Estimates of li(θ(x)) π(x) and the Riemann hypothesis

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1 Esimaes of liθ π he Riemann hypohesis Jean-Louis Nicolas 20 mai 206 Absrac To Krishna Alladi for his siieh birhday Le us denoe by π he number of primes, by li he logarihmic inegral of, by θ = p log p he Chebichev funcion le us se A = liθ π Revisiing a resul of Ramanujan, we prove ha he asserion A > 0 for is equivalen o he Riemann hypohesis Keywords : Chebichev funcion, Riemann hypohesis, Eplici formula 200 Mahemaics Subjec Classificaion : N37, M26, N56 Inroducion Le us denoe by π he number of primes by li he logarihmic inegral of see, below, 22 I has been observed ha, for small, π < li holds, bu Lilewood cf [7] or [5, chap 5] has proved ha, for ending o infiniy, he difference π li oscillaes infiniely many ofen beween posiive negaive values Le us se θ = p log p, he Chebichev funcion, A = liθ π Wha is he behavior of A? In [0, 220, 222, ], under he Riemann hypohesis, Ramanujan proved ha 2 A = 2 + / 2 log 2 + O log 3 where runs over he non-rivial zeros of he Riemann ζ funcion Moreover, in [0, 226], Ramanujan wries under he Riemann hypohesis 2 2 = = + = 2 3 = 2 + γ 0 log4π = 0046 where γ 0 is he Euler consan concludes 4 under he Riemann hypohesis 0 such ha, for 0, A is posiive The aim of his paper is o make hese resuls effecive, in paricular, o show ha Ramanujan s resul 4 is rue for 0 = Research parially suppored by CNRS, Insiu Camille Jordan, UMR 5208

2 Le us se λ = 2 Under he Riemann hypohesis, we have see below λ = We shall prove 2 = = Theorem Under he Riemann hypohesis, we have 6 lim sup A log λ = 2046, 7 lim inf A log 2 2 λ = 953, 8 A is posiive for, 9 A 2 λ log 2 for 37, 0 A M log 2 for 2, where M = A3643log / 3643 = Corollary Each of he five asserions 6 0 is equivalen o he Riemann hypohesis Proof : In 984, Robin cf [9, Lemma 2 8] has shown ha, if he Riemann hypohesis does no hold, here eiss b > /2 such ha A = Ω ± b, ie lim sup A b he five asserions of he heorem are no longer saisfied > 0 lim inf A b < 0 Noaion π = p is he prime couning funcion Π = p k k = κ k= π /k k wih κ = log log 2 θ = log p ψ = κ log p = θ /k p p m k= { log p if = p k Λ = is he von Mangold funcion 0 if no ψ = ψ 2 Λ Π = Π Λ 2 log li denoes he logarihmic inegral of cf below 22 are he Chebichev funcions 2

3 L = li log, L 2 = li log log 2, F = L / log 2, F 2 = L 2 / log 3 > F F 2 are defined below in 36 γ 0 = is he Euler consan λ is defined in 5, cf also 226 f = lim f where f : C C is a comple funcion runs over he non-rivial T I T zeros of he Riemann ζ funcion 2 Plan of he aricle In 2, we shall recall some definiions prove some resuls ha we shall use in he sequel, firs, in 22, abou he logarihmic inegral,, furher, in 23, abou he Riemann ζ funcion eplici formulas of he heory of numbers In 3, he proof of Theorem is given Firs, we wrie A = A + A 2 wih A = liψ Π A 2 = liθ liψ + Π π In 3, under he Riemann hypohesis, an esimae of A is given, by applying he eplici formulas In 32, i is shown ha A 2 depends on he quaniy By = πy θy/ log y which is carefully sudied In 33 resp 34, an effecive lower resp upper esimae for A is given when 0 8 In 35, for < 0 8, esimaes of A are given by numerical compuaion Finally, Theorem is proved in wo seps, depending on he cases 0 8 or > 0 8 The compuaions, boh algebraic numerical, have been carried ou wih Maple On he websie [2], one can find he code a Maple shee wih he resuls We ofen implicily use he following resul : for u v posiive, he funcion logu v is increasing for e u/v decreasing for > e u/v Moreover 2 ma 2 Preliminary resuls 2 Effecive esimaes log u u u v = e v Wihou any hypohesis, Pla Trudgian [8] have shown by compuaion ha 2 θ < for 0 < so improving on resuls of Schoenfeld [] Dusar [3] Under he Riemann hypohesis, for 599, we shall use he upper bounds cf [, 63] 22 ψ log 2 θ log 2 8π 8π 3

4 22 The logarihmic inegral For real >, we define li as cf [, p 228] li = d log = lim ε 0 + We have he following values : 0 ε 0 + +ε d = log 2 d log + li li From he definiion of li, i follows ha 24 d d li = log d 2 d 2 li = log 2 We also have li = γ 0 + loglog + where γ 0 = 0577 is he Euler consan which implies n= log n n n! 25 li = loglog + γ 0 + o, + Le N be a posiive ineger For >, we have cf [2] 26, for, 27 li = N d log N = li k! N! log k N k= k! log k k= + O log N+ Lemma 2 For >, we have 28 L 2 = li log log 2 = F 2 log 3 < 405 log 3 For 0 38, we have 29 L 2 < F 2 0 For > 29, we have log 3 20 L 2 > 2 log 3 Proof : le us se cf he Maple shee [2] f = 3 log li + f 2 = log + 2 log log 2 = 2 F 2 log 2, log log 3 li = f 4

5 f 3 = f 2 = 6 log 4 Since f 2 = f 3 is negaive, f 2 decreases vanishes for 2 = I follows ha f = f 2 / is posiive for < < 2 negaive for > 2 so ha f has a maimum for = 2, f 2 = f vanishes so does F 2 in wo poins 3 = = From 25, we ge lim + F 2 = 0 he variaion of F 2 is given in he following array : F 2 = L2 / log The proof of follows from Array 2 also he proof of 20, afer deducing from f 2 2 = 0 ha F 2 2 = 2 holds In he same way, i is possible o sudy he variaion of he funcion The deails can be found on [2] We have log F = L / log 2 = li / log 2, F = L log Since L =, Array 22 yields 23 > 04 = L = li log > log 2 / log li The derivaive of li/ is = F which, from Array 22, is posiive for < < log 2 negaive for > Therefore, we have 24 > = li li = < 3 4 Lemma 22 Le a be wo real numbers saisfying ep a < a 3 Le κ κ 2 be wo inegers such ha log 2 κ < κ 2 = log a Then we have 25 κ 2 k=κ + k L /k κ3 /κ log 3 L 2a 5

6 Proof : Le us se T = κ 2 k=κ + k L /k I follows from Array 22 ha, for >, L = F / log log 2 holds herefore, T 785 log 2 κ 2 k=κ + k /k Now, as ep >, he funcion / is posiive decreasing for 0 < log so ha T 785 κ2 log 2 / d 785 log log a /κ κ log 2 / d = 785 κ a du log 3 u by he change of variable u = / Finally, by 26 28, we ge /κ T 785 L 2 /κ L 2 a log 3 /κ L 2a which ends he proof of Lemma 22 Lemma 23 Le a 2 a 3 be real numbers κ 2 = log log a Then we have 26 κ 2 k /2k 5 4 /4 Proof : Le us se T = κ 2 k /2k Since a 3 >, he funcion /2 / is posiive decreasing for > 0 so ha T = κ 2 2 /4 + k /2k log log a 2 /4 + 2 k=3 /2 by he change of variable u = /2 Finally, by 26 24, we ge T 2 /4 + li /4 li a 5 4 /4 li a d = /4 du 2 /4 + a log u 26 follows since a 2 > 452 so ha, from Array 23, li a > 0 holds Lemma 24 Under he Riemann hypohesis, for 599, one has 27 θ log 9 log liθ li θ log ψ θ log ψ log 9 log log liψ li ψ log, liψ liθ ψ θ log + 9 log

7 Proof : Le us suppose ha 599 holds From 22, we ge 220 ψ θ log 2 Furher, for h >, Taylor s formula 24 yield 8π = log2 8π log π 599 > li + h = li + h log h 2 2 ξ log 2 ξ, wih ξ min, + h Le us se h = θ ; we have h + = θ θ599 > From 220, we ge ξ b wih b = From 22, i follows ha 0 ξ log 2 ξ b log 2 b = b log 2 + log b 2 log b log 2 + log b log 2 log599 h 2 2 ξ log 2 ξ log 4 28 π 2 ξ log 2 ξ log π 2 = log2 < 9 log which, wih 22, proves 27 In he same way, seing h = ψ yields 28, 29 follows by subsracing 27 from The Riemann ζ funcion We shall use he wo eplici formulas 222 ψ = ψ 2 Λ = log2π 2 log 2, > 223 Π = Π Λ 2 log = li li d log log, >, which can be found in many books in analyic number heory, for insance [5, chap 4] To Formula 223, we prefer he form described in [6, p , wih R = 0] : 224 Π = Π Λ 2 log = li 0 d d log log, > 225 We also have cf [4, p 67] or [2, p 272] = + γ 0 2 log π log 2 = = + = 2 + γ 0 log4π =

8 3 Proof of Theorem 3 Sudy of A = liψ Π Under he Riemann hypohesis, we wrie γ = I ie = 2 + iγ Lemma 3 Under he Riemann hypohesis, we have γ Proof : I is possible o ge beer esimaes for he sum γ 3, bu, for our purpose, he above upper bound will be enough By observing ha 2 = = 4 + γ2 ha he firs zero of ζs is / i cf [4, p 96] or Wikipedia, we ge γ 2 = + /4γ 2 /4 + γ 2 + / /4 + γ Furher, from 226, we ge γ γ = which complees he proof of Lemma 3 Lemma 32 For >, under he Riemann hypohesis, we have 0 d = log + wih 3 K log 3 Proof : By parial inegraion, one has 2 log 2 + K so ha we ge 0 d = 0 log + 3 d I 3 2 K = log 2 3 follows from Lemma log log /2 d = I 3 3 d log 3 d 2 log 3 I 3 8

9 Proposiion 3 Under he Riemann hypohesis, for 599, we have A = liψ Π = 2 log 2 + J wih log log 3 J 300 log 3 + log 2 Proof : Le us wrie liψ = li + ψ log wih, from 28, for 599, + J = li + ψ + Λ/2 log log 2 J 0 Therefore, from , we have A = li + log = 0 + J log2π 2 log Λ + J d + d Λ 2 log 2 + log 2 log li 0 d log + J + J 2 + J 3 wih J 2 = log 2 log2π log Furher, from Lemma 32, one ges J 3 = log /2 2 log d 2 log 34 A = 2 log 2 + J wih 35 J = K + J + J 2 + J 3 K is as in Lemma 32 I remains o bound J 2 + J 3 We have which, for 599, implies 0 J 3 log J 3 = 2 log d log d 2 = log + /2 2 log 2 2 log < log2π log 0 < J 2 + J 3 < log 2 Therefore, 32 resuls from 3, 33,

10 32 Sudy of A 2 = liθ liψ + Π π For y 2, le us se By = πy θy log y = log p log y p y Noe ha By is nonnegaive, nondecreasing coninuous, since for p prime, lim y p, y<p By = πp θp log p log p = Bp In he wo following lemmas, we give esimaes of By Lemma 33 Le y be a real number saisfying y 0 = 83 y We have 36 By L y = liy y log y while, if y y = 599, under he Riemann hypohesis, we have y 37 By L y + 4π Under he Riemann hypohesis, for y y 2 = 2903, we have y 38 By L y 4π Proof : By Sieljes s inegral, one has 39 πy = Furher, we have 30 By = y 2 y d[θ] 2 log θ y0 y log 2 d = + 2 y 0 By 2 26, for y , we ge y y 0 θ log 2 d y y 0 log 2 d = liy y = θy y log y + θ 2 log 2 d θ y log 2 d = By 0 + y 0 θ log 2 d log y liy 0 + y 0 log y 0 = L y L y 0, so ha 30 yields By L y + By 0 L y 0, which proves 36, since By 0 L y 0 = < 0 cf [2] Replacing y 0 by y in 30 yields 3 By = By + wih T y, y = y y y y θ d, from 22, log 2 32 T y, y From 3 32, i follows ha θd log 2 = By L y + L y + T y, y y By L y + y log 2 8π log 2 d = y y 4π y 4π + By y L y 4π y which proves 37, since By L y 4π = < 0 In he same way han he one used o ge 3, for y y 2, we obain y By = By 2 L y 2 + L y + T y, y 2 L y 4π + By 2 L y 2 + 4π 0 y2

11 y2 as By 2 L y 2 + 4π = > 0, his complees he proof of Lemma 33 Le us se εy = { 0 if y if y > I follows from ha, under he Riemann hypohesis, one has y 33 By L y + εy for y 83 4π Proposiion 32 Under he Riemann hypohesis, for 599, we have 34 A 2 = liθ liψ + Π π = κ k B/k + U wih 35 κ := log log 2 Proof : From 29, for 599, we ge U 9 log liθ liψ = θ ψ log From he definiion of ψ Π, his implies A 2 = κ π /k k + U wih U 9 log θ/k + U log which, via he definiion of B, proves 34 I is convenien o inroduce he noaion { 405 if < F2 = F 2 if > 38 F = { 785 if < 95 F if > 95 so ha, from Arrays 2 22, for >, F 2 F are nonincreasing we have 37 L 2 = F 2 log 3 F 2 log 3 L = F log 2 F log 2 Lemma 34 Le us se a = 04 For > 0 8, we se κ = log log 2, κ 2 = log log a le κ be an ineger saisfying 3 κ < κ 2 Then, under he Riemann hypohesis, we have κ B /k k 2 log F log 3 2 κ k /k + F log 2 /k κ3 /κ log 3 k= log 5 Proof : For 2 k κ 2 we have /k /κ2 log a/ log = a,, under he Riemann hypohesis, i follows from 33 ha which implies ha B /k L /k + ε /k /2k 4π κ B /k k T + T 2 + T 3 + T 4 + T 5

12 wih T = 2 L, T 2 = T 4 = κ k=κ 2+ κ k=3 L /k, T 3 = k B /k, T 5 = k κ 2 κ 2 k=κ + ε /k /2k 4kπ L /k, k From he definiion of L, L 2, F, F 2 from 37, one has T = L log 2 = 2 log log 3 F 2 2 log F log 3 2 T 2 = κ k=3 L /k k = κ k=3 k /k log 2 F /k κ k=3 k /k F log 2 /k From Array 2, L 2 04 is posiive, so ha, from Lemma 22 wih a = 04, we have T κ3 /κ log 3 L κ3 /κ log κ3 /κ log 3 For k κ 2 + > log / log a, we have /k < a ; since y By is nondecreasing, we have B /k Ba = B04 = 766 < 72 κ κ log d log 2 d T k k=κ κ 2+ 2 log log a log log/a log a = 72 log log = 72 log log 2 log a log 2 log a log log a 72 log + log 2 log/a = 72 log log 2 log a log a 72 log + log 2 log0 8 = /a log + log log/a + log a log/a Since ε is nondecreasing vanishes for 0 7, from Lemma 23, one ges T 5 = κ 2 ε /k /2k 4kπ ε = 5 6π ε log 5 κ 2 log 5 /4 < 5 6π /2k 4kπ 5 6π ε /4 log 5 log = /4 log 5, which complees he proof of Lemma A lower bound fora Proposiion 33 Under he Riemann hypohesis, for 9 0 6, we have 38 A log 2 2 λ + log 7993 log3 8 8π/ Proof : Since By is nonnegaive, from 34 35, we ge, for 599 A 2 2 B 9 log log 5 2

13 As , we may apply 38 which yields A 2 L /4 2 4π 9 log = 2 Now, as > 29 2, by 20, i follows A 2 2 log log 3 /4 9 log2 4π 0000 = log 2 log 2 + L 2 /4 9 log2 4π 0000 From Proposiion 3, one has : A 2 log log log 3 so ha A = A + A 2 saisfies A log log log2 8π 9 log4 / /300 log2 log 8π 8 log4 / which, via 5, implies 38 Corollary 3 Under he Riemann hypohesis, for 0 8, we have 39 A log 2 2 λ + 52 log Proof : From, he funcions log3 /4 we have cf [2] log5 are decreasing for 0 8 herefore, 7993 log3 8 log log π/ π 8 log = An upper bound fora Proposiion 34 Under he Riemann hypohesis, for 0 8, we have 320 A log λ + Qκ, log where κ is an ineger saisfying 3 κ < log log Qκ, = 4 F log3 + wih F 2 F defined in 36 κ k=3 k F /k log /2 /k κ3 /2 /κ log log Proof : From Proposiion 3 5, for 599, we have A λ log log while, from Proposiion 32, we have A 2 κ k B/k + 9 log Therefore, from Lemma 34, we ge he upper bound 320 for A = A + A 2 3

14 Corollary 32 Under he Riemann hypohesis, for 0 8, we have 322 A log λ log Proof : We choose κ = 5 observe ha, from 36, all he erms of he righ h side of 32 are posiive nonincreasing for 0 8 so ha Q5, Q5, 0 8 = 2529 cf [2] Corollary 33 Under he Riemann hypohesis, for ending o infiniy, we have o o 323 log 2 2 λ + A log log λ + log Proof : The lower bound of 323 follows from Proposiion 33 From Array 2, from 23 from 36, one sees in 32, ha lim F2 = 2 lim F /3 = so ha 32 yields lim Q3, = 8 + 2/300 he upper bound of 323 follows from Proposiion 34 wih κ = 3 35 Numerical compuaion Le us denoe by p p + he primes surrounding he prime p Proposiion 35 For < , A is nondecreasing There eiss infiniely many primes p for which Ap < Ap holds Proof : Le us consider a prime p saisfying 3 p < From 2, one has θp Ap Ap = liθp liθp d = + θp log > + θp θp = log p log θp log θp > 0 From Lilewood cf[7] or [5, chap 5], we know ha here eiss C > 0 a sequence of values of going o infiniy such ha θ + C log log log Le p be he larges prime For p large enough, one has θp = θ + C log log log > p + log p Ap Ap < which complees he proof of Proposiion 35 log p log θp = log p logθp log p < 0 Remark In [8, p 8], Pla Trudgian have proved he eisence of u saisfying 727 < u < 728 θe u e u > 0 52 If P is he larges prime e u, his implies AP < AP + θp = θe u > e u > P + u P + log P log P logθp log P < AP 4

15 Proposiion 36 i For we have 324 A > 0 ii Under he Riemann hypohesis, for 2 we have 325 A 2 + λ log log wih equaliy for = iii Under he Riemann hypohesis, for we have 326 A log λ log iv For we have 327 A log 2 wih equaliy for = 3643 v Under he Riemann hypohesis, for 84 we have 328 A log 2 2 λ + 52 log vi For 37 < 89 we have 329 A log 2 2 λ Proof : Firs, for 2, we define C c by A = log λ + C A = log log 2 2 λ + c log so ha C = log A log2 2 λ c = log A log2 2 + λ i 324 follows from Proposiion 35 A = 030 Noe ha A7 = 054 < 0 cf [2] ii If 0 8, 325 follows from Corollary 32 If 2 < 409, from 2, one has log 2 / 6/e 2, from Proposiion 35, A A so ha C = log A log2 2 λ log e 2 2 λ < 205 which proves 325 If 409 < 0 8, le p be he larges prime As 409 > e 6 holds, from, for [p, p +, he funcion log Ap log2 2 λ is decreasing, which implies 330 C Cp 5

16 , by compuaion, which complees he proof of 325 ma C = ma Cp = C33647 = p<08 iii For 0 8, 326 follows from Corollary 32 We compue p 0 = he larges prime < 0 8 such ha Cp For p + 0 = < 0 8, we denoe by p he larges prime, from 330, one has C Cp < 2522, which implies 326 Then, one calculaes lim p + 0, <p+ 0 C = log p + 0 Ap 0 log2 p + 0 p λ = As he above value is < 2522, we have o solve he equaion C = 2522 for p 0 < p + 0 = find iv For he funcion log 2 / is maimal for = e 4 = 5459 where is value is 6/e 2 = 26 cf 2 As A is nondecreasing, for < 59, we have For p 59 p < p +, one has A log2 6 6 A53 = 55 = 250 e2 e2 A log2 = Ap log2 Ap log2 p p we compue he maimum of Ap log2 p p for 59 p < 0000 which is equal o 5064 for p = 3643 v Le us se f = log 2 2 λ + 52 log For 0 8, A > f follows from Corollary 3 Le p be a prime saisfying e 6 < 409 p < 0 8 For p < p +, one has A = Ap, c = log Ap log2 2 + λ, c = Aplog2 6 log 22 λ < 0 2 3/2 so ha c is decreasing c cp def == lim p +, <p + c = log p+ Ap log2 p + p λ Therefore, for 409 < 0 8 one has c min 409 p<0 8 cp, by compuaion, one ges min cp = c409 = p<08 which implies A > f The funcion f is decreasing on, = 55 ] increasing for cf [2] Therefore, for < a < b, he upper bound of f on he inerval [a, b is mafa, fb We have A84 = A83 < f84 while, for 84 < 89, A = A83 > maf84, f89 f holds For 89 p 40 = 409, one checks ha Ap > mafp, fp + holds which shows ha A > f for 89 < 409 complees he proof of 328 6

17 vi From, he funcion ϕ = log 2 / is increasing for e 4 = decreasing for e 4 so ha, for < a < b, he lower bound of ϕ on he inerval [a, b is minϕa, ϕb Le p be a prime saisfying p 83 From i, one has Ap > 0, for [p, p +, A log2 = Ap log2 Ap minϕp, ϕp + To prove 329, i remains o check ha Ap minϕp, ϕp + > 2 λ holds for 37 p Proof of Theorem Proof : The proof of 6 follows from Corollary 32 while Corollary 3 yields 7 The proof of 8 resuls of Proposiion 36, i v Inequaliy 9 resuls of Proposiion 36, v vi If 0000, Inequaliy 0 follows from Proposiion 36, iv, while for > 0000, Proposiion 36, ii, implies A log λ log which ends he proof of Theorem log λ = log 0000 log 2 Acknowledgemens I am pleased o hank Marc Deléglise for his compuaions for several discussions abou his paper References [] M Abramowiz I A Segun Hbook of Mahemaical Funcions, Dover Publicaions, Inc New-York [2] H Cohen Number Theory Volume II, Analyic modern ools, Springer, 2007 [3] P Dusar Esimaes of some funcions over primes wihou R H, cf hp://arivorg/abs/ v, 200 [4] H M Edwards Riemann s Zea funcion, Academic Press, 974 [5] A E Ingham The disribuion of prime numbers, Cambridge Mahemaical Librairy, Cambridge Universiy Press, Cambridge, 990 Reprin of he 932 original, Wih a foreword by R C Vaughan [6] E Lau Hbuch der Lehre von der Vereilung der Primzahlen, I, 2nd ed, Chelsea, New- York, 953 [7] J E Lilewood Sur la disribuion des nombres premiers C R Acad Sci Paris Sér I Mah, 58, 94, [8] D J Pla T Trudgian On he firs sign change of θ, hp://arivorg/abs/ 40794v, [9] G Robin Sur la différence Liθ π, Annales Faculé des Sciences Toulouse, 6, 984, [0] S Ramanujan Highly composie numbers, Proc London Mah Soc Serie 2, 4, 95, Colleced papers, Cambridge Universiy Press, 927, [] L Schoenfeld Sharper bounds for he Chebyshev funcions θ ψ II, Mah Comp, 30, 976, [2] hp://mahuniv-lyonfr/homes-www/~nicolas/liheahml 7

18 Jean-Louis Nicolas, Univ Lyon, Universié Claude Bernard Lyon, CNRS UMR 5208 Insiu Camille Jordan, Mahémaiques, Bâ Doyen Jean Braconnier, 43 Bd du Novembre 98, F Villeurbanne cede, France hp://mahuniv-lyonfr/homes-www/nicolas/ 8