AN ELEMENTARY PROOF OF THE PRIME NUMBER ...may/REU2017/REUPapers/Choudhary.pdfAN ELEMENTARY PROOF OF THE PRIME NUMBER THEOREM 3 Thus, the mobius and unit functions are inverses of

AN ELEMENTARY PROOF OF THE PRIME NUMBER

THEOREM

ABHIMANYU CHOUDHARY

Abstract. This paper presents an ”elementary” proof of the prime number

theorem, elementary in the sense that no complex analytic techniques areused. First proven by Hadamard and Valle-Poussin, the prime number the-

orem states that the number of primes less than or equal to an integer x

asymptotically approaches the value xln x

. Until 1949, the theorem was con-sidered too ”deep” to be proven using elementary means, however Erdos and

Selberg successfully proved the theorem without the use of complex analysis.

My paper closely follows a modified version of their proof given by NormanLevinson in 1969.

Contents

1. Arithmetic Functions 12. Elementary Results 23. Chebyshev’s Functions and Asymptotic Formulae 44. Shapiro’s Theorem 105. Selberg’s Asymptotic Formula 126. Deriving the Prime Number Theory using Selberg’s Identity 15Acknowledgments 25References 25

1. Arithmetic Functions

Definition 1.1. The prime counting function denotes the number of primes notgreater than x and is given by π(x), which can also be written as:

π(x) =∑p≤x

1

where the symbol p runs over the set of primes in increasing order.

Using this notation, we state the prime number theorem, first conjectured byLegendre, as:

Theorem 1.2.

limx→∞

π(x) log x

x= 1

Note that unless specified otherwise, log denotes the natural logarithm.1

2 ABHIMANYU CHOUDHARY

2. Elementary Results

Before proving the main result, we first introduce a number of foundationaldefinitions and results.

Definition 2.1. An arithmetical function or a sequence, is a function whose domainis the natural numbers, and codomain is either the real numbers or the complexnumbers.

Definition 2.2. We define the divisor sum of an arithmetic function a to be:∑d|n

a(d)

where the symbol d ranges over the set of positive divisors of n.

Definition 2.3. We define the Dirichlet product or Dirichlet Convolution of twoarithmetic functions f, g as:

(f ∗ g)(n) =∑d|n

f(d)g(nd

)Note that the Dirichlet Product is commutative and associative. Moreover, the

set of Arithmetical functions has an identity I over this product, and every arith-metical function with the property that f(1) 6= 0 has an inverse f−1 such thatf ∗ f−1 = I. It is easy to verify that the identity function I is given by:

I(n) =

⌊1

n

⌋=

{1 if n = 1

0 otherwise

Definition 2.4. We define the Mobius function, µ as:

µ(n) =

1 if n = 1

(−1)k if n = p1, ..., pk for primes p1, ..., pk

0 otherwise

Thus, the Mobius function allows us to determine a ”parity” of sorts for anysquarefree integer.

Theorem 2.5. The divisor sum of the mobius function is given by:∑d|n

µ(d) =

⌊1

n

⌋=

{1 if n = 1

0 otherwise= I(n)

This can be verified using the fundamental theorem of arithmetic and the bino-mial theorem. Note that this divisor sum yields the identity function, an importantproperty we will use momentarily.

Definition 2.6. We define the unit function by:

u(n) = 1

for all natural n.

We see that the divisor sum in 2.5 can be rewritten as:∑d|n

µ(d) =∑d|n

µ(d)1 =∑d|n

µ(d)u(nd

)= (µ ∗ u)(n) =

⌊1

n

⌋

AN ELEMENTARY PROOF OF THE PRIME NUMBER THEOREM 3

Thus, the mobius and unit functions are inverses of each other. We can use thisproperty to derive a powerful formula, known as the Mobius inversion formula.

Theorem 2.7 (Mobius inversion formula). If f, g are arithmetical functions and:∑d|n

g(d) = f(n)

then: ∑d|n

f(d)µ(nd

)= g(n)

Proof. We have by 2.3 that:

g ∗ u = f

Taking the convolution of both sides with µ we have:

µ ∗ (g ∗ u) = µ ∗ f

Using associativity and commutativity, we can write the above expression as:

g = µ ∗ f = f ∗ µ

as required. �

Corollary 2.8 (Generalized Mobius Inversion). If f, g are arithmetical functionsand:

g(x) =∑n≤x

f(xn

)then

f(x) =∑n≤x

µ(n)g(xn

)where the symbol n ranges over all integers not greater than x.

We now introduce Von Mangoldt’s function given by the symbol Λ.

Definition 2.9 (Von Mangoldt’s Function). For every integer n ≥ 1 we define:

Λ(n) =

{log(p) if n = pk for some prime p and k ≥ 1

0 otherwise

The above definition is fairly powerful as it turns a multiplication problem (primefactorization), into an addition problem through the use of logarithms. We are alsoprohibited from ”double counting” any prime factors, as we will see in the nexttheorem.

Theorem 2.10 (Divisor sum of the Von Mangoldt Function).∑d|n

Λ(d) = log n

The proof for this result can be derived follows naturally from the fundamentaltheorem of arithmetic. Roughly speaking, the function counts each prime factorof n exactly as many times as it appears in the prime factorization of n. Throughsumming and properties of the logarithm, the result follows.


3. Chebyshev’s Functions and Asymptotic Formulae

Definition 3.1. We say that a function f is ”big-oh g(x) ” for all x ≥ a or writethat:

f(x) = O(g(x))

if there exists a constant M such that for all x ≥ a we have:

f(x) ≤Mg(x)

Corollary 3.2. Let f, g be Riemann Integrable functions such that f(t) = O(g(t))for t ≥ a. Then we have: ∫ x

a

f(t)dt = O(∫ x

a

g(t)dt

)Definition 3.3. We say that f is asymptotic to g or f ∼ g, if:

limx→∞

f(x)

g(x)= 1

Definition 3.4. We define the extension of an arithmetic function a as a mapR+ → R given by:

a(x) = a (bxc) for all x ∈ R+

We now have a suitable way to extend the domain of arithmetic functions to thepositive reals. We now have a new set of tools to our disposal, namely those of cal-culus. We now supply a powerful summation formula that allows us to approximatethe partial sums of arithmetic functions.

Theorem 3.5 (Abel’s Summation Formula). Let f be a real valued function with aRiemann-Integrable derivative for t ≥ 1. Let a(n) be an arithmetical function andlet A(x) be the partial sum of a up to x. Then:∑

n≤x

a(n)f(n) = f(x)A(x)−∫ x

1

f ′(t)A(t)dt

Proof. Taking suitable a, f , we have that:∑1≤n≤x

A(n)−A(n− 1) = [A(1)−A(0)] + ...+ [A(n)−A(n− 1)]

This sum clearly telescopes to the value A(n) − A(0). Because A(0) is an emptysum, we have that: ∑

n≤x

a(n) =∑n≤x

[A(n)−A(n− 1)]

Mutliplying both sides by f(n):∑n≤x

a(n)f(n) =∑n≤x

[A(n)−A(n− 1)]f(n)

Expanding the right hand side:∑n≤x

[A(n)−A(n− 1)]f(n) =∑n≤x

A(n)f(n)−∑n≤x

A(n− 1)f(n)

Reindexing, it follows that:∑n≤x

A(n)f(n)−∑n≤x

A(n− 1)f(n) =∑n≤x

A(n)f(n)−∑

n≤x−1

A(n)f(n+ 1)


Note that: ∑n≤x

A(n)f(n) =∑

n≤x−1

A(n)f(n) +A(x)f(x)

So we have:∑n≤x

A(n)f(n)−∑

n≤x−1

A(n)f(n+1) = A(x)f(x)+∑

n≤x−1

A(n)f(n)−∑

n≤x−1

A(n)f(n+1)

Combining we have:

A(x)f(x) +∑

n≤x−1

A(n)f(n)−∑

n≤x−1

A(n)f(n+ 1) =∑

n≤x−1

A(n)(f(n+ 1)− f(n))

Because f has Riemann integrable derivative, we can apply the fundamental theo-rem of calculus to it and say:∑

n≤x−1

A(n)(f(n+ 1)− f(n)) =∑

n≤x−1

A(n)

∫ n+1

n

f ′(t)dt

Because A(n) is constant on the interval [n, n + 1) and takes on the value A(n)everywhere, we can place it inside the integral. Thus:∑

n≤x−1

A(n)

∫ n+1

n

f ′(t)dt =∑

n≤x−1

∫ n+1

n

A(t)f ′(t)dt

. Summing the integrals, we have:∑n≤x−1

∫ n+1

n

A(t)f ′(t)dt =

∫ x

1

A(t)f ′(t)dt

Thus, in conclusion, we see that∑n≤x

a(n)f(n) = A(x)f(x) +

∫ x

1

A(t)f ′(t)dt

as we need. �

Corollary 3.6 (Euler’s Summation Formula). Let f be a function with Riemann-integrable derivative defined on the interval [1, x]. Then:∑

n≤x

f(n) =

∫ x

1

f(t)dt+

∫ x

1

f ′(t)dt+ f(x)(bxc − x)

Proof. This result follows from the Abel summation formula. �

We will use these results to derive results about the asymptotic behavior of cer-tain arithmetic functions. We first introduce two important functions of Chebyshevwhose asymptotic behavior we will examine.

Definition 3.7. For x > 0 we define the Chebyshev ϑ function by:

ϑ(x) =∑p≤x

log p

where the symbol p runs over all primes not exceeding x.

Definition 3.8. For x > 0 we define the Chebyshev ψ function by:

ψ(x) =∑n≤x

Λ(n)


Lemma 3.9. For x > 1 we have:

0 ≤ ψ(x)

x− ϑ(x)

x≤ (log x)2

2√x log 2

Remark 3.10. Notice by the squeeze theorem that if either quotient ψ(x)x , ϑ(x)x has

a limit to infinity, then the other quotient has the same limit.

Theorem 3.11. The following 3 statements are equivalent:

(1)

limx→∞

π(x) log x

x= 1

(2)

limx→∞

ϑ(x)

x= 1

(3)

limx→∞

ψ(x)

x= 1

(4)

limx→∞

ψ(x)− xx

= 0

Remark 3.12. Throughout the rest of this paper, the function ψ(x)− x is denotedby R(x) and will be known as remainder function.

We prove that 1 is equivalent to 2. The equivalency of 2 and 3 follows from 3.10and the equivalency of 3 and 4 follows trivially. We first use a corollary of 3.5,Abel’s Summation Formula:

Corollary 3.13. We have formulas:

(1)

ϑ(x) = π(x) log x−∫ x

2

π(t)

t

(2)

π(x) =ϑ(x)

log x+

∫ x

2

ϑ(t)

t log2 t

These results can be derived in a straightforward manner using the summationformula. We now use them to prove 3.10.

Proof. We first assume (1) from 3.11. We have then by 3.13 that:

ϑ(x)

x=π(x) log x

x− 1

x

∫ x

2

π(t)

t

It suffices then to show that:

limx→∞

1

x

∫ x

2

π(t)

t= 0

We know by our assumption that:

π(t)

t= O

(1

log t

)


for t ≥ 2. So:1

x

∫ x

2

π(t)

t= O

(1

x

∫ x

2

1

log tdt

)by 3.2. We can bound this integral as follows:∫ x

2

1

log tdt =

∫ √x2

1

log tdt+

∫ x

√x

1

log tdt ≤

√x

log 2+x−√x

log√x

and this approaches 0 as x approaches infinity. So by the squeeze theorem, theoriginal integral approaches 0, and we have that:

limx→∞

ϑ(x)

x= 1

Similarly, to prove the other direction, we now assume that:

limx→∞

ϑ(x)

x= 1

Using our second integral formula, we have:

π(x) log x

x=ϑ(x)

x+

log x

x

∫ x

2

ϑ(t)

t log2 tdt

As we did previously, we must show the integral expression on the right hand sideapproaches 0 as we take a limit to infinity. Our initial assumption implies thatϑ(t) = O(t), so we have that:

log x

x

∫ x

2

ϑ(t)

t log2 tdt = O

(log x

x

∫ x

2

1

log2 tdt

)Again, bounding this integral in a similar manner, we have:∫ x

2

1

log2 tdt ≤

∫ √x2

1

log2 t+

∫ x

√x

1

log2 tdtdt ≤

√x

log2 2+x−√x

log2√xAnd again by the squeeze theorem, we see our limit is 0, as necessary, and thesecond direction is complete. We have proven the equivalency of 3.11.1 and 3.11.2,and the other equivalences follow from this, as stated before. Knowing this, wenow set out to prove the prime number by proving equivalent form 3.11.4, i.e that

limx→∞R(x)x = 0 �

We now examine the asymptotic behavior of Von-Mangoldt’s function and otherarithmetical functions. The following identities will prove useful later in our analysisof Chebyshev’s ϑ and ψ functions. Before we do this however, we will introduce aset of corollaries about partial sums of dirichlet products (see definition 2.3).

Theorem 3.14. Let f, g be arithmetic functions and denote h = f ∗ g. Let H,F,Gdenote the partial sums of their respective functions. Then we have that:

H(x) =∑n≤x

f(n)G(xn

)=

Applying the above to a single arithmetic function, we have that f = f ∗ I andthus:

Theorem 3.15. For F (x) =∑n≤x f(n) we have:∑

n≤x

∑d|n

f(d) =∑n≤x

f(n)⌊xn

⌋=∑n≤x

F(xn

)


We now introduce an identity about the partial sums of the harmonic series:

Theorem 3.16. ∑n≤x

1

n= log x+ γ +O

(1

x

)Where γ is a constant, hereafter called the ”Euler-Mascheroni” constant.

Proof. This is a consequence of 3.6, the Euler summation formula. �

Theorem 3.17. ∑x≤n

Λ(n)⌊xn

⌋= logbnc!

Proof. This follows from a combination of 2.10 and 3.15. �

Lemma 3.18. (Legendre)

bxc! =∏p≤x

pα(p)

Where:

α(p) =

∞∑m=1

⌊x

pm

⌋Proof. This identity is a consequence of 3.17. �

Theorem 3.19. For x ≥ 2, we have:

logbxc! = x log x− x+O(log x)

Proof. We see that:

logbxc! =∑n≤x

log n

Applying Euler’s Summation formula, we have:∑n≤x

log n =

∫ x

1

log t dt+

∫ x

1

(t− btc)t

dt− (x− bxc) log x

= x log x+

∫ x

1

(t− btc)t

dt+O(log x)

We know that:t− btct

= O(

1

t

)So, we have: ∫ x

1

(t− btc)t

dt = O(∫ x

1

1

tdt

)= O(log x)

Thus, ∑n≤x

log n = x log x+O(log x)

�

and an immediate corollary that follows from this:

Corollary 3.20. ∑x≤n

Λ(n)⌊xn

⌋= x log x− x+O(log x)


Proof. This follows from 3.17. �

Corollary 3.21.∑n≤x

Λ(n)⌊xn

⌋=∑n≤x

ψ(xn

)= x log x− x+O(log x)

Proof. This follows from 3.15 and the definition of the ψ function as the partialsums of Von Mangoldt’s function. �

Remark 3.22. Note that we can also use the less precise approximation:∑n≤x

Λ(n)⌊xn

⌋=∑n≤x

ψ(xn

)= x log x+O(x)

as the logarithm is dominated by the linear term. This approximation will be usefullater on, when we discuss Shapiro’s theorem.

The next theorem follows as a consequence of 3.21.

Theorem 3.23. ∑p≤x

⌊x

p

⌋log p = x log x+O(x)

Proof. We know by 3.21 that∑n≤x

Λ(n)⌊xn

⌋= x log x− x+O(log x)

Now, we will reindex the above sum so it can be written in terms of primes lessthan or equal to x. We know that Λ is nonzero for prime powers and 0 otherwise.Thus, we have ∑

n≤x

Λ(n)⌊xn

⌋=∑p≤x

∞∑m=1

Λ(pm)

⌊x

pm

⌋Note that the above ”infinite” sum is indeed a finite sum, as for sufficiently large

m, we have pm ≥ x and thus⌊xpm

⌋= 0. By the definition of Λ we have:

∑p≤x

∞∑m=1

Λ(pm)

⌊x

pm

⌋=∑p≤x

∞∑m=1

log p

⌊x

pm

⌋Decomposing the above sum we have:∑

p≤x

∞∑m=1

log p

⌊x

pm

⌋=∑p≤x

∞∑m=2

log p

⌊x

pm

⌋+∑p≤x

log p

⌊x

p

⌋We now prove that: ∑

p≤x

∞∑m=2

log p

⌊x

pm

⌋= O(x)

We know by definition of the floor function that:∑p≤x

∞∑m=2

log p

⌊x

pm

⌋≤∑p≤x

∞∑m=2

log p

(x

pm

)


Summing the geometric series of the right sum we have:∑p≤x

∞∑m=2

log p

(x

pm

)= x

∑p≤x

log p

(1

(p− 1)p

)Again, we have:

x∑p≤x

log p

(1

(p− 1)p

)≤ x

∞∑n=2

log n

(1

(n− 1)n

)= O(x)

as the series on the right hand side converges by comparison. �

4. Shapiro’s Theorem

We now provide a proof of Shapiro’s theorem, an important theorem whichrelates partial sums of the form

∑n≤x

⌊xn

⌋a(n) to the often more interesting sums

of form∑n≤x a(n). Specifically, we will use this to derive a result about the

behavior of the partial sums of the ψ function.

Theorem 4.1. (Shapiro’s Tauberian Theorem) Let a(n) be a nonnegative sequencesuch that: ∑

n≤x

a(n)⌊xn

⌋= x log x+O(x)

Then the following are true:

(1) For n ≥ 1 we have: ∑n≤x

a(n)

n= log x+O(1)

(2) There exists a constant M such that:∑n≤x

a(n) ≤Mx for all x ≥ 1

(3) There exists a constant m such that:∑n≤x

a(n) ≥ mx for all x ≥ 1

Proof. Define functions S, T by:

S(x) =∑n≤x

a(n), T (x) =∑n≤x

a(n)⌊xn

⌋We first show the inequality:

S(x)− S(x

2

)≤ T (x)− 2T

(x2

)holds. We have:

T (x)− 2T(x

2

)=∑n≤x

a(n)⌊xn

⌋− 2

∑n≤ x

2

a(n)⌊ x

2n

⌋Reindexing we have

T (x)− 2T(x

2

)=∑n≤ x

2

a(n)⌊xn

⌋− 2a(n)

⌊ x2n

⌋+

∑x2<n≤x

a(n)⌊xn

⌋


The sum on the left hand side will be nonnegative because b2xc − 2bxc is alwaysnonnegative (this can be checked by considering the min and max of both functions)and our sequence is nonnegative. Thus, we have:

T (x)− 2T(x

2

)≥

∑x2<n≤x

a(n)⌊xn

⌋For n larger than x

2 , we observe that the term⌊xn

⌋is 1 as 1 is the floor of any

number between 1 and 2. Thus, we have:

T (x)− 2T(x

2

)≥

∑x2<n≤x

a(n)⌊xn

⌋=

∑x2<n≤x

a(n) = S(x)− S(x

2

)as required. We now use this fact to prove statement 2. Recall by our initialassumption that:

T (x) = x log x− x+O(x)

We have then that:

T (x)− 2T(x

2

)= x log x− x+O(x)− 2

(x2

logx

2− x

2+O(x)

)= O(x)

By the previous inequality, we can establish that:

S(x)− S(x

2

)= O(x)

for all x ≥ 1. Thus, we have a constant M ′ such that S(x)−s(x2 ) ≤M ′x. Applyingthis to successive values of x as follows, we can get expressions of the forms:

S(x

2

)− S

(x4

)≤M ′x

2

S(x

4

)− S

(x8

)≤M ′x

4

...

and so on, we see that when we sum these expressions, all terms cancel except S(x)and we are left with:

S(x) ≤M ′x(

1 +1

2+

1

4+ ...

)= 2M ′x

so we have found an appropriate constant, namely 2M ′. We now prove part 1. Wesee that: ⌊x

n

⌋=(xn

)+O(1)

So:

T (x) =∑x≤n

⌊xn

⌋a(n) =

∑x≤n

((xn

)+O(1)

)a(n)

= x∑x≤n

a(n)

n+∑n≤x

a(n)O(1)

= x∑x≤n

a(n)

n+O(x)


If we substitute the asymptotic formula we have for T (x), it follows almost imme-diately that:

x∑x≤n

a(n)

n= log x+O(1)

as needed. �

We can now state a number of corollaries that follow immediately from thisresult:

Corollary 4.2. The following asymptotic formulae hold:

(1)ψ(x) = O(x)

(2) ∑n≤x

Λ(n)

n= log x+O(1)

(3)ϑ(x) = O(x)

(4) ∑p≤x

log p

p= log x+O(1)

All of these corollaries follow from previous asymptotic formulae derived in sec-tion 3.

5. Selberg’s Asymptotic Formula

We now prove a major result, first derived by Atle Selberg in 1949.

Theorem 5.1 (Selberg’s Asymptotic Formula). For x > 0 we have the following:

ψ(x) log x+∑n≤x

Λ(n)ψ(xn

)= 2x log x+O(x)

We first prove a lemma which will help us obtain the final result

Lemma 5.2 (Tatuzawa Iseki Identity). Let F be a real valued function defined onR+ and let G be given by:

G(x) = log x∑n≤x

F(xn

)Then, we have:

F (x) log x+∑n≤x

F(xn

)Λ(n) =

∑d≤x

µ(d)G(xd

)Proof. We first rewrite F (x) log x as a sum. We have:

(5.3) F (x) log x =∑n≤x

⌊1

n

⌋F(xn

)log

x

n=∑n≤x

∑d|n

µ(d)F(xn

)log

x

n

Now, we can use the Mobius inversion formula to say that:

Λ(n) =∑d|n

µ(d) logn

d


So we have:

(5.4)∑n≤x

F(xn

)Λ(n) =

∑n≤x

F(xn

)∑d|n

µ(d) logn

d

Adding (5.3) and (5.4) we have:∑n≤x

F(xn

)∑d|n

µ(d)[log(xn

)+ log

(nd

)]This is equal to: ∑

n≤x

F(xn

)∑d|n

µ(d) logx

d

So, summarizing our results so far, we have:

(5.5) F (x) log x+∑n≤x

F(xn

)Λ(n) =

∑n≤x

F(xn

)∑d|n

µ(d) logx

d

Now, taking the right hand side and writing n = qd we have:

(5.6)∑n≤x

F(xn

)∑d|n

µ(d) logx

d=∑d≤x

µ(d) logx

d

∑q≤ x

d

F

(x

qd

)And by our initial definition of G, we can rewrite the right hand side of (5.6) as:∑

d≤x

µ(d)G(xd

)as needed. �

We now prove 5.1 using this lemma.

Proof. We apply 5.2 to the functions ψ(x) and x − γ − 1, where γ is the EulerMascheroni constant. For ψ(x) we can define the associated Gψ as:

Gψ(x) = log x∑n≤x

ψ(xn

)= log x(x log x− x+O(log x))

= x log2 x− x log x+O(log2 x)

which follows from 3.21. We have for x− γ + 1 that:

Gγ(x) = log x∑n≤x

(xn− γ + 1

)= x log x

∑n≤x

1

n− log x

∑n≤x

(γ + 1)

By 3.16, we have:

x log x∑n≤x

1

n= x log x

(log x+ γ +O

(1

x

))


So:

Gγ(x) = x log x

(log x+ γ +O

(1

x

))− log x

∑n≤x

(γ + 1)

= x log x

(log x+ γ +O

(1

x

))− log x(γ + 1)x

= x log2 x− x log x+O(log x)

We see that the difference between Gψ and Gγ is O(log2 x). We will use the weakerestimate of O(

√x). Now, apply 5.2 (Tatuzawa Iseki) to ψ and x− γ + 1. We have

for ψ that:

(5.7) ψ(x) log x+∑n≤x

ψ(xn

)Λ(n) =

∑d≤x

µ(d)Gψ

(xd

)and for x− γ + 1 we have:

(5.8) (x− γ + 1) log x+∑n≤x

x

n=∑d≤x

µ(d)Gγ

(xd

)Subtracting the RHS of 5.7 from that of 5.8,we have the term:∑

d≤x

µ(d)(Gψ

(xd

)−Gγ

(xd

))Applying our O(

√x) estimate for the difference of Gψ and Gγ with x

d as our argu-ment, we have:

(5.9)∑d≤x

µ(d)(Gψ

(xd

)−Gγ

(xd

))= O

∑d≤x

√x

d

Factoring, we see that 5.9 is equal to:

(5.10) O

√x∑d≤x

√1

d

= O(x)

(√x

∫ x

1

1√tdt

)= O(x)

Using 5.2 (Tatuzwa Iseki), this time on the auxiliary function:

ψ(x)− (x+ γ + 1)

we have:

[ψ(x)−(x+γ+1)] log x+∑x≤n

[ψ(xn

)− x

n− γ − 1

]Λ(n) =

∑d≤x

µ(d)(Gψ

(xd

)−Gγ

(xd

))and we know from 5.10 that the RHS is O(x), so we have:

ψ(x)− (x+ γ + 1)] log x+∑x≤n

[ψ(xn

)− x

n− γ − 1

]Λ(n) = O(x)


Rearranging terms and simplifying we have:

ψ(x) log x+∑n≤x

Λ(n)(xn

)= O(x) + (x+ γ + 1) log x+

∑n≤x

(xn

+ γ + 1)

Λ(n)

= O(x) + x log x+O(log x) + x∑n≤x

Λ(n)

n+ (γ + 1)

∑n≤x

Λ(n)

= O(x) + x log x+O(log x) + x log x+O(1) +O(log x)

= 2x log x+O(x)

which is the Selberg Asymptotic identity. �

We now provide two alternate formulations of Selberg’s Identity. The corollarybelow follows immediately through an application of 3.5 (Abel Summation).

Corollary 5.11.∑n≤x

Λ(n) log n = ψ(x) log x−∫ x

1

ψ(t)

tdt = ψ(x) log x+ O(x)

Definition 5.12. Define:

Λ2(n) = Λ(n) log n+∑d|n

Λ(d)Λ(nd

)= Λ(n) log n+ (Λ ∗ Λ)(n)

By 3.15 the partial sums of Λ2 are given by:∑n≤x

Λ2(n) =∑n≤x

Λ(n) log n+∑n≤x

(Λ ∗ Λ)(n) = ψ(x) log x+O(x) +∑n≤x

Λ(n)ψ(xn

)So an equivalent restatement of Selberg’s identity is:

(5.13)∑n≤x

Λ2(n) = 2x log x+O(x)

Moving the 2x log x to left hand side and applying , we have:

(5.14) Q(x) =∑n≤x

Λ2(n)− 2∑n≤x

log n =∑n≤x

Λ2(n)− 2 log n = O(x)

6. Deriving the Prime Number Theory using Selberg’s Identity

We now move to derive the prime number theorem using Selberg’s identity. Wewill mainlybe working with the function R(x) = ψ(x) − x. We first introducea lemma that allows us to restate Selberg’s identity in terms of the remainderfunction, rather than ψ.

Lemma 6.1. Selberg’s identity can be restated as:

R(x) log x+∑n≤x

Λ(n)R(xn

)= O(x)

Unfortunately, due to the nature of ψ(x), R(x) is also particularly temperamen-

tal. and so is the quotient, R(x)x . Thus, we will use the smoother:

S(x) =

∫ x

2

R(t)

tdt


as our starting point. If we can show that:

limx→∞

S(x)

x= 0

then we can show the same result for the quotient involving R, which is the resultwe need.We now prove two properties of S.

Lemma 6.2. The following is true of S(x):

(1)

S(x) = O(x)

(2)

S(x) is Lipschitz

Proof. We first show 1. Recall that ψ(x) is O(x) for x ≥ 2 and thus we have someconstant M1 such that:

ψ(x) ≤M1x for all x ≥ 2

Subtracting x we have:

R(x) = ψ(x)− x ≤M1x− x = x(M1 − 1) for all x ≥ 2

Dividing by x, we see:R(x)

x≤ (M‘ − 1)

So clearly, R(x)x = O(1). We know then that:

S(x) =

∫ x

2

R(t)

tdt ≤

∫ x

2

O(1)dt = O(∫ x

2

1dt

)= O(x)

as needed. To show the function is Lipschitz, we can use the fact that for anyx1, x2 ≥ 2:

|S(x1)| ≤M1|x1|

|S(x2)| ≤M1|x2|as S(x) = O(x). Subtracting the expressions and applying the triangle inequalitywe have:

|S(x1)− S(x2)| ≤ |S(x1)| − |S(x2)| ≤M1|x1| −M1|x2| ≤M1|x1 − x2|

and thus S is Lipschitz, as required. �

Corollary 6.3.

S′(y) =R(x)

x= O(1)

where S′(y) is defined when y 6= pk for some prime p.

This was proven indirectly in the proof of 6.2. The restriction on y gives us aguarantee of continuity. We now introduce an important corollary:

Corollary 6.4.

S(y) log y +∑n≤y

Λ(n)S( yn

)= O(y)


Proof. Dividing the expression in (6.1) by x and integrating both sides from 2 toy, we have: ∫ y

2

R(x)

xlog xdx+

∫ y

2

∑n≤x

Λ(n)S(xn

)dx = O(1)

Integrating the first expression from the left by parts, we have:∫ y

2

R(x)

xlog x = log y

∫ y

2

R(x)

x−∫ y

2

S(x)1

xdx

= S(y) log y −O(y)

Decomposing the second integral we have:∫ y

2

∑n≤x

Λ(n)S(xn

)dx =

∑n≤x

Λ(n)

∫ y

2

S(xn

)dx

This can be integrated by applying the substitution u = xn and gives us:∫ y

2

∑n≤x

Λ(n)S(xn

)dx =

∑n≤x

Λ(n)S( yn

)Combining, we have:

S(y) log y −O(y) +∑n≤x

Λ(n)S( yn

)= O(1)

and the result follows. �

Lemma 6.5. There exists a constant Z1 such that:

log2 |S(y)| ≤∑m≤x

Λ2

∣∣∣S ( ym

)∣∣∣+ Z1y log y

Proof. We prove this by beginning with (6.3). We have that:

S(y) log y +∑n≤y

Λ(n)S( yn

)= O(y)

We now substitute y for yk , for some positive dummy variable k, and we have:

S(yk

)log(yk

)+∑n≤ y

k

Λ(n)S( y

kn

)= O

(yk

)=O(y)

k

Multiplying both sides by Λ(k) we get:

Λ(k)S(yk

)log(yk

)+ Λ(k)

∑n≤ y

k

Λ(n)S( y

kn

)=O(y)Λ(k)

k

Summing from 1 ≤ k ≤ y we have:∑k≤y

Λ(k)S(yk

)log(yk

)+

∑k≤y

∑n≤ y

k

Λ(k)Λ(n)S( y

kn

) = O(y)∑k≤y

Λ(k)

k

The RHS simplifies to:

O(y)(y log y +O(1)) = O(y log y)


by 6.2. We now simplify the LHS. First, we break up the sum in the first set ofbrackets using properties of logarithms. We have:∑

k≤y

Λ(k)S(yk

)log(yk

)=∑k≤y

Λ(k)S(yk

)log (y)−

∑k≤y

Λ(k)S(yk

)log (k)

Rewriting the above sum and making substitutions, we then have:

∑k≤y

Λ(k)S(yk

)log (y)−

∑k≤y

Λ(k)S(yk

)log (k)+

∑k≤y

∑n≤ y

k

Λ(k)Λ(n)S( y

kn

) = O(y log y)

We turn our attention to:

−∑k≤y

Λ(k)S(yk

)log (k) +

∑k≤y

∑n≤ y

k

Λ(k)Λ(n)S( y

kn

)We first perform a sign change, and see the above is equal to:

−

∑k≤y

Λ(k)S(yk

)log (k)−

∑k≤y

∑n≤ y

k

Λ(k)Λ(n)S( y

kn

)Let m = kn. Because k ≤ y and n ≤ y

k , we know m ≤ y. We can thus reindex theleft hand portion of the above (the symbol changes from k to m but the value ofthe sum does not). We have:

−

∑m≤y

Λ(m)S( ym

)log (m)−

∑k≤y

∑n≤ y

k

Λ(k)Λ(n)S( ym

)Because k ≤ y and n ≤ y

k , we see that the above means nk = m ≤ y, so reindexagain:

−

∑m≤y

Λ(m)S( ym

)log (m)−

∑m≤y

[S( ym

)( ∑kn=m

Λ(k)Λ(n))

)]Combining and applying the definition of Λ2 we have:

−

∑m≤y

S( ym

)Λ2(m)

Now, we reconsider: ∑

k≤y

Λ(k)S(yk

)log (y)

Moving the logarithm to the outside, we have:

log y∑k≤y

Λ(k)S(yk

)By 6.1, this is:

log y(O(y)− S(y) log y)


So in summary, we have:

log y(O(y)− S(y) log y)−

∑m≤y

S( ym

)Λ2(m)

= O(y log y)

And moving the sum to the RHS will give us the result. �

Because we showed that Λ2(m) = 2 logm+O(m) near the end of section 6, ournext lemma will show that we can use weights of 2 logm, making our sum easier towork with.

Lemma 6.6. There exists a constant Z2 such that:

(6.7) log2 y|S(y)| ≤ 2∑m≤y

∣∣∣S ( ym

)∣∣∣ logm+ Z2y log y

Proof. We define a second remainder function C(y), as given by:

(6.8)∑m≤y

∣∣∣S ( ym

)∣∣∣ (Λ2(m)− 2 logm) = C(y)

Now, recall the function Q defined by:

(6.9) Q(y) =∑m≤y

(Λ2(m)− 2 logm)

We see that:

Λ2(m)− 2 logm = Q(m)−Q(m− 1)

Substituting, we have:∑m≤y

∣∣∣S ( ym

)∣∣∣ (Q(m)−Q(m− 1)) = C(y)

We can ignore all terms with m < 2 as S(m) = 0 for those terms. Throughexamination (this becomes exceedingly clear when terms are written out), we canreindex the sum above as:

(6.10) C(y) =∑

2≤m≤y

(∣∣∣S ( ym

)∣∣∣− ∣∣∣∣S ( y

m+ 1

)∣∣∣∣)Q(m)

Applying the reverse triangle inequality (|x− y| ≥ ||x| − |y||), we have:

C(y) ≤∑

2≤m≤y

(∣∣∣∣S ( ym)− S(

y

m+ 1

)∣∣∣∣)Q(m)

Because S is Lipschitz, we have some Z3 such that:(6.11)∑2≤m≤y

(∣∣∣∣S ( ym)− S(

y

m+ 1

)∣∣∣∣)Q(m) ≤∑

2≤m≤y

(Z3

∣∣∣∣( ym)−(

y

m+ 1

)∣∣∣∣)Q(m)

Moreover, we can bound Q(m) by some linear function of m as Q(m) = O(m) by5.14. So, making Z3 large enough and substituting 6.11 we have:

C(y) ≤ Z3m∑

2≤m≤y

(∣∣∣∣( ym)−(

y

m+ 1

)∣∣∣∣)


Factoring a y term and simplifying the right hand side, we have:

C(y) ≤ yZ3m∑

2≤m≤y

(∣∣∣∣( 1

m

)−(

1

m+ 1

)∣∣∣∣) = yZ3

∑2≤m≤y

∣∣∣∣( 1

m+ 1

)∣∣∣∣We see that:

C(y) ≤ yZ3

∑2≤m≤y

∣∣∣∣( 1

m+ 1

)∣∣∣∣ ≤ yZ3

∫ y

1

1

mdm = Z3y log y

Combining the above with our expression for C(y) proves the lemma. �

We now further simplify this inequality by replacing the sum above with anintegral.

Lemma 6.12. There is a constant Z4 such that:

log2 y|S(y)| ≤ 2

∫ y

2

∣∣∣S (yu

)∣∣∣ log u du+ Z4y log y

Proof. Note the following bound on the integral:

logm∣∣∣S ( y

m

)∣∣∣ ≤ ∫ m+1

m

∣∣∣S (yu

)∣∣∣ log u du

As log is an increasing function. Now, by the triangle inequality, we see that:∣∣∣S ( ym

)∣∣∣ ≤ ∣∣∣S (yu

)∣∣∣+∣∣∣S ( y

m

)− S

(yu

)∣∣∣Integrating we have:∫ m+1

m

∣∣∣S ( ym

)∣∣∣ log u du ≤∫ m+1

m

∣∣∣S (yu

)∣∣∣ log u du+

∫ m+1

m

∣∣∣S ( ym

)− S

(yu

)∣∣∣ log u du

Denote the integral furthest to the right as Jm. Now, using the Lipschitz propertyand the appropriate constant M1 from 6.2, we have that:

Jm =

∫ m+1

m

∣∣∣S ( ym

)− S

(yu

)∣∣∣ log u du ≤∫ m+1

m

M1

∣∣∣ ym− y

u

∣∣∣ log u du

We can bound this by:

Jm ≤∫ m+1

m

M1

∣∣∣ ym− y

u

∣∣∣ log u du ≤M1

(y

m− y

m+ 1

)∫ m+1

m

log u du

Simplifying and bounding the final integral on the LHS we have:

Jm ≤M1

(y

m− y

m+ 1

)∫ m+1

m

log u du ≤M1ylog(m+ 1)

m(m+ 1)

Because m ≥ log(m+ 1) we have:

Jm ≤M1y

m+ 1

Now, returning to our original expression, we see:

(6.13) logm∣∣∣S ( y

m

)∣∣∣ ≤ ∫ m+1

m

∣∣∣S (yu

)∣∣∣ log u du+M1y

m+ 1

Using Z4 = Z2 +M1 and applying 6.13 to 6.6, we have the desired result. �


The inequality of 6.12 assumes a simpler form with a change of variables. Lettingx = log y and letting v = log y

u , we can rewrite 6.12 as:

(6.14) x2|S(ex)| ≤ 2

∫ x−log 2

0

|S (ev)| (x− v)e(x−v)dv + Z4xex

Performing another change of variables by defining the function W (x) = e−xS(ex),and applying the transform to 6.14, we have that:

(6.15) x2|W (x)|e−x

≤∫ x−log 2

0

|S (ev)| (x− v)e(x−v)dv + Z4xex

Further simplifying 6.15 we have:

x2|W (x)|e−x

≤ 2

∫ x−log 2

0

|W (v)|ev(x− v)ex

evdv + Z4xe

x

|W (x)|ex ≤ 2

x2

∫ x−log 2

0

|W (v)|(x− v)exdv + Z4xex

x2

(6.16) |W (x)| ≤ 2

x2

∫ x

0

|W (v)|(x− v)dv + Z41

x

The transformations above were done to yield a function W which is essentiallydominated by a weighted average of itself. We now prove two lemmas about |W (x)|.

Lemma 6.17.

lim supx→∞

|W (x)| = α ≤ 1

Proof. By 6.3 we know that:

lim supx→∞

R(x)

x≤ 1

By definition of S, it follows that:

lim supy→∞

|S(y)|y≤ 1

It is clear from this that lim supx→∞ |W (x)| ≤ 1 �

Lemma 6.18. Let

lim supx→∞

1

x

∫ x

0

|W (t)|dt = β

Then β ≥ α = 1

This key result will be proven using lemmas 6.5, 6.6, and 6.12.

Proof. Recall that by 6.16:

|W (x)| ≤ 2

x2

∫ x

0

|W (v)|(x− v)dv + Z41

x

We first decompose the integral on the right hand side by using a dummy variableand rewriting it as an iterated integral. We see that 6.16 can be rewritten as:

(6.19) |W (x)| ≤ 2

x2

∫ x

0

udu

(1

u

∫ u

0

|W (v)|dv)

+ Z41

x


which can be verified by reversing the order of integration. Note that:

2

x2

∫ x

0

udu = 1

Thus we have the integral on the right handside of the form:

1

u

∫ u

0

|W (v)|dv =1

u

∫ u

0

|e−vS(ev)|dv

And this is bounded by M1 (our Lipschitz continuity constant) from lemma 6.2,i.e:

1

u

∫ x

0

|W (v)|dv =1

u

∫ x

0

|e−vS(ev)|dv ≤M1

Thus, if we fix some x1, and take any x > x1, we have:

I(x) =2

x2

∫ x

0

udu

(1

u

∫ x

0

|W (v)|dv)

≤ 2M1

x2

∫ x1

0

udu+2

x2

∫ x

x1

udu

(1

u

∫ x

0

|W (v)|dv)

As we separate the integrals at x1. Given arbitrary ε > 0, if we choose x1 sufficientlylarge, we have by the definition of β (limit supremum) that:

1

u

∫ x1

0

|W (v)|dv ≤ β + ε

for all u ≥ x1. Thus, substituting into our inequality for I(x) we have:

I(x) ≤ M1x21

x2+ (β + ε)

(1− x21

x2

)Thus, for large x, we have by 6.19 that:

|W (x)| ≤ M1x21

x2+ (β + ε)

(1− x21

x2

)+Z4

x

Letting x → ∞, we have that α ≤ β + ε,and the inequality holds as it is true forarbitrary ε. �

We seek to show that α = 0. To do this, we will use two more facts about W ,proving them along the way.

Lemma 6.20. Let k = 2M1 . Then, the following holds:

|W (x1)−W (x2)| ≤ k|x1 − x2|

Proof. We see that by definition:

|W ′(x)| = −e−xS(ex) + S′(ex) ≤ e−x|S(ex)|+ |S′(ex)| ≤ c+ c = 2c

and it follows by the methods used in 6.2 that the condition follows. �

Lemma 6.21. If W (v) 6= 0 for v1 < v < v2 then there exists M2 such that:∫ v2

v1

W (v)dv ≤M2


Proof. We see through an application of Abel summation that:∫ x

2

ψ(t)

t= log x+O(1)

Or, using the remainder function:

(6.22)

∫ x

2

R(t)

t2dt = O(1)

Taking S(y)y2 , we do another double integral decomposition:∫ x

2

S(y)

y2dy =

∫ x

2

dy

y2

∫ y

2

R(t)

t=

∫ x

2

R(t)

t

(∫ x

t

dy

y2

)dt

This simplifies to: ∫ x

2

R(t)

t2dt− 1

x

∫ x

2

R(t)

tdt

Using 6.2 and 6.22, it follows that∫ x2S(y)y2 dy = O(1). Performing change of variables

with y = eu, x = ev we have: ∫ v

log 2

W (u)du = O(1)

Letting v = v1 and v = v2 and subtracting the integrals with these as endpoints,we have that the result integral is bounded, and thus there exists M2 such that:∫ v2

v1

W (u)du ≤M2

�

Lemma 6.23. A function W (x) subject to 6.17, 6.18, 6.20 and 6.21 must haveβ = 0.

Proof. Take ω > α. Then from the definition of α, there exists xω such that for allx ≥ xω we have:

(6.24) |W (x)| ≤ ω

If W (x) 6= 0 for large x, it follows from 6.21 that ω = 0 and thus α = 0. Thus,suppose that W has arbitrarily large zeros. Let a, b be adjacent zeros of W (x) forx > xω. We now have 3 cases.

(1) (b− a) ≥ 2M2

ωBy 6.21, as x 6= 0 for a < x < b we have:∫ b

a

W (u)du ≤M2 ≤1

2(b− a)ω

Thus, the average of |W | on (a, b) is less than 12ω.

(2) (b− a) ≥ 2ωkHere, it follows from 6.20 (Lipschitz property of W )that if the graph of|W (x)| increases as rapidly as possible from x = a to x = b that it cannot

lie above a triangle with height k(b−a)2 ≤ ω and thus:∫ b

a

|W (x)|dx ≤ 1

2(b− a)ω


(3) 2M2

ω ≥ (b− a) ≥ 2ωkWe can use the reasoning as in case 2 for any points a distance ω/k fromeach endpoint. Otherwise, by 6.24 we have:∫ b

a

|W (x)| ≤ w2

k+

(b− a− 2ω

k

)ω

Simplifying the right hand side above we have:

(b− a)ω

(1− ω

k(b− a)

)≤ (b− a)ω

(1− ω2

2M2k

)And this is strictly less than:

(6.25) (b− a)ω

(1− α2

2M2k

)as M2k > 1 and α ≤ 1. If x1 is the first zero of W (x) to the right of xωand x is the largest zero to the left of y then by 6.25 and 6.21 imply that:∫ y

0

|W (x)|dx ≤∫ x1

0

|W (x)|dx+ (x− x1)ω

(1− α2

2M2k

)+M2

Dividing by y and noting that x ≤ y we have:

1

y

∫ y

0

|W (x)|dx ≤ 1

y

∫ x1

0

|W (x)|dx+ (x− x1)ω

(1− α2

2M2k

)+M2

y

Letting y →∞ we see that:

β ≤ ω(1− α2

2M2k)

and because β ≥ α, we see:

α ≤ ω(

1− α2

2M2k

)Since this inequality holds for all ω > α, it must hold for ω = α. Thus,α3 ≤ 0 and since α ≥ 0 it follows α = 0.

It follows then that:

limy→∞

S(y)

y= 0

Thus, for any given ε > 0 we have for large y that:

|S(y)| ≤ 1

3ε2y

Hence, we see:

S(y(1 + ε))− S(y) ≤ 1

3ε2(y(1 + ε) + y) < ε2y

Expanding S, we have: ∫y

y(1 + ε)R(t)

tdt ≤ ε2y

By the definition of R and because ψ is nondecreasing, we have:

ψ(y)

y(1 + ε)

∫ y(1+ε)

y

dt−∫ y(1+ε)

y

dt ≤ ε2y


Hence, we have that ψ(y)y ≤ (1 + ε)2. Similarly, because S(y)− S(y(1− ε)) ≥ −ε2y

for large y leads to ψ(y)y ≥ (1− ε)2. Because ε is arbitrary, it follows that:

limx→∞

ψ(x)

x= 1

proving 1.2. �

Acknowledgments. It is a pleasure to thank my mentor, Karen Butt for herassistance in writing this paper, as well as her having the confidence in me to writeon a topic that most would consider ambitious. I would also like to thank theprogram director, Peter May, for organizing this research experience.

References

[1] Norman Levinson. A Motivated Account of the Elementary Proof of the Prime Number The-

orem. https://www.jstor.org/stable/2316361

[2] Tom M. Apostol. Introduction to Analytic Number Theory.http://plouffe.fr/simon/math/IntrodAnalyticNTApostol.pdf

Documents

AN ELEMENTARY PROOF OF THE PRIME NUMBER ...may/REU2017/REUPapers/Choudhary.pdfAN ELEMENTARY PROOF OF THE PRIME NUMBER THEOREM 3 Thus, the mobius and unit functions are inverses of