Chapter 7 The Prime Number Theorem

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1 Chapter 7 The Prime Number Theorem In this final chapter we will take advantage of an opportunity to apply many of the ideas and results from earlier chapters in order to give an analytic proof of the famous prime number theorem: If π(x) is the number of primes less than or equal to x, then x π(x)lnx asx.that is, π(x) is asymptotically equal to x/ ln x as x. (In the sequel, prime will be taken to mean positive prime.) Perhaps the first recorded property of π(x) is that π(x) as x, in other words, the number of primes is infinite.this appears in Euclid s Elements.A more precise result that was established much later by Euler (737) is that the series of reciprocals of the prime numbers, , is a divergent series.this can be interpreted in a certain sense as a statement about how fast π(x) as x.later, near the end of the 8-th century, mathematicians, including Gauss and Legendre, through mainly empirical considerations, put forth conectures that are equivalent to the above statement of the prime number theorem (PNT).However, it was not until nearly years later, after much effort by numerous 9-th century mathematicians, that the theorem was finally established (independently) by Hadamard and de la Vallée Poussin in 896.The quest for a proof led Riemann, for example, to develop complex variable methods to attack the PNT and related questions. In the process, he made a remarkable and as yet unresolved conecture known as the Riemann hypothesis, whose precise statement will be given later.now it is not clear on the surface that there is a connection between complex analysis and the distribution of prime numbers.but in fact, every proof of the PNT dating from Hadamard and de la Vallée Poussin, up to 949 when P.Erdös and A.Selberg succeeded in finding elementary proofs, has used the methods of complex variables in an essential way.in 98, D.J. Newman published a new proof of the PNT which, although still using complex analysis, nevertheless represents a significant simplification of previous proofs.it is Newman s proof, as modified by J.Korevaar, that we present in this chapter. There are a number of preliminaries that must be dealt with before Newman s method can be applied to produce the theorem.the proof remains far from trivial but the steps

2 2 CHAPTER 7. THE PRIME NUMBER THEOREM along the way are of great interest and importance in themselves.we begin by introducing the Riemann zeta function, which arises via Euler s product formula and forms a key link between the sequence of prime numbers and the methods of complex variables. 7. The Riemann Zeta function The Riemann zeta function is defined by ζ(z) = where n z = e z ln n.since n z = n Re z, the given series converges absolutely on Re z> and uniformly on {z :Rez +δ} for every δ>.let p,p 2,p 3,... be the sequence 2,3,5,... of prime numbers and note that for =, 2,... and Re z>, we have n= /p z =+ p z Now consider the partial product m m p z = + = =( p z n z + p 2z +. + p 2z + ). By multiplying the finitely many absolutely convergent series on the right together, rearranging, and applying the fundamental theorem of arithmetic, we find that the product is the same as the sum n P m n, where P z m consists of along with those positive integers whose prime factorization uses only primes from the set {p,...,p m }.Therefore We now state this formally. = p z 7.. Euler s Product formula = n=, Re z>. nz For Re z>, the Riemann zeta function ζ(z) = n= /nz is given by the product ( ) p z = where {p } is the (increasing) sequence of prime numbers. The above series and product converge uniformly on compact subsets of Re z>, hence ζ is analytic on Re z >.Furthermore, the product representation of ζ shows that ζ has no zeros in Re z> (Theorem 6..7). Our proof of the PNT requires a number of additional properties of ζ.the first result is concerned with extending ζ to a region larger than Re z>.

3 7.. THE RIEMANN ZETA FUNCTION Extension Theorem for Zeta The function ζ(z) /(z ) has an analytic extension to the right half plane Re z>. Thus ζ has an analytic extension to {z :Rez>,z } and has a simple pole with residue at z =. Proof.For Re z>, apply the summation by parts formula (Problem 2.2.7) with a n = n and b n =/n z to obtain Thus But k n= [ n (n +) z n z ] k + (n +) z = n= [ n (n +) z ] n z = nz = k k z (n +) z. k k z n+ n n= [ n n= t z dt = z (n +) z n z n+ where [t] is the largest integer less than or equal to t.hence we have k n= k n z =+ (n +) z = k k z + z n= Letting k, we obtain the integral formula ζ(z) =z = k z + z n= k for Re z>.consider, however, the closely related integral z tt z dt = z Combining this with () we can write ζ(z) z =+z n n+ n [t]t z dt. ]. [t]t z dt [t]t z dt [t]t z dt () t z dt = z z =+ z. ([t] t)t z dt. Now fix k> and consider the integral k ([t] t)t z dt. By (3.3.3), this integral is an entire function of z.furthermore, if Re z>, then k ([t] t)t z dt k t Re(z+) dt t Re z dt = Re z.

4 4 CHAPTER 7. THE PRIME NUMBER THEOREM This implies that the sequence f k (z) = k ([t] t)t z dt of analytic functions on Re z> is uniformly bounded on compact subsets. Hence by Vitali s theorem (5..4), the limit function f(z) = ([t] t)t z dt (as the uniform limit on compact subsets of Re z>) is analytic, and thus the function +z ([t] t)t z dt is also analytic on Re z>.but this function agrees with ζ(z) z for Re z>, and consequently provides the required analytic extension of ζ to Re z >.This completes the proof of the theorem. We have seen that Euler s formula (7..) implies that ζ has no zeros in the half plane Re z>, but how about zeros of (the extension of) ζ in < Re z? The next theorem asserts that ζ has no zeros on the line Re z =.This fact is crucial to our proof of the PNT. 7..3Theorem The Riemann zeta function has no zeros on Re z =,so(z )ζ(z) is analytic and zero-free on a neighborhood of Re z. Proof.Fix a real number y and consider the auxiliary function = h(x) =ζ 3 (x)ζ 4 (x + iy)ζ(x + i2y) for x real and x>.by Euler s product formula, if Re z> then ln ζ(z) = ln p z = Re Log( p z )=Re = = n= n p nz where we have used the expansion Log( w) = n= wn /n, valid for w <.Hence ln h(x) =3ln ζ(x) +4ln ζ(x + iy) +ln ζ(x + i2y) =3Re = = n= +Re = n= n p nx n p nx = n= But p iny = e iny and p i2ny form +4Re n p nx i2ny Re(3 + 4p iny = n= n p nx iny + p i2ny ). = e i2ny.thus Re(3 + 4p iny 3+4cosθ + cos 2θ =3+4cosθ + 2 cos 2 θ = 2( + cos θ) 2. + p i2ny ) has the

5 7.. THE RIEMANN ZETA FUNCTION 5 Therefore ln h(x) and consequently h(x) = ζ 3 (x) ζ 4 (x + iy) ζ(x + i2y). Thus h(x) x = (x )ζ(x) 3 ζ(x + iy) x 4 ζ(x + i2y) x. But if ζ(+iy) =, then the left hand side of this inequality would approach a finite limit ζ ( + iy) 4 ζ( + i2y) as x + since ζ has a simple pole at with residue.however, the right hand side of the inequality contradicts this.we conclude that ζ( + iy). Since y is an arbitrary nonzero real number, ζ has no zeros on Re z =. Remark The ingenious introduction of the auxiliary function h is due to Mertens (898).We now have shown that any zeros of ζ in Re z > must lie in the strip < Re z <. The study of the zeros of ζ has long been the subect of intensive investigation by many mathematicians.riemann had stated in his seminal 859 paper that he considered it very likely that all the zeros of ζ in the above strip, called the critical strip, lie on the line Re z = /2.This assertion is now known as the Riemann hypothesis, and remains as yet unresolved.however, a great deal is known about the distribution of the zeros of ζ in the critical strip, and the subect continues to capture the attention of eminent mathematicians.to state ust one such result, G.H.Hardy proved in 95 that ζ has infinitely many zeros on the line Re z = /2.Those interested in learning more about this fascinating subect may consult, for example, the book Riemann s Zeta Function by H.M. Edwards. Another source is We turn next to zeta s logarithmic derivative ζ /ζ, which we know is analytic on Re z >. In fact, more is true, for by (7..3), ζ /ζ is analytic on a neighborhood of {z :Rez and z }.Since ζ has a simple pole at z =,sodoesζ /ζ, with residue Res(ζ /ζ, ) =. [See the proof of (4.2.7).] We next obtain an integral representation for ζ /ζ that is similar to the representation () above for ζ. [See the proof of (7..2).] But first, we must introduce the von Mangoldt function Λ, which is defined by { if n = p m for some m, Λ(n) = otherwise. Thus Λ(n) islnp if n is a power of the prime p, and is if not.next define ψ on x by ψ(x) = n x Λ(n). (2) An equivalent expression for ψ is ψ(x) = p x m p (x)lnp,

6 6 CHAPTER 7. THE PRIME NUMBER THEOREM where the sum is over primes p x and m p (x) is the largest integer such that p mp(x) x. (For example, ψ(.4) = 3 ln ln [ 3 + ln ] 5 + ln 7.) Note that p mp(x) x iff m p (x)lnp ln x iff m p (x) ln x.thus m ln x p(x) = where as before, [ ] denotes the greatest integer function.the function ψ will be used to obtain the desired integral representation for ζ /ζ Theorem For Re z>, ζ (z) ζ(z) = z ψ(t)t z dt (3) where ψ is defined as above. Proof.In the formulas below, p and q range over primes.if Re z>, we have ζ(z) = p ( p z ) by (7..), hence ζ (z) = p = ζ(z) p = ζ(z) p p z ( p z ) 2 q z q p p z ( p z ) 2 ( p z ) p z p z. Thus ζ (z) ζ(z) = p p z p z = p p nz. n= The iterated sum is absolutely convergent for Re z>, so it can be rearranged as a double sum (p,n),n (p n ) z = k z k where k = p n for some n.consequently, ζ (z) ζ(z) = k z Λ(k) = k z (ψ(k) ψ(k )) k= by the definitions of Λ and ψ.but using partial summation once again we obtain, with a k = k z,b k+ = ψ(k), and b = ψ() = in Problem 2.2.7, k= k= M M k z (ψ(k) ψ(k )) = ψ(m)(m +) z + ψ(k)(k z (k +) z ). k=

7 7.2. AN EQUIVALENT VERSION OF THE PRIME NUMBER THEOREM 7 Now from the definition (2) of ψ(x) wehaveψ(x) x ln x, so if Rez > wehave ψ(m)(m +) z asm.moreover, we can write M M ψ(k)(k z (k +) z )= ψ(k)z k= = = z k= M z k= k M k+ k+ k t z dt ψ(t)t z dt ψ(t)t z dt because ψ is constant on each interval [k, k + ).Taking limits as M, we finally get ζ (z) ζ(z) = z ψ(t)t z dt, Re z>. 7.2 An Equivalent Version of the Prime Number Theorem The function ψ defined in (2) above provides yet another connection, through (3), between the Riemann zeta function and properties of the prime numbers.the integral that appears in (3) is called the Mellin transform of ψ and is studied in its own right.we next establish a reduction, due to Chebyshev, of the prime number theorem to a statement involving the function ψ Theorem The prime number theorem holds, that is, x π(x)lnx, iff x ψ(x) asx. Proof.Recall that ψ(x) = [ ] ln x p x ln x () p x =lnx p x = (ln x)π(x).

8 8 CHAPTER 7. THE PRIME NUMBER THEOREM However, if <y<x, then π(x) =π(y)+ π(y)+ <y+ ln y y<p x y<p x y<p x y + ln y ψ(x). ln y (2) Now take y = x/(ln x) 2 in (2), and we get π(x) x (ln x) 2 + ln x 2lnlnx ψ(x). Thus π(x) ln x x ln x + ln x ψ(x) ln x 2lnlnx x. (3) It now follows from () and (3) that ψ(x) x ln x x π(x) ln x + ln x ψ(x) ln x 2lnlnx x and from this we can see that x ψ(x) iffx π(x)lnx asx. The goal will now be to show that ψ(x)/x asx.a necessary intermediate step for our proof is to establish the following weaker estimate on the asymptotic behavior of ψ(x) Lemma There exists C> such that ψ(x) Cx, x >.For short, ψ(x) =O(x). Proof.Again recall that ψ(x) = p x [ ln x ]lnp, x >.Fix x> and let m be an integer such that 2 m <x 2 m+.then ψ(x) =ψ(2 m )+ψ(x) ψ(2 m ) ψ(2 m )+ψ(2 m+ ) ψ(2 m ) = [ ] ln 2 m + p 2 m Consider, for any positive integer n, n<p 2n =ln 2 m <p 2 m+ n<p 2n p. [ ln 2 m+ ].

9 7.3. PROOF OF THE PRIME NUMBER THEOREM 9 Now for any prime p such that n<p 2n, p divides (2n)!/n! =n! ( ) 2n n.since such a p does not divide n!, it follows that p divides ( ) 2n n.hence ( ) 2n p < (+) 2n =2 2n, n n<p 2n and we arrive at n<p 2n <2n ln 2. Therefore m = m < 2 k ln 2 < 2 m+ ln 2 p 2 m k= 2 k <p 2 k k= and <2 m+ ln 2. 2 m <p 2 m+ [ ] ln x But if p x is such that >, then ln x 2 and hence x p2 so that x p. Thus those terms in the sum [ ] [ ] ln x ln x p x where > occur only when p x, and the sum of terms of this form contribute no more than ln x = π( x)lnx. p x It follows from the above discussion that if 2 m <x 2 m+, then ψ(x) 2 m+ ln2+2 m+ ln 2 + π( x)lnx =2 m+2 ln 2 + π( x)lnx < 4x ln 2 + π( x)lnx 4x ln 2 + x ln x =(4ln2+ ln x)x. x Since x ln x asx, we conclude that ψ(x) =O(x), which proves the lemma. 7.3Proof of the Prime Number Theorem Our approach to the prime number theorem has been along traditional lines, but at this stage we will apply D.J. Newman s method (Simple Analytic Proof of the Prime Number Theorem, American Math.Monthly 87 (98), ) as modified by J.Korevaar (On

10 CHAPTER 7. THE PRIME NUMBER THEOREM Newman s Quick Way to the Prime Number Theorem, Math.Intelligencer 4 (982), 8-5).Korevaar s approach is to apply Newman s ideas to obtain properties of certain Laplace integrals that lead to the prime number theorem. Our plan is to deduce the prime number theorem from a Tauberian theorem (7.3.) and its corollary (7.3.2). Then we will prove (7.3.) and (7.3.2) Auxiliary Tauberian Theorem Let F be bounded and piecewise continuous on [, + ), so that its Laplace transform G(z) = F (t)e zt dt exists and is analytic on Re z>.assume that G has an analytic extension to a neighborhood of the imaginary axis, Re z =.Then F (t) dt exists as an improper integral and is equal to G().[In fact, F (t)e iyt dt converges for every y R to G(iy).] Results like (7.3.) are named for A. Tauber, who is credited with proving the first theorem of this type near the end of the 9th century.the phrase Tauberian theorem was coined by G.H. Hardy, who along with J.E. Littlewood made a number of contributions in this area.generally, Tauberian theorems are those in which some type of ordinary convergence (e.g., convergence of F (t)e iyt dt for each y R), is deduced from some weaker type of convergence (e.g., convergence of F (t)e zt dt for each z with Re z > ) provided additional conditions are satisfied (e.g., G has an analytic extension to a neighborhood of each point on the imaginary axis).tauber s original theorem can be found in The Elements of Real Analysis by R.G. Bartle Corollary Let f be a nonnegative, piecewise continuous and nondecreasing function on [, ) such that f(x) = O(x).Then its Mellin transform g(z) =z f(x)x z dx exists for Re z> and defines an analytic function g.assume that for some constant c, the function g(z) c z has an analytic extension to a neighborhood of the line Re z =.Then as x, f(x) x c. As stated earlier, we are first going to see how the prime number theorem follows from (7.3.) and (7.3.2). To this end, let ψ be as above, namely ψ(x) = [ ] ln x. p x

11 7.3. PROOF OF THE PRIME NUMBER THEOREM Then ψ is a nonnegative, piecewise continuous, nondecreasing function on [, ).Furthermore, by (7.2.2), ψ(x) =O(x), so by (7.3.2) we may take f = ψ and consider the Mellin transform g(z) =z ψ(x)x z dx. But by (7..4), actually g(z) = ζ (z)/ζ(z), and by the discussion leading up to the ζ statement of (7..4), (z) ζ(z) + z has an analytic extension to a neighborhood of each pointofrez =, hence so does g(z) z. Consequently, by (7.3.2), we can conclude that ψ(x)/x, which, by (7.2.), is equivalent to the PNT. Thus we are left with the proof of (7.3.) and its corollary (7.3.2). Proof of (7.3.) Let F be as in the statement of the theorem. Then it follows ust as in the proof of (7..2), the extension theorem for zeta, that F s Laplace transform G is defined and analytic on Re z >.Assume that G has been extended to an analytic function on a region containing Re z.since F is bounded we may as well assume that F (t),t. For <λ<, define G λ (z) = λ F (t)e zt dt. By (3.3.3), each function G λ is entire, and the conclusion of our theorem may be expressed as lim G λ() = G(). λ That is, the improper integral F (t) dt exists and converges to G().We begin the analysis by using Cauchy s integral formula to get a preliminary estimate of G λ () G(). For each R>, let δ(r) > be so small that G is analytic inside and on the closed path. ir γ - R γ + R δ(r).. δ γ γ + R = R + γ - R. -ir Figure 7.3. γ R in Figure (Note that since G is analytic on an open set containing Re z,

12 2 CHAPTER 7. THE PRIME NUMBER THEOREM such a δ(r) > must exist, although it may well be the case that δ(r) asr +.) Let γ + R denote that portion of γ R that lies in Re z>, and γ R the portion that lies in Re z<.by Cauchy s integral formula, G() G λ () = (G(z) G λ (z)) dz. () 2πi γ R z Let us consider the consequences of estimating G() G λ () by applying the usual M- L estimates to the integral on the right hand side of () above.first, for z γ + R and x =Rez, we have G(z) G λ (z) z = R F (t)e zt dt R R λ λ λ e λx F (t) e xt dt e xt dt = R x R x = R Re z. But / Re z is unbounded on γ + R, so we see that a more delicate approach is required to shows that G() G λ () asλ.indeed, it is here that Newman s ingenuity comes to the fore, and provides us with a modification of the above integral representation for G() G λ ().This will furnish the appropriate estimate.newman s idea is to replace the factor /z by (/z)+(z/r 2 ) in the path integral in ().Since (G(z) G λ (z))z/r 2 is analytic, the value of the path integral along γ R remains unchanged.we further modify () by replacing G(z) and G λ (z) by their respective products with e λz.since e λz is entire and has the value at z =, we can write G() G λ () = (G(z) G λ (z))e λz ( 2πi γ R z + z R 2 ) dz. Note that for z = R we have (/z)+(z/r 2 )=(z/ z 2 )+(z/r 2 ) = (2 Re z)/r 2, so that if z γ + R, (recalling (2) above), (G(z) G λ (z))e λz ( z + z R 2 ) Re z e λ Re z λ Re z 2Rez e R 2 = 2 R 2. Consequently, (G(z) G λ (z))e λz ( 2πi γ + z + z dz R 2 ) R R by the M-L theorem.note that this estimate of the integral along the path γ + R is independent of λ.now let us consider the contribution to the integral along γ R of the integral (2)

13 7.3. PROOF OF THE PRIME NUMBER THEOREM 3 along γ R.First we use the triangle inequality to obtain the estimate (G(z) G λ (z))e λz ( 2πi γ z + z dz R 2 ) R G(z)e λz ( 2πi γ z + z R 2 ) dz + G λ (z)e λz ( 2πi R γ z + z dz R 2 ) R = I (R) + I 2 (R). First consider I 2 (R).Since G λ (z) is an entire function, we can replace the path of integration γ R by the semicircular path from ir to ir in the left half plane.for z on this semicircular arc, the modulus of the integrand in I 2 (R) is ( λ F (t)e zt λz 2Rez dt)e R 2 ) 2 Re z Re z R 2 = 2 R 2. (Note that F, we can replace the upper limit of integration by, and e λx for x.) This inequality also holds if Re z = (let z iy).thus by the M-L theorem we get I 2 (R) (/2π)(2/R 2 )(πr) =/R, again. Finally, we consider I (R).This will be the trickiest of all since we only know that on γ R, G is an analytic extension of the explicitly defined G in the right half plane.to deal with this case, first choose a constant M(R) > such that G(z) M(R) for z γ R. Choose δ such that <δ <δ(r) and break up the integral defining I (R) intotwo parts, corresponding to Re z< δ and Re z δ.the first contribution is bounded in modulus by 2π M(R)e λδ ( δ(r) + R )πr = 2 RM(R)( δ(r) + R )e λδ, which for fixed R and δ tends to as λ.on the other hand, the second contribution is bounded in modulus by 2π M(R)( δ(r) + R )2R arcsin δ R, the last factor arising from summing the lengths of two short circular arcs on the path of integration.thus for fixed R and δ(r) we can make the above expression as small as we please by taking δ sufficiently close to.so at last we are ready to establish the conclusion of this theorem.let ɛ> be given.take R =4/ɛ and fix δ(r), <δ(r) <R, such that G is analytic inside and on γ R.Then as we saw above, for all λ, (G(z) G λ (z))e λz ( 2πi γ + z + z dz R 2 ) R = ɛ 4 R and also 2πi γ R (G λ (z)e λz ( z + z R 2 ) dz R = ɛ 4.

14 4 CHAPTER 7. THE PRIME NUMBER THEOREM Now choose δ such that <δ <δ(r) and such that Since 2π M(R)( δ(r) + R )2R arcsin δ R < ɛ 4. 2 RM(R)( δ(r) + R )e λδ < ɛ 4 for all λ sufficiently large, say λ λ, it follows that which completes the proof. Proof of (7.3.2) G λ () G() <ɛ, λ λ Let f(x) and g(z) be as in the statement of the corollary.define F on [, + ) by F (t) =e t f(e t ) c. Then F satisfies the first part of the hypothesis of the auxiliary Tauberian theorem, so let us consider its Laplace transform, G(z) = which via the change of variables x = e t becomes G(z) = = (e t f(e t ) c)e zt dt, ( x f(x) c)x z dx x f(x)x z 2 dx c x z dx = f(x)x z 2 dx c z g(z +) = z + c z = z + [g(z +) c z c]. It follows from the hypothesis that g(z +) (c/z) has an analytic extension to a neighborhood of the line Re z =, and consequently the same is true of the above function G. Thus the hypotheses of the auxiliary Tauberian theorem are satisfied, and we conclude that the improper integral F (t) dt exists and converges to G().In terms of f, this says that (e t f(e t ) c) dt exists, or equivalently (via the change of variables x = e t once more) that ( f(x) x c)dx x

15 7.3. PROOF OF THE PRIME NUMBER THEOREM 5 exists.recalling that f is nondecreasing, we can infer that f(x)/x c as x.for let ɛ> be given, and suppose that for some x >, [f(x )/x ] c 2ɛ.It follows that Hence, f(x) f(x ) x (c +2ɛ) x(c + ɛ) for x x c +2ɛ c + ɛ x. c+2ɛ c+ɛ x x ( f(x) x c)dx x c+2ɛ c+ɛ x x ɛ +2ɛ dx = ɛ ln(c x c + ɛ ). But x 2 x c) dx x asx,x 2, because the integral from to is convergent. Thus for all x sufficiently large, x ( f(x) c+2ɛ c+ɛ x x ( f(x) x c)dx x <ɛln(c +2ɛ c + ɛ ). However, reasoning from the assumption that [f(x )/x ] c 2ɛ, we have ust deduced the opposite inequality.we must conclude that for all x sufficiently large, [f(x )/x ] c <2ɛ. Similarly, [f(x )/x ] c> 2ɛ for all x sufficiently large.[say [f(x )/x ] c 2ɛ. The key inequality now becomes f(x) f(x ) x (c 2ɛ) x(c ɛ) for ( c 2ɛ c ɛ )x x x and the limits of integration in the next step are from c 2ɛ c ɛ x to x.] Therefore f(x)/x c as x, completing the proof of both the corollary and the prime number theorem. The prime number theorem has a long and interesting history.we have mentioned ust a few of the many historical issues related to the PNT in this chapter.there are several other number theoretic functions related to π(x), in addition to the function ψ(x) that was introduced earlier.a nice discussion of some of these issues can be found in Eric W.Weisstein, Prime Number Theorem, from MathWorld A Wolfram Web Resource, This source also includes a number of references on PNT related matters. References.Karl Sabbach, The Riemann Hypothesis-The Greatest Unsolved Problem in Mathematics, Farrar, Strass and Giroux, New York, Julian Havil, Gamma-Exploring Euler s Constant, Princeton University Press, Princeton and Oxford, 23, Chapter 5.

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