D.J. Newman’s Method of Proof for the Prime Number TheoremMatt Rosenzweig

1 Introduction

Define a function π : [0,∞) → N by π(x) = # {primes p ≤ x}. A classic result in analytic number theoryknown as the Prime Number Theorem (PNT) states that

π(x) ∼ x

log x⇐⇒ lim

x→∞

π(x)x

log x

= 1

Equivalently, if we define functions ψ : [0,∞) → C, ψ(x) =∑n≤x Λ(n), where Λ(n) is the von Mangoldt

function defined by log(p) if n is a prime power and 0 otherwise and θ : [0,∞)→ C, θ(x) =∑p≤x log p, then

the PNT states that

ψ(x) ∼ x, θ(x) ∼ x

There are several proofs of the PNT, varying in method and complexity. It is beyond of the scope of thispaper to survey them all, so we briefly mention a few against which to establish the ‘novelty’ of Newman’sproof. We refer the reader interested in a survey of different proofs of the PNT to [2]. The earliest proofs ofthe PNT given by Hadamard and de la Vallee Poussin in 1896 involve complex analysis, the nonvanishing ofthe Riemann zeta function ζ(s) on the closed half plane {Re(s) = σ ≥ 1}, and estimates for the behavior ofζ(s) at ∞. de la Valle Poussin later gave a more precise form showing that

ψ(x) = x+O

(x

exp(c√

log x)

), π(x) = li(x) +O

(x

exp(c√

log x)

),

for some absolute constant c > 0, where li(x) =∫ x

21

log udu is the logarithmic integral. A proof of this result

can be found in Chapter 6 of [7]. To prove this, one first shows that there is a zero-free region of ζ(s).More precisely, there exists an absolute constant c > 0 such that ζ(s) is zero-free for σ ≥ 1− c

log(|t|+4) . One

then uses Perron’s formula with error term, an estimate of the form ζ′

ζ (s) . log(|t| + 4) for the growth at

∞, and the residue calculus to complete the proof. Another class of proofs dispense with complex analysisand instead use Fourier analysis. A relatively simple version of one such proof can be found in [4]. SuchFourier-analytic proofs still use the nonvanishing of ζ(s) for σ ≥ 1 and may or may not need an estimatefor ζ(s) at ∞. The proof in [4] does, whereas the proof based on the Wiener-Ikehara theorem, which canbe found in Section 8.3 of [7], does not. These proofs, though, depend on non-trivial facts about Fouriertransforms, and in that sense, are non-elementary.

Newman’s proof, of which this paper seeks to give an exposition, is in the spirit of asking what is theminimal amount of information needed to show that π(x) ∼ x

log x . It turns out that all we really need is

the nonvanishing of ζ(s) for σ ≥ 1; one does not even need an estimate for ζ(s) at ∞. Then, with someelementary measure theory, the Cauchy integral formula, and a simple estimate for the growth of primesproven by Chebyshev in the early nineteenth century, one can prove the PNT. In exchange, though, for theseweaker hypotheses, we lose the more precise error estimate that one can obtain by using Perron’s formula.

We start by reviewing some basic facts about Dirichlet series, in particular the Riemann zeta function,which we will need to complete the proof. We then prove that ζ(s) and the associated Dirichlet seriesΦ(s) =

∑p(log p)p−s have meromorphic continuations and that ζ(s) 6= 0 for σ ≥ 1. Next, we prove an

equivalent formulation of the basic PNT as stated above using Chebyshev’s first function θ(x) =∑p≤x log p

and show that θ(x) = O(x). We conclude the preliminary results for the PNT with a proof of a Tauberiantheorem for Laplace transforms. Having obtained all the necessary ingredients, we conclude this paper witha very short and very elegant proof of the PNT.

Before diving into the mathematics, we make some comments on notation that will be used in this paper.For complex z ∈ C, we will write x = Re z and y = Im z, and for complex s ∈ C, we will write Re s = σ andIm s = t. We will use the Vinogradov notation f(x) . g(x) to indicate |f(x)| ≤ C |g(x)|, for some constantC > 0, which is equvialent to writing f(x) = O(g(x)). Some authors, such as Montgomery and Vaughan use

� instead of .. We will write f(x) ∼ g(x) if limx→∞f(x)g(x) = 1.

2 Riemann Zeta Function and Friends

2.1 Review of Dirichlet Series

Let f : Z≥1 → C be a complex-valued function, and define an infinite series by F (s) =∑∞n=1 f(n)n−s. We

call the series giving F (s) a Dirichlet series.

1

Proposition 2.1. 1. Suppose that the Dirichlet series F (s) =∑∞n=1 f(n)n−s converges for some s =

s0 and H > 0 is a given constant. Then the series F (s) is uniformly convergent in the sector{s ∈ C : σ ≥ σ0, |t− t0| ≤ H(σ − σ0)}.

2. Any Dirichlet series F (s) =∑∞n=1 f(n)n−s has an abscissa of convergence σc in R ∪ {±∞} with

the property that F (s) converges for all s with σ > σc, and for no s with σ < σc. Moreover, if s0 is apoint with σ0 > σc, then there is a neighborhood of s0 in which α(s) converges uniformly.

Proof. Define R(u) :=∑n>u f(n)n−s0 . For any s, we can express

∑Nn=M+1 f(n)n−s as

N∑n=M+1

f(n)n−s =

N∑n=M+1

f(n)ns−s0(R(n− 1)−R(n)) = −∫ N

M

us0−sdR(u) = −us0−sR(u)|NM +

∫ N

M

R(u)d(us0−s)

= Ms0−sR(M)−Ns0−sR(N) + (s0 − s)∫ N

M

R(u)us0−s−1du

Let M ∈ N be sufficiently large such that |R(u)| ≤ ε,∀u ≥M . If σ > σ0, then∣∣∣∣∣N∑

n=M+1

f(n)n−s

∣∣∣∣∣ ≤ 2ε+ ε |s− s0|∫ ∞M

uσ0−σ−1du ≤(

2 +|s− s0|σ − σ0

)ε

For s in the prescribed region,

|s− s0|+ (σ − σ0) + |t− t0| ≤ (H + 1)(σ − σ0)

We conclude that∑NM+1 f(n)n−s is uniformly small for M < N sufficiently large, and the conclusion follows

by the Cauchy criterion.

Since F (s) =∑∞n=1 f(n)n−s is locally uniformly convergent for σ > σc, the following proposition is a

consequence of Weierstrass’ theorem.

Proposition 2.2. F (s) is analytic for σ > σc, and the differentiated series is locally uniformly convergentto

F ′(s) = −∞∑n=1

f(n)(log n)n−s

Proposition 2.3. Let α(s) =∑∞n=1 ann

−s and β(s) =∑∞n=1 bnn

−s be two Dirichlet series, and set γ(s) =∑∞n=1 cnn

−s, where

cn =∑d|n

adbnd−1 =∑km=n

akbm

If s is a point at which α(s) and β(s) are both absolutely convergent, then γ(s) is absolutely convergent andγ(s) = α(s)β(s).

Proof. This is immediate from the fact that the Cauchy product of two absolutely convergent series isabsolutely convergent, and therefore can be rearranged to obtain γ(s).

A function f is called an arithmetic function if its domain is a subset of Z. An arithmetic functionf(n) is said to be multiplicative if f(1) = 1, f(mn) = f(m)f(n), (m,n) = 1. An arithmetic function f(n)is called totally multiplicative if f(1) = 1 and f(mn) = f(m)f(n) for all m,n.

Proposition 2.4. If f is multiplicative and∑∞n=1 |f(n)|n−σ <∞, then

∞∑n=1

f(n)n−s =∏p

∞∑k=0

f(pk)p−ks

If f is totally multiplicative, then the RHS can be more succinctly expressed as∏p(1− f(p)p−s)−1.

2

Proof. For any prime p,

∞∑k=0

∣∣f(pk)∣∣ p−kσ ≤ ∞∑

n=1

|f(n)|n−σ <∞

by hypothesis. So each sum∑∞k=0 f(pk)p−ks is absolutely convergent. Let y be a positive real number and

let N (y) = {n ∈ N : p|n⇒ p ≤ y}. Since a product of finitely many absolutely convergent series is absolutelyconvergent and therefore may be arbitrarily rearranged, we see that

Πy =∏p≤y

( ∞∑k=0

f(pk)p−ks

)=

∑n∈N (y)

f(n)n−s

Hence, ∣∣∣∣∣Πy −∞∑n=1

f(n)n−s

∣∣∣∣∣ ≤ ∑n/∈N (y)

|f(n)|n−σ

If n ≤ y, then all the prime factors of n are≤ y, hence n ∈ N (y). Consequently, the above≤∑n>y |f(n)|n−σ,

which is the tail of a convergent series and therefore can be made arbitrarily small by taking y large. Weconclude that the partial products Πy →

∑∞n=1 f(n)n−s as y → ∞ and remark that the convergence is

uniform in the half plane {σ′ ≥ σ + ε} for any ε > 0.

One particularly important example of a Dirichet series is the Riemann zeta function ζ(s), is defined forσ > 1 by the absolutely convergent Dirichlet series

ζ(s) =

∞∑n=1

n−s

It should not come as too much of a surprise that a proof of the PNT might involve the Riemann zetafunction, since ζ(s) has absolutely convergent Euler product

ζ(s) =∏p

(1− p−s

)−1, σ > 1

where the infinite product ranges over all primes p.Using the Euler product, we can obtain the following result, which we will need for the proof of the PNT.

Lemma 2.5. ζ(s) 6= 0 for σ > 1.

Proof. We know that ζ(σ) 6= 0 for any σ > 1. Taking the natural logarithm of the Euler product (and usingthe continuity of log) gives

−∑p

log(1− p−σ) =∑p

∞∑k=1

p−kσ

k,

where we use the power series expansion of log z (the principal branch). Since the series on the RHS isconvergent with nonnegative terms, the order of summation is irrelevant. By analytic continuation, we seethat

log ζ(s) =

∞∑n=2

Λ(n)

log nn−s, σ > 1

where Λ(n) is the von Mangoldt function. But∑∞n=1

Λ(n)lognn

−s is a Dirichlet series which converges for σ > 1.

If ζ(s) = 0 for some s = s0, then |log ζ(s)| → ∞, s→ s0. Since the zeroes of an analytic function are isolated,we have

lims→s0

|log ζ(s)| ≤ lims→s0

∞∑n=2

∣∣∣∣Λ(n)

log n

∣∣∣∣n−σ ≤ ∞∑n=1

n−σ0 <∞,

where we use the fact that a Dirichlet series is analytic in its half plane of convergence to obtain thepenultimate inequality. We conclude that ζ(s) 6= 0 in {s ∈ C : σ > 1}.

3

2.2 Zero-Free Region for ζ(s)

By means of a functional equation, it can be shown that ζ(s) extends to an entire meromorphic function witha simple pole of residue 1 at s = 1 (see Corollary 10.3 in [7]) , but for our purposes, it will suffice to showthat ζ(s) has a meromorphic extension to the half plane {σ > 0} with a simple pole at s = 1 of residue 1.

Lemma 2.6. ζ(s)− 1s−1 extends to an analytic function in the half plane {σ > 0}.

Proof. For σ > 1,

ζ(s)− 1

s− 1=

∞∑n=1

n−s −∫ ∞

1

x−sds =

∞∑n=1

n−s −∞∑n=1

∫ n+1

n

x−sdx =

∞∑n=1

[∫ n+1

n

n−sdx−∫ n+1

n

x−sdx

]Fix n ≥ 1. Observe that for x ∈ [n, n+ 1],

∣∣n−s − x−s∣∣ =

∣∣∣∣s∫ x

n

u−s−1du

∣∣∣∣ ≤ σ ∫ n+1

n

u−σ−1du ≤ σn−σ−1

Since∑∞n=1 n

−σ−1 converges for all σ > 0, we see that

∞∑n=1

∫ n+1

n

[n−s − x−s]dx

converges uniformly in any closed half plane {σ ≥ ε}, for ε > 0. Thus, the series defines an analytic functionin the half plane {σ > 0}.

We now show that ζ(s) does not vanish on the line σ = 1. This author knows of two ways of showingthis. The first is by showing that the Dirichlet series

Φ(s) :=∑p

(log p)p−s,

where the summation ranges over all primes p, has meromorphic extension to a domain containing the closedhalf plane {σ ≥ 1} with a simple pole of residue 1 at s = 1, and then applying the argument principle witha little bit of cleverness. The second is by means of an elegant argument given by F. Merten in 1898, whichmakes use of a simple trigonometric identity. Regardless of which argument we use, we will still need thefunction Φ(s) for a key step of the proof of the PNT. We will present both arguments in the order they havebeen listed above.

By comparison with ζ(s), it is immediate that Φ(s) is an absolutely convergent Dirichlet series for σ > 1.As with ζ(s), we can say something about the meromorphic continuation of Φ(s).

Lemma 2.7. The function

Φ(s)− 1

s− 1

extends analytically to Re s ≥ 1 and ζ(s) 6= 0 for Re s = 1.

Proof. Recall the Euler product ζ(s) =∏p(1− p−s)−1, σ > 1. Since ζ(s) 6= 0 for Re s > 1, we can take the

natural logarithm of both sides to obtain

log ζ(s) = −∑p

log(1− p−s)⇒ −ζ′(s)

ζ(s)=∑p

(log p)p−s

1− p−s

Rewriting the RHS,

−ζ′(s)

ζ(s)=∑p

log p

ps − 1= Φ(s) +

∑p

log p

ps(ps − 1)

For σ > 12 , I claim that the series

∑p

log pps(ps−1) is convergent. Indeed, choose δ > 0 such that σ > 1

2 + δ. Since

log x . xδ2 as x ↑ ∞ and ps − 1 . ps, there exists N ∈ N and C > 0 such that∑

p>N

∣∣∣∣ log p

ps(ps − 1)

∣∣∣∣ ≤ C ∑p>N

pδ2

p1+2δ≤∑p

1

p1+δ≤∞∑n=1

1

n1+δ<∞

4

Moreover, we see that the convergence is uniform on the closed half plane{σ ≥ 1

2 + ε}

, for any ε > 0. By

Weierstrass’ theorem,∑p

log pps(ps−1) is analytic in the half plane

{σ > 1

2

}. We see that Φ(s) is a meromorphic

function in the half plane{σ > 1

2

}with poles at s = 1 and the zeroes of ζ(s), and the function Φ(s)− 1

s−1

is analytic at s = 1. If we can show that ζ(s) has no zeroes for σ ≥ 1, then it follows that Φ(s) − 1s−1 is

analytic in an open set containing the closed half plane {σ ≥ 1}. Since we have already shown that ζ(s) 6= 0for σ > 1, it remains to show that ζ(s) 6= 0 for σ = 1, which we now do.

Recall that if a meromorphic function f vanishes with order k at s0, then

lims→s0

(s− s0)f ′(s)

f(s)= Res(

f ′

f; s0) = k,

and if f has a pole of order k at s0, then

lims→s0

(s− s0)f ′(s)

f(s)= Res(

f ′

f; s0) = −k

Suppose that ζ(s) has a zero of order µ ≥ 0 at s = 1 + iα, where α ∈ R and α 6= 0. Since ζ(s) ∈ R for s real,

by the Schwarz reflection principle ζ(s) = ζ(s). Hence, s = 1 − iα is also a zero of order µ of ζ(s). Denote

the multiplicity of the zeroes of ζ(s) at s = 1± i2α (if there are any) by ν ≥ 0. Since Φ(s) differs from ζ′(s)ζ(s)

by an analytic function, we have that

limε↓0

εΦ(1 + ε) = 1, limε↓0

εΦ(1 + ε± iα) = −µ, limε↓0

εΦ(1 + ε± i2α) = −ν

But for ε > 0,

2∑k=−2

(4

k + 2

)Φ(1 + ε+ ikα) =

∑p

2∑k=−2

(4

k + 2

)log p

p1+ε+ikα=∑p

log p

p1+ε

4∑k=0

(4

k

)(pi

α2 )k(p−i

α2 )4−k

=∑p

log p

p1+ε(pi

α2 + p−i

α2 )4 ≥ 0,

since piα2 + p−i

α2 ∈ R. Hence,

limε↓0

2∑k=−2

(4

k + 2

)εΦ(1 + ε+ ikα) = −ν − 4µ+ 6− 4µ− ν = −2ν − 8µ+ 6 ≥ 0,

which implies that µ ≤ 6−2ν8 < 1. We conclude that µ = 0⇐⇒ ζ(1 + iα) 6= 0.

Lemma 2.8.

ζ(σ)3 |ζ(σ + it)|4 |ζ(σ + 2it)| ≥ 1, ∀σ > 1

Note that the inequality, in fact, holds for all σ ≥ 1, by continuity, where we interpret the LHS as a limitin the case σ = 1.

Proof. Recall the trigonometric identity 3 + 4 cos(θ) + cos(2θ) = 2(1 + cos(θ))2. Using the Euler product forζ(s), we have

ζ(s) = exp

(∑p

− log(1− p−s)

)= exp

(∑p

∞∑n=1

p−ns

n

)

= exp

(∑p

∞∑n=1

p−nσ cos(nt log p) + ip−nσ sin(nt log p)

n

)

= exp

(∑p

∞∑n=1

p−nσ cos(nt log p)

n

)exp

(i∑p

∞∑n=1

p−nσ sin(nt log p)

n

)We see that |ζ(s)| = exp(

∑p

∑∞n=1 n

−1p−nσ cos(nt log p)). Hence,

ζ(σ)3 |ζ(σ + it)|4 |ζ(σ + 2it)| = exp

(∑p

∞∑n=1

p−nσ

n[3 + 4 cos(nt log p) + cos(2nt log p)]

)

= exp

(∑p

∞∑n=1

p−nσ

n2(1 + cos(nt log p))2

),

which is ≥ 1 since the argument is ≥ 0.

5

Lemma 2.9. ζ(s) 6= 0 for s = 1 + it, t 6= 0.

Proof. Suppose ζ(s) = 0 for some 1 + it with t 6= 0. Since ζ(s) is analytic at s = 1 + it, we can expand ζ(s)in a power series about s = 1 + it in some open disk D(1 + it; r). So for 0 < |δ| < r

2 ,

|ζ(1 + δ + it)| =

∣∣∣∣∣∞∑n=1

anδn

∣∣∣∣∣ ≤ |δ|∞∑n=1

|an|(r

2

)n−1

= C |δ|

where we use the fact that a power series converges absolutely, strictly within its radius of convergence. Recallthat from the meromorphic extension of ζ(s) to the half plane {σ > 0} we obtain the estimate ζ(1 + δ) =O(δ−1) as δ ↓ 0. Since ζ(s) is analytic at s = 1 + 2it and therefore ζ(1 + δ + 2it) is bounded as δ ↓ 0, weobtain

1 ≤ ζ(1 + δ)3 |ζ(1 + δ + it)|4 |ζ(1 + δ + 2it)| . |δ|−3 |δ|4 = |δ| ,

which yields a contradiction for |δ| sufficiently small.

3 Chebyshev’s Estimate and Equivalent Form of PNT

We define Chebyshev’s first function, denoted by θ, by θ(x) :=∑p≤x log p, where the summation ranges over

all primes p ≤ x. We have the following estimate:

Lemma 3.1. θ(x) = O(x).

Proof. For n ∈ Z≥1, we have by the binomial formula that

22n = (1 + 1)2n =

2n∑k=0

(2n

k

)≥(

2n

n

)≥

∏n<p≤2n

p = eθ(2n)−θ(n)

Taking the natural logarithm of both sides, we obtain 2n log 2 ≥ θ(2n)− θ(n). It follows that for x ∈ R≥0,

θ(x)− θ(x2

) = θ(x)− θ(bx2c) = θ(x)− θ(2bx

2c) + θ(2bx

2c)− θ(bx

2c) ≤ log x+ 2bx

2c log 2 ≤ (1 + log 2)x,

where b·c denotes the floor function. Let r ∈ N be minimal such that 2r > x. Then

θ(x) = θ(x)− θ( x2r

) =

r−1∑j=0

θ(x

2j)− θ( x

2j+1) ≤

r−1∑j=0

(1 + log 2)x

2j≤ 2(1 + log 2)x

Proposition 3.2.

π(x) ∼ x

log x⇐⇒ ψ(x) ∼ x⇐⇒ θ(x) ∼ x

Proof. For a fixed prime p,# {n ≤ x : Λ(n) = log p} is the largest integer jp such that jp log p ≤ log x. Hence,

ψ(x) =∑p≤x

jp log p ≤∑p≤x

log x ≤ π(x) log x

For 0 < δ < 1 and y = x1−δ, we have

π(x) = π(y) +∑

y<p≤x

1 ≤ y +θ(x)

log y= x1−δ +

1

1− δθ(x)

log x

Combing these estimates, we obtain the inequality

θ(x)

x≤ ψ(x)

x≤ π(x)

x/(log x)≤ log x

xδ+

1

1− δθ(x)

x≤ log x

xδ+

1

1− δψ(x)

x

The equivalence follows from noting that log xxδ→ 0, x ↑ ∞ for all δ > 0.

We will show below that θ(x) ∼ x

6

4 Tauberian Theorem for Laplace Transforms

A well-known theorem of Landau (see Theorem 1.7 in [7]) in the elementary theory of Dirichlet seriesstates that if F (s) =

∑∞n=1 ann

−s is a convergent Dirichlet series with finite abscissa of convergence σcand all but finitely many nonnegative coefficients, then s = σc is a singularity of F (s). Equivalently, ifF (s) =

∑∞n=1 ann

−s is analytic in the region {σ > σc} ∪D(σc; r), then∑∞n=1 ann

−s converges for all s withreal part σ ≥ σc. In Newman’s original paper [8], he proves an analogous theorem: if F (s) =

∑∞n=1 ann

−s

is a Dirichlet series with |an| ≤ 1 such that F (s) has analytic continuation to a domain containing {σ ≥ 1},then

∑∞n=1 ann

−s converges for σ ≥ 1. We instead follow Korevaar in [5] in working with Laplace transformsrather than Dirichlet series directly. Newman’s clever contour integration argument, which is the essence ofthis proof of the PNT, is, however, used in both approaches.

Definition 4.1. Let f be a bounded measurable function on [0,∞). The (one-sided) Laplace transformof f is defined by

g(z) :=

∫ ∞0

f(t)e−ztdt, Re z > 0

It is not hard to show that g : {z ∈ C : Re z > 0} → C is analytic. Let ∆ ⊂ C be a triangle, and let ∂∆denote its boundary parametrized by a positively oriented piecewise smooth curve. Since f is bounded and|e−zt| ≤ e−xt, where x = Re(z), we have that∫

∂∆

(∫ ∞0

∣∣f(t)e−zt∣∣ dt) |dz| ≤ ‖f‖L∞ ∫

∂∆

(∫ ∞0

e−xtdt

)|dz| <∞,

since∫∞

0e−xt is a continuous function by the dominated convergence theorem and therefore bounded on the

compact set ∂∆. We can therefore apply Fubini’s theorem to obtain that∫∂∆

(∫ ∞0

f(t)e−zt|dt)dz =

∫ ∞0

f(t)

(∫∂∆

e−ztdz

)dt = 0,

where we the inner integral vanishes by the analyticity of e−zt and Cauchy’s theorem. We conclude fromMorera’s theorem that g is analytic on the half plane {Re z > 0}.

Suppose the Laplace transform of f has an analytic continuation to an open set containing the closedhalf plane {Re z ≥ 0}. Given that if we plug z = 0 into the integrand of the Laplace transform, we obtain∫∞

0f(t)dt, we ask if this hypothesis tells us anything about the integrability of the original function f? Such

results are called Tauberian theorems, and it turns out that if the Laplace transform has such an analyticcontinuation, then the ‘improper’ Lebesgue integral of f exists on the domain [0,∞). We make this precisewith the following theorem.

Theorem 4.2. Let f be a bounded measurable function on [0,∞). Suppose the Laplace transform of f

g(z) =

∫ ∞0

f(t)e−ztdt,

which is analytic in the right half plane H = {z ∈ C : Re z > 0}, extends to analytic function in an open set

containing H = {z ∈ C : Re z ≥ 0}. Then limT→∞∫ T

0f(t)dt exists, and

limT→∞

∫ T

0

f(t)dt = g(0),

the value of the analytic extension at z = 0.

Proof. Let M > 0 such that |f(t)| ≤ M for all t ≥ 0. For T > 0 fixed, the function gT (z) =∫ T

0f(t)e−zT is

entire by the same argument used to show that the Laplace transform is analytic in {Re z > 0}. I claim that

limT→∞

gT (0) = g(0)

We show this by countour integration. Let R > 0 be large and δ > 0 be sufficiently small so that g(z) isanalytic in a domain containing D = {z ∈ C : |z| ≤ R,Re z ≥ −δ}. Let Γ = ∂D, positively oriented. By theCauchy integral formula,

g(0)− gT (0) =1

2πi

∫Γ

[g(z)− gT (z)]

(1 +

z2

R2

)ezT

dz

z

7

Let x = Re z. For x > 0,

|g(z)− gT (z)| =∣∣∣∣∫ ∞T

f(t)e−ztdt

∣∣∣∣ ≤M ∫ ∞T

e−xtdt =Me−Tx

x.

and for |z| = R,∣∣∣∣ezT (1 +z2

R2

)1

z

∣∣∣∣ = exT∣∣∣∣ 1

Re− arg z +

1

Rearg z

∣∣∣∣ =exT

R2

∣∣Re− arg z +Rearg z∣∣ =

2exT |x|R2

From the continuity of g(z)− gT (z), we conclude that∣∣∣∣[g(z)− gT (z)]ezT(

1 +z2

R2

)1

z

∣∣∣∣ ≤ 2M

R2, ∀z ∈ Γ+ = Γ ∩ {Re z ≥ 0} ,

and therefore, ∣∣∣∣∣ 1

2πi

∫Γ+

[g(z)− gT (z)]ezT(

1 +z2

R2

)1

zdz

∣∣∣∣∣ ≤ M

R

We now consider the integal over Γ− = Γ∩{Re z < 0} by considering g and gT separately. Since gT is entire,by Cauchy’s theorem we may replace the contour Γ− by the semi-circle of radius R, Γ′− = {|z| = R,Re z < 0}.For Re z = x < 0,

|gT (z)| =

∣∣∣∣∣∫ T

0

f(t)e−ztdt

∣∣∣∣∣ ≤M∫ T

0

e−xtdt ≤M∫ T

−∞e−xtdx =

Me−xT

|x|

Using our previous estimate for ezT (1 + z2

R2 ) 1z , we obtain∣∣∣∣∣ 1

2πi

∫Γ′−

gT (z)ezT(

1 +z2

R2

)1

zdz

∣∣∣∣∣ ≤ M

R

Since g is analytic on Γ−, there exists K = K(R, δ) > 0 such that∣∣∣∣g(z)

(1 +

z2

R2

)1

z

∣∣∣∣ ≤ K, ∀z ∈ Γ−

Since ezT is bounded on Γ− and converges to 0 on compact subsets of Γ−, as T ↑ ∞, it follows from thedominated convergence theorem that

limT→∞

∣∣∣∣∣ 1

2πi

∫Γ−

g(z)ezT(

1 +z2

R2

)1

zdz

∣∣∣∣∣ = 0

Hence,

lim supT→∞

|g(0)− gT (0)| ≤ 2M

R

Letting R→∞, we conclude that limT→∞ gT (0) = g(0).

5 Main Argument

The argument for the proof of the PNT is relatively simple and boils down to showing that the integral∫ ∞1

θ(x)− xx2

dx

converges as an improper Lebesgue integral. From there, the rest of the proof is almost a triviality. We use

Chebyshev’s estimate to show that the function θ(x)−xx2 is bounded, and we use the analyticity of the function

Φ(s) − 1s−1 in the closed half plane {σ ≥ 1} to show that the Laplace transform has analytic continuation

to a domain containing the closed right half plane {σ ≥ 0}. The reader will recall from the proof of Lemma

2.7 that the analyticity of Φ(s) − 1s−1 is where we need the nonvanishing of ζ(s) for σ ≥ 1. Hence, θ(x)−x

x2

satisfies the hypotheses of Theorem 4.2, and the conditional convergence of the integral follows immediately.

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Lemma 5.1. The integral ∫ ∞1

θ(x)− xx2

dx

converges.

Proof. Consider the function f : [0,∞)→ R, given by

f(t) := θ(et)e−t − 1

As a consequence of Chebyshev’s estimate, f is a bounded measurable function on [0,∞). I claim that theone-sided Laplace transform of f ,

g(z) =

∫ ∞0

f(t)e−ztdt, Re z > 0

extends to an analytic function on an open set containing {Re z ≥ 0}. Since θ(x) is increasing and right-continuous, θ(x) induces a Lebesgue-Stieltjes measure. Integrating by parts, for σ > 1, we have

Φ(s) =∑p

log p

ps=

∫ ∞1

dθ(x)

xs= s

∫ ∞1

θ(x)dx

xs+1= s

∫ ∞0

θ(et)e−stdt,

where we make the change of variable x = et to obtain the last equality. Hence,

g(s) =

∫ ∞0

f(t)e−stdt =

∫ ∞0

θ(et)e−(s+1)tdt−∫ ∞

0

e−stdt =Φ(s+ 1)

s+ 1− 1

s,

which extends analytically to {Re s ≥ 0} by Lemma 2.7. Thus, f satisfies the hypotheses of Theorem 4.2,and we conclude that

limT→∞

∫ T

0

f(t)dt = limT→∞

∫ T

0

[θ(et)e−t − 1

]dt = lim

T→∞

∫ T

0

[θ(x)

x− 1

]dx

x=

∫ ∞0

θ(x)− xx2

dx

Theorem 5.2. θ(x) ∼ x.

Proof. We prove the statement by contradiction. First, suppose that there exists c > 1 such that θ(x) ≥ cxfor all x sufficiently large (say ≥ x0). Then since θ is nondecreasing, for all x ≥ x0,∫ cx

x

θ(t)− tt2

dt ≥∫ cx

x

cx− tt2

dt =

∫ c

1

c− tt2

dt > 0

But since∫∞

1θ(t)−tt2 dt is convergent by the preceding lemma and the integrand is nonnegative,

∫ cxx

θ(t)−tt2 dt→

0, x→∞, which is a contradiction. Now suppose that there exists c < 1 such that θ(x) ≤ cx for all x ≥ x0.Then for x ≥ c−1x0, ∫ x

cx

θ(t)− tt2

dt ≤∫ x

cx

ct− tt2

dt <

∫ 1

c

c− tt2

dt < 0,

which by the same argument, contradicts the convergence of∫∞

1θ(t)−tt2 dt.

References

[1] Ahlfors, Lars V. Complex Analysis (Third Edition). McGraw-Hill, 1979.

[2] Bateman, Paul T. and Harold G. Diamond. A Hundred Years of Prime Numbers, The American Mathe-matical Monthly 103 (1996), no. 9, 729-741.

[3] Bogachev, V.I. Measure Theory, Vol. I. Springer, 2007.

[4] Dym, H. and H.P. McKean, Fourier Series and Integrals. Academic Press, 1972.

[5] Korevaar, J. On Newman’s Quick Way to the Prime Number Theorem, The Mathematical Intelligencer4 (1982), no. 3, 108-115.

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[6] Lax, Peter D. and Lawrence Zalcman, Complex Proofs of Real Theorems. American Mathematical Society,2012.

[7] Montgomery, Hugh and Robert Vaughn, Multiplicative Number Theory: I. Classical Theory. CambridgeUniversity Press, 2006.

[8] Newman, D.J. Simple Analytic Proof of the Prime Number Theorem, The American MathematicalMonthly 87 (1980), no. 9, 693-696.

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