Sharper bounds for ψ , θ , π , p k

ESA - CNRS 6090

Sharper bounds for ψ , θ , π , p _k

Pierre Dusart

Rapport de recherche n° 1998-06

Université de Limoges, 123 avenue Albert Thomas, 87060 Limoges Cedex Tél. 05 55 45 73 23 - Fax. 05 55 45 73 22 - laco@unilim.fr

http://www.unilim.fr/laco/

PIERRE DUSART

Abstract. In default of a proof of theRiemannhypothesis, the best estimates forψ(x) andθ(x) and hence ofπ(x), p_k and other functions of the primes, depend on the current state of knowledge of the zeros ofζ(s). With a better knowledge about the zeros of theRiemannζfunction, we can show sharper bounds forψ(x), θ(x), π(x) and primesp_k.

1. Introduction

In many applications it is useful to have explicit error bounds in the prime num- ber theorem. RosserandSchoenfelddeveloped an analytic method which com- bines a numerical verification of theRiemann hypothesis with a zero-free region.

The aim of this paper is to find sharper bounds for the Chebyshev’s functions ψ(x), the logarithm of the least common multiple of all integers not exceedingx, andθ(x), the product of all primes not exceedingx:

θ(x) =X

p6x

lnp, ψ(x) = X

pν6xp,ν

lnp

where sum runs over primespand respectively over powers of primesp^ν. The Prime Number Theorem could be written as follows:

ψ(x) =x+o(x), x→+∞.

An equivalent formulation of the above theorem should be: for all ε > 0, there existsx₀=x₀(ε) such that

|ψ(x)−x|< ε x forx>x0

or

|θ(x)−x|< ε x forx>x0.

This article hangs up on some known results: the most important works on effective results have been shown by Rosser & Schoenfeld [13, 14, 16], Robin [9] &

Massias[6] andCosta Pereira[4].

The proofs for estimates of ψ(x) in [14] are based on the verification of Rie- mannhypothesis to a given height and an explicit zero-free region forζ(s) whose form is essentially that the classical one ofDe la vall´ee Poussin. Rosser &

Schoenfeld have shown that the first 3 502 500 zeros ofζ(s) are on the critical strip. Van de Lune et al [17] have shown that the first 1 500 000 000 zeros are

Date: June 11, 1998.

1991Mathematics Subject Classification. Primary 11A25, 11A40; Secondary 11L20, 11-04.

Key words and phrases. Estimates of elementary functions.

1

on the critical strip. We will improve bounds forψ(x) andθ(x). We will prove the following results:

|ψ(x)−x| 6 0.006409 x

lnx forx>exp(22),

|θ(x)−x| 6 0.006788 x

lnx forx>10 544 111,

|θ(x)−x| 6 0.2 x

ln²x forx>3594641,

|θ(x)−x| 6 515 x

ln³x forx >1,

|θ(x)−x| 6 1717433 x

ln⁴x forx >1.

We apply these results onp_k, thek^thprime, andθ(p_k). Let’s denote by ln₂xfor ln lnx.

The asymptotic expansion of p_k is well known; Cesaro [2] then Cipolla [3]

expressed it in 1902:

pk =k (

lnk+ ln2k−1 + ln2k−2

lnk −ln²₂k−6 ln2k+ 11 2 ln²k +O

õln2k lnk

¶3!) .

A more precise work about this can be find in [10, 15]. We group together the results onp_k and onθ(p_k):

θ(p_k) > k µ

lnk+ ln2k−1 +ln2k−2.0553 lnk

¶

fork>exp(22), (1)

θ(pk) 6 k µ

lnk+ ln2k−1 +ln2k−2 lnk

¶

fork>198, (2)

p_k > k(lnk+ ln₂k−1) fork>2, (3)

pk 6 k(lnk+ ln2k−0.9484) fork>39017, (4)

p_k 6 k µ

lnk+ ln₂k−1 +ln2k−1.8 lnk

¶

fork>27076, (5)

pk > k µ

lnk+ ln2k−1 +ln2k−2.25 lnk

¶

fork>2.

(6)

The formula (2) has been proved byRobinin [9].

We use the above results to prove that, forx>3275, the interval

£x, x+x/(2 ln²x)¤

contains at least one prime. Let’s denote byπ(x) the number of primes not greater thanx. We show that

x lnx

µ 1 + 1

lnx

¶

x>5996 π(x)_x>1⁶ x lnx

µ

1 + 1.2762 lnx

¶ .

More precise results onπ(x) are also shown:

π(x) > x

lnx−1 forx>5393,

π(x) 6 x

lnx−1.1 forx>60184, π(x) > x

lnx µ

1 + 1

lnx+ 1.8 ln²x

¶

forx>32299, π(x) 6 x

lnx µ

1 + 1

lnx+ 2.51 ln²x

¶

forx>355991.

We gave in§7.1 some others inequalities.

2. Results of Rosser & Schoenfeld

We rewrite here some results of [14] which we use. We want to boundψ(x)−x for “moderate” values ofx. Remember the used notations:

we want to findεsuch that we have, forx>exp(b)

|ψ(x)−x|< εx.

Let

R= 9.645908801, X =p

ln(x)/R, Kν(z, x) =1

2 Z _∞

x

t^ν−1exp{(−1/2)z(t+ 1/t)}dt forz >0, x>0, realν, φ_m(y) =y^−m−1exp(−X²/ln(y/17)),

Rm(δ) =¡

(1 +δ)^m+1+ 1¢m

, D= 158.84998, T₁=1

δ

µ2R_m(δ) 2 +mδ

¶1/m

.

LetN(T) the number of zeros such that 0< γ6T and its approximation F(T) = T

2πln T 2π− T

2π+7 8 and let

R(T) = 0.137 lnT + 0.443 ln₂T+ 1.588.

LetAsuch all zerosρ=β+iγof theRiemannζfunction in the critical strip are such that the real partβ= 1/2 for 0< γ6A. The numberA, used byRosser&

Schoenfeld, is defined as the unique solution of the equationF(A) = 3,502,500, which yieldsA= 1,894,438.51224· · · andF(A) =N(A). We use a new value ofA, defined as the unique solution of the equation F(A) = 1,500,000,001. According to [17], this yieldsA= 545,439,823.215· · · andF(A) =N(A).

Proposition 1([14], Theorem 4). Let T1>D,mbe a positive integer and Ω₁ = 2 +mδ

4π (µ

lnT₁ 2π+ 1

m

¶2

+ 0.038207 + 1

m²− 2.82m (m+ 1)T1

)

Ω2 = (0.159155)Rm(δ)z 2m²17^m

½

zK2(z, A⁰) + 2mln µ17

2π

¶

K1(z, A⁰)

¾ +R_m(δ){2R(Y)φ_m(Y)−R(A)φ_m(A)},

wherez= 2X√

m= 2p

mb/R,A⁰ = (2m/z) ln(A/17)and Y = max{A, 17 expp

b/((m+ 1)R)}. If b >1/2 0< δ <(1−exp(−b))/mthen, for allx>exp(b),

|ψ(x)−x|< εx

whereε= Ω1exp(−b/2) + Ω2δ^−m+mδ/2 + exp(−b) ln(2π).

As we use this theorem, we write explicitly the proof which doesn’t appear in Rosser&Schoenfeld’s article; They gave only indications.

Proof: In all this proof, the numbering refers to [14] and some non-defined nota- tions too. AsT2= 0, S3(m, δ) = 0.

S₄(m, δ) = 2 X

β>1/2 γ>0

Rm(δ) exp(−X²ln(γ/17))

δ^m|ρ(ρ+ 1)· · ·(ρ+m)| < Rm(δ) δ^m

X

γ>A

φ(γ) by writingφ(y) :=φ_m(y). By using corollary of lemma 7, (φ(y) :=φ_m(y))

X

γ>A

φ(γ)6{ 1

2π+q(y)} Z _∞

A

φ(y) ln

³ y 2π

´

dy+E₀ where q(y) = ^{0.137 lny+0.443}

ylnyln(2π^y ) and E0 = 2R(Y)φ(Y)−R(A)φ(A) because N(A) = F(A) with Y = max(A, W) whereW satisfies (W −y)φ⁰(y)>0 fory > A. Let’s computeW.

φ⁰(y) =φ⁰_m(y) =y^−m−2exp(−X²/ln(y/17)) µ

−(m+ 1) + X² ln²(y/17)

¶ . Henceφ⁰_m(y)>0 fory>17 exp

³q X² m+1

´

. According to (3.16), Z +∞

A

φm(y) ln

³ y 2π

´

dy= z²

2m²+ 17^mK2(z, A⁰) +zln(17/(2π))

m17^m K1(z, A⁰) where z= 2X√

m andA⁰ = (2m/z) ln(A/17). Moreover, q(Y)6q(A)67·10⁻⁹. So forx>exp(b),

S4(m, δ)<Rm(δ) δ^m ( 1

2π+7·10⁻⁹) Ã

z²

2m²+ 17^mK2(z, A⁰) +zln¡₁₇

2π

¢

m17^m K1(z, A⁰)

! +E0

wherez= 2 qmb

R ,A⁰= (2m/z) ln(A/17).

We deduce that

S₃(m, δ) +S₄(m, δ)6Ω₂δ^−m.

By using the upper bound ofS1(m, δ) +S2(m, δ) given in this previous lemma, we conclude by lemma 8 that

1

x|ψ(x)−(x−ln(2π)−1/2·ln(1−1/x²))|6Ω₁/√

x+ Ω₂δ^−m+mδ/2 thus

|ψ(x)−x|6(Ω1exp(−b/2) + Ω2δ^−m+mδ/2 + exp(−b) ln(2π))x.

2

3. Results on ψ(x) andθ(x)

We group here together some known results which we will use in our proofs.

Proposition 2. (i) -θ(x)6ψ(x)for allx.

(ii) -ψ(x)−θ(x)<√ x+⁶₅√³

xfor10⁸6x610¹⁶. (iii) -ψ(x)−θ(x)<1.43√

xfor allx >0.

(iv) - θ(x)< x for0< x610¹¹. (v) -θ(x)<1.000081xforx >0.

(vi) - |θ(x)−x|<0.0077629_ln^x_x forx >1.04·10⁷. Proof: The first inequality can be easily deduced from

ψ(x) = X∞ k=1

θ(x^1/k).

The second is given in [4]. The next one is theorem 13 of [13]. The three last ones

are taken from [16] p. 360. 2

For “large” values ofx, we use the following theorem (theorem 11 p. 342 of [16]):

Theorem 1. Let R= 9.645908801andX = qlnx

R .

|ψ(x)−x|< xε0(x) forx>17,

|θ(x)−x|< xε0(x) forx>101 whereε₀(x) =p

8/(17π)X^1/2exp(−X).

Theorem 2. Let bbe a positive real. For x>exp(b),

|ψ(x)−x|6εx, whereε is given in table 1.

Corollary 1. Forx>exp(50), we have

|ψ(x)−x|60.905·10⁻⁷x.

Proof: We make no mathematical change of theRosser&Schoenfeld’s method.

We only change the value A by a bigger one. Apply proposition 1 with A = 545,439,823.215· · · as defined in section 2. Evaluation of the quantities of propo- sition 1, for example with the computer algebra Maple ¹, gives new bounds given in table 1.

If we takeA= 1,894,438.51224· · ·, this implementation permits to recover the results found byRosserandSchoenfeld.

In particular, we compute the value of εforx>e⁵⁰. Chooseδ= 0.947265625· 10⁻⁸andm= 18. By computation, we find a value forεlightly less than 0.905·10⁻⁷.

This value has been verified withPari. 2

We need afterwards to have bounds for θ(x) to 10¹¹. The first method is to computeθ(x) until 10¹¹ (we need a powerful computer). The second method uses the remark that the table of [1], which gives the upper and the lower bounds of Li(x)−π(x) by intervals up to 10¹¹, permits to find precise bounds forθ(x) up to 10¹¹. The results are in table 2.

We construct this table in the following manner:

Letρ₁(a, b) = min_pk∈[a,b](Li(p_k)−k) andR₁(a, b) = max_pk∈[a,b](Li(p_k)−k). Write

1We thank theM´edicisgroup for the use of their computers.

r(a, b) = min_x∈[a,b](Li(x)−π(x)) andR(a, b) = max_x∈[a,b](Li(x)−π(x)). Note that r(a, b) > min(ρ1(a, b)−1/2,Li(a)−π(a)) and R(a, b) 6 max(R1(a+ 1, b) + 1 + 1/2,Li(b)−π(b)). The interval [0,10¹¹] is split into intervals [ai, bi].

θ(x) = π(x) ln(x)− Z x

2

π(y) y dy

= π(x) ln(x)− Z 100

2

π(y) y dy−

Z x 100

π(y)−Li(y)

y +Li(y)

y dy

> x−(Li(x)−π(x)) lnx+C+ X

[ai,bi] 1006ai,bi6b

r(ai, bi) ln µbi

a_i

¶

+r(b, x) ln

³x b

´ (7)

whereC= Li(100) ln(100)−100−R100 2

π(y)

y dyandbis the upper bound of the last interval which doesn’t containx. If we don’t know the value ofπ(x), we can bound the difference Li(x)−π(x) byR(a, x).

We have an upper bound of θ(x) by replacing in formula (7) all r by R and converselyRbyr.

Theorem 3. 1. |ψ(x)−x|60.006409_ln^x_x forx>exp(22).

2. |θ(x)−x|60.006788_ln^x_x forx>10 544 111.

Proof: We use table 1 by intervals:

if x > exp(22) then |ψ(x)−x| 6 2.78652·10⁻⁴x 6 23∗2.78652·10⁻⁴_ln^x_x 6 6.409·10⁻³_ln^x_x forx6exp(23).

Ifx>exp(23) then|ψ(x)−x|61.8436·10⁻⁴x65.5308·10⁻³_ln^x_x forx6exp(30).

Ifx>exp(30) then |ψ(x)−x|60.978·10⁻⁵x65.868·10⁻³_ln^x_x forx6exp(600).

Ifx>exp(600) then|ψ(x)−x|60.75·10⁻⁷x61.5·10⁻⁴_ln^x_x forx6exp(2000).

Moreover,

|ψ(x)−θ(x)|<√ x+⁶₅√³

xfor 10⁸6x610¹⁶ thus

|ψ(x)−θ(x)|<0.00037871_ln^x_x for exp(22)6x6exp(30) and

|ψ(x)−θ(x)|<1.43√

xforx >0 thus

|ψ(x)−θ(x)|<1.32·10⁻⁵_ln^x_x forx>exp(30).

Forx>exp(2000), we use theorem 1 to conclude that

|ψ(x)−x|and|θ(x)−x|60.00164 x

lnx forx>exp(2000).

With Table 2 and a computer verification, we extend the result onθ, showed until now forx>exp(22), until 10 544 111. This shows that the result of proposition 2(vi) ofRosser&Schoenfeldis almost best possible. 2 The above result gives a formula with order 1 for power of logarithm. We need sometimes the next orders. We obtain better results thanRosser&Schoenfeld ones.

Theorem 4. Let η2= 3.965,η3= 515andη4= 1717433.

Forx >1,

|θ(x)−x|< ηk

x ln^kx. (We can chooseη2= 0.2 forx>3594641.)

Proof: In all cases, we use table 1 by intervals. Starting withk= 2, we show that

|θ(x)−x|<0.2 x ln²x.

Forx>exp(3220), according to theorem 1 we have

|θ(x)−x|< xε(x)< η2

x ln²x withη₂= 0.19923 forx>exp(3220).

Forx>1.04·10⁷, |θ(x)−x|60.0077629_ln^x_x 60.1941_ln^x2x forx6e²⁵. Forx>e²⁵, |ψ(x)−x|610⁻⁴x 60.09_ln^x2x forx6e³⁰. Forx>e³⁰, |ψ(x)−x|610⁻⁵x 60.1_ln^x2x forx6e¹⁰⁰. Forx>e¹⁰⁰, |ψ(x)−x|60.9·10⁻⁷x 60.1521_ln^x2x forx6e¹³⁰⁰. Forx>e¹³⁰⁰, |ψ(x)−x|60.6·10⁻⁷x 60.1944_ln^x2x forx6e¹⁸⁰⁰. Forx>e¹⁸⁰⁰, |ψ(x)−x|60.42·10⁻⁷x 60.168_ln^x2x forx6e²⁰⁰⁰. Forx>e²⁰⁰⁰, |ψ(x)−x|60.37·10⁻⁷x 60.196_ln^x2x forx6e²³⁰⁰. Forx>e²³⁰⁰, |ψ(x)−x|60.292·10⁻⁷x 60.1825_ln^x2x forx6e²⁵⁰⁰. Forx>e²⁵⁰⁰, |ψ(x)−x|60.244·10⁻⁷x 60.18453_ln^x2x forx6e²⁷⁵⁰. Forx>e²⁷⁵⁰, |ψ(x)−x|60.19·10⁻⁷x 60.197_ln^x2x forx6e³²²⁰. Moreover,

for exp(25)6x6exp(30), |ψ(x)−θ(x)| <√ x+⁶₅√³

x <0.0023725_ln^x2x, and forx>exp(30), |ψ(x)−θ(x)| <1.43√

x <0.0004_ln^x2x. Now we are interested in casek= 3. As|θ(x)−x|60.0077629_ln^x_x, we have

|θ(x)−x|6310.516 x

ln³x forx6exp(200).

Note that forx>exp(200), the difference betweenθ andψ is negligible since, for x>exp(200),

|ψ(x)−θ(x)|<1.43√

x60.5·10⁻³⁶ x ln³x.

Forx>e²⁰⁰, |θ(x)−x|60.8561317·10⁻⁷x 6500_ln^x3x forx6e¹⁸⁰⁰. Forx>e¹⁸⁰⁰, |θ(x)−x|60.419134·10⁻⁷x 6510_ln^x3x forx6e²³⁰⁰. Forx>e²³⁰⁰, |θ(x)−x|60.2917036·10⁻⁷x 6456_ln^x3x forx6e²⁵⁰⁰. Forx>e²⁵⁰⁰, |θ(x)−x|60.243946·10⁻⁷x 6508_ln^x3x forx6e²⁷⁵⁰. Forx>e²⁷⁵⁰, |θ(x)−x|60.1877·10⁻⁷x 6507_ln^x3x forx6e³⁰⁰⁰. Forx>e³⁰⁰⁰, |θ(x)−x|60.137602·10⁻⁷x 6514_ln^x3x forx6e³³⁴¹. Now apply theorem 1 forx>exp(3341); we find

|θ(x)−x|6514.826 x ln³x.

We use the same methods for casek= 4. As|θ(x)−x|60.0077629_ln^x_x, we have

|θ(x)−x|61676786.4 x

ln³x forx6exp(600).

Note that, for x >exp(600), the difference between θ and ψ is negligible seeing that, forx>exp(600),

|ψ(x)−θ(x)|<1.43√

x610⁻¹¹⁹ x ln⁴x.

Forx>e⁶⁰⁰, |θ(x)−x|60.744205·10⁻⁷x 6 ^1190728x_ln4x forx6e²⁰⁰⁰. Forx>e²⁰⁰⁰, |θ(x)−x|60.3675·10⁻⁷x 6 ^1435547x_ln⁴_x forx6e²⁵⁰⁰. Forx>e²⁵⁰⁰, |θ(x)−x|60.243946·10⁻⁷x 6 ^1395162x_ln4x forx6e²⁷⁵⁰. Forx>e²⁷⁵⁰, |θ(x)−x|60.1877·10⁻⁷x 6 ^1520370x_ln4x forx6e³⁰⁰⁰. Forx>e³⁰⁰⁰, |θ(x)−x|60.137602·10⁻⁷x 6 ^1716527x_ln4x forx6e³³⁴².

Now apply theorem 1 forx>exp(3342): we find

|θ(x)−x|61717433 x ln⁴x.

We verify by computer forx61,04·10⁷. We find that|θ(x)−x|60.2_ln^x2x for x>3594641 and|θ(x)−x|63.9648085· · ·_ln^x2_x forx >1 (the value 3.9648085· · ·

is choosen forp₁₇). 2

4. Results onpk andθ(pk)

For some steps of demonstrations, we use the following results:

Lemma 1.

pk > k(lnk) fork>2, pk 6 k(lnk+ ln2k) fork>6, pk 6 klnpk fork>4, pk > k(lnpk−2) fork>5.

Proof: These inequalities have been proved by Rosser. The first inequality can be found in [11]; the next one in [12]. The last ones can be easily deduced from

x

lnx < π(x)< x lnx−2

(see [12]). 2

Rossershowed that pk >klnk and improved his result withSchoenfeld by showing thatpk >k(lnk+ ln2k−3/2) (see [14]). En 1983,Robin[9] succeeded to prove that

p_k >k(lnk+ ln₂k−1.0072629).

Massias&Robin[6] are able to show that

pk>k(lnk+ ln2k−1) fork6exp(598) andk>exp(1800).

We have show that for allk>2 in [5].

Theorem 5. Fork>2, we have

p_k>k(lnk+ ln₂k−1) wherepk isk^thprime number.

First method to find a lower bound forθ(pk).

Proposition 3. Let

fβ(k) :=k µ

lnk+ ln2k−1 + ln2k−β lnk

¶ . Assume that fork>k₀

|θ(p_k)−p_k|6ck.

Let β be a real satisfying

β(lnk0−1) > lnk0(ln2k0+ 1) + 1−ln2k0

−ln²k₀ln µ

1 +ln₂k₀−1 lnk0

+ln₂k₀−β−clnk₀ ln²k₀

¶ . If θ(pk0)>fβ(k0)then

θ(pk)>fβ(k) fork>k0.

Proof: Assume that

pk >ha(k) :=k µ

lnk+ ln2k−1 + ln₂k−a lnk

¶ . Hence

θ(p_k)−θ(p_k0) = Xk

n=k0+1

lnp_n >

Xk

n=k0+1

lnh_a(n).

Moreover

f_β⁰(k) = lnk+ ln₂k+ln₂k−β+ 1

lnk −ln₂k−(β+ 1) ln²k and

lnpk >lnha= lnk+ ln2k+ ln µ

1 +ln2k−1

lnk +ln2k−a ln²k

¶ . It’s sufficient to show whenf_β⁰ 6lnha because

fβ(k)−fβ(k0) = Z k

k0

f_β⁰(x)dx6 Xk

n=k0+1

lnha(n)6θ(pk)−θ(pk0), equivalently

ln₂k+ 1−β

lnk −ln₂k−(β+ 1) ln²k 6ln

µ

1 +ln₂k−1

lnk +ln₂k−a ln²k

¶

β(lnk−1)>lnk(ln2k+ 1)−ln²kln µ

1 + ln2k−1

lnk +ln2k−a ln²k

¶

+ 1−ln2k.

Sincepk>θ(pk)−ck, we can choosea=β+clnk. Then we find that β(lnk−1) > (lnk)(ln₂k+ 1)−(ln²k) ln

µ

1 +ln₂k−1

lnk +ln₂k−β−clnk ln²k

¶

+1−ln2k.

2 We apply proposition 3. Forc= 0.007, the computed values ofβ are:

lnk0 β 22 2.0553 23 2.0532 25 2.04975 30 2.04397 100 2.02961 1000 2.0156 2000 2.0128 Corollary 2. Forp_k>10¹¹,

θ(pk)>k µ

lnk+ ln2k−1 + ln2k−2.0548 lnk

¶ .

Proof: We takek₀= 4118054814 andc= 0.006788. We obtainβ= 2.0548. Using

table 2, we checkθ(10¹¹)>f_2.0548(k₀). 2

The major drawback of the above method is that we must check if θ(p_k0) >

f_β(k₀) which needs a computation ofθ(x) for large x. The next method doesn’t need this hypothesis.

Second method to find a lower bound forθ(pk).

Proposition 4. Let k0>exp(exp(3)). Assume that

|θ(x)−x|6c x

lnx forx>pk0. Then

θ(pk)>k µ

lnk+ ln2k−1 +ln₂k−β lnk

¶

fork>k0

withβ solution of the following equality

β = lnk0(1 +α)−ln2k0+ 1 lnk0−1

α = 1

2

(ln₂k₀−a)² lnk0

+a a = 1−ln₂k₀−β

lnk0

+c

Proof: Applying lemma 1 of [9] fora= 1 and lnk0>exp(a+ 2) in the same way than theorem 7 of [9], we find a first value ofβ writtenβ0. Since|θ(pk)−pk|6ck, we can apply lemma 1 one more time with the new value ofa: we choose

a= 1−ln2k0−β0

lnk₀ +c.

In fact the series{β_k}k is a convergent one. We determine the limit by solving the equation of the fixed point fork=k₀

β = lnk

µ

1 +¹₂^{(ln2k−1+ln2}

k−β lnk −c)²

lnk + 1−^ln2_ln^k_k^−β +c

¶

−ln2k+ 1

lnk−1 .

Whenk₀ grows to infinity,β decreases to 2 +c. 2

Applying the previous proposition forc= 0.007, the computed values ofβ are:

lnk0 β exp(3) 2.0675

25 2.056 30 2.04938 1000 2.01356 2000 2.0128 Corollary 3. Fork>exp(30),

θ(p_k)>k µ

lnk+ ln₂k−1 +ln₂k−2.05 lnk

¶ . Theorem 6.

pk>k µ

lnk+ ln2k−1 +ln2k−2.25 lnk

¶

fork>2.

Proof: Sinceθ(x)< xfor 0< x <10¹¹, we deduce immediately p_k >θ(p_k)>k

µ

lnk+ ln₂k−1 +ln2k−2.1454 lnk

¶ . Forp_k >exp(4000), it follows from theorem 1 that

|θ(x)−x|< xε(x)< η₂ x ln²x

withη2= 0.040033 forx>exp(4000). Thus, forpk >exp(4000), p_k>θ(p_k)−η₂ p_k

ln²pk

>k µ

lnk+ ln₂k−1 +ln2k−2.1−0.040033 lnk

¶ . For 10¹¹6pk6exp(4000),

p_k > θ(p_k) µ 1

1 +ε

¶

>(1−ε)θ(p_k)

> k µ

lnk+ ln2k−1 +ln2k−β lnk −ε

µ

lnk+ ln2k−1 +ln2k−β lnk

¶¶

. withβ= 2.0548. Let’s study the map

f(ε, k) :=−β−ε µ

lnk+ ln2k−1 + ln2k−β lnk

¶ lnk.

We choose at the first timeε=ε(10¹¹) = 0.00008. We verify thatf(ε,exp(30))>

−2.1319 and so on, by small intervals up to exp(4000). We deduce thatf(ε, k)>

−2.25 forp_k between exp(22) and exp(4000). 2

Theorem 7.

pk6k µ

lnk+ ln2k−1 + ln2k−1.8 lnk

¶

fork>27076.

Proof: Assume first that (see lemma 1) pk

ln²pk

6 k

lnk. Forx>3594641, we have establish that

|θ(x)−x|< xε(x)< η2

x ln²x withη₂= 0.2. Thus, forp_k >3594641,

pk 6θ(pk) +η2

p_k ln²p_k 6k

µ

lnk+ ln2k−1 + ln₂k−2 + 0.2 lnk

¶ .

We terminate the proof by a computer check. 2

Theorem 8. Fork>39017, we have

pk6k(lnk+ ln2k−0.9484).

Proof: As (see theorem 7) p_k 6k

µ

lnk+ ln₂k−1 + ln2k−1.8 lnk

¶ , we have, forx>exp(33),

pk6k(lnk+ ln2k−0.9484).

¿From

|θ(p_k)−p_k|6ε(p_k)p_k, we deduce that

p_k6θ(p_k) +ε(p_k)p_k. As

θ(pk)6k µ

lnk+ ln2k−1 + ln2k−2 lnk

¶ ,

it remains to show that

g(k) :=−1 +ln2k−2

lnk +ε(pk)pk

k <−0.9484 for 41180548136k6exp(30). If we choose (cf. [6])

p_k6k(lnk+ ln₂k−0.9427) and ε(p_k) = 0.000079989 + 1/√

10¹¹ for 10¹¹ 6 p_k 6 exp(30) and next ε(p_k) = 10⁻⁵+ 1.43/√

exp 30 forp_k >exp(30), we obtain the result forp_k >10¹¹. Now we study if the result is true forp_k 610¹¹. Writeα= 0.9484 and

f(k) =k(lnk+ ln2k−α).

Thanks toBrent [1], we have the lower and the upper bounds of the difference of Li(x)−π(x) into various intervals up to 10¹¹. Hence if, fork∈[k0, k1],

Li(f(k))−π(p_k)>M_pk

0,pk1 := max

k∈[k0,k1](Li(p_k)−π(p_k)) we could deduce that

pk6f(k).

We considerg(k) = Li(f(k))−k.

g⁰(k) =1 + 1/lnk−α−ln(1 + (ln2k−α)/lnk) lnk+ ln₂k+ ln(1 + (ln₂k−α)/lnk)

admits a minimum for lnk≈exp(2 +α+ 2/exp(2 +α)). Forα= 0.9484, this lower bound is positive. We show, thanks to the following table, that

Li(f(k))−π(p_k)>M_pk0,pk1

for 170368 6 k 6 4118054813 (2312573 6 pk 6 10¹¹). The values of π(x) are extracted from table 3 ofRiesel[8] (Ex.: π(10⁷) = 664579)

k pk 6· Li(f(k))−k R1=Mpk0,pk1

170368 2.315·10⁶ 261.0004 M_2·10⁶_.5·10⁶=261 348512 5·10⁶ 414.1091 M_5·10⁶_.10⁷=346 664578 10⁷ 634.5851 M₁₀⁷_,2·10⁷=435 1270606 2·10⁷ 983.965 M_2·10⁷_,5·10⁷=692 3001133 5·10⁷ 1788.864 M_5·10⁷_,5·10⁸=1724 26355866 5·10⁸ 8890.45 M_5·10⁷_,10¹⁰=7048 455052510 10¹⁰ 93238.1 M10¹⁰,10¹¹=17065

Fork= 39017..170367, a check has been made thanks toParisystem. 2 5. Interval which contains at least one prime

We already know the result ofSchoenfeld[16] showing that, forx>2010759.9, the interval ]x, x+x/16597[ contains at least one prime. We improve this result with the following proposition. You can see also [7].

Proposition 5. Fork>463, p_k+16p_k

µ

1 + 1

2 ln²p_k

¶ .

Proof: Assume that, fork>k0, p_k >k

µ

lnk+ ln₂k−1 +ln₂k−α₀ lnk

¶

and

pk6k µ

lnk+ ln2k−1 + ln2k−α1

lnk

¶ .

pk

µ 1 + γ

ln²pk

¶

−pk+1=pk−pk+1+γ pk

ln²pk

> k µ

lnk+ ln₂k−1 + ln₂k−α₀ lnk

¶

+γ p_k ln²pk

−(k+ 1) µ

ln(k+ 1) + ln2(k+ 1)−1 + ln₂(k+ 1)−α₁ ln(k+ 1)

¶

= k

µ

lnk−ln(k+ 1) + ln₂k−ln₂(k+ 1) + ln₂k−α₀

lnk −ln₂(k+ 1)−α₁ ln(k+ 1)

¶

− µ

ln(k+ 1) + ln₂(k+ 1)−1 +ln₂(k+ 1)−α₁ ln(k+ 1)

¶

+γ p_k ln²pk

But,

ln(k)−ln(k+ 1) =−ln(1 + 1/k), ln₂(k)−ln₂(k+ 1) =−ln

µ

1 + ln(1 + 1/k) lnk

¶ , ln₂k−α₀

lnk −ln₂(k+ 1)−α₁

ln(k+ 1) >(α₁−α₀)/ln(k+ 1)

because ^ln_x^x is decreasing for x>1/e hencef(k)>f(k+ 1) where f(x) := ^ln_ln²_x^x. We chooseγ such that

γ pk

ln²pk

> k µ

ln(1 + 1/k) + ln µ

1 +ln(1 + 1/k) lnk

¶

+ (α0−α1)/ln(k+ 1)

¶

+ µ

ln(k+ 1) + ln₂(k+ 1)−1 + ln2(k+ 1)−α1

ln(k+ 1)

¶

; (8)

We must chooseγ > α0−α1to make the inequality true for large k. As pk

ln²pk

> klnk (lnk+ ln2k)², the inequality (8) is satisfied withγ= 1/2 if

ln(k+ 1)/2>(α0−α1)(lnk+ ln2k).

Forα0= 2.25 andα1= 1.8, this holds fork>exp(82). Theorem 12 of [16] showed that the interval ]x, x+x/16597[ contains at least one prime for x> 2010759.9.

Apply it forx=pk, this yields the result for 2010759.96pk6exp(91). According to [16] p. 355,

pn+1−p_n6652 forp_n62.686·10¹², and so

p_k+16p_k µ

1 + 1

2 ln²p_k

¶

when

p_k

2 ln²p_k > k

2(lnk+ ln2k) >652

that is to say for k > 2·10⁴. For k = 463..20000, a direct check with computer

conclude the proof. 2

Theorem 9. For allx>3275, there exists a primepsuch that x < p6x

µ

1 + 1 2 ln²x

¶ .

This result is better thatRosser&Schoenfeld’s one forx>3·10³⁹. Proof: Letx>2. There existsk∈N^∗ such that

pk6x < pk+1. As the mapx7→x(1 + 1/(2 ln²x)) is increasing,

pk 6x⇒pk

µ

1 + 1

2 ln²pk

¶

6x

µ

1 + 1 2 ln²x

¶ . According to proposition 5, fork>463,

p_k+16p_k µ

1 + 1

2 ln²pk

¶ . We deduce that, forx>p463= 3299,

x < pk+16x µ

1 + 1 2 ln²x

¶ . Moreover, forx>3274.0111,

3299< x µ

1 + 1 2 ln²x

¶ .

2 6. Results on π(x)

Remember that

π(x) = x lnx

µ 1 + 1

lnx+ 2

ln²x+O( 1 ln³x)

¶ . Theorem 10. 1. - _ln^x_x¡

1 +_ln¹_x¢

6π(x)forx>599.

2. -π(x)6 _ln^x_x¡

1 +^1.2762_lnx ¢

forx >1(the value 1.2762 is chosen forx=p₂₅₈= 1627).

3. - π(x)6 _ln^x_x¡

1 +^1.0992_lnx ¢

forx>1.332·10¹⁰. 4. - π(x)6 _lnx−1.1^x forx>60184.

5. - π(x)> _lnx−1^x forx>5393.

6. - _ln^x_x¡

1 + _ln¹_x+_ln^1.82x

¢6π(x)forx>32299.

7. - π(x)6 _ln^x_x¡

1 +_ln¹_x+_ln^2.512x

¢forx>355991.

Let’s start with the first three inequalities.

Lower bound ofπ(x).•We began with x>10⁸. π(x) =π(p)− θ(p)

ln(p)+ θ(x) ln(x)+

Z x p

θ(y)dy yln²(y).

Butθ(x)>x−0.024_ln(x)^x forx>758711 ([16], Th.7 p.357). Hence, forp>758711, this yields

π(x)> θ(x) ln(x)+ x

ln²(x)+π(p)− θ(p) ln(p)− p

ln²(p)+ (2−0.024) Z _x

p

dy ln³(y). Sinceθ(x)>x−0.0077629_ln(x)^x forx>1.04·10⁷,

π(x)> x lnx

µ

1 +1−0.008 ln(x)

¶

forx>10⁸ because

π(p)− θ(p) ln(p)− p

ln²(p)+ (2−0.024) Z x

p

dy

ln³(y) >0 forx>10⁸, by choosing

p=p61000= 760267 andθ(p)< p.

•According to [13](Th 16, p. 72), we have for 11< x610⁸ , Li(x)−Li(√

x)< π(x).

For 18596x610⁸,

Li(x)−Li(√

x)> x lnx

µ

1 +0.992 lnx

¶ . Fork= 1..284 (p284= 1861), we check by computer that

π(pk−1/2) =k−1> pk

lnp_k µ

1 + 0.992 lnp_k

¶ , inequality true fork>110. Henceπ(x)> _ln^x_x¡

1 + ^0.992_lnx ¢

forx>p₁₀₉= 599.

We will obtain a better result with an upper bound forθ in order 2 in lnx.

Upper bound forπ(x). According to [16] (p. 360), we know that : θ(x)< x for 0< x610¹¹,

θ(x)<1.000081x forx >0, and that

|θ(x)−x|60.0077926 x

lnx forx>1.04·10⁷.

Letb= 0.0077926,c= 1.000081 andK= 10¹¹. π(x) = θ(x)

lnx + Z _x

2

θ(y)dy yln²y

= θ(x) lnx +

Z K 2

θ(y)dy yln²y +

Z x K

θ(y)dy yln²y

< x lnx

µ 1 + b

lnx

¶ +

Z _K

2

dy ln²y +

Z _x

K

dy ln²y +b

Z _x

K

dy

ln³y forx>K

= x

lnx µ

1 + b lnx

¶ +

· y ln²y

¸x 2

+ 2 Z _x

2

dy ln³y +b

Z _x

K

dy ln³y

< x lnx

µ

1 + b+ 1 lnx

¶ + 2 x

ln³x+ 6 Z _x

2

dy ln⁴y +b

· y ln³y

¸x K

+ 3b Z _x

K

dy ln⁴y ButZ _x

K

dy ln⁴y =

Z ^√_x

K

dy ln⁴y +

Z _x

√x

dy ln⁴y <(√

x−K)/ln⁴(K) +x−√ x ln⁴√

x if√ x>K.

To have

π(x)6 x lnx

µ 1 + β

lnx

¶

we choose β > b+1+2 +b

lnx +ln²x x

à 6

Z K 2

dy

ln⁴x−b K

ln³K + (6 + 3b) µ√

x−K

ln⁴K +x−√ x ln⁴√

x

¶!

. We obtainβ >1.03 whenx>exp(100).

π(x) = π(K)−θ(K) lnK +θ(x)

lnx + Z x

K

θ(y)dy yln²y

< π(K)−θ(K) lnK +c x

lnx+c Z x

K

dy ln²y

= π(K)−θ(K) lnK +c K

lnK +c(Li(x)−Li(K))

= M +c·Li(x) whereM is a constant.

Write

∆(x) = x lnx

µ 1 + β

lnx

¶

−(M+c·Li(x)).

∆⁰(x) =¡

(1−c) ln²(x) + (β−1) ln(x)−2β¢ /ln³x.

∆⁰ equals zero when

lnx= −(β−1)±p

(β−1)²+ 8β(1−c)

2(1−c) .

We can takeπ(K) = 4118054813 and^θ(K)_lnK > _ln^K_K¡

1−^0.007_lnK¢

hence, forβ = 1.0992, the local minimumx0≈exp(22.5775) (x0 < K) is negative but ∆(K) is positive.

Hence, forx∈[K,exp(100)],

π(x)< x lnx

µ

1 +1.0992 lnx

¶ .

By consulting table given in [1], we show that the result remains true for x >

1.332·10¹⁰. According to [13] (Th. 16 p. 72) and [1], we have forx6K, π(x)<Li(x).

Let’s study the difference

∆(x, β) = x lnx

µ 1 + β

lnx

¶

−Li(x).

This function admits a minimum at point x = exp(2β/(β −1)). For β₀ :=

1.276103273, the value of this minimum equals to−9.972985. We consult the table ofBrent[1]. This yields the lower difference between π(x) and Li(x) in different intervals [10ⁿ,10ⁿ⁺¹] withn= 0..10. We deduce that

x lnx

µ 1 + β₀

lnx

¶

−π(x) = x lnx

µ 1 + β₀

lnx

¶

−Li(x) + (Li(x)−π(x))>0 forx>10⁴. Fork= 1..1230, we verify that

π(p_k) =k6 p_k lnpk

µ 1 + β₀

lnpk

¶ .

We choose the value forβ0 such that it holds for all integersk, includingk= 258.

Next we consider the others formulas. Let

x0= 1.04·10⁷, K=π(x0)−θ(x0) lnx0

.

Numericallyπ(x0) = 689382 andθ(x0) = 10395445.63690637· · · Write J(x;a) =K+ x

lnx+a x ln³x+

Z _x

x0

µ 1

ln²y+a 1 ln⁴y

¶ dy Since

π(x) =π(x0)−θ(x0) lnx₀ +θ(x)

lnx + Z x

x0

θ(y)dy yln²y and|θ(x)−x|60.2_ln^x2x forx > x0, we have, forx>x0,

J(x;−0.2)6π(x)6J(x; 0.2).

Write M(x;c) = _ln^x_x¡

1 + _ln¹_x+_ln^c2x

¢ for upper bound’s function for π(x). Let’s write the derivatives ofJ(x;a) and ofM(x;c) with respect tox:

J⁰(x;a) = 1 lnx+ a

ln³x−2 a ln⁴x, M⁰(x;c) = 1

lnx+c−2 ln³x − 3c

ln⁴x.

We must choose c > 2 +a to have J⁰ < M⁰ when lnx > (3c−2a)/(c−2−a).

Choosec = 2.51. We verify by computer thatJ(10¹¹; 0.2)< M(10¹¹; 2.51). Since Li(x)< M(x; 2.51) forx>10⁷, we verify by direct computation for small values of xto obtain

π(x)< x lnx

µ 1 + 1

lnx+ 2.51 ln²x

¶

forx>355991.

Now write

m(x;d) = x lnx

µ 1 + 1

lnx+ d ln²x

¶ .

We study the derivatives: we must choosed62−ato haveJ⁰> m⁰.