Sharper bounds for ψ, θ, π, p k

Transcription

1 Laboratoire d Arithmétique, de Calcul formel et d Optimisation ESA - CNRS 6090 Sharper bounds for ψ, θ, π, p k Pierre Dusart Rapport de recherche n Université de Limoges, 23 avenue Albert Thomas, Limoges Cede Tél Fa laco@unilimfr

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3 SHARPER BOUNDS FOR ψ, θ, π, p k PIERRE DUSART Abstract In default of a proof of the Riemann hypothesis, the best estimates for ψ and θ and hence of π, p k and other functions of the primes, depend on the current state of knowledge of the zeros of ζs With a better knowledge about the zeros of the Riemann ζ function, we can show sharper bounds for ψ,θ,π and primes p k Introduction In many applications it is useful to have eplicit error bounds in the prime number theorem Rosser and Schoenfeld developed an analytic method which combines a numerical verification of the Riemann hypothesis with a zero-free region The aim of this paper is to find sharper bounds for the Chebyshev s functions ψ, the logarithm of the least common multiple of all integers not eceeding, and θ, the product of all primes not eceeding : θ = ln p, p ψ = p,ν p ν where sum runs over primes p and respectively over powers of primes p ν The Prime Number Theorem could be written as follows: ψ = + o, + An equivalent formulation of the above theorem should be: for all ε>0, there eists 0 = 0 ε such that ψ <ε for 0 or θ <ε for 0 This article hangs up on some known results: the most important works on effective results have been shown by Rosser & Schoenfeld [3, 4, 6], Robin [9] & Massias [6] and Costa Pereira [4] The proofs for estimates of ψ in [4] are based on the verification of Riemann hypothesis to a given height and an eplicit zero-free region for ζs whose form is essentially that the classical one of De la vallée Poussin Rosser & Schoenfeld have shown that the first zeros of ζs are on the critical strip Van de Lune et al [7] have shown that the first zeros are ln p Date: June, Mathematics Subject Classification Primary A25, A40; Secondary L20, -04 Key words and phrases Estimates of elementary functions

4 2 PIERRE DUSART on the critical strip We will improve bounds for ψ and θ We will prove the following results: ψ ln θ ln θ 02 ln 2 θ 55 ln 3 θ ln 4 for ep22, for 0 544, for , for >, for > We apply these results on p k, the k th prime, and θp k Let s denote by ln 2 for ln ln The asymptotic epansion of p k is well known; Cesaro [2] then Cipolla [3] epressed it in 902: { p k = k +ln 2 k + ln 2 k 2 ln2 2 k 6ln 2 k + 2ln 2 k + O ln2 } 3 k A more precise work about this can be find in [0, 5] We group together the results on p k and on θp k : θp k k +ln 2 k + ln 2 k for k ep22, θp k k +ln 2 k + ln 2 k 2 for k 98, p k k +ln 2 k for k 2, p k k +ln 2 k for k 3907, p k k +ln 2 k + ln 2 k 8 for k 27076, p k k +ln 2 k + ln 2 k 225 for k 2 The formula 2 has been proved by Robin in [9] We use the above results to prove that, for 3275, the interval [, + /2 ln 2 ] contains at least one prime Let s denote by π the number of primes not greater than We show that + ln ln 599 π > ln ln

5 SHARPER BOUNDS FOR ψ, θ, π, p k 3 More precise results on π are also shown: π for 5393, ln π for 6084, ln π + ln ln + 8 ln 2 for 32299, π + ln ln + 25 ln 2 for We gave in 7 some others inequalities 2 Results of Rosser & Schoenfeld We rewrite here some results of [4] which we use We want to bound ψ for moderate values of Remember the used notations: we want to find ε such that we have, for epb ψ <ε Let R = , X = ln/r, K ν z, = t ν ep{ /2zt +/t}dt for z>0, 0, real ν, 2 φ m y =y m ep X 2 / lny/7, R m δ = + δ m+ + m, D = , T = /m 2Rm δ δ 2+mδ Let NT the number of zeros such that 0 <γ T and its approimation F T = T 2π ln T 2π T 2π and let RT =037 ln T ln 2 T +588 Let A such all zeros ρ = β +iγ of the Riemann ζ function in the critical strip are such that the real part β =/2 for 0 <γ A The number A, used by Rosser & Schoenfeld, is defined as the unique solution of the equation F A =3, 502, 500, which yields A =, 894, and F A =NA We use a new value of A, defined as the unique solution of the equation F A =, 500, 000, 00 According to [7], this yields A = 545, 439, and F A =NA Proposition [4], Theorem 4 Let T D, m be a positive integer and { Ω = 2+mδ ln T 4π 2π + } m m 2 282m m +T Ω 2 = 05955R { } mδz 7 2m 2 7 m zk 2 z,a +2mln K z,a 2π +R m δ{2ry φ m Y RAφ m A},

6 4 PIERRE DUSART where z =2X m =2 mb/r, A =2m/z lna/7 and Y = ma{a, 7 ep b/m +R} If b>/2 0<δ< ep b/m then, for all epb, ψ <ε where ε =Ω ep b/2+ω 2 δ m + mδ/2 + ep bln2π As we use this theorem, we write eplicitly the proof which doesn t appear in Rosser & Schoenfeld s article; They gave only indications Proof: In all this proof, the numbering refers to [4] and some non-defined notations too As T 2 =0,S 3 m, δ =0 S 4 m, δ =2 R m δ ep X 2 lnγ/7 δ m < R mδ ρρ + ρ + m δ m φγ γ>a β>/2 γ>0 γ>a by writing φy :=φ m y By using corollary of lemma 7, φy :=φ m y φγ { y 2π + qy} φyln dy + E 0 2π 037 ln y+0443 y ln y ln y where qy = and E 2π 0 =2RY φy RAφA because NA = F A with Y = maa, W where W satisfies W yφ y 0 for y>a Let s compute W φ y =φ my =y m 2 ep X 2 X 2 / lny/7 m ++ Hence φ my 0 for y 7 ep + X 2 m+ A According to 36, ln 2 y/7 y z 2 φ m yln dy = A 2π 2m 2 +7 m K 2z,A + z ln7/2π m7 m K z,a where z =2X m and A =2m/z lna/7 Moreover, qy qa So for epb, S 4 m, δ < R mδ δ m 2π z 2 2m 2 +7 m K 2z,A + z ln 7 2π m7 m K z,a +E 0 mb where z =2 R, A =2m/zlnA/7 We deduce that S 3 m, δ+s 4 m, δ Ω 2 δ m By using the upper bound of S m, δ+s 2 m, δ given in this previous lemma, we conclude by lemma 8 that ψ ln2π /2 ln /2 Ω / +Ω 2 δ m + mδ/2 thus ψ Ω ep b/2+ω 2 δ m + mδ/2 + ep bln2π

7 SHARPER BOUNDS FOR ψ, θ, π, p k 5 3 Results on ψ and θ We group here together some known results which we will use in our proofs Proposition 2 i - θ ψ for all ii - ψ θ < for iii - ψ θ < 43 for all >0 iv - θ <for 0 < 0 v - θ < for >0 vi - θ < ln for >04 07 Proof: The first inequality can be easily deduced from ψ = θ /k k= The second is given in [4] The net one is theorem 3 of [3] The three last ones are taken from [6] p 360 For large values of, we use the following theorem theorem p 342 of [6]: ln Theorem Let R = and X = R ψ <ε 0 for 7, θ <ε 0 for 0 where ε 0 = 8/7πX /2 ep X Theorem 2 Let b be a positive real For epb, ψ ε, where ε is given in table Corollary For ep50, we have ψ Proof: We make no mathematical change of the Rosser & Schoenfeld s method We only change the value A by a bigger one Apply proposition with A = 545, 439, as defined in section 2 Evaluation of the quantities of proposition, for eample with the computer algebra Maple, gives new bounds given in table If we take A =, 894, , this implementation permits to recover the results found by Rosser and Schoenfeld In particular, we compute the value of ε for e 50 Choose δ = and m = 8 By computation, we find a value for ε lightly less than This value has been verified with Pari We need afterwards to have bounds for θ to0 The first method is to compute θ until 0 we need a powerful computer The second method uses the remark that the table of [], which gives the upper and the lower bounds of Li π by intervals up to 0, permits to find precise bounds for θ upto 0 The results are in table 2 We construct this table in the following manner: Let ρ a, b = min pk [a,b]lip k k and R a, b = ma pk [a,b]lip k k Write We thank the Médicis group for the use of their computers

8 6 PIERRE DUSART ra, b = min [a,b] Li π and Ra, b = ma [a,b] Li π Note that ra, b minρ a, b /2, Lia πa and Ra, b mar a +,b++ /2, Lib πb The interval [0, 0 ] is split into intervals [a i,b i ] πy θ = π ln y dy 7 = π ln πy y dy Li π ln + C + where C = Li00 ln πy Liy + Liy dy y y bi ra i,b i ln + rb, ln b [a i,b i ] 00 a i,b i b πy y dy and b is the upper bound of the last interval which doesn t contain If we don t know the value of π, we can bound the difference Li π byra, We have an upper bound of θ by replacing in formula 7 all r by R and conversely R by r a i Theorem 3 ψ ln for ep22 2 θ ln for Proof: We use table by intervals: if ep22 then ψ ln ln for ep23 If ep23 then ψ ln for ep30 If ep30 then ψ ln for ep600 If ep600 then ψ ln for ep2000 Moreover, ψ θ < for thus ψ θ < ln for ep22 ep30 and ψ θ < 43 for >0thus ψ θ < ln for ep30 For ep2000, we use theorem to conclude that ψ and θ for ep2000 ln With Table 2 and a computer verification, we etend the result on θ, showed until now for ep22, until This shows that the result of proposition 2vi of Rosser & Schoenfeld is almost best possible The above result gives a formula with order for power of logarithm We need sometimes the net orders We obtain better results than Rosser & Schoenfeld ones Theorem 4 Let η 2 =3965, η 3 = 55 and η 4 = For >, θ <η k We can choose η 2 =02 for Proof: In all cases, we use table by intervals Starting with k = 2, we show that θ < 02 ln 2

9 SHARPER BOUNDS FOR ψ, θ, π, p k 7 For ep3220, according to theorem we have θ <ε <η 2 ln 2 with η 2 =09923 for ep3220 For , θ ln 094 for e 25 ln 2 For e 25, ψ for e 30 ln 2 For e 30, ψ for e 00 ln 2 For e 00, ψ for e 300 ln 2 For e 300, ψ for e 800 ln 2 For e 800, ψ for e 2000 ln 2 For e 2000, ψ for e 2300 ln 2 For e 2300, ψ for e 2500 ln 2 For e 2500, ψ for e 2750 ln 2 For e 2750, ψ for e 3220 ln 2 Moreover, for ep25 ep30, ψ θ < < ln 2, and for ep30, ψ θ < 43 < ln 2 Now we are interested in case k =3 As θ ln, we have θ 3056 ln 3 for ep200 Note that for ep200, the difference between θ and ψ is negligible since, for ep200, ψ θ < ln 3 For e 200, θ for e 800 ln 3 For e 800, θ for e 2300 ln 3 For e 2300, θ for e 2500 ln 3 For e 2500, θ for e 2750 ln 3 For e 2750, θ for e 3000 ln 3 For e 3000, θ for e 334 ln 3 Now apply theorem for ep334; we find θ ln 3 We use the same methods for case k =4 As θ ln, we have θ ln 3 for ep600 Note that, for ep600, the difference between θ and ψ is negligible seeing that, for ep600, ψ θ < ln 4 For e 600, θ for e 2000 ln 4 For e 2000, θ for e 2500 ln 4 For e 2500, θ for e 2750 ln 4 For e 2750, θ for e 3000 ln 4 For e 3000, θ for e 3342 ln 4

10 8 PIERRE DUSART Now apply theorem for ep3342: we find θ ln 4 We verify by computer for, We find that θ 02 ln 2 for and θ for > the value ln 2 is choosen for p 7 4 Results on p k and θp k For some steps of demonstrations, we use the following results: Lemma p k k for k 2, p k k +ln 2 k for k 6, p k k ln p k for k 4, p k kln p k 2 for k 5 Proof: These inequalities have been proved by Rosser The first inequality can be found in []; the net one in [2] The last ones can be easily deduced from ln <π < ln 2 see [2] Rosser showed that p k k and improved his result with Schoenfeld by showing that p k k +ln 2 k 3/2 see [4] En 983, Robin [9] succeeded to prove that p k k +ln 2 k Massias & Robin [6] are able to show that p k k +ln 2 k for k ep598 and k ep800 We have show that for all k 2 in [5] Theorem 5 For k 2, we have p k k +ln 2 k where p k is k th prime number First method to find a lower bound for θp k Proposition 3 Let f β k :=k +ln 2 k + ln 2 k β Assume that for k k 0 Let β be a real satisfying θp k p k ck β 0 0 ln 2 k 0 ++ ln 2 k 0 ln 2 k 0 ln + ln 2 k 0 + ln 2 k 0 β c 0 0 ln 2 k 0 If θp k0 f β k 0 then θp k f β k for k k 0

11 Proof: Hence Moreover Assume that SHARPER BOUNDS FOR ψ, θ, π, p k 9 p k h a k :=k θp k θp k0 = +ln 2 k + ln 2 k a k n=k 0+ ln p n f βk =lnk +ln 2 k + ln 2 k β + k n=k 0+ ln h a n ln 2 k β + ln 2 k and ln p k ln h a =lnk +ln 2 k +ln + ln 2 k + ln 2 k a ln 2 k It s sufficient to show when f β ln h a because f β k f β k 0 = equivalently ln 2 k + β k k 0 f βd ln 2 k β + ln 2 k β ln 2 k + ln 2 k ln k n=k 0+ ln h a n θp k θp k0, ln + ln 2 k + ln 2 k a ln 2 k + ln 2 k + ln 2 k a ln 2 k + ln 2 k Since p k θp k ck, we can choose a = β + c Then we find that β ln 2 k + ln 2 kln + ln 2 k + ln 2 k β c ln 2 k + ln 2 k We apply proposition 3 For c =0007, the computed values of β are: 0 β Corollary 2 For p k 0, θp k k +ln 2 k + ln 2 k Proof: We take k 0 = and c = We obtain β =20548 Using table 2, we check θ0 f k 0 The major drawback of the above method is that we must check if θp k0 f β k 0 which needs a computation of θ for large The net method doesn t need this hypothesis

12 0 PIERRE DUSART Second method to find a lower bound for θp k Proposition 4 Let k 0 epep3 Assume that θ c for p k0 ln Then θp k k +ln 2 k + ln 2 k β for k k 0 with β solution of the following equality β = 0 + α ln 2 k α = ln 2 k 0 a 2 + a 2 0 a = ln 2 k 0 β + c 0 Proof: Applying lemma of [9] for a = and 0 epa + 2 in the same way than theorem 7 of [9], we find a first value of β written β 0 Since θp k p k ck, we can apply lemma one more time with the new value of a: we choose a = ln 2 k 0 β 0 + c 0 In fact the series {β k } k is a convergent one We determine the limit by solving the equation of the fied point for k = k ln 2 k + ln 2 k β c 2 ln2 k β + + c ln 2 k + β = When k 0 grows to infinity, β decreases to 2 + c Applying the previous proposition for c = 0007, the computed values of β are: 0 β ep Corollary 3 For k ep30, θp k k +ln 2 k + ln 2 k 205 Theorem 6 p k k +ln 2 k + ln 2 k 225 for k 2 Proof: Since θ <for 0 <<0, we deduce immediately p k θp k k +ln 2 k + ln 2 k 2454 For p k ep4000, it follows from theorem that θ <ε <η 2 ln 2

13 SHARPER BOUNDS FOR ψ, θ, π, p k with η 2 = for ep4000 Thus, for p k ep4000, p k θp k η 2 ln 2 k p k For 0 p k ep4000, p k θp k εθp k +ε k p k +ln 2 k + ln 2 k β ε +ln 2 k + ln 2 k ln 2 k + ln 2 k β with β =20548 Let s study the map fε, k := β ε +ln 2 k + ln 2 k β We choose at the first time ε = ε0 = We verify that fε, ep30 > 239 and so on, by small intervals up to ep4000 We deduce that fε, k > 225 for p k between ep22 and ep4000 Theorem 7 p k k +ln 2 k + ln 2 k 8 for k Proof: Assume first that see lemma p k ln 2 k p k For , we have establish that θ <ε <η 2 ln 2 with η 2 =02 Thus, for p k , p k p k θp k +η 2 ln 2 k +ln 2 k + ln 2 k 2+02 p k We terminate the proof by a computer check Theorem 8 For k 3907, we have p k k +ln 2 k Proof: As see theorem 7 p k k +ln 2 k + ln 2 k 8, we have, for ep33, p k k +ln 2 k From θp k p k εp k p k, we deduce that p k θp k +εp k p k As θp k k +ln 2 k + ln 2 k 2,

14 2 PIERRE DUSART it remains to show that gk := + ln 2 k 2 for k ep30 If we choose cf [6] + εp k p k k < p k k +ln 2 k and εp k = / 0 for 0 p k ep30 and net εp k = / ep 30 for p k ep30, we obtain the result for p k 0 Now we study if the result is true for p k 0 Write α =09484 and fk =k +ln 2 k α Thanks to Brent [], we have the lower and the upper bounds of the difference of Li π into various intervals up to 0 Hence if, for k [k 0,k ], Lifk πp k M pk0,p k := ma k [k 0,k ] Lip k πp k we could deduce that We consider gk = Lifk k p k fk g k = +/ α ln + ln 2 k α/ +ln 2 k + ln + ln 2 k α/ admits a minimum for ep2 + α +2/ ep2 + α For α =09484, this lower bound is positive We show, thanks to the following table, that Lifk πp k M pk0,p k for k p k 0 The values of π are etracted from table 3 of Riesel [8] E: π0 7 = k p k Lifk k R = M pk0,p k M = M = M 0 7,2 0 7= M 2 0 7,5 0 7= M 5 0 7,5 0 8= M 5 0 7,0 0= M 0 0,0 =7065 For k = , a check has been made thanks to Pari system 5 Interval which contains at least one prime We already know the result of Schoenfeld [6] showing that, for , the interval ], + /6597[ contains at least one prime We improve this result with the following proposition You can see also [7] Proposition 5 For k 463, p k+ p k + 2ln 2 p k

15 SHARPER BOUNDS FOR ψ, θ, π, p k 3 Proof: Assume that, for k k 0, p k k +ln 2 k + ln 2 k α 0 and p k + γ But, k = k p k k +ln 2 k + ln 2 k α ln 2 p k+ = p k p k+ + γ p k p k ln 2 p k +ln 2 k + ln 2 k α 0 k + + γ p k ln 2 p k lnk ++ln 2 k + + ln 2k + α lnk + lnk ++ln 2 k ln 2 k ++ ln 2 k α 0 lnk ++ln 2 k + + ln 2k + α lnk + γ ln 2 k p k lnk lnk += ln + /k, + ln 2 k ln 2 k += ln ln 2 k α 0 ln 2k + α lnk + ln + /k ln 2k + α lnk + + γ p k ln 2 p k, α α 0 / lnk + because ln ln2 is decreasing for /e hence fk fk + where f := We choose γ such that p k ln + /k ln + /k+ln + +α 0 α / lnk lnk ++ln 2 k + + ln 2k + α lnk + ; ln We must choose γ>α 0 α to make the inequality true for large k As p k k ln 2 p k +ln 2 k 2, the inequality 8 is satisfied with γ =/2 if lnk +/2 α 0 α +ln 2 k For α 0 =225 and α =8, this holds for k ep82 Theorem 2 of [6] showed that the interval ], + /6597[ contains at least one prime for Apply it for = p k, this yields the result for p k ep9 According to [6] p 355, p n+ p n 652 for p n , and so p k+ p k + 2ln 2 p k

16 4 PIERRE DUSART when p k k 2ln p k 2 +ln 2 k that is to say for k For k = , a direct check with computer conclude the proof Theorem 9 For all 3275, there eists a prime p such that <p + 2ln 2 This result is better that Rosser & Schoenfeld s one for Proof: Let 2 There eists k N such that p k <p k+ As the map +/2 ln 2 is increasing, p k p k + 2ln 2 p k According to proposition 5, for k 463, p k+ p k + 2ln 2 p k We deduce that, for p 463 = 3299, Moreover, for 32740, <p k < + 2ln 2 + 2ln 2 + 2ln 2 6 Results on π Remember that π = + ln ln + 2 ln 2 + O ln 3 Theorem 0 - ln + ln π for π ln ln for > the value 2762 is chosen for = p258 = π ln ln for π ln for π ln for ln + ln + 8 ln π for π ln + ln + 25 ln for Let s start with the first three inequalities

17 SHARPER BOUNDS FOR ψ, θ, π, p k 5 Lower bound of π We began with 0 8 π =πp θp lnp + θ ln + θydy p y ln 2 y But θ 0024 this yields ln for 7587 [6], Th7 p357 Hence, for p 7587, π θ ln + ln 2 + πp θp lnp p ln 2 p Since θ ln for 04 07, because by choosing π πp θp lnp ln ln p ln 2 p dy p ln 3 y for 0 8 dy p ln 3 y 0 for 08, p = p 6000 = and θp <p According to [3]Th 6, p 72, we have for < 0 8, Li Li <π For , Li Li ln ln For k =284 p 284 = 86, we check by computer that p k πp k /2 = k ln p k , ln p k inequality true for k 0 Hence π ln ln for p09 = 599 We will obtain a better result with an upper bound for θ in order 2 in ln Upper bound for π According to [6] p 360, we know that : θ < for 0 < 0, θ < for > 0, and that θ ln for

18 6 PIERRE DUSART Let b = , c =00008 and K =0 π = θ ln + θydy 2 y ln 2 y = θ K ln + θydy 2 y ln 2 y + θydy K y ln 2 y < + b K dy + ln ln 2 ln 2 y + dy K ln 2 y + b dy K ln 3 for K y = + b [ ] y dy + ln ln ln 2 +2 y 2 2 ln 3 y + b dy K ln 3 y < + b + +2 [ ] ln ln ln 3 +6 dy y 2 ln 4 y + b dy ln 3 +3b y K ln 4 y But K dy ln 4 y = K dy ln 4 y + dy ln 4 y < K/ ln 4 K+ ln 4 if K To have π + β ln ln we choose β>b++ 2+b ln + ln2 K dy 6 2 ln 4 b K K ln 3 K +6+3b ln 4 K + ln 4 We obtain β 03 when ep00 π = πk θk ln K + θ ln + θydy K y ln 2 y dy K ln 2 y = πk θk ln K + c K + c Li LiK ln K = M + c Li where M is a constant < πk θk ln K + c ln + c Write = + β M + c Li ln ln = cln 2 +β ln 2β / ln 3 equals zero when ln = β ± β 2 +8β c 2 c We can take πk = and θk ln K > K ln K 0007 ln K hence, for β =0992, the local minimum 0 ep <K is negative but K is positive Hence, for [K, ep00], π < ln ln K

19 SHARPER BOUNDS FOR ψ, θ, π, p k 7 By consulting table given in [], we show that the result remains true for According to [3] Th 6 p 72 and [], we have for K, π < Li Let s study the difference, β = + β Li ln ln This function admits a minimum at point = ep2β/β For β 0 := , the value of this minimum equals to We consult the table of Brent [] This yields the lower difference between π and Li in different intervals [0 n, 0 n+ ] with n =00 We deduce that + β 0 π = + β 0 ln ln ln ln for 0 4 Fork =230, we verify that p k Li + Li π > 0 + β 0 ln p k πp k =k ln p k We choose the value for β 0 such that it holds for all integers k, including k = 258 Net we consider the others formulas Let 0 =04 0 7, K = π 0 θ 0 ln 0 Numerically π 0 = and θ 0 = Write J; a =K + ln + a ln ln 2 y + a ln 4 dy y Since π =π 0 θ 0 + θ ln 0 ln + θydy 0 y ln 2 y and θ 02 ln 2 for > 0,wehave,for 0, J; 02 π J;02 Write M; c = ln + ln + c ln for upper bound s function for π Let s 2 write the derivatives of J; a and of M; c with respect to : J ; a = ln + M ; c = a ln 3 2 a ln 4, ln + c 2 ln 3 3c ln 4 We must choose c>2+a to have J <M when ln >3c 2a/c 2 a Choose c =25 We verify by computer that J0 ;02 <M0 ;25 Since Li <M;25 for 0 7, we verify by direct computation for small values of to obtain π < + ln ln + 25 ln 2 for Now write m; d = + ln ln + d ln 2 We study the derivatives: we must choose d 2 a to have J >m

20 8 PIERRE DUSART Choose d =8 As m 0 ;8 <J 0 ; 02 and by direct computation for small values, we obtain π > + ln ln + 8 ln 2 for To show that π, ln b we proceed in the same way By comparing with the derivative of J; a, we show that we must choose b For b = and a = 02, J ; 02 is greater that the derivative of /ln for > Moreover, for = 0, 0 ln 0 <J 0; 02 We verify by computer that, for , π ln Now we show that we have, for 6084, π ln Since + a < ln ln ln a for >epa, an above result shows the result for Now we study if π ln for 0 Write k = ln and compare k with Li The derivative of k / ln is 200 ln ln ln 3 ln 2 So we deduce that the difference k / ln is positive for belonging to [0 6, 0 ] Since Li0 6 <k0 6, we deduce that, for 0 6 0, Li <k A computer check concludes the demonstration up to 0 6 We show that + π for 599 ln ln by using inequality 6 of theorem 0 Net we verify by computer the lower bound is true from = Applications 7 Other inequalities Let γ be Euler s constant γ Theorem Let B = γ + p ln /p+/p For >, p ln 2 B 0 ln ln 3 p For 0372, p ln 2 B 0 ln ln 3 p

21 SHARPER BOUNDS FOR ψ, θ, π, p k 9 Proof: By 420 of [3], p =ln 2 + B + θ ln p p θy y+lny y 2 ln 2 dy y Hence p ln θ θy y +lny 2 B + ln p y 2 ln 2 dy y As θ 02/ ln 2 Theorem 4 and +lny y ln 4 y dy = 2ln 2 + 3ln 3, we have the result for We conclude by computer s check Theorem 2 Let E = γ n=2 p ln p/pn For >, ln p 02 ln E p ln + 02 ln 2 For 2974, p p ln p p ln E 02 ln + 02 ln 2 Proof: By 42 of [3], ln p θ =ln + E + p θy y y 2 dy Hence ln p θ θy y ln E p p y 2 dy As dy y ln 2 y = ln, theorem 4 yields the result for We conclude by computer s check Theorem 3 For >, and and For 2973, p p p p < e γ + 02 p ln ln 2 ln p p >eγ ln 02 ln 2 > e γ 02 p ln ln 2 ln p p <eγ ln + 02 ln 2

22 20 PIERRE DUSART Proof: By theorem and by definition of B, wehave 0 ln ln 3 γ ln 2 p ln /p 0 ln ln 3 p> p Let S = p> ln /p+/p = n=2 p> p We have n γ ln 2 p Take the eponential of both sides to obtain p p ln /p S 0 ln ln 3 e γ 0 ep S + ln ln ln 3 We use lower bound for S given in [3] p 87: S> 02 ln Hence, for 00, e γ p ln ep0/ ln2 We have also In the same way, as p p p p e γ ln ep 0/ ln 2 γ ln 2 ln /p S 0 ln ln 3, p we obtain the others inequalities since S 0 Theorem 4 For >, π2 π ln For 3285, π2 π ln 07 ln 2 Remark: As π >/ln for 7 corollary of [3], we have π2 < 2π for real 3 Proof: By theorem 0, we have for 6084, 2 ln 2 2 <π2 π < ln ln 2 ln Let f = 2 ln 2 ln ln As+u< u < +u +2u2 for u /2, f 2 ln 2 ln ln In a same way, if we write g = 2 ln 2 ln ln,

23 SHARPER BOUNDS FOR ψ, θ, π, p k 2 we have g ln 22 2 ln 2 ln 2 4 ln 72 Mandl s conjecture We want to show the conjecture of Robert Mandl written in the article of Rosser & Schoenfeld [4] p 243 Theorem 5 Let Then, 9 0 Proof: Write S n = n p k k= S n n2 2 ln n +ln 2 n 3/2 for n , S n n2 2 ln n +ln 2 n 463 for n 779 sn = n2 2 ln n +ln 2 n 3/2 fn = sn n ln n Hence f n = n ln n +ln 2 n + ln n 2lnn f n = lnn +ln 2 n + 3 2lnn ln 2 n n We take the Taylor series epansion of f between n and n +: fn + fn =f n+ f n with n<n <n+ 2 Since f increases,we obtain fn + fn f n+ f n + f n+lnn 2 By theorem 6, p n n ln n +ln 2 n + ln ln n 225, ln n this yields n ep ep275, p n fn + fn For n 0 =0 6, S n0 fn 0 Suppose that S n fn upton Then S n = S n + p n fn+p n fn + Thus, for all n n 0, S n fn which implies S n sn since p n n ln n We verify that S n sn for n

24 22 PIERRE DUSART by a computer check Inequality 0 can be find in [6] We have n p k = n2 n 2 k= ln n +ln 2 n 32 + o 2 p n = n2 2 ln n +ln 2 n +o Theorem 6 Mandl For n 9, we have p + p p n /n < 2 p n Proof: By 0, we have for n 779, S n n n 2 ln n +ln 2 n 463 Now, theorem 5 yields the result G Robin gave me a conjecture concerning a lower bound of the above quantity: Proposition 6 For n 2, we have We have n n k= p k = n2 2 p [ n 2 ] n n p i i= ln n +ln 2 n 32 + o np [n/2] = n2 2 ln n +ln 2 n ln 2 + o Proof: According to theorem 7, we have for k 27076, So, for n 5452, np n/2 n2 2 p k k +ln 2 k + ln 2 k 8 ln n +ln 2 n ln 2 + ln 2 n 8 S n ln n ln 2 We make a direct computer check to show that np n/2 S n pour k Proposition 7 For a 2 and b 2, or for b =and a 5, we have p ab >ap b For a and b positive integers, we have ap b+ + bp a+ <p ab+2 2 If a and b are integers greater than 9, we have p ab ap b + bp a

25 SHARPER BOUNDS FOR ψ, θ, π, p k 23 b m δ ε / / Table ψ ε for epb For a and b positive integers, we have p ab+2 ap b+ + bp a+ Proof: Use the inequalities 3 and 4 References [] R P Brent, Irregularities in the Distribution of Primes and Twin Primes, Math Of Computation, Vol 29, Number 29 January 975 pp UMT, Math Of Computation, Vol 30, Number 34 April 976 p 379 [2] E Cesaro, Sur une formule empirique de M Pervouchine, Comptes rendus hebdo des séances de l académie des sciences, Vol CXIX, 895 pp [3] M Cipolla, La determinazione assintotica dell n imo numero primo, Matematiche Napoli, Vol 3, 902 pp [4] N Costa Pereira, Estimates for the Chebyshev Function ψ θ, Math Of Computation, Vol 44, Number 69 January 985 pp 2-22 [5] P Dusart, The k th prime is greater than k +lnlnk for k 2, Math Of Computation, to appear [6] Jean-Pierre Massias & Guy Robin, Bornes effectives pour certaines fonctions concernant les nombres premiers, Publication Département Math de Limoges and Journal Th Nombres de Bordeau, Vol pp [7] O Ramaré, Short effective intervals containing primes, Journal Number Th, to appear [8] Hans Riesel, Prime Numbers and Computer Methods for Factorization, Birkhäuser 985 [9] Guy Robin, Estimation de la fonction de Tchebychef θ sur le k ième nombre premier et grandes valeurs de la fonctions ωn, nombre de diviseurs premiers de n, Acta Arithmetica, vol 42, numéro pp

26 24 PIERRE DUSART θ θ Table 2 Bounds for θ [0] Guy Robin, Permanence de relations de récurrence dans certains développements asymptotiques, Publications de l institut mathématique, tome 43 57, 988 pp 7-25 [] J Barkley Rosser, the n-th prime is greater than n log n, Proc London Math Soc 2, Vol 45, 939 pp 2-44 [2] J Barkley Rosser, Eplicit Bounds for some functions of prime numbers, Amer J Math, Vol 63, 94 pp [3] J Barkley Rosser & L Schoenfeld, Approimate Formulas for Some Functions of Prime Numbers, Illinois Journal Math pp [4] J Barkley Rosser & L Schoenfeld, Sharper Bounds for the Chebyshev Functions θ and ψ, Math Of Computation, Vol 29, Number 29 January 975 pp [5] Bruno Salvy, Fast computation of some asymptotic functional inverses, J Symbolic Computation, Vol 7, 994 pp [6] Lowell Schoenfeld, Sharper Bounds for the Chebyshev Functions θ and ψii, Math Of Computation, Vol 30, Number 34 April 976 pp [7] J van de Lune, H J J te Riele & DT Winter, On the Zeros of the Riemann Zeta Function in the Critical StripIV Math Of Computation, Vol 46, Number 74 April 986 pp

27 SHARPER BOUNDS FOR ψ, θ, π, p k 25 Département de Math, LACO, UPRES A 6090, 23 avenue Albert THOMAS, LIMOGES cede address: dusart@unilimfr