) ≥ α log(x)π(x).

Now, by diving by x ψ(x) α log(x) α log(x)π(x) + ≥ . x x1−α x

Next, by taking the lim sup on both sides and using the fact that ψ(x) ∼ x as x → ∞ and 0 < α < 1 log(x)π(x) 1 ≥ α lim sup . x→∞ x

Then, letting α → 1 log(x)π(x) 1 ≥ lim sup . x→∞ x

22 This shows that statement 2 implies statement 1.

Next, we show that statement 1 implies statement 2. This proof is very similar to the proof that statement 2 implies statement 1 and we will show it for completeness. We already know ψ(x) log(x)π(x) ≤ . x x

By taking the lim sup on both sides and using π(x) ∼ x/ log(x) as x → ∞

ψ(x) log(x)π(x) lim sup ≤ lim sup = 1. x→∞ x x→∞ x

We already know for 0 < α < 1,

ψ(x) α log(x) α log(x)π(x) + ≥ . x x1−α x

Hence, ψ(x) α log(x)π(x) α log(x) ≥ − x x x1−α

Now by taking the lim inf on both sides and using the fact π(x) ∼ x / log(x) as x → ∞ and letting α → 1 ψ(x) 1 ≤ lim inf . x→∞ x

This shows that statement 1 implies statement 2.

To show that statement 2 and statement 3 are equivalent, we ﬁrst show statement 2 implies statement 3. In order to show ψ(x) ∼ x as x → ∞ it sufﬁces to show

ψ(x) ψ(x) 1 ≤ lim inf and lim sup ≤ 1. x→∞ x x→∞ x

We start by getting an upper bound and a lower bound for ψ(x).

23 Let α < 1 < β. Since ψ(x) is increasing, it follows that

βx x Z Z (βx − x)ψ(x) ≤ ψ(u) du and ψ(u) du ≤ (x − αx)ψ(x).

x αx

So we see x βx R ψ(u)du R ψ(u)du αx ≤ ψ(x) ≤ x . x − αx βx − x

ψ(x) Now we’ll show lim sup x ≤ 1. Clearly, x→∞

βx R ψ(u)du ψ(x) ≤ x βx − x ψ (βx) − ψ (x) ψ(x) ≤ 1 1 . βx − x

First, by dividing both sides by x

ψ(x) ψ (βx) − ψ (x) ≤ 1 1 x βx2 − x2 ψ(x) 1 ψ (βx) − ψ (x) ≤ 1 1 x β − 1 x2 ψ(x) 1 2β2ψ (βx) ψ (x) ≤ 1 − 1 . x β − 1 2β2x2 x2

x2 Second, by taking the lim sup on both sides and using the fact ψ1(x) ∼ 2 as x → ∞

ψ(x) 1 β2 1 lim sup ≤ − x→∞ x β − 1 2 2 ψ(x) β + 1 lim sup ≤ . x→∞ x 2

Lastly, by letting β → 1 ψ(x) lim sup ≤ 1. x→∞ x

24 By using the lower bound for ψ(x) we can similarly show that 1 ≤ lim inf ψ(x) . Thus state- x→∞ x ment 2 implies statement 3. It can similarly be shown that statement 3 implies statement 2.

3.2 Estimates of the Zeta functions

We will now show an easier way to look at the analytic continuation of the zeta function, which quickly moves forward its extension in the half-plane <(s) > 0. This procedure will help in analyzing the growth properties of s near the line <(s) = 1 and these will be ∞ helpful as we move to the next section. The focus of this is to compare the sum P n−s n=1 ∞ with the integral R x−sdx. 1

∞ Proposition 3.2. There is a sequence of entire functions {δn(s)}n=1 that satisfy the estimate

σ+1 |δn(s)| ≤ |s|/n , where s = σ + it, and such that

N X 1 Z dx X − = δ (s) ns xs n 1≤n

whenever N is an integer > 1.

This proposition has the following consequence.

Corollary 3.3. For <(s) > 0 we have

1 ζ(s) − = H(s) s − 1

∞ P where H(s) = δn(s) is holomorphic in the half-plane <(s) > 0. n=1

25 n+1 Proof. Compare P n−s with P R x−s dx, and set 1≤n

n+1 Z 1 1 δ (s) = − dx. n ns xs n

The mean-value theorem applied to f(x) = x−s yields

1 1 |s| − ≤ , whenever n ≤ x ≤ n + 1. ns xs nσ+1

σ+1 Therefore |δn(s)| ≤ |s|/n and since

N n+1 Z dx X Z dx = xs xs 1 1≤n

Remark. This idea can be replicated to yield the continuation of s into the entire complex plane.

The proposition when apply shows that the growth of ζ(s) near the line <(s) = 1 is well ∞ behave. Recall that when <(s) > 1 we have |ζ(s)| ≤ P n−σ, and so ζ(s) is bounded in n=1 any half-plane <(s) ≥ 1 + δ, with δ > 0. We shall see that on the line <(s) = 1, |ζ(s)| is majorized by |t| for every > 0, and that the growth near the line is not substandard. The approximations below are not favorable. In truth, they are somewhat rough, but they work for the plan laid out here.

Proposition 3.4. Suppose s = σ + it with σ, t ∈ R. Then for each σ0, 0 ≤ σ0 ≤ 1, and

every > 0, there exists a constant c such that

1−σ0+ 1. |ζ(s)| ≤ c|t| , if σ0 ≤ σ and |t| ≥ 1.

0 2. |ζ (s)| ≤ c|t| , if 1 ≤ σ and |t| ≥ 1.

26 The proposition that follows is actually a quantitative version of Theorem 2.12.

Proposition 3.5. For every > 0, we have 1/|ζ(s)| ≤ c|t| when s = σ + it, σ ≥ 1 and |t| ≥ 1.

Proof. By Corollary 2.14, if σ > 1 and t is real, then log |ζ3(σ)ζ4(σ + it)ζ(σ + i2t)| ≥ 0. This allows us to see

|ζ4(σ + it)| ≥ |ζ−3(σ)||ζ−1(σ + i2t)| for σ ≥ 1 and |t| ≥ 1.

Also, we know that s = 1 is a simple pole for ζ(s), so that

|ζ(σ)| ≤D(σ − 1)−1 as σ → 1, for some D > 0

|ζ−3(σ)| ≥D(σ − 1)3 and by using the estimate for ζ(s) in Proposition 3.4

|ζ(σ + i2t)| ≤ c|t| if σ ≥ 1 and |t| ≥ 1.

We can put this all together to see

1 |ζ4(σ + it)| ≥ |ζ−3(σ)||ζ−1(σ + i2t)| ≥ D(σ − 1)3 |t|− for all σ ≥ 1 and |t| ≥ 1. c

So we can conclude

|ζ(σ + it)| ≥ k(σ − 1)3/4|t|−/4 for all σ ≥ 1 and |t| ≥ 1.

We now consider two separate cases, depending on whether the inequality σ − 1 ≥ A|t|−5 holds, for some appropriate constant A (whose value we choose later).

27 Case 1: If σ − 1 ≥ A|t|−5 then

|ζ(σ + it)| ≥ k(σ − 1)3/4|t|−/4 ≥ kA|t|−4 and it sufﬁces to replace 4 by to conclude the proof of the desired estimate. Case 2: If σ − 1 < A|t|−5 then

We ﬁrst choose σ0 > σ with σ0 − 1 = A|t|−5. By the triangle inequality

|ζ(σ0 + it)| − |ζ(σ0 + it) − ζ(σ + it)| ≤ |ζ(σ + it)|.

Apply the mean value theorem and the estimate for the derivatives of ζ(s) in Prop 3.4:

|ζ(σ0 + it) − ζ(σ + it)| ≤ c0|σ0 − σ||t| ≤ c0|σ0 − 1||t|.

By using |ζ(σ + it)| ≥ k(σ − 1)3/4|t|−/4 where we select σ = σ0,

|ζ(σ + it)| ≥ |ζ(σ0 + it)| − |ζ(σ0 + it) − ζ(σ + it)| ≥ k(σ0 − 1)3/4|t|−/4 − c0(σ − 1)|t|.

k 4 0 −5 0 3/4 −/4 Let A = ( 2c0 ) and recalling that σ − 1 = A|t| . We need to show k(σ − 1) |t| = 2c0(σ0 − 1)|t|,

k(σ0 − 1)3/4|t|−/4 = k(A|t|−5)3/4|t|−/4 = kA3/4|t|−5|t|.

28 −5 σ0−1 Now by solving for |t| = A ,

(σ0 − 1) (σ0 − 1) kA3/4|t|−5|t| = k |t| = k = 2c0(σ0 − 1)|t|. A1/4 k 2c0

So we get

|ζ(σ + it)| ≥ k(σ0 − 1)3/4|t|−/4 − c0(σ − 1)|t| = c0(σ0 − 1)|t| = A0|t|−4.

By replacing 4 by we get the desired inequality.

Proposition 3.6. For all c > 1

c+i∞ 1 Z xs+1 −ζ 0 (s) ψ (x) = ds. 1 2πi s(s + 1) ζ(s) c−i∞

In order to arrive for the proof of this formula, we isolate the necessary contour integrals in a lemma and useful identity.

Lemma 3.7. If c > 0, then

 c+i∞  1 Z as 0 if 0 < a ≤ 1, ds = 2πi s(s + 1)  1 c−i∞ 1 − a if 1 ≤ a.

Here, the integral is over the vertical line <(s) = c.

Proof. Case 1: Suppose 1 ≤ a, deﬁne

es log(a) f(s) = . s(s + 1)

Our contour Γ(T ) consists of two contours. The ﬁrst is the vertical segment S(T ) from c − iT to c + iT . The second is the half-circle C(T ) centered at c of radius T , lying

29 to the left of the vertical segment. We equip Γ(T ) with the positive (counterclockwise) orientation. Where we choose T so large that 0 and −1 are contained in the interior of Γ(T ).

S(T )(t) = c + it − T ≤ t ≤ T π 3π C(T )(t) = c + T eit ≤ t ≤ 2 2

−1 Clearly, Res(f, 0) = 1, Res(f, −1) = a . By the residue theorem we can say

Z 1 f(s) ds = 2πi 1 − . a Γ(T )

Since

Z Z as Z as Z as f(s) ds = ds = ds + ds s(s + 1) s(s + 1) s(s + 1) Γ(T ) Γ(T ) S(T ) C(T )

30 c+i∞ R 1 R in order to show f(s) ds = 1 − a it sufﬁces to show f(s) ds → 0 as T → ∞. c−i∞ C(T ) This follows immediately by ﬁrst noticing three inequalities:

1. |s| = |T eit + c| ≥ |T eit| − |c| = T − c.

2. |s + 1| = |T eit + c + 1| ≥ |T eit| − |c + 1| = T − c − 1.

3. If s = σ + it then |as| = |es log(a)| ≤ ec log(a) since σ ≤ c .

Now, by putting all three inequalities together we get.

3π/2 Z as Z |as| it ds ≤ |iT e | dt s(s + 1) |s||s + 1| C(T ) π/2 3π/2 Z ec log(a) ≤ T dt (T − c)(T − c − 1) π/2 ec log(a) = π T. (T − c)(T − c − 1)

Now, by letting T → ∞ we see

Z f(s) ds → 0 as T → ∞.

C(T )

Case 2: Suppose 0 < a ≤ 1.

Let’s define a contour similar to case one. We define a half circle on the right hand side centered at c with radius T and counter clockwise direction and a vertical line from c − iT to c + iT . Notice that there are no poles in this region. So, by the Cauchy Integral Formula we can say that the integral will have a value of zero. Similarly, we can show that the integral along the half circle will go to zero as T goes to infinity.

31 Proposition 3.8. If <(s) > 1, then

0 ∞ ζ (s) X Λ(n) − = . ζ(s) ns n=1

By Lemma 2.12 X p−ms log(ζ(s)) = . m p,m Now by differentiating both sides

0 ∞ ζ (s) X p−ms log(p)m X X Λ(n) = − = − log(p)p−ms = − . ζ(s) m ns p,m p,m n=1

Now we are ready to prove Proposition 3.6, which follows immediately from two u First, observe we can write ψ(u) = P Λ(n) as n=1

 ∞  X 1 if n ≤ u, ψ(u) = fn(u)Λ(n) where fn(u) =  n=1 0 otherwise.

Next, observe

x Z ψ1(x) = ψ(u) du 0 x ∞ Z X = fn(u)Λ(n) du 0 n=1 ∞ x X Z = Λ(n) fn(u) du n=1 0 x x X Z = Λ(n) du

n=1 n x X = Λ(n)(x − n). n=1

32 Now, by using Proposition 3.8

c+i∞ 1 Z xs+1 −ζ(s) ds 2πi s(s + 1) ζ(s) c−i∞

c+i∞ ∞ 1 Z xs+1 X Λ(n) = ds 2πi s(s + 1) ns c−i∞ n=1

∞ c+i∞ X 1 Z ( x )s = xΛ(n) n ds 2πi s(s + 1) n=1 c−i∞

Next, by using Lemma 3.7 where a = x/n and using the fact that the integral will be nonzero for 1 ≤ x/n we conclude that the integral is nonzero for n ≤ x

∞ c+i∞ x x X 1 Z ( x )s X n X xΛ(n) n ds = xΛ(n)(1 − ) = Λ(n)(x − n) = ψ (x). 2πi s(s + 1) x 1 n=1 c−i∞ n=1 n=1

This completes the proof.

3.3 Proof of the Prime Number Theorem

Theorem 3.9.

2 ψ1(x) ∼ x /2 as x → ∞

This section is wholly focused on proving the above theorem, the consequence of which is that we will have proved the Prime Number Theorem. The main factors in the argument are:

• the formula in Proposition 3.6 connecting ψ1 and ζ

c+i∞ 1 Z xs+1 −ζ0(s) ψ (x) = ds for c > 1 1 2πi s(s + 1) ζ(s) c−i∞

33 • the non-vanishing of the zeta function on <(s) = 1

• the estimates for ζ near the line given in Proposition 3.4 together with Proposition 3.5.

Let us now look closer at the strategy:

• In the mention integral, for ψ1(x) , change the line of integration <(s) = c with c > 1 to <(s) = 1.

• If we could accomplish that, then the size of the factor xs+1 in the integrand would be of order x2 which is near to the answer what we are looking for, as oppose to xc+1 , c > 1 which is much too big.

However, this would leave us with two problems that we have to address:

• The pole of ζ(s) at s = 1 but when taken into account will contribute the main term

2 x /2 of the asymptotic of ψ1(x).

• What remains must be proven smaller than this term. Therefore, we must correct the inaccurate estimate of order x2 when integrating on the line <(s) = 1.

Our plan is as follow.

Proof. Fix c > 1 and x ≥ 2. We will denote the integrated by F (s)

xs+1 −ζ0(s) F (s) = . s(s + 1) ζ(s)

34 First we deform the vertical line from c − i∞ to c + i∞ to the path γ(T ). (The segments of γ(T ) on the line <(s) = 1 consist of T ≤ t ≤ ∞, and −∞ < t ≤ −T ) . Here T ≥ 3 and T will be chosen appropriately later. By using Cauchy’s theorem allow us to see that

c+i∞ 1 Z 1 Z F (s) ds = F (s) ds. 2πi 2πi c−i∞ γ(T )

Next, we pass from the contour γ(T ) to the contour γ(T, δ). For ﬁxed T , we choose δ > 0 small enough so that ζ has no zeros in the box

s = {σ + it | 1 − δ ≤ σ ≤ 1, |t| ≤ T }

Such a choice can be made since ζ does not vanish on the line σ = 1 . In fact, by Corollary 3.3, we know ζ(s) = 1/(s − 1) + H(s), where H(s) is holomorphic near s = 1. Therefore

35 −ζ(s)/ζ(s) = 1/(s − 1) + h(s) where h(s) is holomorphic near s = 1 , and so the residue of F (s) at s = 1 equals x2/2. So we can say

1 Z x2 F (s) ds = 2πi 2 γ(T )−γ(T,δ) which implies that

1 Z x2 1 Z F (s) ds = + F (s) ds. 2πi 2 2πi γ(T ) γ(T,δ)

We now decompose the contour γ(T, δ) as γ1 + γ2 + γ3 + γ4 + γ5 and estimate each of the integrals R F (s) ds, j = 1, 2, 3, 4, 5. γj First we argue that there exists T so large that

Z 2 Z 2 x x F (s) ds ≤ and F (s) ds ≤ . 2 2 γ1 γ5

To see this, we ﬁrst note that for s ∈ γ5 one has

|xs+1| = x1+σ = x2.

Then, by Proposition 3.4 and 3.5 we have, for example, that |ζ0(s)/ζ(s)| ≤ A|t|1/2,

∞ Z Z 1 2 |t| 2 F (s) ds ≤ cx dt. t2 γ5 T

Since the integral converges, we can make the right-hand side ≤ x2/2 upon taking T sufﬁciently large. The argument for the integral over γ1 is the same.

36 Having now ﬁxed T , we choose δ appropriately small. On γ3, note that

|xs+1| = x1+1−δ = x2−δ

from which we conclude that there exists a constant CT (dependent on T ) such that

Z 2−δ F (s) ds ≤ CT x .

γ3

Finally, on the small horizontal segment γ4 (and similarly on γ2), we can estimate the integral as follows

1 Z Z 2 0 1+σ 0 x F (s) ds ≤ C x dσ ≤ C . T T log(x) γ4 1−δ

0 We conclude that there exist constant CT and CT such that

x2 Z x2 2 2−δ 0 ψ1(x) − = F (s) ds ≤ x + CT x + CT . 2 log(x) Γ(T,δ)

Multiplying through by 2/x2 , we see that

2ψ1(x) −δ 0 1 − 1 ≤ 2 + 2CT x + 2C x2 T log(x) and therefore, for all large x we have

2ψ1(x) − 1 ≤ 4. x2

37 Therefore, since we have proved

2 ψ1(x) ∼ x /2 as x → ∞ the proof of the Prime Number Theorem is completed.

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