Lectures # 5 and 6: The Prime Number Theorem.

Transcription

1 Lecures # 5 and 6: The Prime Number Theorem Noah Snyder July 8, 22 Riemann s Argumen Riemann used his analyically coninued ζ-funcion o skech an argumen which would give an acual formula for π( and sugges how o prove he prime number heorem This argumen is highly unrigorous a poins, bu i is crucial o undersanding he developmen of he res of he heory Noice ha log ζ(s = p n n p ns for Re(s > Leing J( = p k k, noice ha log ζ(s = s dj( again for Re(s > Now use inegraion by pars o ge log ζ(s = s J( s d Now his is a Mellin ransform, so, assuming some echnical resuls, we should be able o use Melin inversion Thus, J( = σ+i log ζ(s s ds 2πi σ i s This converges when σ > Thus in order o find a formula for J( we need only ge a beer formula for log ζ(s Riemann claimed ha ξ(s = ξ( ( ρ s ρ, where he produc is aken over all roos of he ξ funcion (ha is, over all nonrivial zeroes of he ζ-funcion This produc does no converge absoluely, and we should pair any erms wih im(ρ posiive wih a corresponding erm wih negaive imaginary par o ge a convergen produc The proof of his produc formula basically depends on geing nice bounds on he growh of he number of zeroes Now we noice ha, ζ(s = 2 s(s πs/2 Γ(s/2 ξ(s = 2 s(s πs/2 Γ(s/2 ξ( ρ ( s ρ log ζ(s = log 2 log s log(s + s 2 log π log Γ(s/2 + log ξ( + ρ ( log s ρ We wan o subsiue his ino our inegral formula and evaluae ermwise, however doing so would lead o divergen inegrals (for eample in he s 2 log π erm Thus Riemann firs inegraed by pars o ge, J( = 2πi σ+i ( d log ζ(s s ds log σ i ds s Now we can subsiue our formula for ζ(s and evaluae erm by erm Wih a good bi of work, Riemann evaluaed hese inegrals and go he formula, J( = Li( ρ Li( ρ + ( 2 d log 2 log

2 Noice ha J( = n= n π( n We can inver his formula o ge, π( = n= µ(n n J(/n This gives us a formula for π( Is dominan erm is n= µ(n n Li(/n This would show he prime number heorem if we could acually prove ha his erm was dominan The key o proving his is o show ha he ρ Li(ρ erms are each smaller, ha is o say we need o show ha Re(ρ < 2 Chebyshev s Funcions Before Riemann s work he only significan progress owards he prime number heorem was made by Chebyshev who proved ha, for sufficienly large and some consans c < < c 2, c log π( c 2 log To prove his he inroduced wo funcions which are crucial in laer proofs of prime number heory Recall ha we conjecure ha he chances ha a number n is prime is roughly log n Thus, if we couned each prime as log p insead of as, hen we would ge a beer behaved funcion Definiion 2 Le θ(n = p n log p (where, as usual, a jumps we define he funcion o be halfway in beween he wo values As we ve seen from Riemann s argumen i is ofen simpler o coun prime powers insead of primes Definiion 22 Le ψ(n = p k n log p There is anoher way of wriing ψ in erms of Von Mangold s Λ funcion Definiion 23 Le { log n if n is a prime power Λ(n = else Clearly ψ( = n Λ(n Firs we noice ha one can epress each of he funcions ψ, θ, π, and J in erms of any of he ohers Proposiion 24 J( = π( = n= n π(/n µ(n n π(/n n= ψ( = θ( = θ( /n n= µ(nψ( /n n= Proof We ve already shown he firs wo, and he proof of he second wo are eacly he same Proposiion 25 π( = θ( log + θ( (log 2 d J( = ψ( log + ψ( = J( log θ( = π( log ψ( (log 2 d J( d π( 2

3 Proof Noice ha π( = log dθ( The heorem follows from inegraion by pars Similarly J( = log dψ(, and we inegrae by pars again Conversely, θ( = log dπ( and ψ( = log dj( Inegraing hese by pars gives he second wo equaions Since θ and ψ are rivially O( log and π and J are rivially O( we can rewrie hese equaions in erms of error esimaes The long and shor of all of his is ha o prove he prime number heorem i is enough o prove any of π Li(, J( Li(, θ(, or ψ( Furhermore, given any eplici error erms in he above approimaions we can find eplici error erms for all of he oher approimaions As i urns ou ψ is he easies funcion o deal wih Proposiion 26 For any n, d n Λ(d = log n Proof Noice ha n = p n pk where k is he larges number such ha p k n Thus, n = p k np Taking he logarihm shows ha log n = d n Λ(d There is anoher way of looking a his ideniy In shorhand his proposiion claims Λ = log n Thus i is equivalen o some ideniy involving Dirichle series Noice ha n= Λ(nn s = p (log pp ms = ζ (s ζ(s m= Also log n = ζ (s f(s, Λζ(s = f(s, log n eacly as we had hoped o show n= Before leaving his las proof we noice ha one of he equaions can be rewrien 3 Chebyshev s Theorem ζ (s ζ(s = s dψ( Theorem 3 For sufficienly large and some consans c < < c 2, c ψ( c 2 Proof Chebyshev noiced ha if we sum d n Λ(d = log n over all n, hen T ( = Λ(m = m log n = log! m n By Sirling s formula Noice ha by Möbius inversion, T ( = log! = log + O(log T ( = Λ(m = Λ(m = ( ψ n m n m n m n n ψ( = µ(nt n= ( n This suggess ha finie epressions which have several erms from n= µ(nt ( n will give good approimaions o ψ Bu we also wan good cancellaions when we plug in he approimaion from Sirling s formula For eample, i would be informaive o look a epressions of he following form: ( ( T ( T T, 2 2 3

4 ( ( ( T ( T T + T, ( ( ( ( T ( T T T + T, ec We will look a he firs epression T ( 2T ( 2 Chebyshev looked a he hird epression and was able o ge consans c and c 2 closer o Noice, ( T ( 2T = ( Λ(m 2 m 2m m The lefhand side is log 2 + O(log The righhand side is ( Λ(m m Λ(m = ψ( 2m m m for large and any consan ε >, In paricular, we can ake c = 69 Similarly, he righhand side is Λ(m m ( m log 2 ε ψ( ( Λ(m = ψ( ψ 2m 2 2 m ψ( ψ ( 2 log 2 + O(log Summing hese esimaes yields, In paricular, we can ake c 2 = 38 ψ( 2 log 2 + O(log 2 By our previous resuls relaing ψ and π, we also ge ha 79 π( 38 log log 4 Reducing he Prime Number Theorem o Facs Abou ζ(s Recall ha Inegrae by pars o see ha ζ (s ζ(s = ζ (s ζ(s = s s dψ( ψ( s d Our general mehod of aack is o rewrie his as a Mellin ransform and hen use Mellin inversion o rerieve ψ( in erms of ζ(s However, o make cerain inegrals behave well laer on, we firs make a sligh change Inegraes by pars again o noice ha Definiion 4 Le φ( = ζ (s ζ(s = s2 ψ( d ( ψ( d s d 4

5 Therefore we have ζ (s ζ(s = s2 φ( s d To wrie his as a Mellin ransform we make he change of variables s s ζ ( s ζ( s ( s 2 = φ( s d In order o apply Mellin inversion we mus check o see ha he echnical condiions of ha heorem are saisfied Noice ha since ψ( = O( log we have φ( = O(log Therefore he inegrand in he Mellin ransform converges absoluely for R(s < Also, ζ (s ζ(s (log nn σ Thus, for any posiive ε, in he region Re(s + ε, he funcion ζ (s ζ(s consan he inegral σ+i ζ ( s ζ( s ( s 2 d σ i n= is bounded by an absolue converges absoluely for any σ < Therefore he condiions of Mellin inversion are saisfied and, φ( = σ+i 2πi σ i ζ ( s ζ( s ( s 2 s ds, for any Re(s < Now we can change variables back s s and muliply boh sides by o ge, Proposiion 42 For any s wih Re(s > he following inegral converges absoluely and φ( = σ+i 2πi σ i ζ (s ζ(s s 2 s d Noice ha hus far we could have gone hrough he argumen wih ψ( insead of φ( and he resuling formula would have a /s insead of /s 2 Our argumen from here on in consiss of several pars Firs we will assume ha here are no zeroes of he ζ funcion on he line Re(s = We will prove his in he ne secion Thus he only pole of he inegrand in he halfplane Re(s is s = We can subrac off his pole o ge a erm which conribues he dominan erm The remaining inegral we can move all he way o he line Re(s = Then we will ge an eplici bound on his inegral This will give us an approimaion for φ( Finally we will need o erac an esimae for ψ( from our knowledge concerning ψ( So noice ha φ( = 2πi σ+i σ i s s 2 s d 2πi σ+i σ i ( ζ (s ζ(s + s s 2 s d The firs inegral can be wrien as he limi of an inegral abou he recangle wih corners +/T ±it and T ± it The inegrals along all bu he righ side die very quickly Thus our inegral is he sum of he residues o he lef of Re(s = 2 The only poles are a s = and s = To his end epand s = e s log = + s log + s 2 log 2 + Thus he residue a s = is log A s = he residue is Therefore his inegral conribues he erm log (The noes ha I am basing his on say ha his inegral is log I canno find ou where he comes from, bu I do no rus my abiliy o do comple analysis very well, and so ha is probably righ Noneheless since we are only ineresed in approimaion he will no maer given our assumpion ha ζ( + i, we have proved: 5

6 Proposiion 43 φ( = log 2πi i ( ζ ( + i ζ( + i + i ( + i 2 ei log d In order o esimae his las inegral we will need a few esimaes on he size of ζ(s and ζ (s These will be proved in he ne secion Thus we will make he following assumpions: Proposiion 44 Leing s = σ + i as usual, we have he bound ζ ( k(s = O(log k in he region σ > log and > 2 Also we have ζ(s = O(log7 in he region σ and > 2 ( Proposiion 45 For any ineger k, φ( = + O Proof Le (log k f( = ( ζ ( + i 2πi ζ( + i + i ( + i 2 Recall ha φ( = R f(ei log Since he second erm is rapidly oscillaing, if we can ge a decen bound on f( we should ge a very good bound on φ( From our esimaes concerning ζ and is derivaives, ( log f ( k( = O ( + 2 Therefore for each k here is a consan C(k wih f ( k( d C(k Now we inegrae by pars k imes o see ha, f(e i log d = ( i log k R R f(e i log C(k R log k R f ( k(e i log Combining his wih our earlier resuls yields our required resuls Noice ha had we aemped o run hrough he above argumen wih ψ he final inegral would no have converged absoluely One would sill epec he oscillaory erm o cancel hings ou, bu proving his would be more difficul All ha remains o do (oher han he analyic resuls pu off ill ne secion is o urn his esimae for φ ino an esimae for ψ I is perhaps surprising ha one can do his, since we are essenially differeniaing an approimaion Bu since ψ behaves so nicely we can in fac do his Theorem 46 For any ineger k, ψ( = + O( log k/2 Proof Suppose he ε( is any funcion saisfying < ε( 2 Le g k( = ha for all sufficienly large and some consan C, since g k (2 g k (, Cg k ( φ( Cg k ( φ( + ε( φ( ε( + Cg k ( + ε( + Cg k ( ε( + 3Cg k ( We have proved log k 6

7 On he oher hand, since ψ is an increasing funcion, φ( + ε( φ( = Combining hese wo equaions shows ha +ε( ψ( + ε( + 3Cg k ( + ε( ε( ψ( Considering φ( φ( ε( in he same way yields φ( 2Cg k( ε( ε( d ψ( + ε( + ε( + 6Cg k( ε( Now we can choose ε( in such a way o minimize he error erm The bes such choice is ε( = c g k ( where we choose c small enough so ha we sill have ε( 2 Plugging his epression ino our previous resuls yields he heorem This is equivalen o he prime number heorem Plugging our esimae for ψ ino our previous relaions, π( = ( log + 2 log 2 d + O log k However, by inegraion by pars, Li( = log + 2 ( π( = Li( + O log k d + O( we have log 2 ( Noice ha he approimaion π( = log only holds, a priori, up o O 5 Some Facs Abou ζ(s Proposiion 5 For any real, ζ( + i log 2 Proof Throughou his proof any ime we use he symbol c i means a paricular consan which may change from equaion o equaion Recall ha log ζ(s p s + c p Reζ(s p cos log p p σ + c If s = + i were a zero of he zea funcion, hen lim σ + log ζ(σ + i = cos log p lim σ + p σ = This implies ha he vas majoriy of numbers cos log p are near nearly all he numbers log p would lie near he poins of he arihmeic progression (2n + π This is impossible because his regulariy would sugges ha cos(2 log p were nearly for he vas majoriy of primes This in urn suggess ha ζ(s has a pole a s = + 2i Now we make his argumen rigorous Suppose ζ(s had a zero a s = + i, hen ζ(s/(s i would be analyic near s = + i In paricular, aking he real par of log of ζ(s/(s i, we see ha cos( log p p σ < log(σ + c p 7

8 Le δ > be some small posiive number Le S be he sum of p σ over all primes which saisfy (2n + π log p < δ for some ineger n, and le S 2 be he sum over primes which do no saisfy his condiion For erms in he second sum cos( log p > cos δ S (cos δs 2 < log(σ + K On he oher hand, since here is a simple pole a, we have S + S 2 < log(σ + c S (cos δs 2 < S S 2 + c S 2 < c cos δ However, since +2πi is no a pole of ζ(s, he real par of log ζ(s is bounded above near s = +2i Therefore cos 2 log p p σ < c p Again we can spli his sum up over he wo ses of primes For primes of he firs ype cos(2 log p > cos 2δ > S cos 2δ S 2 < c S < c ( cos δ cos 2δ Hence, for some consan depending on δ, S + S 2 < C(δ Leing σ approach makes he lefhand side blow up which is a conradicion For a more clever bu perhaps less informaive proof ha a zero a ζ( + i would force a pole a ζ( + 2i look a he proof of his resul on one of he ne few copied pages The proofs of he following wo resuls are on he ne few phoocopied pages Proposiion 52 Leing s = σ + i as usual, we have he bound ζ ( k(s = O(log k in he region σ > log and > 2 Proposiion 53 Leing s = σ + i as usual, we have he bound ζ(s = O(log7 in he region σ and > 2 8