The Prime Number Theorem

Transcription

1 he Prime Number heorem Yuuf Chebao he main purpoe of thee note i to preent a fairly readable verion of a proof of the Prime Number heorem (PN, epanded from Setion 7-8 of Davenport tet [3]. We intend to make thi epoition a elf-ontained a poible. However, ome reult from earlier etion in the tet, a well a ome priniple from omple analyi, are aumed and lited without proof in Setion below. In Setion, we derive a formula for the um of the von Mangoldt funtion, whih i a key ingredient of our proof. he final tep of the proof are arried out in Setion 3, where ome onnetion between the PN and the Riemann Hypothei (RH i alo diued. Let u firt reall a few definition: he prime ounting funtion π( i the number of all prime le than or equal to a poitive real number. he logarithmi integral funtion i li( = log t dt. Note that ome author denote thi funtion a Li(. onvention by Davenport. Here we follow the he big O notation: Let f, g and h be omple-valued funtion whoe domain ontain a ommon infinite interval [a,. he relation f( = O(g( mean there eit ome ontant M and 0 uh that f( M g( for all 0. he relation f( = h( + O(g( mean f( h( = O(g(. hi i to be undertood that f( an be approimated by h(, where g( i an error term. We oaionally write f( g( for f( = O(g(. he form of the PN that we will be proving i ( π( = li( + O e log where i a poitive ontant. We will alo how that an immediate onequene of ( i another form of the PN: π( = ( log + O log. ( (

2 Several proof of the PN have been diovered. Signifiant ontribution that lead to the firt ueful proof ould be traed bak at leat to the 8th entury, ine the time of Legendre and Gau. Below we reord a number of notoriou ahievement and ome well-known proof of the PN, lited by year. We do not intend to produe an ehautive timeline nor a omplete hitorial aount here. 85 Pafnuty Chebyhev howed that there eit ontant 0.9 and. uh that π( log for all large. he preeding log inequalitie are abbreviated a π(, and an be read a π( log ha the ame order of magnitude a. hu Chebyhev wa the firt log to demontrate the orret order of magnitude for π(. See Setion 3. of [6] for a detailed diuion of Chebyhev etimate. 859 Bernhard Riemann, in hi influential memoir On the number of prime le than a given magnitude, ued the zeta funtion ζ( whih wa etended analytially throughout the omple plane (eept for a pole at =, to invetigate π(. Among other thing, he etablihed an important funtional equation for ζ(, namely, ξ( = ξ(, where ξ( = π Γ( ζ( and Γ i the gamma funtion. In addition, he howed ome onnetion between π( and the zero of ζ(. hi paved a way to the firt omplete proof of the PN a few deade later. 896 Jaque Hadamard and Charle Jean de la Vallée-Pouin independently gave the firt omplete proof of the PN. he proof ue tehnique from omple analyi, partiularly the Argument Priniple and the Reidue heorem, and i baed on ome areful analyi of the zero of ζ(, a uggeted earlier by Riemann. hi proof i ometime referred to a an analyti proof of the PN. 93 Shikao Ikehara, in [7], ued Wiener auberian heorem to derived the PN. An outtanding apet of thi proof i that it only require ζ( to be zero-free on the line Re( =, rather than a larger zero-free region a in the previou proof by Hadamard and de la Vallée-Pouin. 949 Paul Erdő in [4] and Atle Selberg in [9] publihed eentially the ame elementary proof of the PN. It i elementary in the ene that no ophitiated tehnique from omple analyi wa ued, although the argument wa quite intriate. Unfortunately, there wa a bitter diagreement between the two mathematiian a to whom the redit of the proof hould be given. See [5] for an aount of the dipute. 980 Donald J. Newman found a imple proof of the PN. It i analyti, but imple in a ene that it hardly ue anything beyond Cauhy Integral Formula, and ue laial reult on θ( = p log p by Chebyhev. It an alo be viewed a a impler verion of the auberian argument by Ikehara. hi i pereived by ome a the eaiet proof the PN o far. See [] or [8] for Newman proof. Our proof of the PN will follow eentially a laial approah by Hadamard and de la Vallée-Pouin.

3 he following diagram illutrate logial dependene of the main reult in thee note. An arrow from A to B indiate that we ue A (poibly along with ome other reult to prove B. heorem - heorem 3 heorem 4 heorem 5-6 Remark heorem 8 heorem 0 heorem Corollary 4 Corollary 9 heorem 3 heorem 4 3 heorem and heorem are bai priniple about ontour integral that we will ue to derive two formula for the um of von Mangoldt funtion in heorem 0 and heorem. Notie that heorem i a entral reult that i alo baed on heorem 3 and heorem 4, whih are ome fat about the gamma funtion and the zeta funtion. heorem 3 i a ruial etimate for the um of von Mangoldt funtion that will eventually yield heorem 4, whih i the PN. Preliminarie In thi etion, we ollet ome reult (motly without proof that are required for the net etion. Firt of all, in the derivation of the error term, we almot alway rely on the following fat about upper bound for the moduli of ontour integral: heorem. Let C := z(t, t [a, b] be a mooth ar and let f(z be a ontinuou funtion on C. hen b f(dz f(z(t dz(t ML C where M := ma f(z(t and L = b t [a,b] a z (t dt i the length of C. a he relation f(zdz ML i alled the ML inequality. Another important C priniple that we will be uing repeatedly in Setion i: heorem. (he Reidue heorem Let C be a imple loed urve in a region A, travered one in the ounterlokwie diretion. Let f be a omple-valued funtion whih i analyti on and inide C, eept at ome iolated ingularitie z,..., z k inide C. hen πi C f(zdz = k Re(f, z j. For the proof of the two theorem above, ee, for eample, [] or []. Further, we will etenively ue many fat about the Riemann zeta funtion ζ(, whih i loely related to the gamma funtion Γ(. j=i

4 4 he gamma funtion an be defined on C 0,,,... } by [ ( Γ( = e γ + e /n] n n= ( where γ = lim n n k= log n i the Euler ontant. It i known that Γ( i k zero-free and analyti in the omple plane, eept at imple pole = 0,,, 3,.... In Setion, we will need a fat about the logarithmi derivative of the gamma funtion that Γ ( Γ( = log + O( (3 when and π + δ < arg( π δ for any δ > 0. hi fat i taken from Setion 0 of [3]. he Riemann zeta funtion i defined for Re( > by ζ( = n= /n, and an be etended to the right half-plane Re( > 0 by ζ( = } d + where } = i the frational part of. One way to etend ζ( to the whole omple plane i to ue the following form of Riemann funtional equation: ( π ζ( = π ζ( Γ( in. (4 We know that ζ( i analyti in C, eept for a imple pole at =. It an be een from (4 that the only zero of ζ( on the left half-plane are =, 4, 6,... ; thee are alled the trivial zero of the Riemann zeta funtion, and eah of them ha order one. All the remaining zero of ζ( are in the ritial trip C : 0 < Re( < }, and are alled nontrivial zero. he Riemann Hypothei (RH aert that all nontrivial zero of ζ( are on the ritial line Re( =. See hapter VII of Conway tet [] for a thorough introdution to the gamma funtion and the zeta funtion. he following theorem i drawn from Setion of [3], and will be very ueful for our work in Setion. heorem 3. (Partial fration deompoition of ζ ( ζ( he zeta funtion ha infinitely many zero in the ritial trip. Let Z be the et of all uh nontrivial zero. hen for any we have ζ ( ζ( = B + log π Γ ( + Γ ( + + ( ρ + (5 ρ where B Γ ( + Γ ( + = + log 4π + n= ρ Z ( + n. (5 n

5 From (5 and (5 we immediately have ζ (0 ζ(0 = log π + log 4π = log(4π = log π. (6 Alo in Setion, we are going to ue the following reult from page 99 in Setion 5 of [3]: heorem 4. (i Let > 0 and let Ω := = σ + it : t < }. Let N Ω ( be the number of all nontrivial zero of ζ( inide Ω. hen N Ω ( = O(log. (7 (ii Let > 0. Fi = σ + i, where σ and ζ( 0. Let ρ denote any nontrivial zero of ζ(. hen ζ ( ζ( = + O(log. (8 ρ ρ Ω heorem 5. here eit a ontant > 0 uh that ζ( ha no zero in the region } = σ + it : σ. log( t + heorem 6. For > 0, let N( be the number of all zero of ζ( in the retangle = σ + it : 0 < σ <, 0 < t < }. hen 5 N( = π log π π + O(log. For the proof of heorem 5 and heorem 6, ee Setion 3 and 5, repetively, of [3]. Remark 7. (i From heorem 5, it an be inferred that if ρ = σ + it i a zero of ζ( uh that σ > 0, then σ < log( t +. Moreover, for uffiiently large, there eit ome ontant > 0 uh that if ρ = σ + it i a zero of ζ( with σ > 0 and t <, then σ < log. (9 Indeed, for t <, we have log( t + < log( + log for large, o that / log( t + < /( log, and now it i lear that (9 hold with = /. (ii From heorem 6, it i eay to ee that we have a rough etimate N( = O( log. (0 he impliation above, peifially (9 and (0, will be ued in our proof of the PN in Setion 3.

6 6 An epliit formula for ψ( In thi etion, we will derive a ueful formula (ee heorem for the funtion ψ( := Λ(n n where > 0 and Λ(n i the von Mangoldt funtion defined by log p, when n = p Λ(n = m, p i a prime and m ; 0, ele. Our target formula for ψ( will involve nontrivial zero of ζ( in the ritial trip, whih i a onequene of an appliation of the Reidue heorem. An intermediate tep that will lead to the deired formula i the Perron integral formula with error term (heorem 0. o prepare for our verion of Perron formula, we firt prove the following formula for a line integral: heorem 8. Let > 0 be a fied ontant. For y > 0 and > 0, let I(y, := πi +i i y d. hen I(y, = ( } O y min,, if 0 < y < ; log y ( + O, if y = ; ( } + O y min,, if y >. log y ( Before proving heorem 8, we note that the following formula for lim I(y, i immediate: Corollary 9. Let > 0 be a fied ontant. For y > 0, we have 0, if 0 < y < ; +i y πi i d =, if y = ;, if y >. Proof. (of heorem 8 Firt, we onider the ae y =. Uing the parametrization L(t = + it, t for the line egment L from i to + i, we have I(, = πi +i i = πi d + it idt

7 7 Hene, = π [ = π = π 0 = [ π 0 it + t dt + t dt i + t dt + t dt = ( t π artan 0 π = π I(, = π hi how I(, = + O (. + t dt. ] t + t dt [ i 4π log( + t + t dt π + t dt ] + t dt ] ( t dt = O. Net, we onider the ae 0 < y <. Let Γ = L L L L 3 be a retangular ontour oniting of the line egment L, L, L, and L 3 with the orientation a hown in Figure. Figure. Line egment and irular ar ued to verify ( Sine the funtion y / i analyti on and inide Γ, by Cauhy heorem it follow that y πi Γ d = 0. So from Γ = L L L L 3 we have +i y πi i d = ] y πi [L L d + y L3 d + y d. (.

8 8 Now we determine ome bound for eah of the integral on the RHS of (.. On L, we have = σ + i, where 0 < σ d. / / beaue on L. Alo reall that y σ+it = y σ y it = y σ e it log t = y σ It i trivial that beaue y i = o + i in = for all R, and y > 0. hu, y L d y d d y σ dσ y σ dσ = y σ log y = L y log y. In the lat equation above, we ue the fat that y σ 0 a σ beaue 0 < y <. hi how ( y L y d = O. (. log y A imilar onideration a above alo yield L3 y ( y d = O. (.3 log y On L, we have = d+it, where t ; o y = y d and / /d on thi egment. herefore, y L d y d y d yd dt = d d. L By letting d (and hene moving L very far to the right, we ee that y d L would be a mall a we wih. So the bound from thi integral i negligible when ompared to (. and (.3. hu we an infer from (.-(.3 that ( y I(y, = O. (.4 log y On the other hand, onider Γ = L C, where C i an ar of a irle with radiu R = +, entered at the origin, a hown in Figure. Again, by Cauhy heorem we have y πi Γ d = 0. Note that y / y /R on C. So by the ML inequality, it follow that +i y πi d = y πi C d π y R πr = y. hi how i I(y, = O(y. (.5 In light of (.4 and (.5, we would take the minimum value of y and y /( log y a our bet bound, depending on the value of /( log y. herefore, a deired. ( I(y, = O y min, } log y

9 he ae y > an be treated imilarly a in the previou ae. Firt let Γ 3 be the ontour L L 4 L 5 L 6, where L 4, L 5, and L 6 are line egment a hown in Figure. hi time the funtion y / ha a imple pole at = 0 inide Γ 3, and i ( y Re, 0 Γ 3 y = lim y 0 =. So the Reidue heorem yield d =. A a reult, πi +i y πi d = ] y πi [L4 L5 d + y L6 d + y d. (.6 By eentially the ame method that we ued to how (.4, the um of three integral on the RHS of (.6 will be bounded by y /( log y. Notie that the aumption y > give y y on Γ 3. Moreover, let Γ 4 = L C, where C i the irular ar on the left ide of L a hown in Figure. hen y / alo ha a imple pole inide Γ 4 at = 0 with reidue. herefore, πi +i i y d = y πi C d. (.7 By the ame tehnique that we ued to how (.5, the integral on the RHS of (.7 will be bounded by y. hu we have ( y I(y, = + O and I(y, = + O(y log y ( } and hene I(y, = + O y min,. hi omplete the proof. log y It i time now to turn our attention to ψ(. Oberve that ψ( ha a jump at eah point when i a prime power, that i, when = p m for ome prime p and m N. If we modify ψ( at eah of thee diontinuou point by taking the average on the left and the right ide of the point, then it turn out that formula ( will be effetive in approimating the value of thi modified funtion. More preiely, we will be able to handle the funtion ψ 0 ( defined by ( ψ 0 ( := Λ(n + Λ(n n< n ψ( Λ(, if i a prime power; = ψ(, otherwie. Indeed, we an ue ( to epre thi funtion a ( ψ 0 ( = Λ(n I n, + R(, = n= Λ(n n= = πi n= ( +i πi i ( +i i Λ(n n (/n d d + R(, where >, and R(, ollet all error term a given by (. 9 + R(, (

10 0 where Reall that the Dirihlet erie Λ(n n= onverge abolutely for Re( >, n n= Λ(n n = ζ ( ζ(. (3 Sine >, the erie Λ(n n= onverge abolutely and uniformly to ζ ( n ζ( on the ompat line egment from i to + i. herefore, we an interhange the um and the integral in ( to obtain ψ 0 ( = πi hi give part of our net reult: heorem 0. For > we have where +i i n= +i Λ(n n ψ 0 ( = πi i ( log R(, = O + (log min d + R(, (4 ζ ( ζ( d + R(, (5, } and i the ditane between and the nearet prime power other than. Proof. Formula (5 follow readily from (3-(4. o prove (6, firt note that by ( we have ( } R(, Λ(n min, n log + Λ(. (6. n n=,n WLOG, we may aume > e and hooe = + log = + log = log = e. Sine < and Λ( log, we ee that Λ( log R(, n=,n Λ(n n min, log n >. hen (6. hu by (6. we have } + log. (6. Net, we partition the et of integer n in (6. into two et, namely, A = n : n 3 4 or n 4 } and B = n : < n < 4 } 3 and etimate the um on eah et eparately. For n A, we have log > log 4. So / log i bounded above. Alo reall n 3 n that ζ ( ζ( herefore, ha a Laurent erie repreentation + k=0 k( k, o that ζ (σ ζ(σ σ a σ +. (6.3

11 n A Λ(n n } min, log n n= Λ(n n = ζ ( ζ( = ζ ( + log ζ ( + log log. (6.4 Notie that we ue (6.3 in the etimate (6.4. For n B, we have 3 4 < n < 4 3 and n. We write B = B B where B = n : 3 4 < n < } and B = n : < n < 4 3 }. Sine eah of thee et will ontribute nothing to the um in (6. if it doe not ontain a prime power, we may aume that eah of B and B ontain ome prime power. Let n B, and let q be the bigget prime power in B. Oberve that q < 3 4 = 4 <. And from log( t = t t t3 3 for t <, we have log q = log q = log ( q Sine q B, we know that < < 4. herefore, by (6.5 q q 3 Λ(q q min, } log (log min, q q. (6.5 }. (6.6 ( q For other n B whih are prime power, we an write n = q m for 0 < m <. So 4 log n log q q m = log q m ( = log m m q q q. Sine m 4 it follow that, and 4 m= m d = log t

12 n B,n q Λ(n n min, } 4 log (log n (log m= 4 m= min, m q } m log. (6.7 Let n B, and let r be mallet prime power in B. hen by a imilar argument that we ue to obtain (6.6 and (6.7, it an be hown that } } Λ(r min, r log (log min, (6.8 (r r and n B,n r Λ(n n min, } log n log. (6.9 Let be the ditane between and the nearet prime power other than. hen = min q, r }. By referring bak to (6. and olleting all bound from (6.4 and (6.6-(6.9 we an now onlude hi verifie (6. R(, log + (log min, }. Now we are ready to etablih the main reult in thi etion: heorem. For > 0 we have ψ 0 ( = ρ ρ log( log π + E(, (7 where t < ( log ( E(, = O + (log min, and ρ = σ + it i any zero of ζ( in the ritial trip. } Proof. We may aume, WLOG, a in the proof of heorem 0 that > e. Chooe = +. hen < <. We will be uing formula (5-(6 from heorem log 0. o verify (7-(8, we will evaluate πi +i i ζ ( ζ( d by uing the Reidue heorem. In doing o, we will ue the line egment from i to + i a part of a loed retangular ontour C that enloe everal pole of the funtion g( := ζ ( ζ(. (8

13 3 Figure. he ontour C = L L L L 3 and ome pole of g( Let L, L, L, and L 3 be the path a in Figure, and let C := L L L L 3. hen C enloe ome pole of g(. Notie that all zero of ζ( are pole of g(, and ine ζ( ha infinitely many zero, we mut enure that C doe not pa through any zero of ζ(. Here i why it i poible to do o: o hooe L that avoid the zero of ζ(, imply reall that the zero on the left half-plane are only the trivial zero at even negative integer. So we may hooe U to be an odd negative integer. o hooe L and L 3, reall from (7 in heorem 4 that for large enough, there eit a ontant M > 0 uh that the number N Ω ( of zero of ζ( in the trip Ω = σ + it : < t < + } i at mot M log. If we ubdivide Ω into k := M log + horizontal trip, ay Ω,..., Ω k, where eah trip ha the ame height h = /k, then it i impoible that eah trip ontain at leat one zero; otherwie, N Ω ( k > M log, a ontradition. So there eit ome trip Ω j that ontain no zero of ζ(, and we an move L into that trip. Sine the zero of ζ( i ymmetri about the real ai, it follow immediately that L 3 pae through no zero of ζ(. Let u now alulate the reidue of the pole of g inide C. Reall a general fat that if f( ha a zero or a pole at 0, then f /f alway ha a imple pole at 0, and ( f Re f, 0 = lim ( 0 f ( m, if 0 f( = 0 i a zero of order m; (7. m, if 0 i a pole of order m. In partiular, if ρ i a zero or pole of ζ(, then ρ 0 and ρ i a imple pole of g( = ζ ( (note that ζ( i an entire funtion. So Re (g, ρ = lim ρ ( ρ ζ ( ζ(. (7.

14 4 Sine ζ( ha a imple pole at =, it follow from (7.-(7. that Re(g, = lim( ζ ( ζ( = ( =. (7.3 If ρ = n i a trivial zero of ζ(, then by the partial fration deompoition of ζ (/ζ( (equation (5 and (5 in heorem 3 we have Re(g, n = lim ( + ( n nζ ζ( = n n. (7.4 If ρ = σ + it i a nontrivial zero of ζ(, then (5 and (5 again imply Re(g, ρ = lim( ρ ζ ( ρ ζ( = ρ ρ. (7.5 For the imple pole = 0 of g(, we have by (6 after heorem 3. Re(g, 0 = lim ζ ( 0 ζ( = ζ (0 ζ(0 By the Reidue heorem and (7.3-(7.6 we now have +i g(d = ρ πi i ρ + t < [ + πi U n= n n log π = log π (7.6 ] g(d + g(d + g(d. (7.7 L L L 3 Letting U in (7.7, and realling that log( t = t t t3 for 3 t <, o that n n = n n = log( n= n= we obtain from (7.7 all the main term a laimed in (7. o verify (8 for the error term E(,, we find a bound for eah integral on the RHS of (7.7. On L, we have = σ + i. We firt find a bound for ζ ( on the egment ζ( L, := L : σ } of L. Reall from (8 in heorem 4 that for any uh, ζ ( ζ( = + O(log (8. ρ ρ Ω where ρ = α + it i any zero of ζ( in Ω. Oberve that ρ = (σ α + ( t t.

15 So ρ t for all ρ Ω. Moreover, a a reult from our hoie of, we know that and hene t > h = t M log + 5 (8. = O(log. (8.3 Sine the number of all ρ in Ω i N Ω ( = O(log by (7 in heorem 4, it follow from (8.-(8.3 that ρ t = N Ω( t = O(log. (8.4 ρ Ω ρ Ω Now by (8. and (8.4 we have on L,. herefore, g(d L, (log ζ ( ζ( = O(log (8.5 σ+i σ + i dσ log = log log log σ dσ σ log (8.6 ine = + log = e. Net, we find a bound for the integral on L, := L : U σ }. Reall from (4 that ζ( atifie a funtional equation ( π ζ( = π ζ( Γ( in. Letting = w, we obtain = w and ( πw ζ( = ζ( w = w π w ζ(wγ(w o (8.7 beaue in( π θ = o θ. We will onider (8.7 for L,. Uing the produt rule for differentiation, we obtain ζ ( ζ( = (w ζ ζ(w = log π π ( πw tan + ζ (w ζ(w + Γ (w Γ(w. (8.8

16 6 By our hoie of L and U, it i eay to ee that πw tay away from the ingularitie of tan z. In fat, ine tan z ha ingularitie at z = (k+π where k Z, and ine any L, tay outide any irle of radiu entered at any trivial zero ρ = k of ζ, it follow that w (k + = ( w k = ( k >. hi implie the eond term π tan ( πw in (8.8 i bounded by a ontant. In addition, for L,, we have α := Re(w. Sine ζ (w ζ(w for Re(w >, we have ζ (w ζ(w n= Λ(n n α n = n= n α n= n α ( = ζ α ( 3 ζ = Λ(n n= n w beaue ζ i dereaing on the real interval (,. hi how the thrid term in (8.8 i alo bounded by a ontant. o find a bound for the lat term, we refer bak to (3: Γ ( Γ( = log + O(. Sine w =, we ee that Γ (w Γ(w Our analyi proeeding (8.8 now yield ζ ( ζ( log w log. herefore, for any poitive odd integer U, ζ ( g(d = L, L, ζ( d = O(log. (8.9 (log L, d σ (log U dσ log = log σ dσ σ log = log log. (8.0

17 However, the bound in (8.0 i dominated by the one in (8.6 on L,. So we onlude g(d log L log. (8. Similarly, we have 7 g(d log L 3 log. (8. On L, we have = U + it, t. Note that by (8.9 we have ζ ( g(d = L 3 L ζ( d (log U L (log U d U U dt log U =. (8.3 U U But the bound in (8.3 an be made a mall a we wih when U, o it i negligible. herefore, from (5-(6, (7.7 and (8.-(8., we an now infer that the error term for (7 i E(, } log + (log min, + log log log log ( + log + (log min hi verifie (8, and hene onlude our proof. + (log min,, }. We finih thi etion by the following formula for ψ 0 (. hey are immediate onequene of (5 and (7 when. Corollary. We have for > ; and ψ 0 ( = πi ψ 0 ( = ρ +i i ζ ( ζ( d (9 ρ ρ log( log π. (0 he inde ρ of the um in (0 range over all nontrivial zero of the zeta funtion in the ritial trip. hi formula might be alled an epliit formula for ψ 0 (. }

18 8 3 A proof of the PN We are now about to finih a proof of the PN. heorem enable u to take a big tep forward. heorem 3. here eit a ontant > 0 uh that ( ψ( = + O. ( e log Proof. Sine we are interet in the aymptoti behavior ψ(, we may aume WLOG that i an integer whih i not a prime power, o that ψ 0 ( = ψ(. hen (7 give ψ( = t < ρ ρ log( log π + E(,. (. A loer look at (. ugget that we need to etimate the um t <, where ρ = σ + it i any nontrivial zero of ζ( in the retangle ρ ρ for large R := σ + it : 0 < σ <, t < }. It i time now to ue our knowledge about a zero-free region of ζ(. Reall from (9 in Remark 7 that for large enough, any zero ρ in R mut atify for ome ontant > 0. A a reult, σ < log ρ = σ < log = log = (e log log A /ρ /t, it follow from (. that t < ρ ρ e log log 0<t< = e log log. (. t. (.3 So it remain to etimate 0<t<. Here we ue (0 from Remark 7 that the t number of nontrivial zero of ζ( in R i o utilize thi, oberve that for any t > 0, dn(t = lim h 0 [N(t N(t h] N( = O( log. (.4 = the number of zero ρ of ζ with Im(ρ = t. It i known that ζ ha no zero for mall t, e.g. when t <. So we may onider only t >. Hene,

19 9 0<t< t = = N( t dn(t + log + = log + log N(t t dt (integration by part t log t t dt (by (.4 log. (.5 hu by (.3 and (.5, t < ρ log ρ (log e log. (.6 Sine i an integer whih i not a prime power, we know that the ditane between and the nearet prime power i at leat. herefore, } (log min, log < log (. So the eond term for E(, in (8 i negligible. Further, we hooe = e log o that log = log and log = log. Note that log ( = log ( e log = (log + log (log. (.7 From (8, (., and (.6-(.7 we now have ψ( log ( + (log e log log e log (log + (log e ( = (log e log + (log e log log e log (.8 where 0 < < min, }. Notie that the lat etimate in (.8 i valid beaue u e u e u and ue u e u. hi omplete the proof.

20 0 heorem 4. (he Prime Number heorem here eit a ontant > 0 uh that ( π( = li( + O Moreover, Proof. Let π ( := n Equivalently, π ( = p m = p π( = e log. ( ( log + O log. (3 Λ(n. hi funtion i related to π( by log n log p m log p = π( + log p log p + p p log p log p + p p 3 + = π( + π( + 3 π( 3 + = π( + O(. log p 3 log p + π( = π ( + O(. (. On the other hand, π ( i related to ψ( by π ( = Λ(n log n n = [ ( Λ(n log + + ] log n log Λ(n n = [ ] Λ(n n n t log t dt + Λ(n log = he lat tep that yield (. an be jutified by n ψ(t ψ( t log dt + t log. (.

21 [ ] Λ(n n n t log t dt = [ ] Λ(n n n t log t dt = [ ] m+ Λ(n t log t dt n = m< = = m< n m< [ n m m m+ m ] Λ(n t log t dt [ ] m+ Λ(n t log t dt m ψ(t t log t dt. n m In the lat equality above, we ue an obervation that for m t < m +, hu, by (.-(., π( = = log + n m Λ(n = n t ψ(t ψ( t log dt + t log + O( log t dt + o introdue li( into our formula, oberve that herefore, by (.3-(.4, π( = li( + li( = log t dt = t log t + = log + Λ(n = ψ(t. ψ(t t ψ( t log dt + + O(. (.3 t log log t dt o ue heorem 3, let C > 0 be a ontant uh that log t dt log. (.4 ψ(t t ψ( t log dt + + O(. (.5 t log ( ψ( = + O e C log. (.6

22 hen, by (.5-(.6 we an dedue π( li( = e C log t log t log dt + e C + log e C log t dt + e C log e C log t dt + e 4 4 e C C log dt + e C log 4 + e C log + e C log e log log t dt + e C log where = C > 0. We have ued an obervation that for t ( 4,, log t > log 4 = log. hi jutifie (. o prove (3, reall that li( an be written a in (.4, where log t dt = and by the ubtitution u = log t we have log log 3 t dt = log log log + e u log du elog u3 log log 3 t dt u 3 du log. Finally, note that e log /(log. hi onlude our proof. After all, we ee that π( and ψ( are loely related. In fat, there i a relation π( = ψ( ( log + O log. (4 hi i a onequene of Chebyhev etimate. See hapter 3 of [6] for a proof of (4. It i obviou from (4 that ψ( if and only if π( /(log. We onlude our note by howing ome onnetion between the RH and an improvement of the error term for π(. Namely, he Riemann Hypothei hold π( = li( + O( log. (5 Notie that the error term for π( in (5 i better than thoe in (, beaue e log = e log log e log log = log.

23 o prove (5, firt uppoe that the RH hold. hen eah zero ρ = σ + it of ζ( in the ritial trip ha the real part σ =. hu, (. beome and (.6 beome t < ρ = σ = ρ ρ (log. Chooing =, and uing (8 and (., we obtain Referring bak to (.5: we then have ψ( log ( π( = li( + log. π( li( + (log ψ(t t ψ( t log dt + + O( t log log. t dt + log + Converely, uppoe π( = li( + O( log. hen it follow, for eample, by (4, that ψ( = + O( log. Oberve that for any fied ε > 0, beaue log log log log ε 0 a. hi implie ψ( = + O ( +ε. Now, by Mellin tranform (ee, for eample, page 4 of [6] we have, for Re( >, ζ ( ζ( = = ψ( d + ψ( d ε d +. d + Sine onverge abolutely for Re( > + ε, it follow from the lat + ε relation that ζ ( an be etended analytially to the half-plane Re( > + ε (of ζ( oure, with an eeption for a pole at =. In partiular, ζ( annot have any zero in thi half-plane. Sine ε > 0 i arbitrary and the nontrivial zero of ζ( are ymmetri about the ritial line, it follow that all nontrivial zero of ζ( mut be on thi line. herefore, the Riemann Hypothei hold. 3

24 4 Referene. J. Bak, and D. J. Newman, Comple Analyi., 3rd edition, Springer-Verlag, New York, 00.. J. B. Conway, Funtion of One Comple Variable I, nd Edition, Springer- Verlag, New York, H. Davenport, Multipliative Number heory, 3rd Edition, Springer-Verlag, New York, P. Erdő, On a new method in elementary number theory whih lead to an elementary proof of the prime number theorem, Pro. Nat. Aad. Si. U.S.A., 35 (949, D. M. Goldfeld, he Elementary Proof of the Prime Number heorem: An Hitorial Perpetive, goldfeld/erdoselbergdipute.pdf, Retrieved Marh 7, A. J. Hildebrand, Introdution to Analyti Number heory Math 53 Leture Note, Fall 005, hildebr/ant/, Verion , Retrieved Marh 7, S. Ikehara, An etenion of Landau theorem in the analyti theory of number, Journal of Mathemati and Phyi of the Maahuett Intitute of ehnology, 0 (93, D. J. Newman, Simple analyti proof of the prime number theorem, Amerian Mathematial Monthly, 87 (980, A. Selberg, An elementary proof of the prime-number theorem, Annal of Mathemati, 50 (949,