TWO PROOFS OF THE PRIME NUMBER THEOREM. 1. Introduction

Transcription

1 TWO PROOFS OF THE PRIME NUMBER THEOREM PO-LAM YUNG Introduction Let π() be the number of primes The famous prime number theorem asserts the following: Theorem (Prime number theorem) () π() log as + (This means lim + (π() log )/ = ) It has been known since Euclid that there are infinitely many primes Euler gave an alternative proof of the infinitude of primes based on the divergence of /p But it seems that Gauss and Legendre were the first to consider distributions of primes By studying large tables of primes (primes up to millions!), Gauss noted that the density of primes is approimately / log Chebyshev made some important progress in the 850 s The landmark paper of Riemann [0] (and his only one on this subject) made clear the connection of the asymptotics of π() to the ζ function that now bears his name Subsequently, in 896, the first complete proof of the prime number theorem was given independently by Hadamard [5] and de la Vallée Poussin [] About 50 years later, elementary approaches to the prime number theorem were also discovered, most notably by Erdös [3] and Selberg [8] Our goal in this article is to elucidate a comple analytic proof of the prime number theorem, given in Chapter 7 of [9] We will also give a variant of that proof based on the work of D J Newman [6] (but proceeds via Chebyshev s ψ function instead of ϕ; see also the eposition in [] for another account of Newman s proof) Before we begin, we note here that Li() = log t dt also satisfies Li() / log So the prime number theorem can also be written as π() Li() as + Indeed, Li() satisfies the following asymptotics: for any N > 0, one has Li() = ( ) log + (log ) +! + + (N )! (log ) 3 (log ) N + O (log ) N+ Li() turns out to be a better approimation of π() than / log Date: October 6, 06

2 Chebyshev s ψ function The proofs of the prime number theorem we will give proceeds via Chebyshev s ψ function: ψ() := [ ] log log p, > 0 log p p The following proposition is well known: Proposition The prime number theorem is equivalent to the assertion that () ψ() Proof Indeed, assume for the moment that () holds Then since ψ() p dividing both sides by and letting +, we get Also, for any α (0, ), we have ψ() log p p α <p lim inf log = π() log, π() log() log p (π() π( α )) log( α ) α(π() α ) log Hence if () holds, then dividing the above inequality by, and letting, we get that α lim sup + (π() α ) log = α lim sup + Letting α, we get π() log lim sup + Together we obtain (), and the prime number theorem holds π() log The converse implication, namely that () implies (), is not much harder Since we do not need this direction of the implication, we leave this verification to the interested reader Note that ψ can be rewritten as ψ() = p m N: p m log p = n Λ(n) where Λ is the von Mangoldt function, defined for n N by { log p, if n = p m for some prime p and some positive integer m Λ(n) = 0, otherwise We are interested in the asymptotics of ψ() as + We will see, in the net section, that one can study this by considering the Dirichlet series corresponding to {Λ(n)} n=, namely Λ(n) n s n= See also Proposition of Chapter 7 of [9]

3 This is in turn basically the logarithmic derivative of the Riemann zeta function, and it is ultimately why the Riemann zeta function makes its appearance in this approach towards the prime number theorem 3 The Mellin transform We digress a little to discuss three important integral transforms: the Mellin transform, the Laplace transform and the Fourier transform Suppose f : (0, ) C is a measurable function that vanishes on (0, ) Suppose further that there eists some a R, A > 0 such that f() A a for all [, ) Let a 0 be the infimum of all a R, for which there eists A > 0 such that the above estimate holds Then the Mellin transform of f is defined by Mf(s) = 0 f() s d for all s C with Re s > a 0 ; indeed the integral defining Mf(s) converges absolutely there, and defines a holomorphic function of s in that half plane The Mellin transform is really the Laplace transform in disguise Indeed, suppose F : R C is a measurable function that vanishes on (, 0) Suppose further that there eists some a R, A > 0 such that F (t) Ae at for all t [0, ) Let a 0 be the infimum of all a R, for which there eists A > 0 such that the above estimate holds Then the Laplace transform of F is defined by LF (s) = F (t)e st dt for all s C with Re s > a 0 ; indeed the integral defining LF (s) converges absolutely there, and defines a holomorphic function of s in that half plane If F (t) := f(e t ) where f is as in the above definition of the Mellin transform, then Mf(s) = LF (s) whenever they are defined Recall also the Fourier transform on R If G L (R), then its Fourier transform is defined by Ĝ(τ) = G(t)e itτ dt for all τ R If F is as in the above definition of the Laplace transform, then for all c > a 0 and all τ R, we have LF (c + iτ) = F c (τ) where F c (t) := F (t)e ct Our goal was to understand asymptotics of ψ() = n Λ(n) as + The strategy we will follow is indeed fairly general, and the initial steps works perfectly well when the sequence {Λ(n)} n= is replaced by any sequence of comple numbers {a n} n=, as long as a n = n o() as n We phrase it in the following proposition: 3

4 Proposition 3 Suppose {a n } n= is a sequence of comple numbers satisfying a n = n o() as n Let f : (0, ) C be defined by f() = n a n Then the Mellin transform of f is defined for all s C with Re s >, and is given by for all such s, where Mf(s) = s D a(s) D a (s) := is the Dirichlet series corresponding to the sequence {a n } n= (also defined for Re s > ) In particular, the Mellin transform of ψ is defined for all s C with Re s >, and is given for all such s by Mψ(s) = Λ(n) s n s n= n= a n n s Proof Suppose a n = n o() as n, and f() = n a n Then for any a >, there eists A > 0 such that f() A a Furthermore, we can rewrite f(), as f() = a n χ [n, ) () n= where χ [n, ) is the characteristic function of the interval [n, ) Hence the Mellin transform of f is defined for all s C with Re s >, and is given by Mf(s) = a n s d = s n= n n= a n n s = s D a(s) for all such s (The interchange of the sum with the integral can be justified using Fubini s theorem) Since Λ(n) = n o() as n, applying the result to a n = Λ(n) yields the desired conclusion for Mψ(s) The proposition suggests that in order to understand ψ() = n Λ(n) (or more generally f() = n a n where {a n } is as in the proposition), it may be helpful to study the corresponding Dirichlet series D Λ (s) (or D a (s)); indeed, if we can invert the Mellin transform, then we can hope to convert information about the Dirichlet series D Λ (s) (or D a (s)) into information about ψ() (or f()) The success of this approach ultimately lies with our ability to invert the Mellin transform; we study the latter, by studying how one could invert the Fourier and the Laplace transforms First, recall that the Fourier transform can be inverted by the following formula under suitable hypothesis on G For instance, if both G and Ĝ are in L (R), then for all t R G(t) = π Ĝ(τ)e itτ dτ We will need a slightly different form of the Fourier inversion formula, when Ĝ is not necessarily integrable: 4

5 Proposition 4 Suppose G L (R), and t 0 R is a point where the following limits all eist: G (t + 0 G(t + 0 ) := lim t t + 0 ) := lim G(t), t t + 0 G(t 0 ) := lim G(t), t t 0 G(t) G(t + 0 ), G (t 0 ) := lim t t 0 t t 0 G(t) G(t 0 ) t t 0 Then T lim Ĝ(τ)e it0τ G(t + 0 dτ eists, and equals ) + G(t 0 ) T + π T In particular, if G L (R) is piecewise C (meaning that there eists a strictly increasing sequence {t n } n= such that for all n Z, G is differentiable on (t n, t n+ ), and both G(t) and G (t) has a limit as t t + n and t t n+ ), then for all t R lim T + π T T Ĝ(τ)e itτ dτ = G(t+ ) + G(t ) The proof uses the famous lemma of Riemann-Lebesgue: Lemma 5 (Riemann-Lebesgue) If H L (R), then Ĥ(τ) 0 as τ ± In particular, if H L (R), then (3) H(t) sin(tt )dt = as T + R Ĥ( T ) Ĥ(T ) i 0 Proof of Lemma 5 If h is a smooth function with compact support on R, then ĥ is rapidly decreasing at infinity; indeed ĥ(τ) = h(t)e itτ dt = ( R (iτ) N h(t) d ) N e itτ dt = d N h R dt (iτ) N R dt N e itτ dt for all N N, so ĥ(τ) C N τ N for all N N, where C N := dn h L In particular then ĥ(τ) 0 as τ ± Now if dt N (R) H L (R), then we approimate H by a smooth function with compact support; indeed for any ε > 0, there eists a smooth function h with compact support on R, such that It follows that H h L (R) ε Ĥ(τ) Ĥ h(τ) + ĥ(τ) H h L (R) + ĥ(τ) ε + ĥ(τ) Since ĥ(τ) 0 as τ ±, we conclude that lim sup Ĥ(τ) ε τ ± Since this is true for all ε > 0, we conclude that Ĥ(τ) 0 as τ ±, as desired 5

6 Proof of Proposition 4 Suppose t 0 is as in the proposition For any T > 0, we have T Ĝ(τ)e it0τ dτ = T ( ) G(t)e itτ dt e it0τ dτ π T π T R = T G(t) e i(t0 t)τ dτdt π R T = G(t) sin((t 0 t)t ) dt π R t 0 t = G (t 0 t) sin(tt ) dt π t R (The interchange of the integrals in the second equality is justified by Fubini s theorem since G L (R)) We also recall the well-known fact that (4) 0 sin t dt = π t (This can be proved, for instance, using contour integrals) Thus and It follows that π = π T T 0 G(t + 0 ) = π G(t 0 ) = π 0 Ĝ(τ)e it 0τ dτ G(t+ 0 ) + G(t 0 ) G (t 0 t) G(t + 0 ) t 0 sin(tt )dt + π G(t + 0 )sin(tt ) dt, t G(t 0 )sin(tt ) dt t 0 G (t 0 t) G(t 0 ) t sin(tt )dt Now since G (t + 0 ) and G (t 0 ) both eist, there eists some > 0 such that [G(t 0 t) G(t + 0 )]/t is bounded for t (, 0), and [G(t 0 t) G(t 0 )]/t is bounded for t (0, ) We define [G (t 0 t) G(t + 0 )]/t if t (, 0) H(t) := [G (t 0 t) G(t 0 )]/t if t (0, ) G(t 0 t)/t if t Then H L (R) (because H(t) is bounded when 0 < t <, and H(t) G(t 0 t) when t ), and π = π T R T Ĝ(τ)e it0τ dτ G(t+ 0 ) + G(t 0 ) H(t) sin(tt )dt + G(t+ 0 ) π sin(tt ) dt + G(t 0 ) t π sin(tt ) dt t As T +, the first term on the right hand side tends to zero by the lemma of Riemann-Lebesgue (see (3)) The second and the third terms tend to zero as well, since sin(tt ) dt = t sin(tt ) dt = t as T + (see (4)) This concludes the proof of Proposition 4 6 T sin(t) dt 0 t

7 In view of the connection of the Laplace transform to the Fourier transform, we obtain the following corollary of Proposition 4: Proposition 6 Suppose F : R C is a piecewise C function that vanishes on (, 0) Suppose further that there eists some a R, A > 0 such that F (t) Ae at for all t [0, ) Let a 0 be the infimum of all a R, for which there eists A > 0 such that the above estimate holds Then for all c > a 0, and all t R, we have (5) πi c+i c i LF (s)e st ds = F (t+ ) + F (t ), in the sense that if γ c,t is the vertical contour joining c it to c + it, then lim LF (s)e st ds T + πi γ c,t eists, and is equal to (F (t + ) + F (t ))/ The integral on the left-hand side of (5) is called the Bromwich integral The proposition gives a precise set of conditions under which the Bromwich integral inverts the Laplace transform of a function Proof of Proposition 6 Suppose F and c are as above Then F c (t) := F (t)e ct is piecewise C, and is in L (R) since there eists A > 0 such that F c (t) Ae (c a0)t/ for all t [0, ) Now LF (c + iτ) = F c (τ) for all τ R Thus Proposition 4 applied to F c shows that πi c+i c i LF (s)e st ds = πi = ect π lim T T + T T lim T + T = e ct F c(t + ) + F c (t ) = F (t+ ) + F (t ) LF (c + iτ)e (c+iτ)t idτ F c (τ)e itτ dτ By a change of variable = e t, we obtain the following corollary for the inverse of the Mellin transform as well Proposition 7 Suppose f : (0, ) C is a piecewise C function that vanishes on (0, ) Suppose further that there eists some a R, A > 0 such that f() A a for all [, ) Let a 0 be the infimum of all a R, for which there eists A > 0 such that the above estimate holds Then for all c > a 0, and all > 0, we have πi c+i c i Mf(s) s ds = f(+ ) + f( ), 7

8 in the sense that if γ c,t is the vertical contour joining c it to c + it, then lim Mf(s) s ds T + πi γ c,t eists, and is equal to (f( + ) + f( ))/ Proof Apply Proposition 6 to F (t) := f(e t ), noting that Mf(s) = LF (s) The following proposition then follows: Proposition 8 Suppose {a n } n= and D a(s) are as in Proposition 3 Then for all c >, and all > 0, we have { c+i (6) D a (s) s πi s ds = n< a n if / N a + n< a n if N c i In particular, for all c >, and all > 0, we have { c+i Λ(n) s (7) πi n s s ds = ψ() if / N (ψ( + ) + ψ( ))/ if N c i n= Equation (6) is sometimes known as Perron s formula Proof of Proposition 8 To prove (6), it suffices to apply Proposition 7 to f() := n a n, since f() is piecewise constant, and Proposition 3 shows that Mf(s) = D a (s)/s whenever Re s > Equation (7) then follows from (6) by setting a n = Λ(n) 4 Relation to the Riemann ζ function We now continue to prove the prime number theorem Our strategy was to prove the asymptotics () where ψ() := n Λ(n) In the previous section, we have epressed ψ in terms of the Dirichlet series of {Λ(n)} n=, namely n= Λ(n) n s This is intimately connected to the Riemann ζ function, which we will see as follows Recall that the Riemann ζ function is defined by ζ(s) =, valid for Re s > ns n= We note here that Proposition 7 also provides an alternative proof of Lemma 4 in Chapter 7 of [9] Indeed, if we take f to be such that f() = (/) for, and f() = 0 for (0, ), then f is bounded, continuous, piecewise C, and the Mellin transform of f is Mf(s) = /s(s + ) for all s C with Re s > 0 Thus Proposition 7 implies Lemma 4 of Chapter 7 of [9] Similarly, instead of doing a contour integration, one can work out Eercise 6 of Chapter 7 of [9] by interpreting it as an appropriate instance of Proposition 7 We leave the details to the interested readers 8

9 We will assume known the following product factorization of ζ over all primes: ζ(s) =, valid for Re s > p s p Taking logarithmic derivative, we get ζ (s) ζ(s) = p p s log p p s = p valid for Re s >, so in view of the definition of Λ(n), we see that (8) Thus (7) can now be rewritten as (9) πi c+i c i ζ (s) ζ(s) = n= ζ (s) s ζ(s) s ds = m log p p ms, Λ(n), valid for Re s > ns { ψ() if / N (ψ( + ) + ψ( ))/ if N We may thus hope to obtain asymptotics of ψ(), by studying bounds for ζ and ζ Observe that s = c if s is on the contour of integration in the integral in (7) or (9) Thus to show that ψ() remains small as +, we should shift the contour of integration {Re s = c} to the left as far as Λ(n) s n= n s s possible A technical point arises here: the integrand in (7) or (9), namely, is O(/ s ) only on the contour of integration So the integrals in (7) or (9) may not converge absolutely, and this is inconvenient when we shift the contour integrals As a result, in the net section, we show that instead of studying asympotics of ψ(), it suffices to study the asymptotics of a smoothed out version of ψ(), that we denote by ψ () This ψ () has a Mellin transform that decays more rapidly at infinity, and the analog of (7) or (9) for ψ would converge absolutely, making it easier to deal with 5 A technical point Continuing from the last section, let ψ : (0, ) R be defined by for all > 0 ψ () = Proposition 9 ψ() if and only if ψ () / 0 ψ(y)dy Thus in view of Proposition, to prove the prime number theorem, it suffices to prove that (0) ψ () Proof of Proposition 9 Indeed, suppose (0) holds Then for any α (0, ), we have so ψ () ψ (α) = ψ() α ψ () ψ (α) ( α) = ψ(y)dy ( α)ψ(), 9 ψ () ψ (α) (α) α α

10 Letting +, we see that Letting α, we see that Similarly, for any β (, ), so ψ() Letting +, we see that lim sup + ψ() ψ (β) ψ () = lim sup + α α = + α β ψ() ψ (β) ψ () (β ) = ψ(y)dy (β )ψ(), ψ (β) β ψ () (β) β ψ() lim inf + β β = β + Letting β +, we see that ψ() lim inf + Together we see that ψ(), as desired The converse implication is similar Since we do not need this direction of the implication, we leave this verification to the interested reader So from now on, we concentrate on proving asymptotics (0) for ψ () Note that for any a >, there eists a constant A > 0 such that ψ ()/ A a Thus the Mellin transform of ψ ()/ is defined for all s C with Re s >, and is given for all such s by 0 ψ () d s = s + = 0 s + Mψ(s) = s(s + ) n= ψ () d d (s+) d Λ(n) n s ζ (s) = s(s + ) ζ(s) Now since ψ() = n Λ(n) is piecewise constant on (0, ), ψ () = 0 ψ(y)dy is continuous and piecewise linear there Hence ψ ()/ is continuous and piecewise C on (0, ) Proposition 7 then shows that for all c > and all > 0, we have 3 ie ψ () = πi () ψ () = πi c+i c i c+i c i 3 This is precisely Proposition 3 of Chapter 7 of [9] 0 ζ (s) s(s + ) ζ(s) s ds ζ (s) s(s + ) ζ(s) s+ ds

11 ζ We note that the integrand above, namely (s) s(s+) ζ(s) s+, is O(/ s ) on the contour of integration, thanks to the appearance of the quadratic factor s(s + ) in the denominator (contrary to the linear factor s in the integrand of (7) or (9)) This makes it easier for us to shift the contour of integration {Re s = c} in a moment To accomplish the latter, we will need to know that ζ continues meromorphically past the line {Re s = } We summarize in the net section the facts we will need about the continuation of ζ 6 Analytic continuation of ζ We will assume known that ζ has a meromorphic continuation to the half-space S η := {s C: Re s > η} for some η <, so that the only singularity of ζ in this strip is a simple pole at s = In other words, there eists η <, and a holomorphic function h(s) on S η, such that () ζ(s) = h(s) s on S η We will also assume known the following upper bound for ζ : for any ε > 0, there eists A > 0, such that (3) ζ (s) A Im s ε/ whenever s C with Re s and Im s All these can be proved, for instance, by comparing ζ(s) = n= n s to the corresponding integral s d = /(s ) (and using Cauchy s estimate for the bound on ζ ) 4 Indeed, with more work (for instance by establishing the functional equation of ξ(s) := π s/ Γ(s/)ζ(s)), one can take η all the way to in the above claims 7 Non-vanishing of ζ on Re s = Recall that our strategy towards proving the prime number theorem is to establish asymptotics (0) for ψ () We have represented ψ () as a contour integral in () With the analytic continuation of ζ in the above section in mind, we would like to shift the contour of integration to the left as far as possible (just like what is typically done when one computes the Bromwich integral, in inverting the Laplace transform) This relies on knowing where the zeroes of ζ(s) are, since every zero of ζ contributes a pole in s of the integrand in () We now prove the following theorem Theorem 0 ζ has no zeroes on the vertical line where Re s = Proof We need three observations First, note that ζ is real-valued on {s R: s > } Hence ζ(s) = ζ(s) for all s C This shows that the (non-real) zeroes of ζ comes in conjugate pairs: if s is a zero of ζ, then so is s, and ζ vanishes to the same order at both s and s 4 See Proposition 5, Corollary 6, Proposition 7 in Chapter 6 of [9]

12 Net, recall that if f is a meromorphic function near a point z 0, then the order of of f at z 0 (which is positive if f vanishes there, negative if f has a pole there) can be computed via the residue of the logarithmic derivative of f at z 0 In particular, for any t R, the order of ζ at + it is ɛ ζ ( + it + ɛ) (4) ord +it ζ = lim ɛ 0 ζ( + it + ɛ) Finally, recall the logarithmic derivative of ζ, given by (8) What will be important for us is that Λ(n) is real and non-negative for all positive integers n Now we are ready to put all these together First, since ζ has a pole at s =, it cannot have a zero at s = Net, suppose ζ is zero at s = + it for some t R We show that this is impossible by considering the orders of ζ at ± it, ± it and : Combining (4) and (8), we see that ord +it ζ = lim ɛ ɛ 0 + n= ord +it ζ = lim ɛ ɛ 0 + n= Λ(n) n it n+ɛ Λ(n) n it n+ɛ ord ζ = lim ɛ Λ(n) ɛ 0 + n +ɛ n= ord it ζ = lim ɛ ɛ 0 + n= Λ(n) nit n+ɛ ord it ζ = lim ɛ Λ(n) nit ɛ 0 + n+ɛ n= We now multiply these five equations by, 4, 6, 4, respectively, and add them all up Observe that (5) n it + 4n it n it + n it = (n it/ + n it/ ) 4 = ( cos(t log n/)) 4 0 Since Λ(n) 0 for all n, we then see that ord +it ζ + 4 ord +it ζ + 6 ord ζ + 4 ord it ζ + ord it ζ 0 But ord ζ =, ord +it ζ = ord it ζ, and ord +it ζ = ord it ζ 0 Hence 8 ord +it ζ 6 0, which contradicts our assumption that ζ( + it) = We remark that (5) is really Lemma 4 of Chapter 7 of [9] in disguise Also, by rewriting ζ /ζ as the derivative of log ζ, and undoing the derivative, the above argument essentially gives Corollary 5 of Chapter 7 of [9]

13 By making the above argument more quantitative, we can establish 6 the following lower bound of ζ: for any ε > 0, there eists a constant B > 0, such that (6) ζ(s) B Im s ε/ whenever s C with Re s and Im s In particular, combining with our earlier bound (3) for ζ, we see that for any ε > 0, there eists a constant C > 0, such that (7) ζ (s) ζ(s) C Im s ε whenever s C with Re s and Im s 8 The Proof of the prime number theorem in [9] We can now finish the proof of the prime number theorem as in Chapter 7 of [9] Fi some c > Then () says that ψ () = c+i ζ (s) πi c i s(s + ) ζ(s) s+ ds, where the integration is along the vertical contour {c + iτ : τ R} The integrand is a holomorphic function of s on an open half-space {Re s > η} for some η < In view of estimate (7) for ζ /ζ, and that s+ Re s+ s(s + ) Im s, we can shift the contour of integration, and obtain, for any T > 0, that ψ () = ζ (s) πi s(s + ) ζ(s) s+ ds, γ(t ) where γ(t ) is the contour consisting of 5 straight line segments, joining the following points in order: i, it, c it, c + it, + it, and + i (See p 95 of [9] for a picture of γ(t )) Suppose now ε > 0 is given In view of estimate (7) for ζ /ζ again, we may choose T > 0 large enough, so that ( it +i ) ζ (s) + π i +it s(s + ) ζ(s) ds < ε Then since s+ = on the contours of integration in the above two integrals, we get ( it +i ) ζ (s) + πi s(s + ) ζ(s) s+ ds < ε i Hence for this choice of T, we have (8) ψ () = πi ( c it it +it c+it + + c it +it c+it ) ζ (s) s(s + ) ζ(s) s+ ds + O(ε ) for all > 0, where O(ε ) is a term bounded in absolute value by ε Having chosen T, let (0, min{ η, }) be sufficiently small, where η is as in the above, so that ζ has no zeroes in the closed rectangle { Re s, Im s T } Such eists because ζ etends meromorphically on S η as in the description just before () (so the zeroes of ζ has no accumulation points in S η ), and because ζ has no zeroes on the line segment {Re s =, Im s T } Let γ(t, ) be the contour consisting of 5 straight line segments, joining the following points in order: i, it, it, 6 See Proposition 6 in Chapter 7 of [9] 3

14 + it, + it, and + i (See p 95 of [9] for a picture of γ(t, )) Then γ(t ) γ(t, ) is a rectangular contour with vertices it, c it, c + it, and + it On and inside this rectangular contour, ζ has a pole at s =, and no zeroes anywhere Thus ζ /ζ is meromorphic on and inside this rectangular contour, and has a residue at s = It follows that for each > 0, the function s ζ (s) s(s+) ζ(s) s+ has a simple pole at s =, and nowhere else inside the rectangle whose vertices are it, c it, c + it, and + it The residue of this function at s = is just ( + ) + = Thus from (8), we see that (9) ψ () = πi ( it it ζ (s) +it +it ) + + it +it ζ (s) s(s + ) ζ(s) s+ ds + O(ε ) Now the function s π s(s+) ζ(s) is continuous on the above 3 contours of integration Hence its modulus is bounded above by some constant C T, there Also, similarly Furthermore, All in all, we see that it it s+ ds = +it +it +it it σ+ dσ s+ ds log s+ ds = T log ; ψ () C T, log + T + ε Dividing by /, we see that ψ () / 4C T, log + 4T + ε Now T and are fied once we fi ε If we pick sufficiently large, the right hand side can be made smaller than 3ε This proves that ψ () /, as desired in (0), and completes the proof of the prime number theorem in Chapter 7 of [9] 9 A variant of Newman s proof The above proof of the prime number theorem is based on analysis of the Chebychev s ψ function: ψ() = log p p m N: p m Some work was needed in obtaining quantitative estimates of ζ near the line Re s = (more precisely, an upper bound for ζ (s)/ζ(s) ) On the other hand, in [6], Newman gave another comple analytic proof of the prime number theorem, using only the vanishing of ζ on the line Re s = (and no asymptotics of ζ there), by considering the ϕ function: ϕ() = p 4 log p

15 Below we try to combine the two approaches, and adapt Newman s argument so that it works through Chebychev s ψ function (rather than the ϕ function) Recall that by Proposition, to prove the prime number theorem, it suffices to verify asymptotics () for ψ It may help to first verify a weaker statement, namely that ψ()/ remains bounded as + This is what we are going to do net, via an essentially elementary argument Proposition ψ() remains bounded as + Proof First, we claim that there eists a constant C, such that for any positive integer n, we have (0) ψ(n) ψ(n) Cn To prove this claim, note that () ψ(n) ψ(n) = p {m: n<p m n} log p = log m= {p: n<p m n} p In the product inside the logarithm, consider first the term corresponding to m = We have ( ) n () p n Indeed {p: n<p n} ( ) n (n)(n ) (n + ) = n n! is an integer, so that n! is a factor of (n)(n ) (n + ); also each prime p with n < p n is a factor of (n)(n ) (n + ) Since each such prime p is relatively prime with n!, we see that (n!) {p: n<p n} p divides (n)(n ) (n + ), ie {p: n<p n} p divides ( ) n n In particular, () holds This further implies (3) p n, since ( ) n n {p: n<p n} n k=0 ( ) n = ( + ) n = n k This completes our estimate of the product inside the logarithm on the right hand side of () Net, we consider those terms in the same product corresponding to m If m, and n < p m n, then p n Also, for each prime p, there is at most one power of p that lies in (n, n] Hence p p ( n n) m= {p: n<p m n} Hence, together with (3), we obtain ψ(n) ψ(n) log This establishes our claim (0) p n ( n ( ) n) n n log(n) n log + 5

16 Now that we have the claim (0), we see that there eists a constant C such that ψ() ψ() C for all In fact it suffices to prove this for large To do so, take n to be the integer closest to Then the sum defining ψ() and ψ(n) differ in at most one term, and ψ() ψ(n) C log C Similarly, ψ() ψ(n) C Hence together with the bound for ψ(n) ψ(n) we already established, we see that ψ() ψ() C, as desired Now we just iterate this estimate: ψ() ψ(/) C (/) ψ(/) ψ(/4) C (/4) and sum up a geometric series on the right Then ψ() C as +, as desired Now recall that the Mellin transform of ψ() was given in Proposition 3, which in view of (8) can be written as ψ() d (4) s = ζ (s), valid for Re s > sζ(s) We are interested in showing ψ() 0 as + Hence we are led to consider the following identity: Proposition ( ) ψ() d s = ζ (s) sζ(s), valid for Re s > s Proof This follows from (4) by simply noting that d s =, valid for Re s > s (This could be interpreted as the Mellin transform of χ [, ) ()) From the meromorphic continuation () of ζ to some half-space {Re s > η} with η <, we see that ζ (s) ζ(s) + s etends to a homormorphic function on the same half-space Hence the right hand side of the identity in Proposition etends to a holomorphic function on an open set containing the closed half plane {Re s } This shows that the following Tauberian theorem applies: Proposition 3 Let f() be a bounded function on [, ), and define g(s) = f() d s for Re s > Then g is holomorphic on Re s > If g etends to a holomorphic function on an open set containing the closed half plane Re s, then f() d eists, and is equal to g() 6

17 Indeed f() := ψ() is a bounded function by Proposition, and the integral f() d s = ( ψ() etends to a holomorphic function on an open set containing the closed half plane {Re s } by Proposition Hence assuming Proposition 3 for the moment, we obtain the following proposition: ( ) ψ() d Proposition 4 The improper integral converges The prime number theorem would then follow Proof of Theorem As observed before, it suffices to verify () We argue by contradiction Suppose () is false Then either there eists α > such that ψ( n ) > α n for a sequence { n } with n +, or there eists β < such that ψ(y n ) < βy n for a sequence {y n } with y n + In the first case, since ψ is an increasing function, we have ψ() ψ( n ) α n whenever n In particular, αn ( ) ψ() d αn ( (5) n αn ) d α ( α ) d n =, the last integral being strictly positive, and independent of n This contradicts Proposition 4: in fact, Proposition 4 implies that αn ( ) ψ() d = n αn ) d s ( ) ψ() d n ( ) ψ() d 0 as n, and this is not compatible with the lower bound we have obtained in (5) Similarly, in the second case, we use ψ() ψ(y n ) < βy n whenever y n, to conclude that yn ( ) ψ() d yn ( ) βyn d ( ) β d = < 0 βy n βy n independent of n This contradicts Proposition 4 β It remains to prove Proposition 3 Proof of Proposition 3 Suppose f is bounded, say f() M for all Suppose also that g(s) := f() d s etends holomorphically to an open set containing the closed half plane {Re s } Let g t (s) = t f() d s Then g t is entire for all t, and our goal is to show that g t () converges to g() as t + For ε > 0 and > 0, let Λ ε, be the positively oriented closed contour, given by (6) Λ ε, = C ε + L () + L () + L (3) 7

18 where C ε be the semicircle in the right half plane {Re s > }, that is centered at and of radius /ε; L () is the horizontal straight line joining + iε to + iε ; L () is the vertical straight line joining + iε to iε ; and L (3) is the horizontal straight line joining iε to iε Then for any ε > 0, as long as is sufficiently small, we have, by Cauchy integral formula, that g t () g() = ds [g t (s) g(s)] πi Λ ε, s For various technical reasons, we will actually use the following identity instead (which also follow from Cauchy s integral formula, since the etra factor t s ( + ε (s ) ) is entire in s, and equals when s = ): (7) g t () g() = πi Λ ε, [g t (s) g(s)]t s ( + ε (s ) ) ds s Now we decompose the above path integral into 4 parts, according to (6) For s C ε, we have Hence (8) L () g t (s) g(s) t s t t Re s d f() Re s Mt Re s Re s, + ε (s ) = s ( iε ) s ( + iε ) Re s ε C ε s = ε +L() +L(3) [g t (s) g(s)]t s ( + ε (s ) ) ds πi C ε s CMε Net, let C ε be the semi-circle in the left half plane {Re s < }, that is centered at and of radius /ε Then we integrate the part concerning g in (7), over C ε instead of over L () + L () + L (3) (This is possible because g t is entire) Hence g t (s)t s ( + ε (s ) ) ds s = g t (s)t s ( + ε (s ) ) ds C ε s But on C ε, we have Similarly as before, on C ε, we have Hence (9) πi L () g t (s) t d f() Re s Mt Re s Re s t s ( + ε (s ) ) C tre s ( Re s) ε +L() +L(3) g t (s)t s ( + ε (s ) ) ds s CMε 8

19 Finally, the contribution of g(s) to the contour integral over L () + L (3) is given by (30) g(s)t s ( + ε (s ) ) ds πi s Cε L () +L(3) where g C on L () L (3) This is because L () and L (3) both have lengths, and that / s ε on L () L (3) (Note also that t s since Re s <, and + ε (s ) C on L () L (3) ) Now the contribution of g(s) to the contour integral over L () is given by (3) πi L () g(s)t s ( + ε (s ) ) ds s C t ε This is because the length of L () is /ε, the function g is bounded by C on L (), and t s = t on L () ; also, + ε (s ) C on L (), and / s / on L () Altogether, by (7), (8), (9), (30) and (3), we see that for any ε > 0, there eists a small > 0, such that g t () g() CMε + Ct ε Letting t +, we see that lim sup g t () g() CMε t + Since ε > 0 is arbitrary, this shows that g t () g() as t +, as desired 0 Concluding remarks We end by mentioning some comparisons of the two proofs of the prime number theorem given above The Tauberian proof based on the work of Newman is shorter, and does not involve the use of any quantitative estimates of ζ on the line {Re s = }, whereas the proof given in [9] requires knowing such estimates Nevertheless, the proof given in [9] works in more general contet, when one wants to obtain asymptotics for n a n for any appropriate sequences a n ; also, the proof is more powerful, in the sense that if we had known further information about the zeroes of ζ (say we know that ζ has no zeroes on {Re s > η} for some particular η > 0), then we can use that to our advantage, and obtain lower order correction terms to the asymptotics of π() (This would be hard to do with the Tauberian argument of Newman) Hence it is useful to know the meromorphic continuation of ζ to a region in the comple plane that is as large as possible, and to understand its zeroes there Indeed, one of the famous eplicit formulas in the theory of primes says ψ () = ρ ρ ρ(ρ + ) E() where the sum is taken over all zeroes ρ of the ζ function in the critical strip {0 Re s }, and E() = O() is an error term 7 (There is also a corresponding eplicit formula for ψ(), ecept that that sum does not converge absolutely, contrary to the one for ψ () given above) This highlights, for instance, the importance of the famous Riemann hypothesis, that all zeroes of ζ on the critical 7 See Problem in Chapter 7 of [9] 9

20 strip {0 < Re s < } is on the line Re s = / It has been shown in [7] that the Riemann hypothesis implies that ψ() < (log ) 8π for all 73, and that π() Li() < log 8π for all 657 (Note that the power of on the right hand side is essentially /) Also, the Riemann hypothesis can be shown [] to be equivalent to the estimate π() Li() C log for some constant C There are actually many other equivalent forms of Riemann hypothesis, which is beyond our scope of discussion here References [] Tom M Apostol, A centennial history of the prime number theorem, Number theory, Trends Math, Birkhäuser, Basel, 000, pp 4 MR [] C-L de la Vallée-Poussin, Recherches analytiques sur la théorie des nombres premiers, Annales de la Société Scientifiques de Bruelles 0B (896), (French) [3] P Erdös, On a new method in elementary number theory which leads to an elementary proof of the prime number theorem, Proc Nat Acad Sci U S A 35 (949) [4] P Garrett, Simple Proof of the Prime Number Theorem, online lecture notes (January 0, 05) [5] J Hadamard, Sur la distribution des zéros de la fonction ζ(s) et ses conséquences arithmétiques, Bull Soc Math France 4 (896), 99 0 (French) MR50464 [6] D J Newman, Simple analytic proof of the prime number theorem, Amer Math Monthly 87 (980), no 9, [7] Lowell Schoenfeld, Sharper bounds for the Chebyshev functions θ() and ψ() II, Math Comp 30 (976), no 34, MR [8] Atle Selberg, An elementary proof of the prime-number theorem, Ann of Math () 50 (949), [9] Elias M Stein and Rami Shakarchi, Comple analysis, Princeton Lectures in Analysis, vol, Princeton University Press, Princeton, NJ, 003 [0] B Riemann, Bestimmung einer Function einer veränderlichen compleen Größe durch Grenz- und Unstetigkeitsbedingungen, J Reine Angew Math 54 (857), 4, DOI 055/crll85754 (German) MR [] Helge von Koch, Sur la distribution des nombres premiers, Acta Math 4 (90), no, 59 8, DOI 0007/BF (French) MR55496 [] D Zagier, Newman s short proof of the prime number theorem, Amer Math Monthly 04 (997), no 8,