ANALYTIC PROOF OF THE PRIME NUMBER THEOREM

ANALYTIC PROOF OF THE PRIME NUMBER THEOREM RYAN SMITH, YUAN TIAN Conens Arihmeical Funcions Equivalen Forms of he Prime Number Theorem 3 3 The Relaionshi Beween Two Asymoic Relaions 6 4 Dirichle Series and Euler Producs 7 5 Conour Inegral Reresenaion of ψ )/ 9 6 ζs) near and on he Line σ = 7 Comleion of he Proof of he Prime Number Theorem 6 References 8 Le πn) be he rime couning funcion, ha is, he funcion ha gives he number of rimes less han or equal o n The analyic roof of he rime number heorem can hen be summarized as follows: ψ ) πn) log n as ψ) as lim = n n where ψ ) = Λn) = ψ)d, ψ) = n Λn) and We firs rove he las ar, namely { log if n = m for some rime and some m, 0 oherwise πn) log n ) ψ) as lim = n n Arihmeical Funcions We begin by defining arihmeical funcions Definiion A real- or comle-valued funcion defined on he osiive inegers is called an arihmeical funcion or a number-heoreic funcion The funcions πn) and Λn) referred o above are eamles of arihmeical funcions We now formally define he funcions Λn) and ψn) Definiion Mangold funcion) For every ineger n we define { log if n = m for some rime and some m, Λn) = 0 oherwise

SMITH & TIAN Definiion Chebyshev s ψ-funcion) For > 0 we define Chebyshev s ψ-funcion by he formula ψ) = n Λn) The following heorem will be useful laer in our roof of he heorem Theorem Abel s ideniy) For any arihmeical funcion an) le A) = n an) where A) = 0 if < Assume f has a coninuous derivaive on he inerval [y, ], where 0 < y < Then y ) an)fn) = A)f) Ay)fy) A)f )d y<n Proof Le k = [] and m = [y], so ha A) = Ak) and Ay) = Am) Then k k an)fn) = an)fn) = {An) An )} fn) y<n Now, observe ha = = = = n=m+ k n=m+ k n=m+ k An)fn) n=m+ k n=m An)fn + ) An) {fn) fn + )} + Ak)fk) Am)fm + ) An) n=m+ n k n+ n=m+ n n+ f )d + Ak)fk) Am)fm + ) A)f )d + Ak)fk) Am)fm + ) Ak)fk) = Ak)fk) + Ak)f) Ak)f) = A)f) A) f) fk)) Similarly, = A)f) A) y<n k f )d = A)f) Am)fm + ) = Ay)fy) + m+ y k A)f )d k an)fn) = A)f )d + A)f) m+ Ay)fy) m+ =A)f) Ay)fy) y A)f )d y A)f )d k A)f )d A)f )d Now we inroduce anoher arihmeical funcion which we will use in he roof

ANALYTIC PROOF OF THE PRIME NUMBER THEOREM 3 Definiion Chebyshev s ϑ-funcion) If > 0 we define Chebyshev s ϑ-funcion by he equaion ϑ) = log, where runs over all rimes less han or equal o Also if > 0 we le π) denoe he number of rimes no eceeding We will now eamine he relaionshi beween ϑ) and π) Theorem For we have 3) ϑ) = π) log and 4) π) = ϑ) log + π) d ϑ) log d Proof Le an) denoe he characerisic funcion of he rimes; ha is { if n is rime, an) = 0 oherwise Then we have π) = = <n an), ϑ) = n log = <n an) log n Now we ake f) = log wih y = in Abel s ideniy in Theorem o ge ϑ) = n log = <n which roves 3) since π) = 0 for < Ne, le bn) = an) log n and wrie an) log n = π) log π) log π) d, π) = 3/<n bn) log n, ϑ) = n bn) Takeing f) = / log wih y = 3/ in Abel s ideniy we obain π) = ϑ) log ϑ3/) log 3/ + which roves 4) since ϑ) = 0 if < 3/ ϑ) log d, Equivalen Forms of he Prime Number Theorem We will now show some equivalen forms of he rime number heorem Before acually showing hese equivalencies, we will firs inroduce a new noaion

4 SMITH & TIAN Definiion The big oh noaion) If g) > 0 for all a, we wrie f) = O g)) read: f) is big oh of g) ) o mean ha he quoien f)/g) is bounded for a; ha is, here eiss a consan M > 0 such ha An equaion of he form f) Mg) for all a f) = h) + O g)) means ha f) h) = O g)) We noe ha f) = O g)) for a imlies ) f)d = O g)d for all a a a Theorem 3 The following relaions are logically equivalen: π) log 5) lim = ϑ) 6) lim = ψ) 7) lim = Proof We show his by firs roving ha 5) and 6) are equivalen and hen ha 6) and 7) are equivalen Recall equaions 3) and 4) - from hese we obain, resecively and ϑ) π) log = π) d π) log = ϑ) + log ϑ) log d To show ha 5) imlies 6) we need only show ha 5) imlies Bu 5) imlies π) Now lim ) = O log for so π) d = O π) d = 0 ) d log d log = d log + d log d log + log + log log + log so This shows ha 5) imlies 6) d 0 as log d log

ANALYTIC PROOF OF THE PRIME NUMBER THEOREM 5 To show ha 6) imlies 5) we need only show ha 6) imlies log ϑ) lim log d = 0 Bu 6) imlies ϑ) = O) so Now hence log d log = log ϑ) log log d = O d log + d log d log d log 0 as ) log + log This roves ha 6) imlies 5), so 5) and 6) are equivalen To show ha 6) and 7) are equivalen, we firs noice ha 8) ψ) = m log ϑ /m ) Hence we have ψ) ϑ) = ϑ /m ) 0 m log Bu from he definiion of ϑ) we have he rivial inequaliy ϑ) log log so 0 ψ) ϑ) m log = log log /m log /m ) log ) log log = log log Now we divide all erms in he inequaliy by o obain 9) 0 ψ) ϑ) log log Bu 9) imlies ha ψ) lim ϑ) ) = 0 which imlies ha 6) and 7) are equivalen

6 SMITH & TIAN 3 The Relaionshi Beween Two Asymoic Relaions Now we will show ha he asymoic relaion 0) ψ ) as imlies he asymoic relaion ψ) as We will laer show why 0) is rue using roeries of he Riemann zea funcion We sar wih roving he following lemma Lemma For any arihmeical funcion an) le An) = n an), where A) = 0 if < Then ) n)an) = n A)d Proof We aly Abel s ideniy Theorem ) wih y = o obain ) an)fn) = A)f) A)f )d n if f has a coninuous derivaive on [, ] Taking f) = we have n an)fn) = n nan) and A)f) = n so ) reduces o ) Lemma Le A) = n an) and le A ) = an) 0 for all n If we have he asymoic formula 3) A ) L c as for some c > 0 and L > 0, hen we also have 4) A) cl c as In oher words, formal differeniaion of 3) gives a correc resul an) A)d Assume also ha Proof The funcion A) is increasing since he an) are nonnegaive Choose any β > and consider he difference A β) A ) We have This gives us or A β) A ) = βu u Au)du βu A)du = A)β ) A) β {A β) A )} A) c { A β) β β) c βc A } ) c

ANALYTIC PROOF OF THE PRIME NUMBER THEOREM 7 Kee β fied and le in his inequaliy By 3) we find lim sua) c β Lβc L) = L βc β Now le β + The quoien on he righ is he difference quoien for he derivaive of c a = and has he limi c Therefore A) 5) lim su cl c Now we consider any α wih 0 < α < and consider he difference A ) A α) An argumen similar o he above shows ha lim A) αc inf L c α As α he righ member ends o cl This, ogeher wih 5) shows ha A)/ c ends o cl as When an) = Λn), an) 0 holds and we have A) = ψ) and A ) = ψ ) Therefore, we can aly Lemma and Lemma o obain: Theorem 4 6) ψ ) = n n)λn) Also, he asymoic relaion ψ ) / imlies ψ) as 4 Dirichle Series and Euler Producs In his secion, we will inroduce some basic ideas in Dirichle series and Euler roducs which we will use laer in our roof of he rime number heorem A Dirichle series is a series of he form fn) n s n= where fn) is an arihmeical funcion We call he fn) he coefficiens of he corresonding Dirichle series We inroduce he Riemann zea funcion here as an eamle of a dirichle series Definiion Riemann zea funcion) For any s C, we define ζs) = n s We will now rove a coule of lemmas n= Lemma 3 Le s 0 = σ 0 + i 0 and assume ha he Dirichle series fn)n s0 has bounded arial sums, say fn) n s0 M n

8 SMITH & TIAN for all Then for each s wih σ > σ 0 we have 7) fn) n s Maσ0 σ + s s ) 0 σ σ 0 a<n b Proof Le an) = fn)n s0 and le An) = n an) Then fn)n s0 = an)n s0 s so we can aly Theorem wih f) = s0 s ) o obain a<n b fn) n s = Ab)b s0 s Aa)a s0 s + s s 0 ) b a A) s0 s d Since A) M his gives us fn) b n s a<n b Mbσ0 σ + Ma σ0 σ + s s 0 M σ0 σ d a Ma σ0 σ b σ0 σ aσ 0 σ + s s 0 M σ 0 σ Ma σ0 σ + s s ) 0 δ δ 0 Lemma 4 Le {f n } be a sequence of funcions analyic on an oen subse S of he comle lane, and assume ha {f n } converges uniformly on every comac subse of S o a limi funcion f Then f is analyic on S and he sequence of derivaives {f n} converges uniformly on every comac subse of S o he derivaive f Proof Since f n is analyic on S we have Cauchy s inegral formula f n a) = f n z) πi z a dz where D is any comac disk in S, D is is osiively oriened boundary, and a is any inerior oin of D Because of uniform convergence we can ass o he limi under he inegral sign and obain fa) = πi D D fz) z a dz which imlies ha f is analyic inside D For he derivaive we have f na) = πi D f n z) z a) dz and f a) = πi D fz) z a) dz from which i follows easily ha f na) f a) uniformly on every comac subse of S as n Now we are ready o rove he following heorems Theorem 5 A Dirichle series fn)n s converges uniformly on every comac subse lying inerior o he half-lane of convergence σ > σ 0

ANALYTIC PROOF OF THE PRIME NUMBER THEOREM 9 Proof I suffices o show ha fn)n s converges uniformly on every comac recangle R = [α, β] [c, d] wih α in he half-lane of convergence he half-lane of convergence is simly he half-lane in which he series converges) To do his we use he esimae obained in Lemma 3 8) a<n b fn) n s Maσ0 σ + s s ) 0 σ σ 0 where s 0 = σ 0 + i 0 is any oin in he half-lane where of convergence We choose s 0 = σ 0 where σ 0 < α Then if s C we have σ σ 0 α σ 0 and s 0 s < C, where C is a consan deending on s 0 and R bu no on s Then 8) imlies a<n b fn) n s Maσ0 σ + C ) = Ba σ0 σ α α 0 where B is indeenden of s Since a σ0 α 0 as a + he Cauchy condiion for uniform convergence is saisfied Theorem 6 The sum funcion F s) = fn)n s of a Dirichle series is analyic in is half-lane of convergence, and is derivaive F s) is reresened in his halflane by he Dirichle series 9) F s) = obained by differeniaing erm by erm n= fn) log n n s, Proof We aly Theorem 5 and Lemma 4 o he sequence of arial sum We now aly he revious heorem o ζs) Differeniaing ζs) erm by erm and summing over all n will give us ζ log n s) = n s n= Alying Theorem 6 o ζs) and ζ s) will give us 0) ζ s) ζs) = n= Λn) n s 5 Conour Inegral Reresenaion of ψ )/ Our ne goal is o rove he asymoic relaion in 0) by reresening ψ )/ as a conour inegral For he conour inegral reresenaion of ψ )/, we need some knowledge abou he gamma funcion Γs) defined as ) Γs) = 0 s e d for s = σ + i where σ > 0 A aricularly useful roery of Γs) is given by he following funcional equaion: ) Γs + ) = sγs)

0 SMITH & TIAN Anoher useful roery of he gamma funcion is ha he gamma funcion has simle oles a non-osiive inegers wih residue ) n /n! a n Wih hese observaions, we can now roceed o rove he following lemma Lemma 5 If c > 0 and u > 0, hen for every ineger k we have c+ i u z πi zz + ) z + k) dz = k! u)k if 0 < u, 0 if u >, he inegral being absoluely convergen Proof Firs we noe ha he inegrand is equal o u z Γz)/Γz + k + ) This follows by reeaed use of he funcional equaion in ) To rove he lemma we aly Cauchy s residue heorem o he inegral πi CR) u z Γz) Γz + k + ) dz, where CR) is he conour shown in he following grah a) if 0 < u and b) if u > The radius R of he circle is greaer han k + c so ha all he oles a z = 0,,, k lie inside he circle k R c R a) 0 < u b) u > Now we show ha he inegral along each of he circular arcs ends o 0 as R If z = + iy and z = R he inegrand is dominaed by u z zz + ) z + k) = u z z + z + k u c R z + z + k The inequaliy u u c follows from he fac ha u is an increasing funcion of if 0 < u and a decreasing funcion if u > Now if n k we have c z + n z n = R n R k R since R > k Therefore he inegral along each circular arc is dominaed by πru c R R) k = O R k ) and his ends o 0 as R since k

ANALYTIC PROOF OF THE PRIME NUMBER THEOREM If u > he inegrand is analyic inside CR) hence C R) = 0 Leing R we find ha he lemma is roved in his case If 0 < u we evaluae he inegral around CR) by Cauchy s residue heorem The inegrand has oles a he inegers n = 0,,, k, hence u z Γz) πi CR) Γz + k + ) dz = = k Res z= n n=0 k n=0 = k! Leing R we obain he lemma u z Γz) Γz + k + ) u n Γk + n) Res z= n Γz) = k n=0 k ) u) n = n u)k k! Now we are ready o reresen ψ )/ as a conour inegral Theorem 7 If c > and we have 3) ψ ) = πi c+ i s ss + ) ) ζ s) ds ζs) k n=0 u n ) n k n)!n! Proof From 6) we have ψ )/ = n n/)λn) Now use Lemma 5 wih k = and u = n/ If n we obain n = πi c+ i /n) s ss + ) ds Mulilying his relaion by Λn) and summing over all n we find ψ ) = n c+ i Λn)/n) s ds = πi ss + ) n= since he inegral vanishes if n > This can be wrien as 4) ψ ) = n= c+ i c+ i Λn)/n) s ds πi ss + ) f n s) ds, where f n s) = πi Λn)/n)s ss + ) Ne we wish o inerchange he sum and inegral in 4) For his i suffices o rove ha he series 5) n= c+ i f n s) ds is convergen The arial sums of his series saisfy he inequaliy N n= c+ i Λn)/n) c s s + ds = N n= Λn) n c c+ i c s s + ds A n= Λn) n c,

SMITH & TIAN where A is a consan, so 5) converges Hence we can inerchange he sum and inegral in 4) o obain ψ ) c+ i = f n s) ds = c+ i s Λn) πi n= ss + ) n s ds n= = c+ i s ) ζ s) ds πi ss + ) ζs) by 0) Now divide by o obain 3) Theorem 8 If c > and we have ψ ) 6) ) = πi where 7) hs) = c+ i s hs) ds, ζ s) ss + ) ζs) ) s Proof This ime we use Lemma 5 wih k = o ge ) = c+ i s πi ss + )s + ) ds, where c > 0 Relace s by s in he inegral keeing c > ) and subrac he resul from 3) o obain Theorem 8 Subsiuing s wih c + i will yield which convers 7) ino 8) ψ ) s = c i = c i log e ) = c c+ i hc + i)e i log d Our ne ask is o show ha he righ member in 8) ends o 0 as We wan o firs show ha we can simly u c = in 8) For his we need o sudy ζs) in he neighborhood of he line σ = 6 ζs) near and on he Line σ = In his secion, we will base our work on he following wo facs We will no rove hese facs, alhough he roofs can be easily found in books on analyic number heory, comle analysis, or Fourier series see references) Firs of all, for all s = σ + i wih σ > 0, we have 9) ζs) = N n= n s s N [] s+ s N d + s Differeniaing each member of 9) will give us he second fac ha N ζ log n []) log [] s) = n s + s n= N s+ d N s+ d 30) N s log N N s s s )

ANALYTIC PROOF OF THE PRIME NUMBER THEOREM 3 Now we are ready o obain uer bounds for ζs) and ζ s) Theorem 9 For every A > 0 here eiss a consan M deending on A) such ha 3) ζs) M log and ζ s) M log for all s wih σ / saisfying 3) σ > A log and e Proof If σ we have ζs) ζ) and ζ s) ζ ) and he inequaliies in 3) are rivially saisfied Therefore we can assume σ < and e We have s σ + + < and s so / s / Esimaing ζs) by using 9) we find ζs) N n= σ n σ + N d + N σ+ = N n= n σ + σn σ + N σ Now we make N deend on by aking N = [] Then N < N + and log n log if n N The inequaliy 3) imlies σ < A/ log so n σ = n σ n = n e σ) log n < n ea log n/ log ) n ea = O n σn σ N + N = O) and N σ ζs) = O N n= = N ) N σ O = O), N ) + O) = Olog N) + O) = Olog ) n This roves he inequaliy for ζs) in 3) To obain he inequaliy for ζ s) we aly he same ye of argumen o 30) The only essenial difference is ha an era facor log N aears on he righ Bu log N = Olog ) so we ge ζ s) = Olog ) in he secified region Theorem 0 If σ > we have 33) ζ 3 σ) ζσ + i) 4 ζσ + i) Proof Using he Euler roduc sanning over all rimes and he Taylor series eansion of log, we ge { } { ζs) = e log n s = e log } { } ks = e log / s n= k=0 { = e log ) } { } s = e m ms m=

4 SMITH & TIAN ζσ + i) = e = e = e { { { m mσ+i) m= m= e m= im m σ } { } = e m mσ im m= } { } log im e = e im log m mσ ζ 3 σ) = e ζσ + i) 4 = e ζσ + i) = e } { { { Since m mσ > 0, i suffices o show ha = e m= m= m= m= { } 3 m mσ m mσ m= 4 cosm log ) m mσ cosm log ) m mσ 3 + 4 cosm log ) + cosm log ) 0 cosm log ) m mσ 3 + 4 cosm log ) + cosm log ) = cos m log ) + 4 cosm log ) + Hence he heorem is roved Theorem R, ζ + i) 0 } } = cosm log ) + ) 0 Proof We only need o consider 0 Rewrie he revious heorem by dividing boh sides by σ 4 34) {σ )ζσ)} 3 ζσ + i) σ ) ζσ + i) σ This is valid if σ > Now le σ + in 34) The firs facor aroaches since ζs) has residue a he ole s = The hird facor ends o ζ + i) If ζ + i) were equal o 0, he middle facor could be wrien as ζσ + i) ζ + i) σ 4 ζ + i) 4 as σ + Therefore, if for some 0 we had ζ + i) = 0, he lef member of 34) would aroach he limi ζ + i) 4 ζ + i) as σ + Bu he righ member ends o as σ + and his gives a conradicion Theorem There is a consan M > 0 such ha ζs) < M log7 whenever σ and e and ζ s) ζs) < M log9 }

ANALYTIC PROOF OF THE PRIME NUMBER THEOREM 5 Proof For σ we have ζs) = and by 0) we have n= µn) n s ζ s) ζs) = n= n= Λn) n, n ζ) so he inequaliies hold rivially if σ Suose, hen, ha σ and e Rewrie Theorem 0 by dividing boh sides by ζσ + i) and ake he fourh roo of boh sides ζσ + i) ζσ)3/4 ζσ + i) /4 Now σ )ζσ) is bounded by, say, M in he inerval σ, where M is an absolue consan ζσ) M if < σ σ Also, ζσ + i) = Olog ) if σ by Theorem 9 So for < σ we have ζσ + i) M 3/4 log ) /4 Alog )/4 = σ ) 3/4 σ ), 3/4 where A is an absolue consan Therefore for some consan B > 0 we have 35) ζσ + i) > Bσ )3/4 log ) /4, if < σ and e This also holds rivially for σ = Le α be any number saisfying < α < Then if σ α, e, we may use Theorem 9 o wrie ζσ + i) ζα + i) α Hence, by he riangle inequaliy, ζσ + i) ζα + i) ζσ + i) ζα + i) σ ζ u + i) du α σ)m log α )M log ζα + i) α )M log Bα )3/4 log ) /4 α )M log This holds if σ α, and by 35) i also holds for α σ since σ ) 3/4 α ) 3/4 In oher words, if σ and e we have he inequaliy ζσ + i) Bα)3/4 log ) /4 α )M log for any α, ) Now we make α deend on and choose α so ha he firs erm on he righ is wice he second This requires ) 4 B α = + M log ) 9 Clearly α > and also α < if 0 for some 0 Thus, if 0 and σ we have ζσ + i) α )M log = C log ) 7

6 SMITH & TIAN This inequaliy also holds wih erhas) a differen C if e 0 This roves ha ζs) C log 7 for all σ, e, giving us a corresonding uer bound for /ζs) To ge he inequaliy for ζ s)/ζs) we aly Theorem 9 and obain an era facor of log Now we are ready o finish he analyic roof of he rime number heorem, which we do by roving ha he asymoic relaion in 0) holds We sar by roving he following lemma 7 Comleion of he Proof of he Prime Number Theorem Lemma 6 If fs) has a ole of order k a s = α hen he quoien f s)/fs) has a firs order ole a s = α wih residue k Proof We have fs) = gs)/s α) k, where g is analyic a α and gα) 0 Hence for all s in a neighborhood of α we have f s) = g s) s α) k kgs) { } gs) k = s α) k+ s α) k s α + g s) gs) f s) fs) = k s α + g s) gs) This roves he lemma since g s)/gs) is analyic a α Theorem 3 The funcion is analyic a s = F s) = ζ s) ζs) s Proof By Lemma 6, ζ s)/ζs) has a firs order ole a wih residue, as does /s ) Hence heir difference is analyic a s = The final ar of he roof makes use of he Riemann-Lebesgue lemma from Fourier analysis, which we do no rove here I saes ha if f) d converges, hen lim r herefore, so oo does lim r f) sin r d = 0 and lim r f)e ir d = 0 Theorem 4 For we have ψ ) ) = π where he inegral lemma we have 36) ψ ) f) cos r d = 0 hold, and h + i)e i log d, h+i) d converges Therefore, by he Riemann-Lebesgue and hence ψ) as

ANALYTIC PROOF OF THE PRIME NUMBER THEOREM 7 Proof In Theorem 8 we roved ha if c > and we have ψ ) ) = c+ i s hs) ds, πi where ζ s) hs) = ss + ) ζs) s Our firs ask is o show ha we can move he ah of inegraion o he line σ = To do his we aly Cauchy s heorem o he recangle R shown below ) T 0 c σ T The inegral of s hs) around R is 0 since he inegrand is analyic inside and on R Now we show ha he inegrals along he horizonal segmens end o 0 as T Since he inegrand has he same absolue value a conjugae oins, i suffices o consider only he uer segmen, = T On his segmen, we have he esimaes ss + ) T and ss + )s ) T 3 T Also, here is a consan M such ha ζ s)/ζs) M log 9 if σ and e Hence if T e we have hs) M log9 T T so ha c c s hs) ds c M log9 T T dσ = M c log9 T T c )

8 SMITH & TIAN Therefore he inegrals along he horizonal segmens end o 0 as T, and hence we have c+ i s hs) ds = + i i On he line σ = we wrie s = + i o obain πi Now we noe ha For so e e inegral h + i) d = + i i e e h + i) d we have s hs) ds = π h + i) d + h + i) d converges Similarly, e h + i) M log9 e s hs) ds h + i)e i log d h + i) d + e h + i) d h + i) d converges, so he enire h+i) d converges By he Riemann-Lebesgue lemma, his imlies ha lim h + i)e i log d = 0 This gives us he asymoic relaion: ψ ) ψ) as Hence he rime number heorem is roved References [] Aosol, T M, Inroducion o Analyic Number Theory, Sringer-Verlag, New York, 976 [] Brown, J W, Churchill, R V, Fourier Series and Boundary Value Problems, McGraw-Hill, Columbus, OH, 000 [3] Conway, J B, Funcions of One Comle Variable, Sringer, New York, 986