Asymptotic properties of weighted Goldbach representation function, and its connection with zeta function

Asymptotic properties of weighted Goldbach representation function, and its connection with zeta function

Yiyu Tang June 14, 2020

Sous la direction de Gautami Bhowmik Laboratoire Paul Painlev´e, Universit´e Lille 1

Resum´e

La conjecture de Goldbach est l’un des plus vieux probl`emes non r´esolus des math´ematieques, qui s’´enonce comme suit: Tout nombre entier pair sup´erieur `a 3 peut s’´ecrire comme la somme de deux nombres premiers.

En d’autres termes, sinest un nombre naturel et nous posons (petqsont des nombres premiers) g(n) = X

p+q=n

1, (0.1)

alorsg(n)6= 0 si n est pair. Dans cet article, au lieu d’´etudierg(n) directement, nous consid´erons une version lisse deg(n):

r₂(n) = X

m+k=n

Λ(m)Λ(k), (0.2)

o`u Λ(n) est la fonction de Von Mangoldt. Notre principal int´erˆet est la propri´et´e asymptotique de X

n≤X

r2(n), (0.3)

l’estimation du terme d’erreur a un lien avec des z´eros de fonction de zeta, qui fait le lien entre le probl`eme original avec l’hypoth`ese de Riemann.

En outre, nous consid´erons le cas g´en´eral:

rk(n) = X

m1+...+m_k=n

Λ(m1)Λ(m2)· · ·Λ(mk), (0.4) et la propri´et´e asymptotique deP

n≤Xrk(n), certains r´esultats ont ´et´e prouv´es, mais une g´en´eralisation du cas particulierk= 2 reste `a ´etudier.

L’auteur tient `a remercier le Laboratoire Paul Painlev´e, qui offre une bourse pour l’auteur de Master 2. En particulier, l’auteur remercie Gautami.Bhowmik, qui donnent ce sujet int´eressant et de nombreuses suggestions. L’auteur remercie ´egalement Bruno.Martin, qui enseigne le cours de la th´eorie des nombres analytiques, l’´etude erait beaucoup plus difficile sans ces connaissances de base.

L’auterur souhaite remercies finalement le jury de soutenance.

Contents

1 Introduction 3

1.1 Definition and Notation . . . 4

1.1.1 Some notations . . . 4

1.1.2 Some arithmetic functions . . . 4

1.1.3 Gamma function and Riemann zeta function . . . 6

1.2 Main Results . . . 8

1.3 Further Remarks . . . 9

2 Proof of Theorem 1.5, Part 1 9 3 Proof of Theorem 1.5, Part 2 17 3.1 Proof of Theorem 1.5, Part 2 . . . 17

3.2 Proof of Step 1 in Theorem 1.5 . . . 19

3.2.1 Estimate for Σ3x^2BT¹²log⁴x . . . 22

3.2.2 Estimate for Σ₄^x_T²log⁴x . . . 24

3.3 Proof of Lemma 3.1 . . . 26

4 Proof of Theorem 1.6, Part 1 29 4.1 Proof of the main theorem. . . 29

4.2 Proof of the Lemma 4.1 . . . 31

1 Introduction

In mathematics, one of the most famous and oldest problem is Goldbach’s conjecture, which was proposed in 1742, during the communications of German mathematician Christian Goldbach and Leonhard Euler. In the language of modern mathematics, the Goldbach’s conjecture sates that

Every even integer greater than 2 can be written as the sum of two primes. (1.1) With the help of computers, people verified that for allnless than 4×10¹⁸, Goldbach’s conjecture is true (in 2014). People believe this conjecture is true, just as Euler replied to Goldbach in 1742:

T hat ... every even number is a sum of two primes, I consider it as a completely certain theorem, but I am not able to prove it.

(1.2) In this article, instead of studying Goldbach counting function g(n) = P

p+q=n1, where p and q are prime numbers, we would like to treat a smooth and moderate increasing version, says the weighted Goldbach representation function, by using logarithms:

r2(n) = X

m+k=n

Λ(m)Λ(k), (1.3)

where Λ is the Von Mangoldt function to be defined later, we will see that Λ has logarithmic increase, andr2(n) behaves very similar tog(n).

Our main work in this article is to give an asymptotic formula like the following form:

X

n≤X

r₂(n) =f(X) +O(g(X)), (1.4)

two termsf(X) andg(X) are called themain termand theerror termrespectively. Furthermore, we will also consider the general caserk(n) =P

m₁+...+m_k=nΛ(m1)Λ(m2)· · ·Λ(mk), and study some asymptotic properties ofP

n≤Xrk(n).

In general, main term is easy to obtain, however, if we want to get some good estimates for the error term, then we must deal with non-trivial zeros of the Riemann zeta function. When we talk about non-trivial zeros of zeta function, theRiemann Hypothesis is inevitable. This is an interesting point, it connects two most famous unsolved problems in mathematics, Goldbach’s conjecture and Riemann Hypothesis, together. Moreover, we will see that some good error term estimates are equivalent to Riemann Hypothesis, this reveals the connection between these two problems to a certain extent.

1.1 Definition and Notation

1.1.1 Some notations

Letf, g : R⁺ →Cbe two functions, we use the notation f =O(g) to mean that there exists a constantC so that for allx,

|f(x)| ≤C|g(x)|. (1.5)

We will also denotef g to mean thatf =O(g).

Similarly, we denotef =o(g) to mean that

x→∞lim f(x)

g(x) = 0. (1.6)

Moreover, we use the notation

f g (1.7)

to mean thatf gandgf simultaneously, and f ∼g is to say that lim_x→∞^f(x)_g(x) = 1.

1.1.2 Some arithmetic functions

TheVon Mangoldt functionis the arithmetic function defined by Λ(n) =

(logp, ifn=p^k, k≥1, pis prime

0, otherwise (1.8)

If we decomposenas sum of two integersn=m+k, then Λ(m)Λ(k)6= 0 if and only ifmandkare the power of primes, so we get a seriesr2(n) =P

m+k=nΛ(m)Λ(k). Moreover, when we consider the character functionp(n) of primes, i.e. p(n) = 1 ifnis prime andp(n) = 0 otherwise, then Goldbach counting functiong(n) =P

p+q=n1 can be written as g(n) = X

m+k=n

p(m)p(k).

We notice that, although not the same thing, the general term Λ(m)Λ(k) and p(m)p(k) are very similar, they are 0 whenmork is not prime. Therefore, if we know some properties ofr2(n), then we can get some informations ofg(n). By this idea, in 1937, Vinogradov considered the partial sum

r(n) = X

k₁+k₂+k₃=n

Λ(k₁)Λ(k₂)Λ(k₃),

and after an elaborated estimate ofr(n), he proved a weak form of Goldbach’s conjecture for large integersn, now named as Vinogradov’s Theorem:

Theorem 1.1 (Vinogradov). Any sufficiently large odd integer can be written as a sum of three prime numbers.

For the proof, see Chapter 26 of [1]. However, we would like to explain how Vinogradov recovered the information of g3(n) = P

k+l+m=np(k)p(l)p(m) from r(n). In fact, Vinogradov proved the following estimate:

r(N) = 1

2G(N)N²+OA

N² log^AN

, (1.9)

whereAis any positive integer, and G(N) =Y

p|N

1− 1

(p−1)² Y

p-N

1 + 1

(p−1)³

. (1.10)

We notice that ifN is odd, then G(N)≥Q

p6=2(1−(p−1)⁻²)>0 (For some properties of infinite product, see [20], Chapter 5.) We call the infinite product C2 = 2Q

p6=2(1−(p−1)⁻²)>0 is the Twin Prime Constant. Now by Vinogradov’s estimate, we have (N is odd andpi’s are prime)

N²r(N) = X

p₁+p₂+p₃=N

logp₁logp₂logp₃+ X

(k₁,k₂,k₃)6=(1,1,1)

X

p^k₁¹+p^k₂²+p^k₃³=N

logp₁logp₂logp₃ log³N X

p₁+p₂+p₃=N

1 + X

(k₁,k₂,k₃)6=(1,1,1)

X

p^k₁¹+p^k₂²+p^k₃³=N

logp₁logp₂logp₃

=g3(N) log³N+ X

(k₁,k₂,k₃)6=(1,1,1)

X

p^k₁¹+p^k₂²+p^k₃³=N

logp1logp2logp3.

(1.11) For the second term, by symmetry, we may assumek1>1, we notice that the sum

X

(k₁,k₂,k₃)6=(1,1,1)

X

p^k₁¹+p^k₂²+p^k₃³=N

(1.12)

is finite. Moreover, consider the sumP

p^k1≤N1, firstly we must havep≤√

N, byPrime Number Theorem, there are at most

√ N log√

N such primes (For the prime number theorem, see subsection 1.1.3). Secondly, ifp^k1 ≤N holds, thenk1≤log₂N logN. Therefore

X

p^k1≤N

1logN

√N log√

N √

N . (1.13)

According to the above estimate, we get X

(k1,k2,k3)6=(1,1,1)

X

p^k₁¹+p^k₂²+p^k₃³=N

logp₁logp₂logp₃N³²log³N. (1.14)

This inequality implies g₃(N) N²log⁻³N +O(log₂N), which is positive when N sufficiently large. So any large odd integer can be decomposed as a sum of three primes.

TheFirst Chebyshev functionϑ(x) is defined by ϑ(x) =X

p≤x

logp, (1.15)

and theSecond Chebyshev functionψ(x) is defined by the sum of Λ(n):

ψ(x) =X

n≤x

Λ(n). (1.16)

We notice thatψ(x) =P

p^k≤xlogp=P

p≤x[log_px] logp, wherek∈N\ {0}. Therefore, we have the following relationship ofϑandψ:

ψ(x) =X

k≥1

ϑ(x^1k), (1.17)

for eachx, the sum is finite because the general term is 0 when 2^k > x.

1.1.3 Gamma function and Riemann zeta function

We will use zeta function frequently, and in subsection3.2.1, some properties of complex Gamma function are necessary, so we give some elementary properties of zeta and Gamma function, all of these contents can be found in analytic number theory or complex analysis textbook, like the Chapter 6, 7 of [20], so we omit the proofs and details.

Complex Gamma function In elementary calculus, we have learned the Gamma function Γ(s) =

Z ∞ 0

x^s−1e^−xdx, s >0. (1.18) This definition can be extended to {s ∈ C : Re(s) > 0} directly, just notice that the integral R∞

0 x^s−1e^−xdxconverges absolutely whenRe(s)>0.

Analytic continuation of Γ(s) The definition of Γ(s)can be extended to {s ∈C: Re(s)>0}, and the integral representation of Γ(s) is analytic in this half plane. Moreover, ifRe(s)>0, then Γ(s+ 1) =sΓ(s). As a consequence,Γ(n+ 1) =n!forn= 0,1,2...

By the identity Γ(s+ 1) =sΓ(s), we can extend Γ to a meromorphic function on all ofC. For example, on the half-plane{<(s)>−1}, we define

F₁(s) = Γ(s+ 1)

s . (1.19)

Since Γ(s+ 1) is holomorphic inRe(s)>−1, we see thatF1is meromorphic in that half-plane, with only singularity a simple pole at s= 0 with residue 1. Moreover, ifRe(s)>0, then F1(s) = Γ(s) by previous theorem. SoF1 extends Γ to a meromorphic function on the half-plane{Re(s)>−1}.

Similarly, we can extend Γ(s) to the half-plane{<(s)>−m} withm∈N.

Theorem 1.2. Γ(s)can be extended to a meromorphic function on Cwhose only singularities are simple poles at the non-positive integerss= 0,−1,−2.... The residue ofΓ at−n is(−1)ⁿ/n!.

TheRiemann zeta functionis the function defined by ζ(s) =X

n≥1

1

n^s, <(s)>1. (1.20) Analytic continuation of ζ(s) For<(s)>1, we have

π^−s/2Γ(s/2)ζ(s) = 1 2

Z ∞ 0

u^(s/2)−1[θ(u)−1]du, (1.21)

where θ(t) =P

n∈Ze^−πn2t(t >0) is the Theta function. Using thePoisson summation formula in Fourier analysis, we get

θ(t) = 1

√tθ 1 t

. (1.22)

By this functional equation, π^−s/2Γ(s/2)ζ(s) = 1

s−1−1 s +1

2 Z ∞

1

u^−(s+1)/2+u^(s/2)−1

θ(u)−1

du, <(s)>1. (1.23) Notice that the integral above defines an entire function ins, becauseθ(u) has exponential decay at infinity. Moreover, the right-hand side above remains unchanged if we replacesby 1−s, soξ(s) = π^−s/2Γ(s/2)ζ(s) has an analytic continuation toC with simple poless= 0,1 and ξ(s) =ξ(1−s) for alls∈C.

Now we define

ζ(s) =π^s/2 ξ(s)

Γ(s/2), (1.24)

then ζ(s) is an analytic continuation of P

n≥1 1

n^s. Moreover, by Theorem 1.2, 1/Γ(s/2) is entire with simple zeros at 0, −2,−4, ..., so the simple pole ofξ(s) at 0 is canceled with the simple zero of Γ(s/2) at 0. In conclusion, we have

ζ(s) has a meromorphic continuation in to C, whose only singularity is a simple pole s = 1.

Moreover ζ(−2n) = 0 for alln∈N\ {0}, we say that−2,−4,−6... are thetrivial zeros ofζ.

Prime Number Theorem, Riemann Hypothesis We have known that ζ(−2n) = 0 for all n ∈ N\ {0}, however, the negative even integers are not the only zeros for ζ. The Riemann Hypothesisstates that:

The real part of every non-trivial zero of ζ is 1

2. (1.25)

Like Goldbach’s conjecture, people believe that the Riemann Hypothesis is true. Using computers, more than million of non-trivial zeros has been found, and all of them are on the line <(s) = 1/2.

We denote ρbe the non-trivial zeros of ζ, the summation P

ρ means take sum over all non-trivial zeros ofζ. Furthermore, considering the zeros of zeta function, we have

Theorem 1.3. ζ(s)never vanishes in the half-plane

{s=σ+ it:σ≥1}, (1.26)

and{−2n}n≥1 are the only zeros in the half-plane{s=σ+ it:σ≤0}. So all non-trivial zeros lies in the band{s=σ+ it:σ∈(0,1), t∈R}, we call this band thecritical band.

Moreover, there exists a constant c >0, such thatζ never vanishes in the region (See the figure below)

{s=σ+ it:σ≥1− c

log(2 +|t|)}. (1.27)

For the proof, see Chapter II.3 of [21]. A very simple but interesting observation is that non- trivial zeros are symmetric about the critical line {<(s) = 1/2}, this is obvious from the functional equationζ(s) =π^s/2_Γ(s/2)^ξ(s) and ξ(s) =ξ(1−s). Morevover,non-trivial zeros of zeta function come in conjugate pairs {ρ,ρ}, this is due to the following theorem:¯

Theorem 1.4 (Schwarz reflection principle). Let Ω be an open set of C that is symmetric with respect to the real line, that is

z∈Ω if and only if z¯∈Ω (1.28)

Let Ω⁺ denote the part of Ω that lies in the upper half-plane and Ω⁻ that part that lies in the lower half-plane. Also, let I = Ω∩R. If f⁺ and f⁻ are holomorphic functions in Ω⁺ and Ω⁻ respectively, that extend continuously to I and

f⁺(x) =f⁻(x) f or all x∈I, (1.29)

then the functionf defined on Ωby f(z) =







f⁺(z) z∈Ω⁺, f^±(z) z∈I, f⁻(z) z∈Ω⁻

(1.30) is holomorphic inΩ.

The proof of this theorem can be found in many complex analysis textbook, for example, Chapter 2 of [20], so we omit it.

Figure 1: Zero-free region of zeta function in the critical band, notice 1−_log(2+|t|)^c goes to 1 as|t|

tends to infinity.

If we denoteπ(x) be the number of primes less than or equal tox, thePrime Number Theorem states that:

π(x)∼ x

logx, (1.31)

and it has a more precise form:

There exists a positive constant c >0, such thatψ(x) =x+O(e^−c

√logx). (1.32) Finally, we give a functional equation which would be used in the proof of Lemma 4.1(For the proof of this equation, see [20], Chapter 7.)

−ζ⁰(s) ζ(s) =X

n

Λ(n)n^−s, <(s)>1, (1.33)

summing by parts, we get

−ζ⁰(s) ζ(s) =X

n

Λ(n)n^−s= Z ∞

1⁻

u^−sdψ(u) =s Z ∞

1

u^−s−1ψ(u)du, <(s)>1. (1.34) Notice that we have connectedζ(s) with Λ(n) andψ(n). As we mentioned before, instead of studying Goldbach’s counting function P

p₁+p₂=n1 directly, we would like to deal withP

m+k=nΛ(m)Λ(k).

So by using function Λ, now we connect Goldbach’s problem with Riemann zeta function. We will see this point later in the proof of our main theorem.

1.2 Main Results

Theorem 1.5. Let

r₂(n) = X

m+k=n

Λ(m)Λ(k), (1.35)

then Riemann Hypothesis is true, if and only if the following asymptotic property holds:

X

n≤X

r₂(n) = X²

2 +O(X^3/2). (1.36)

Then we consider the general case, assumek≥2 be an integer.

Theorem 1.6. Let

rk(n) = X

m₁+...+m_k=n

Λ(m1)Λ(m2)· · ·Λ(mk), (1.37) if Riemann Hypothesis is true, then the following asymptotic property holds:

X

n≤X

r_k(n) =X^k

k! +O_k(X^−1/2+klog^kX), (1.38) hereO_k means that the constant C in section 1.1.1is depending onk.

1.3 Further Remarks

Remark 1. We must point out that the main work is on the sumP

n≤Xrk(n), rather thanrk(n) itself. But forrk(n), there are also some estimates can be obtained from thegeneralized Riemann HypothesisofL(s, χ), for this point, one may see [5,14].

Remark 2. Theorem1.5has a more precise form:

X

n≤X

r(n) =X²

2! −2X

ρ

X^ρ+1 ρ(ρ+ 1)+O

(XlogX)⁴³

, (1.39)

the error term was reached by A. Fujii in [6]-[8]. In [3], Bhowmik and Schlage-Puchta improved the error term toO(Xlog⁵X).

Remark 3. Like the case ofk= 2, Theorem1.6has a more precise form:

X

n≤X

rk(n) =X^k

k! −kX

ρ

X^ρ+k−1

ρ(ρ+ 1)· · ·(ρ+k−1) +Ok(X^k−1log^kX), (1.40) for this point, see [15], Theorem 1.3.

Remark 4. Theorem1.6only consider one direction. The author believe that, just like Theorem 1.5, the converse of Theorem1.6 is also true. This is to say if we admit the asymptotic formula of P

n≤Xrk(n) in Theorem1.6, then the Riemann Hypothesis is true. One may follow the idea in the section 3, however, some estimates (without Riemann Hypothesis) for

Sk(x) =X

n≤x

rk(n) (1.41)

is necessary, just like the step 1 in subsection 3.1, and this is not a trivial work, further study for this is needed and the author is still considering this.

2 Proof of Theorem 1.5, Part 1

In this section, we assume Riemann Hypothesis is always true.

For convenience, we write down the proposition which would be proved in this section:

Assuming Riemann Hypothesis, thenX

n≤X

r2(n) =X²

2 +O(X^3/2). (2.1) We use the method in A.Fujii’s paper [6], and complete some details for it. Firstly, it suffices to consider the caseX =N ∈N. The idea of proving the estimate is not difficult: We decompose the sum in several parts time after time, and on each small parts, we get estimates of error termsN³². The idea is simple, but some estimates are not so trivial and we need more elaborated work, so we will proceed these decompositions by several steps.

Step 1: Getting the main term Let

R(y) =−y+X

n≤y

Λ(n), y >0. (2.2)

Then X

n≤N

r2(n) = X

m≤N

Λ(m) X

k≤N−m

Λ(k)

= X

m≤N

Λ(m)

N−m+R(N−m)

= X

m≤N

Λ(m)(N−m) + X

2≤m≤N−2

Λ(m)R(N−m)−Λ(N−1).

(2.3)

For the third term,|Λ(N−1)|=O(logN). Summing by parts, we have X

m≤N

mΛ(m) =ψ(x)x

N 1

− Z N

1

ψ(t)dt

=N ψ(N)− Z N

1

ψ(t)dt

(2.4)

therefore, for the first term, we have

X

m≤N

Λ(m)(N−m) =N ψ(N)−

N ψ(N)− Z N

1

ψ(t)dt

= Z N

1

ψ(t)dt= N²

2 +O(N³²), (2.5) the last equality holds due to the following lemma:

Lemma 2.1. Let ψ(x)be the Chebyshev function, then Z x

2

ψ(t)dt= x²

2 +O(x³²), (2.6)

and letψ0(x) =¹₂(ψ(x+ 0) +ψ(x−0)), thenψ0(x) =ψ(x)whenx6=p^mfor somem≥1, moreover, we have

ψ0(x) =x−X

ρ

x^ρ

ρ −ζ⁰(0) ζ(0) −1

2log(1−x⁻²), x >1. (2.7) Nextly we come to the error term estimate, forT ≥2, let

X

ρ

x^ρ

ρ = X

|=(ρ)|≤T

x^ρ

ρ +R(x, T) (2.8)

where

R(x, T) x

T log²xT+ logx. (2.9)

For the proof, see [18], Chapter VII, Lemma 8.6 and [12], Chapter 12, Theorem 12.1.

Using the first estimate of Lemma2.1, X

n≤N

r2(n) = X

m≤N

Λ(m)(N−m) + X

2≤m≤N−2

Λ(m)R(N−m)−Λ(N−1)

=N²

2 +O(N³²) + X

2≤m≤N−2

Λ(m)R(N−m)

=N²

2 +O(N³²) +I.

(2.10)

Notice that we have finished the first decomposition and get the main termN²/2 and the error term O(N³²). Nextly, according to our idea, now we decompose the last term as some parts, with each partsN³²

Step 2: The estimate of IN√

N Write the last termI as X

2≤m≤N−2

Λ(N−m)R(m) = X

2≤m≤N−2

Λ(N−m)

−m+ψ(m)

, (2.11)

and using the error term estimate of Lemma2.1, we get

I= X

2≤m≤N−2

Λ(N−m)

− X

|=(ρ)|≤T

m^ρ

ρ +Om

T log²(mT)

+O(logm)

=− X

2≤m≤N−2

Λ(N−m) X

|=(ρ)|≤T

m^ρ ρ +

X

2≤m≤N−2

mΛ(N−m)

O1

T log²(N T)

+ψ(N−2)O(logN)

=− X

2≤m≤N−2

Λ(N−m) X

|=(ρ)|≤T

m^ρ

ρ +ON²

T log²(N T)

+O(NlogN),

(2.12) the last equality holds because we use Prime Number Theorem ψ(x) ∼ x, and the following estimates:

X

2≤m≤N−2

mΛ(N−m) =O X

m≤N

mΛ(N−m)

=OZ N 2

ψ(t)dt

=ON² 2

, (2.13)

In conclusion, we writeI=P

2≤m≤N−2Λ(N−m)

−m+ψ(m)

as the form I1+ON²

T log²(N T)

+O(NlogN), (2.14)

whereI₁=−P

2≤m≤N−2Λ(N−m)P

|=(ρ)|≤T m^ρ

ρ . Here we finished the decomposition second time, with two error termN³² for someT to be chosen.

Step 3: estimate of I₁ N√

N Now we continue our decomposition third time for I₁. From now on, not only in this step, we always assume that1T N. By Riemann hypothesis, we can writeρ=¹₂+ iγ, then

I1=− X

2≤m≤N−2

√mΛ(N−m) X

|γ|≤T

m^iγ

1 2+ iγ

=−2=

( X

2≤m≤N−2

√mΛ(N−m) X

0≤γ≤T

m^iγ γ

)

+O X

2≤m≤N−2

√mΛ(N−m) X

0≤γ≤T

m^iγ γ²

!

=−2=(I2) +O(I3).

(2.15)

We considerI₃ at first, summing by parts, and using√

a+ 1−√ a= √

a+ 1 +√ a−1

X

m≤N

√N−mΛ(m) = X

m≤N

√N−m

ψ(m)−ψ(m−1)

= X

m≤N−1

ψ(m) √

N−m−√

N−m−1

X

m≤N−1

ψ(N)

√N−m ψ(N)√

N=O(N

√ N).

(2.16)

For the part P

0≤γ≤T m^iγ

γ² , we need a lemma about the number of zeros of ζ(s) in a prescribed domain.

Lemma 2.2. ForT ≥1, denote N(T)be the number of zeros of ζ(x+ iy)in the rectangle

{(x, y)∈C: x∈(0,1), y ∈(0, T)}, (2.17) then the following estimate holds

N(T) = T 2πlog T

2π− T

2π +O(logT). (2.18)

In fact, we have a more elaborated estimate:

N(T) = T 2πlog T

2π− T 2π+7

8+ 1

πargζ1 2 + iT

+O1 T

. (2.19)

For the proof, see Chapter IX of [22]. We would like to give some explanation about the term argζ ¹₂+ iT

, which is defined by continuity along the polygonalLjoining the point 2,2 + iT, ¹₂+ iT. According to this lemma, if we denote {ρn =βn+ iγn}n be the zeros of ζ in critical band with γn>0, and enumerate it with the increasing orderγ1≤γ2..., then

γn∼ 2πn

logn. (2.20)

So

X

0≤γ≤T

m^iγ γ² X

n

log²n

n² <∞. (2.21)

We can also derive this estimate by Lemma3.2.

Combining these two estimates, we get I3=O(N√

N). Nextly, we deal with the termI2. Step 4: Estimate ofI2N³² We write downI2 again for convenience

I2= X

2≤m≤N−2

√mΛ(N−m) X

0≤γ≤T

m^iγ

γ (2.22)

Firstly, using Lemma2.2, we have X

0<γ≤Y

1

γ =log²Y

4π −log(2π)

2π logY + Z ∞

1

S(t)

t² dt−1 + log(2π)

2π +7

8+ Z ∞

1

η(t)

t² dt+B(Y)

=A(Y) +B(Y)

(2.23)

whereS(t) = _π¹argζ(¹₂+ it),η(t) varifiesη(t) =O(¹_t) fort > t0, and B(Y) = S(Y)

Y −

Z ∞ Y

S(t)

t² dt+η(Y)

Y −

Z ∞ Y

η(t)

t² dt (2.24)

Secondly, letube an integer, if 1≤u < N, then X

n≤u

√nΛ(N−n) = X

n≤u−1

√nΛ(N−n) +√

uΛ(N−u)

= X

N−u<n≤N−1

√N−nΛ(n) +√

uΛ(N−u)

= Z N−1

N−u

pN−yd(y+R(y)) +√

uΛ(N−u),

(2.25)

here we use the Riemann-Stieltjes integral, integrating by parts, we get:

Z N−1 N−u

pN−ydR(y) =p

N−yR(y)

N−1

N−u

+1 2

Z N−1 N−u

√R(y) N−ydy

=R(N−1)−√

uR(N−u) +1 2

Z N−1 N−u

√R(y) N−ydy,

(2.26)

therefore,

X

n≤u

√nΛ(N−n) = Z N−1

N−u

pN−ydy+R(N−1)−√

uR(N−u)

+1 2

Z N−1 N−u

√R(y)

N−ydy+√

uΛ(N−u).

(2.27)

We put

C(u) = Z N−1

N−u

pN−ydy= 2

3(u³² −1) D(u) =R(N−1) +√

u

Λ(N−u)−R(N−u) +1

2 Z N−1

N−u

√R(y) N−y dy.

(2.28)

Ifu≥N, similarly X

n≤u

√nΛ(N−n) = Z N−1

1

pN−ydy+√

N−1 +R(N−1) +1 2

Z N−1 1

√R(y)

N−ydy, (2.29) again, we put

C(u) = 2

3(u³² −1) D(u) = 2

3

(N−1)³² −u³² +√

N−1 +R(N−1) + 1 2

Z N−1 1

√R(y) N−ydy.

(2.30)

Now, by Riemann-Stieltjes integral, we writeI₂as I2=

Z T 1

Z N 1

u^itd(C(u) +D(u))d(A(t) +B(t))

= Z T

1

Z logN 0

e^itxd(C(e^x) +D(e^x))d(A(t) +B(t))

= Z T

1

Z logN 0

−e^itxdC(e^x)dA(t) + Z T

1

Z logN 0

e^itxdC(e^x)d(A(t) +B(t)) +

Z T 1

Z logN 0

e^itxd(C(e^x) +D(e^x))dA(t) + Z T

1

Z logN 0

e^itxdD(e^x)dB(t)

=I4+I5+I6+I7.

(2.31)

We deal withI4,I5 andI6at first, I7 is more difficult, so we put it in the next step. Firstly,I4 is

− Z T

1

Z logN 0

logt−log(2π)

2πt e^itxe^32xdxdtN³² Z T

1

logt

t² dtN³². (2.32) Secondly,I₅ is

Z T 1

Z logN 0

e^itxdC(e^x)d(A(t) +B(t))N³² 1 t

X

0<γ<t

γ⁻¹

T

1

+ Z T

1

1 t²

X

0<γ<t

γ⁻¹dt

!

N³².

(2.33)

Finally, by Fubini’s theorem, we writeI6 as Z logN

0

Z T 1

e^itxdA(t)d(C(e^x) +D(e^x)) = Z logN

0

Z T 1

e^itxlogt−log(2π)

2πt dtd(C(e^x) +D(e^x)). (2.34) Then, by the following estimate

Z T 1

e^itxlogt−log(2π)

2πt dtmin(1

x,log²T), (2.35)

I₆is bounded by Z 1

0

log²Td(C(e^x) +D(e^x)) + Z logN

1

1

xd(C(e^x) +D(e^x))N³² (2.36) Step 5: Estimate ofI7N³² We use the result in [9] (see [9], Lemma 1)

I7TlogN max

0<δ_T¹

Z logN 0

D(e^y+δ)−D(e^y)2

dy¹₂Z T 1

B t+ (2 logN)⁻¹

−B(t)² dt¹₂

,

(2.37) we denote the two integrals in the right-hand site asI8andI9. I9 is easy, we just notice that by the definition ofB(y)

B(y)

t+(2 logN)⁻¹ t

= S(y) +η(y) y

t+(2 logN)⁻¹ t

−

Z t+(2 logN)⁻¹ t

S(y) +η(y)

y² dy. (2.38)

Notice thatS(y) +η(y) =O(1), soB(y)

t+(2 logN)⁻¹ t

¹_t, henceI9RT 1

1

t²dt=O(1).

For I₈, we need a more finer estimate than I₉, because the expressions of D(u) are different in when u < N and u ≥ N. When y ∈ (0,logN), e^y ∈ (1, N), so we needn’t pay much attention on D(e^y). However, if we solve the inequality e^y+δ < N, then y < logN−δ, this suggests us to decomposeI8as RlogN−δ

0 +RN

logN−δ =I10+I11, where I10=

Z logN−δ 0

√

e^yR(N−e^y)−√

e^y+δR(N−e^y+δ) +1 2

Z N−e^y N−e^y+δ

√R(u) N−udu

!2

dy

I11= Z logN

logN−δ

2 3

(N−1)^3/2−e^3(y+δ)/2 +1

2 Z N−e^y

1

√R(u)

N−udu+√

N−1 +√

e^yR(N−e^y)

!² dy.

(2.39)

We deal withI11 firstly, I10 is more difficult so we put it in the next step. There is an integral termRN−e^y

1 inI11, which may cause troubles, so we try to control it firstly. We notice thatN−e^y∈ 0, N(1−e^−δ)

wheny ∈(logN−δ,logN), moreover, by the inequality e^x≥1−xandδ1/T, we have 1−e^−δ ≤δ1/T, henceN−e^yN/T. RememberR(u) =ψ(u)−u, we obtain

Z N−e^y 1

√R(u)

N−udu 1 pN−N/T

Z ^N_T

1

ψ(u)−u

du(N/T)³²

√

N , (2.40)

we use Lemma2.1in the last equality.

Therefore, we have (Using (a+b+c)²a²+b²+c²) I₁₁

Z logN logN−δ

2 3

(N−1)^3/2−e^3(y+δ)/2

+(N/T)³²

√

N +√

N−1 +√

e^yR(N−e^y)

!² dy 1

T N³

T² +N² T log⁴N

,

(2.41)

some explanation for the last termRlogN

logN−δe^yR²(N−e^y)dy is necessary, after a change of variable, it’s RN(1−e^−δ)

0 R²(t)dt, and then we use the estimate R(t) =O(t¹²log²t), see Lemma 2.3 for this point.

Step 5: Estimate ofI₁₀N³² Firstly, we write√

e^yR(N−e^y)−√

e^y+δR(N−e^y+δ) as following

√

e^yR(N−e^y)−√

e^yR(N−e^y+δ) +√

e^yR(N−e^y+δ)−√

e^y+δR(N−e^y+δ)

=√ e^y

R(N−e^y)−R(N−e^y+δ)

−R(N−e^y+δ)(

√

e^y+δ−√

e^y). (2.42)

Moreover,√

e^y+δ−√ e^y=√

e^y(e^δ2−1)√

e^y(exp_2T¹ −1), and we have e^x−1xwhenxis small, hence

√

e^y+δ−√ e^y

√e^y

T . (2.43)

Then using the inequality (a+b+c)²a²+b²+c², we get I10

Z logN−δ 0

e^y

R(N−e^y)−R(N−e^y+δ)² dy+

Z logN−δ 0

Z N−e^y N−e^y+δ

√R(u) N−udu

!² dy +

Z logN−δ 0

e^y

T²R(N−e^y+δ)²dy.

(2.44)

We denote

I12=

Z logN−δ 0

e^y

R(N−e^y)−R(N−e^y+δ)2

dy, I13=

Z logN−δ 0

Z N−e^y N−e^y+δ

√R(u) N−udu

!² dy, I₁₄=

Z logN−δ 0

e^y

T²R(N−e^y+δ)²dy.

(2.45)

I₁₂ is more difficult, so we put it in the next step. To deal withI₁₃ and I₁₄, firstly, we need a lemma concerning the growth ofR(u).

Lemma 2.3. Let θ∈[¹₂,1) fixed, then

ψ(x) =x+O(x^θlog²x) (2.46)

if and only if

ζ(s)6= 0, for all<(s)> θ. (2.47)

For the proof, one may see [12], Chapter 12, Theorem 12.3. In particular, this lemma shows that the Riemann Hypothesis is equivalent to

ψ(x) =x+O(x¹²log²x), (2.48) however, this will never be easier than proving the original form of Riemann Hypothesis.

Using this lemma, I₁₃ andI₁₄can be estimated directly:

I13

Z logN−δ 0

Z N−e^y N−e^y+δ

√ulog²u

√N−udu

!² dy

Nlog⁴N

Z logN−δ 0

Z N−e^y N−e^y+δ

√ du N−u

!² dy

=Nlog⁴N

Z logN−δ 0

√

e^y+δ−√ e^y2

dy Nlog⁴N

T²

Z logN−δ 0

e^ydy N²log⁴N T² .

(2.49)

After a change of variable,

I14= 1 T²δ

Z N−e^δ 0

R²(t)dt 1

T²e^δ

Z N−e^δ 0

tlog⁴tdt N²log⁴N T² .

(2.50)

Step 6: Estimate ofI12N³² After a change of variable, I12=

Z N−1 N(1−e^−δ)

R(ye^δ+N−Ne^δ)−R(y)2

dy

Z N−1 N(1−e^−δ)

R(ye^δ−N(e^δ−1))−R(ye^δ))2

dy+ Z N−1

N(1−e^−δ)

R(ye^δ)−R(y)2

dy

=

Z (N−1)e^δ N(e^δ−1)

R(x−N(e^δ−1))−R(x)² dx+

Z N−1 N(1−e^−δ)

R(ye^δ)−R(y)² dy,

(2.51)

the last two integrals R(N−1)e^δ

N(e^δ−1) and RN−1

N(1−e^−δ) have been treated in Goldston and Montgomery’s article, see [10,19]. Hence

I12N² max

0<δ_T¹

δlog²1

δ N²log²N

T . (2.52)

Conclusion Until now, we have provedI8=I10+I11=I11+I12+I13+I14, where I111

T N³

T² +N² T log⁴N

, I12 N²log²N

T ,

I13N²log⁴N

T² , I14 N²log⁴N T² ,

(2.53)

so

I₈ N³ T³ +N²

T² log⁴N+N²

T log²N. (2.54)

In step 5, we have provedI9=O(1), andI7=TlogNmax_0<δ¹

T

√I8·√

I9, hence I₇TlogNN³²

T³² +N

T log²N+ N

√

T logN

. (2.55)

In step 4, we decomposeI2as I4+I5+I6+I7, and we have also provedI4, I5, I6N³², so

I₂I₇+N³². (2.56)

In step 3, we writeI₁=−2=(I2) +O(I₃) and proveI₃N³², so

I₁I₇+N³². (2.57)

In step 2, we writeI=I₁+O

N²

T log²(N T)

+O(NlogN), and notice thatT N, therefore I N²

T log²(N T) +N³² +TlogNN³² T³² +N

T log²N+ N

√

T logN

, (2.58)

now letT =√

Nlog²N, then 1T N andIN³². Finally, we get X

n≤X

r₂(n) = X

n≤N

r₂(n) = N²

2 +O(N^3/2). (2.59)

3 Proof of Theorem 1.5, Part 2

3.1 Proof of Theorem 1.5, Part 2

We recall the part 2 of Theorem1.5:

Assuming X

n≤X

r2(n) = X²

2 +O(X^3/2), then Riemann Hypothesis is true. (3.1) Step 1 To prove part 2, we need an asymptotic formula ofS(x) =P

n≤xr₂(n) S(x) =x²

2 −X

ρ

2x^ρ+1

ρ(ρ+ 1)+E(x), (3.2)

whereB= sup{<ρ:ζ(ρ) = 0},E(x) =O x^2+4B3 log⁴(2x)

. We have known thatB≤1 by Theorem 1.3, andB ≥1/2 is already known, because there does exist zeros on the line{<(s) = 1/2}, if we consider the positive imaginary parts, then the first non-trivial zero on that line has imaginary parts approximately 14.135.

Step 1 a little bit technical, so we put it in the subsection3.2.

Step 2 Assuming step 1, we define the corresponding Dirichlet series ofr₂(n) F(s) =X

n≥1

r₂(n)

n^s , (3.3)

we notice that

r₂(n) = X

m+k=n

Λ(m)Λ(k)nlogn, (3.4)

so when<(s)>2, the series converges uniformly and absolutely andF(s) has the integral form F(s) =s

Z ∞ 1

S(u)u^−s−1du, (3.5)

therefore,F(s) is analytic in {s∈C: <(s)>2}and by step 1 F(s) = s

2(s−2)+X

ρ

r(ρ)s

(ρ+ 1)(s−ρ+ 1) +s Z ∞

1

Er(u)u^−s−1du+1 2+X

ρ

r(ρ) ρ+ 1

= 1

s−2 +X

ρ

r(ρ) s−ρ−1 +s

Z ∞ 1

Er(u)u^−s−1du+1 2+X

ρ

r(ρ) ρ+ 1

.

(3.6)

wherer(ρ) =−²_ρ. We must point out that due to Lemma3.2, the seriesP

ρ r(ρ)

ρ+1 is convergent.

Moreover,F(s) can be extended meromorphically to the half plane{s∈C: <(s)> ^2+4B₃ }, this is becauseEr(u)u^2+4B3 log⁴(2u).

Step 3 IfB <1, then

1 +B= inf{σ0≥3

2 :F(s)− 1

s−2 is analytic on<(s)> σ0}. (3.7) Denote the set on the right hand is A. By step 2, infA is at most ^2+4B₃ , however, B ≤1, so infA ≤ ^2+4B₃ ≤B+ 1.

For the inverse inequality B + 1 ≤ infA, we know that this is trivially true when B = 1/2, because σ0≥ ³₂, so we may assume that 1/2< B ≤1. Now max{^2+4B₃ ,³₂}<1 +B, there exist an >0 such that max{^2+4B₃ ,³₂}<1 +B−, by the definition ofB, there exists a non-trivial zeroρ with ¹₂ < B− <<(ρ). By the formula ofF(s), in the half plane

{s∈C:<(s)>1 +B− >3

2} (3.8)

F has a pole atρ+ 1 with residue−²_ρ, and we conclude that 1 +B−≤inf{σ₀≥3

2 :F(s)− 1

s−2 is analytic on<(s)> σ₀}, (3.9) finally, let→0.

Step 4 Now we letD(x) =P

n≤xr2(n)−^x₂², then D(x)x³²⁺ for all >0. Consequently F(s)− 1

s−2 =s Z ∞

1

D(u)u^−s−1du+1

2, <(s)>2, (3.10) where the right hand side is analytic on{s∈C:<(s)> ³₂}, since D(u) u³²⁺. So we conclude B≤ ¹₂ by step 3, hence the Riemann Hypothesis holds.

Step 5 Notice that in Step 3, we assume that B <1, however we only have B ≤1 by Theorem 1.3, so we need to exclude the caseB= 1. For this, we need a lemma:

Lemma 3.1. If for some 0< δ <1, the following asymptotic formula holds:

S(X) =X²

2 +O(X^2−δ), (3.11)

then there exists 0< δ⁰<1 such that for all non-trivial zerosρofζfunction,ρsatisfies

<(ρ)<1−δ⁰. (3.12) The proof of this Lemma will be treated in subsection3.3. By our assumption:

X

n≤X

r2(n) = X²

2 +O(X^3/2), (3.13)

just letδ=¹₂, then there exists aδ⁰ such that<(ρ)<1−δ⁰ <1, so we exclude the caseB = 1.

3.2 Proof of Step 1 in Theorem 1.5

We follow the method in [4]. [4] considers the general case ofS(x) forL(s, χ), but we only need to deal with ζ, so some proofs in [4] can be simplified. In order to prove the asymptotic formula of S(x), we need some estimates for the non-trivial zeros of zeta function.

Lemma 3.2. Let ρ=β+ iγ be non-trivial zeros of ζ(s), then (i) For any T ≥1

X

|γ|≤T

1

|ρ| log²(2T). (3.14)

(ii) For any T ≥1

X

|γ|>T

1

|ρ|² log(2T)

T , (3.15)

particularly,P

ρ 1

|ρ|² <∞.

Proof. (i). We may use Lemma2.2N(T)∼cTlogT withc >0, remove finite zeros with 0< γ≤1, we have

X

1<γ≤T

1

|ρ| = X

1≤n≤T

X

n<γ≤n+1

1

|ρ| X

1≤n≤T

logn

n log²(T), (3.16)

so (i) is true by the fact that non-trivial zeros of zeta function come in conjugate pairs.

(ii). Similarly, (ii) is followed by X

|γ|>T

1

|ρ|² X

n≥T

X

n<γ≤n+1

1

|ρ|² X

n≥T

logn

n² logT

T . (3.17)

Recall thatS(x) =P

n≤xr₂(n), so we can writeS(x) as S(x) =X

l≤x

Λ(x)ψ(x−l), (3.18)

however,ψ(x) =x−P

|=(ρ)|≤T x^ρ

ρ +O(_T^xlog²x), whereT ≤x(see Lemma2.1 or2.3), hence S(x) =X

l≤x

Λ(l)

(x−l)− X

|=(ρ)|≤T

(x−l)^ρ ρ

+X

l≤x

Λ(l)O x−l

T log²(x−l)

. (3.19)

For the second term, X

l≤x

Λ(l)O

(x−l)

T log²(x−l)

x

T log²xX

l≤x

Λ(l) x²

T log²x. (3.20) For the first term, we denote

Σ1=X

l≤x

Λ(l)(x−l) Σ2=X

l≤x

Λ(l) X

|=(ρ)|≤T

(x−l)^ρ ρ ,

(3.21)

then the first term is Σ1−Σ2, by Fubini’s theorem, it’s easy to verify that Σ1=

Z x 0

ψ(u)du. (3.22)

Once again, byψ(x) =x−P

|=(ρ)|≤T x^ρ

ρ +O(_T^xlog²x), we have Σ1=

Z x 0

u− X

|=(ρ)|≤T

u^ρ

ρ +Ou

T log²u du

=x²

2 − X

|=(ρ)|≤T

x^ρ+1

ρ(ρ+ 1)+O(x²

T log²x).

(3.23)

Σ₂ is much more difficult than Σ₁, we change the double sum at first, then

Σ2= X

|=(ρ)|≤T

1 ρ

X

l≤x

Λ(l)(x−l)^ρ (3.24)

for the general term ¹_ρP

l≤xΛ(l)(x−l)^ρ, by Riemann-Stieltjes integral 1

ρ X

l≤x

Λ(l)(x−l)^ρ=1 ρ

Z x 0

(x−u)^ρdψ(u)

=1 ρ

Z x 0

(x−u)^ρd

A(u, T) +B(u, T) ,

(3.25)

whereA(u, T) =u−P

|=(ρ)|≤T u^ρ

ρ , andB(u, T) =O _T^ulog²u . Then 1

ρ Z x

0

(x−u)^ρdψ(u) =1 ρ

Z x 0

(x−u)^ρdu−1 ρ

X

|=(ρ⁰)|≤T

Z x 0

(x−u)^ρu^ρ0−1du+1 ρ

Z x 0

(x−u)^ρdB(u, T)

=I1−I2+I3,

(3.26) where

I₁= x^ρ+1 ρ(ρ+ 1) I2= X

|=(ρ⁰)|≤T

Γ(ρ)Γ(ρ⁰) Γ(1 +ρ+ρ⁰)x^ρ+ρ0 I₃=1

ρ(x−u)^ρB(u, T)

u=x u=0

+ Z x

0

(x−u)^ρ−1B(u, T)du O 1

|ρ|

x²

T log²x +

Z x 0

(x−u)^ρ−1B(u, T)du.

(3.27)

Here, Γ(ρ) is the complex Gamma function, we would like to say something about calculating I2. After a change of variableu7→xv,

Z x 0

(x−u)^ρu^ρ0−1du=x^ρ+ρ0 Z 1

0

(1−v)^ρv^ρ0−1dv=x^ρ+ρ0B(ρ⁰, ρ+ 1). (3.28) B(ρ⁰, ρ+ 1) is theBeta Functiondefined by

B(x, y) = Z 1

0

t^x−1(1−t)^y−1dt <x,<y >0. (3.29) Furthermore, we have the following relationship:

B(x, y) =Γ(x)Γ(y)

Γ(x+y), (3.30)