Bent E. Petersen
Contents
1 Introduction 1
2Asymptotics 6
3 The Logarithmic Integral 9
4TheCebyˇˇ sev Functions θ(x) and ψ(x) 11
5M¨obius Inversion 14
6 The Tail of the Zeta Series 16
7 The Logarithm log ζ(s) 17
< s 8 The Zeta Function on e =1 21
9 Mellin Transforms 26
10 Sketches of the Proof of the PNT 29 10.1 Cebyˇˇ sev function method ...... 30 10.2 Modified Cebyˇˇ sev function method ...... 30 10.3 Still another Cebyˇˇ sev function method ...... 30 10.4 Yet another Cebyˇˇ sev function method ...... 31 10.5 Riemann’s method ...... 31 10.6 Modified Riemann method ...... 32 10.7 Littlewood’s Method ...... 32 10.8 Ikehara Tauberian Theorem ...... 32
11ProofofthePrimeNumberTheorem 33
References 39
1 Introduction
Let π(x) be the number of primes p ≤ x. It was discovered empirically by Gauss about 1793 (letter to Enke in 1849, see Gauss [9], volume 2, page 444 and Goldstein [10]) and by Legendre (in 1798 according to [14]) that x π(x) ∼ . log x
This statement is the prime number theorem. Actually Gauss used the equiva- lent formulation (see page 10) Z x dt π(x) ∼ . 2 log t
1 B. E. Petersen Prime Number Theorem
For some discussion of Gauss’ work see Goldstein [10] and Zagier [45].
In 1850 Cebyˇˇ sev [3] proved a result far weaker than the prime number theorem — that for certain constants 0
Note X∞ ψ(x)= θ(x1/n) n=1 where the sum is finite for each x since θ(x1/n)=0ifx<2n.aCebyˇˇ sev proved that the prime number theorem is equivalent to either of the relations θ(x) ∼ x, ψ(x) ∼ x. ˇ θ(x) In addition Cebyˇsev showed that if limx→∞ x exists, then it must be 1, which then implies the prime number theorem. He was, however, unable to establish the existence of the limit.
Like Gauss, Riemann formulated his estimate of π(x) in terms of the logarithmic integral Z x dt Li(x)=(PV) ,x>1. 0 log t In his famous 1859 paper [33] he related the relative error in the asymptotic approximation π(x) ∼ Li(x) to the distribution of the complex zeros of the Riemann zeta function X∞ Y 1 −1 ζ(s)= = 1 − p−s . (1) ns n=1 p prime The Riemann zeta function was actually introduced by Euler as early as 1737. It was used of by Cebyˇˇ sev (in the real domain) prior to Riemann’s use of it. Euler also discovered the functional equation πs ζ(s)=2(2π)s−1 sin Γ(1 − s) ζ(1 − s)(2) 2
Seminar Lecture Notes 2 Version: May 2 1996 B. E. Petersen Prime Number Theorem which he published in 1749. The functional equation was proved by Riemann in [33].
Riemann does not prove the prime number theorem in his 1859 paper. His object was to find an explicit analytic expression for π(x), and he does so. He does comment that π(x)isaboutLi(x)andthatπ(x)=Li(x)+O(x1/2). This would imply π(x) =1+O(x−1/2 log x)=1+o(1), Li(x) which gives the prime number theorem. Thus Hardy’s comment, [14], page 352, that Riemann does not even state the prime number theorem, is not strictly accurate. On the other hand, Riemann’s assertion about the order of the error is so much stronger than what is required for the prime number theorem, that one could maintain that he does not state the weaker theorem.
In 1896 the prime number theorem was finally proved by Jacques Hadamard [12] and also by Charles–Jean de la Vall´ee Poussin [6]. The first part of the proof
is to show that ζ(s) =0if6 Towards the end of his 1859 paper [33] Riemann asserts that π(x) < Li(x). Gauss [9], volume 2, page 444, makes the same assertion. This is known to be true for all x ≤ 108 but was proved false in general in 1914 by Littlewood, [27]. Littlewood showed that π(x) − Li(x) changes sign infinitely often. Indeed Littlewood showed there is a constant K>0 such that (π(x) − Li(x)) log x x1/2 log log log x is greater that K for arbitrarily large x and less than −K for arbitrarily large x. Littlewood’s methods yield no information on where the first sign change occurs. In 1933 Skewes [35] showed that there is at least one sign change at x for some 1034 x<1010 . Skewes proof required the Riemann hypothesis. In 1955 [36] he obtained a bound without using the Riemann hypothesis. This new bound was 10964 1010 . Skewes large bound can be reduced substantially. In 1966 Sherman Lehman [25] showed that between 1.53 × 101165 and 1.65 × 101165 there are more than 10500 successive integers x for which π(x) > Li(x). Lehman’s work suggests there is no sign change before 1020. Perhaps we will see a sign change soon! In 1987 Seminar Lecture Notes 3 Version: May 2 1996 B. E. Petersen Prime Number Theorem te Riele [37] showed that between 6.62 × 10370 and 6.69 × 10370 there are more than 10180 successive integers x for which π(x) > Li(x). In Ramanujan’s second letter to Hardy (in 1913, see [2], page 53) he estimates π(x)by ∞ X µ(n) π(x) ≈ Li(x1/n)(3) n n=1 where µ(n)istheM¨obius function. This expression was obtained by Riemann in 1859, except Riemann has additional terms, arising from the complex zeros of ζ(s). Littlewood points out in a letter to Hardy (discussing Ramanujan’s letter, see [2], page 68) that 1 x1/2 π(x) − Li(x)+ Li(x1/2) =6 O . 2 log x It follows that 1 π(x) − Li(x)+ Li(x1/2) =6 O Li(x1/2) . (4) 2 Thus it is clear that equation (3) can not be interpreted as an asymptotic series for π(x) (though it is an asymptotic series). Ramanujan says to truncate the series at the first term less than one. This gives an excellent approximation to π(x), but it is empirical. The actual expression obtained by Riemann is ∞ X µ(n) π(x)= J(x1/n) n n=1 Z X ∞ dt J(x)=Li(x) − Li(xρ) − log 2 + t(t2 − 1) log t ρ x where ρ runs over the complex roots of the zeta function. The first sum here is actually finite for each x since X 1 J(x)= π(x1/n) n is 0 for x<2. The complete proof of Riemann’s formula (in a different form) was given by von Mangoldt ([43]) in 1895. In connection with equation (3) we now have XN µ(n) XN X π(x) − Li(x1/n)= Li(xρ/n) + “other terms” n n=1 n=1 ρ Seminar Lecture Notes 4 Version: May 2 1996 B. E. Petersen Prime Number Theorem where the omitted terms are not particularly significant. The terms in the double sum are Riemann’s “periodic” terms. Individually they are quite large, but there must be a large amount of cancellation to account for the fact that equation (3) gives a very close estimate of π(x). In the following table the column labelled “Riemann” was computed using the first 6 terms in the series (3). The column labelled “Li(x)” couldR just as well x be labelled “Gauss” as Li(x)differsbyabout1.045 ... from 2 dt/ log t.The table shows why Li(x) is preferred to the asymptotically equivalent expression x/log x. x π(x) x/log x Li(x) Riemann 500 95 80.4 101.7 94.3 1,000 168 144.7 177.6 168.3 2,000 303 263.1 314.8 303.3 500,000 41,638 38,102.8 41,606.2 41,529.4 1,000,000 78,498 72,382.4 78,627.5 78,527.3 1,500,000 114,155 105,477.9 114,263.0 114,145.7 2,000,000 148,933 137,848.7 149,054.8 148,923.4 2,500,000 183,072 169,700.9 183,244.9 183,101.4 3,000,000 216,816 201,151.6 216,970.5 216,816.2 The prime counts in the table above are taken from Edwards [8] and are due to D. N. Lehmer [26]. The other numbers were computed using Maple V3. While the “Riemann” column looks better it turns out that in the long run Li(x)is just as good – see Littlewood [27] and the discussion in Edwards [8], page 87. Here is some additional numerical evidence for the prime number theorem: x π(x) x/log x Li(x) 104 1,229 1085.7 1246.1 108 5,761,455 5.42 × 106 5.762 × 106 1012 37,607,912,018 3.61 × 1010 3.760795 × 1010 1016 279,238,341,033,925 2.71 × 1014 2.79238344 × 1014 1018 24,739,954,287,740,860 2.41 × 1016 2.4739954309 × 1016 The prime counts in the table above are taken from Crandal [5] and are due to Reisel [32] and Odlyzko. The other numbers were computed using Maple V3. Here’s a strong version of the prime number theorem (see Ivi´c [22]) expressed in terms of the Cebyˇˇ sev psi function ψ. Seminar Lecture Notes 5 Version: May 2 1996 B. E. Petersen Prime Number Theorem Theorem 1.1. There exists a constant C>0 such that 3/5 −1/5 ψ(x)=x + O xe−C(log x) (log log x) . With regard to the connection between the complex zeros of the zeta function and the estimate of the error in the prime number theorem (see [22]) we have: Theorem 1.2. 1 ≤ Let 2 α<1.Then ψ(x)=x + O xα(log x)2 if and only if ζ(s) =06 for 2Asymptotics Let f and g be functions defined in a neighborhood of a. The notation f(x)=o(g(x)),x→ a means lim f(x)/g(x)=0. x→a The notation f(x)=O(g(x)),x→ a means there is a constant M such that | f(x) |≤M | g(x) | for all x near a. The notation f(x) ∼ g(x),x→ a means lim f(x)/g(x)=1, that is f(x)=g(x)+o(g(x)),x→ a. x→a We frequently omit the expression “x → a”, especially if a = ∞ or if a may be inferred from the context. Sometimes we want more detailed information about the remainder o(g(x)) above. Let (g(x))n≥1 be a sequence of functions such that gn+1(x)=o(gn(x)) Seminar Lecture Notes 6 Version: May 2 1996 B. E. Petersen Prime Number Theorem for each n ≥ 1. Then X∞ f(x) ∼ cn gn(x)(5) n=1 means Xn f(x) − ck gk(x)=o(gn(x)) (6) k=1 for each n ≥ 1. By “going out one more term” we see that (6) is equivalent to Xn f(x) − ck gk(x)=O(gn+1(x)) (7) k=1 for each n ≥ 1. The series (5) need not converge. Taking longer partial sums need not improve the approximation to f(x), unless we also take larger x.Eveniftheseries(5) converges, it need not converge to the function f(x). The next result is a useful Tauberian result for estimating an integrand. Lemma 2.1. Let f be a function on [2, ∞) and suppose xf(x) is monotone nondecreasing on [2, ∞).Letm and n be real numbers, n =6 −1.If Z x n+1 ∼ x →∞ f(t)dt m ,x , 2 (log x) then (n +1)xn f(x) ∼ ,x→∞. (log x)m Proof. Let x ≥ 2, >0andx(1 − ) ≥ 2. Then Z Z Z x(1+) x(1+) dt 1+ dt f(t)dt = tf(t) ≥ xf(x) = xf(x)log(1+) x x t 1 t and Z Z Z x x dt 1 dt f(t)dt = tf(t) ≤ xf(x) = −xf(x) log(1 − ). x(1−) x(1−) t 1− t Seminar Lecture Notes 7 Version: May 2 1996 B. E. Petersen Prime Number Theorem Let >0. By hypothesis there exists A ≥ 2 such that x ≥ A implies Z n+1 − 2 x n+1 2 x (1 ) ≤ ≤ x (1 + ) m f(t)dt m . (log x) 2 (log x) ≥ 2 Thus if 0 <<1andx max 1− ,A then Z x(1+) xf(x)log(1+) ≤ f(t)dt Zx Z x(1+) x = f(t)dt − f(t)dt 2 2 n+1 n+1 2 n+1 − 2 ≤ x (1 + ) (1 + ) − x (1 ) (log x(1 + ))m (log x)m Z x −xf(x)log(1− ) ≥ f(t)dt x(1−) Z Z x x(1−) = f(t)dt − f(t)dt 2 2 n+1 − 2 n+1 − n+1 2 ≥ x (1 ) − x (1 ) (1 + ) (log x)m (log x(1 − ))m It follows if 0 <<1then m n+1 2 − − 2 (log x) f(x) ≤ (1 + ) (1 + ) (1 ) lim sup n x→∞ x log(1 + ) (log x)mf(x) (1 − 2) − (1 − )n+1(1 + 2) lim inf ≥ . x→∞ xn − log(1 − ) Now the two expressions in above have limit n +1as → 0. Compare lemma 2.1 with Rademacher [31], page 102, Edwards [8], page 82, and Grosswald [11], page 175. Note that asymptotic estimates are not preserved by exponentiation. For ex- ample, we have Lemma 2.2. If m>0 then there exists a continuous increasing function g on [1, ∞) such that 1. 0 Seminar Lecture Notes 8 Version: May 2 1996 B. E. Petersen Prime Number Theorem 2. g(x)=o x ,x→∞ (log x)m 3. log g(x) ∼ log x, x →∞ Proof. x/(log x)m+1 has its minimum (m +1)−(m+1)em+1 at x = em+1. Hence if we define ( x(m +1)−(m+1) if 1 ≤ x ≤ em+1 g(x)= x ≥ m+1 (log x)m+1 if x e then g is continuous and the first two conclusions hold. Now note if x ≥ em+1 then log g(x)=logx − (m + 1) log(log x). Since log(log x) lim =0 x→∞ log x the last conclusion holds. Lemma 2.2 is adapted from [31], page 96. 3 The Logarithmic Integral The logarithmic integral Li(x), x>1, is defined as the Cauchy principal value R ∞ of the divergent integral 0 dt/log t. Explicitly Z Z 1− dt x dt Li(x) = lim + . → 0 0 log t 1+ log t Z − 1 1 1 = lim − dt +log →0 log t t − 1 Z0 x 1 1 + − dt + log(x − 1) − log log t t − 1 1+ Z x − 1 − 1 = log(x 1) + − dt. 0 log t t 1 We can avoid the singularity in the integrand by noting that for any µ>1we have Z x dt Li(x)=Li(µ)+ . µ log t Seminar Lecture Notes 9 Version: May 2 1996 B. E. Petersen Prime Number Theorem Thus Z x dt Li(x)=1.045163780 ···+ 2 log t which shows that estimate of Gauss for π(x) does not differ by more than about 1 from the estimate given by Li(x). It is fairly easy to see that Li(x) has exactly one root µ>1 and therefore Z x dt Li(x)= µ log t where µ =1.45136923488338105 .... This expression is the one Ramanujan used for Li(x). Proposition 3.1. ∞ X x Li(x) ∼ (n − 1)! ,x→∞. (log x)n n=1 Proof. If we integrate by parts n times we have Z x x ··· − x dt Li(x)= + +(n 1)! n + cn + n! n+1 log x (log x) 2 (log t) where cn is a constant. It now suffices to show Z (log x)n x dt lim c + n! =0. →∞ n n+1 x x 2 (log t) Dividing the interval of integration at x1/2 we have Z (log x)n x dt lim →∞ n+1 x x 2 (log t) (log x)n x1/2 − 2 (log x)n x − x1/2 ≤ + 2n+1 x (log 2)n+1 x (log x)n+1 log x n 2n+1 ≤ + x1/2n log x whichconvergesto0asx →∞. The idea of dividing the interval of integration at x1/2 is a standard trick. See Edwards [8], page 85. In particular Li(x) ∼ x/log x so that π(x) ∼ Li(x)andπ(x) ∼ x/log x are equivalent formulations of the prime number theorem. Seminar Lecture Notes 10 Version: May 2 1996 B. E. Petersen Prime Number Theorem 4TheCebyˇˇ sev Functions θ(x) and ψ(x) The Cebyˇˇ sev theta function θ is defined by X θ(x)= log p p≤x where p runs over primes. There is a simple relationship between θ(x)andπ(x): Proposition 4.1. Z θ(x) x θ(t) π(x)= + 2 dt. (8) log x 2 t (log t) Proof. Let p1 =2,p2 =3,p3 =5,··· be the sequence of primes and let Xk ck = log pj so ck − ck−1 =logpk. j=1 Note if x ∈ [pk,pk+1)thenπ(x)=k and θ(x)=ck.Thus Z − Z Z x kX1 pj+1 x θ(t) cj ck 2 dt = 2 dt + 2 dt 2 t (log t) j=1 pj t (log t) pk t (log t) − kX1 1 1 1 1 = c − + c − j log p log p k log p log x j=1 j j+1 k Xk c c − c − c = 1 + j j 1 − k log p log p log x 1 j=2 j Xk θ(x) =1+ 1 − log x j=2 θ(x) = π(x) − . log x If we work with Stieltjes integrals we see easily that Z Z x dθ(t) x dθ(t) π(x)=1+ = 2 log t µ log t Seminar Lecture Notes 11 Version: May 2 1996 B. E. Petersen Prime Number Theorem where 1 <µ<2. Then an integration by parts (see Kestelman [23]) yields the proposition. The striking resemblance to the logaritmic integral is very suggestive. The integral in equation (8) is actually o(x/log x). This follows immediately from the proof of proposition 3.1 once we know θ(x)=O(x). The only obvious estimate for θ(x) however is X θ(x)= log p ≤ π(x)logx ≤ x log x (9) p≤x x Corollary 4.2. If θ(x) ∼ x then π(x) ∼ . log x Proof. As pointed out above, if θ(x)=O(x)then π(x) θ(x) = + o(1). x/log x x Let ( log p if n = pk,pprime Λ(n)= 0otherwise. Then the Cebyˇˇ sev psi function ψ is defined by X X ψ(x)= Λ(n)= log p n≤x pn≤x where, as usual, p runs over primes. We note X∞ ψ(x)= θ(x1/n) (10) n=1 where the sum is actually finite for each x since θ(x)=0ifx<2. If n is the largest integer suchh thatipn ≤ x then n log p is the contribution of the ≤ log x powers of p to ψ(x)andn log p ,where[z] is the greatest integer in z.Thus X log x X ψ(x) ≤ log p ≤ log x 1=π(x)logx. log p p≤x p≤x Seminar Lecture Notes 12 Version: May 2 1996 B. E. Petersen Prime Number Theorem Lemma 4.3. ψ(x)=θ(x)+O x1/2(log x)2 . (11) log x Proof. In equation (10) there are at most log 2 nonzero terms and they decrease in magnitude. Thus θ(x) ≤ ψ(x) ≤ θ(x)+θ(x1/2)logx/log 2. Now apply (9). Corollary 4.4. We have θ(x) ∼ x if and only if ψ(x) ∼ x. Proof. ψ(x) θ(x) θ(x) = + O x−1/2(log x)2 = + o(1). x x x Lemma 4.5. If ψ(x) is the Cebyˇˇ sev psi function then x π(x) ∼ if and only if ψ(x) ∼ x. log x Proof. We have already shown ψ(x) π(x) ≤ . (12) x x/log x Now suppose π(x) ∼ x/log x. Then by (12) we have ψ(x)=O(x). It follows that θ(x)=ψ(x)+O(x1/2(log x)2)=O(x) and therefore by (8) we have π(x) θ(x) = + o(1). x/log x x Then it follows that θ(x) ∼ x and so by corollary 4.4 we have ψ(x) ∼ x. Conversely if ψ(x) ∼ x then by corollary 4.4 we have θ(x) ∼ x and so by corollary 4.2 we are done. Alternate proof: Choose an increasing continuous function g on [1, ∞)such that 0 Seminar Lecture Notes 13 Version: May 2 1996 B. E. Petersen Prime Number Theorem have π(x)=π(g(x)) + (π(x) − π(g(x))) X ≤ g(x)+ 1 g(x) Thus π(x) g(x) log x ψ(x) ≤ + . x/log x x/log x log g(x) x If ψ(x) ∼ x then it follows that lim sup π(x)/(x/log x) ≤ 1. If π(x) ∼ x/log x then it follows that lim inf ψ(x)/x ≥ 1. In view of equation (12) the proof is complete. The alternate proof of Lemma 4.5 is adapted from [31], page 96. The lemma shows we can express the prime number theorem as ψ(x) ∼ x.Noteatthis point the existence of either of the limits lim π(x)/(x/log x) and lim ψ(x)/x is not clear. Exercise 4.6. Prove the estimates θ(x)=O(x) and ψ(x)=O(x) by elemen- tary means (see Cebyˇˇ sev’s work) and then use them to simplify all of arguments above. Note we would also have ψ(x)=θ(x)+O(x1/2 log x) which is somewhat better than equation 11. 5M¨obius Inversion We define the M¨obius function µ(n)by 1ifn =1 µ(n)= (−1)m if n is a product of m distinct primes 0 in all other cases. Note that µ(n)=0ifn is not square–free. An important property of µ(n)is ( X 1ifn =1 µ(d)= 0otherwise d|n Seminar Lecture Notes 14 Version: May 2 1996 B. E. Petersen Prime Number Theorem 1 Let a =(a1,a2, ···) ∈ ` and define X bn = amn. m≥1 ∞ 1 Then b =(b1,b2, ···) ∈ ` . Suppose we have b ∈ ` . Then we can consider X µ(n)bmn. n≥1 Substituting the definition of bmn and making a change of variable we obtain X X X X µ(n)amnk = µ(d)amh = am. n≥1 k≥1 h≥1 d|h Note if an =0forn>N then bn =0forn>N and all the sums are finite. In this case the formula X am = µ(n)bmn n≥1 is called the M¨obius inversion formula. For a discussion of the extension to `1 sequences see Hartman and Wintner [18]. → R Proposition 5.1. Let h : N be totally multiplicative in the sense that h(mn)=h(m)h(n) and h(1) = 1.Letf be a function on [1, ∞) such that f(x)=0if x<2.If X∞ F (x)= h(n)f(x1/n),x≥ 1 (13) n=1 then X∞ f(x)= µ(n)h(n)F (x1/n). (14) n=1 Proof. Fix x and let 1/m am = h(m)f(x ). log x Note that am =0ifm> log 2 .Nowlet X bn = amn ≥ mX1 = h(mn)f(x1/mn) ≥ m 1 X = h(n) h(m)f(x1/mn) m≥1 = h(n)F (x1/n). Seminar Lecture Notes 15 Version: May 2 1996 B. E. Petersen Prime Number Theorem Then by the M¨obius inversion formula X X 1/n f(x)=h(1)f(x)=a1 = µ(n)bn = µ(n)h(n)F (x ). n≥1 n≥1 Note h(n)=nk satisfies the multiplicative hypothesis. The cases k =0and k = −1 occur frequently. Recall if θ(x)istheCebyˇˇ sev theta function and ψ(x) is the Cebyˇˇ sev psi function then X∞ ψ(x)= θ(x1/n). n=1 Thus we now have X∞ θ(x)= µ(n)ψ(x). n=1 Likewise if we introduce ∞ X 1 J(x)= π(x1/n) n n=1 then ∞ X µ(n) π(x)= J(x1/n). n n=1 Note Riemann denoted π(x)byF (x)andJ(x)byf(x). 6 The Tail of the Zeta Series To estimate the zeta function we will need an expression for the tail of the zeta series, that is, ∞ X 1 . ns n=k Consider first a partial sum of the tail of the series Xm 1 ζ(s, k, m)= . ns n=k+1 Clearly Xm n − (n − 1) ζ(s, k, m)= ns n=k+1 − mX1 1 1 = −k1−s + m1−s + n − . ns (n +1)s n=k Seminar Lecture Notes 16 Version: May 2 1996 B. E. Petersen Prime Number Theorem For the summands here we have Z Z n+1 n+1 1 − 1 −s−1 −s−1 n s s = sn x dx = s [x]x dx n (n +1) n n where [x] is the greatest integer in x. It follows that Z m ζ(s, k, m)=−k1−s + m1−s + s [x]x−s−1dx. k Trivially, if s =1,wehave6 Z m −s s 1−s − 1−s s x dx = − m k k 1 s and therefore Z m [x] − x 1 ζ(s, k, m)=s dx + m1−s − k1−s . s+1 − k x 1 s Letting m →∞we obtain the tail of the zeta series ∞ Z ∞ X [x] − x k1−s ζ(s, k)= n−s = s dx − xs+1 1 − s n=k k for < e s>1. It follows that Z Xk ∞ x − [x] k1−s ζ(s)= n−s − s dx + (15) xs+1 s − 1 n=1 k by uniqueness of analytic continuation we see that (15) holds for all ∞ x − [x] − 7 The Logarithm log ζ(s) Recall ζ(s) is meromorphic in the plane with a simple pole at s = 1 (with residue 1) and Y −s −1 ζ(s)= 1 − p , Seminar Lecture Notes 17 Version: May 2 1996 B. E. Petersen Prime Number Theorem Since ζ(s) has no zeros in log ζ(s) defined in the half space For the principal branch of the logarithm we have ∞ X 1 log(1 − z)=− zn, | z | < 1. n n=1 Therefore (16) yields ∞ X X 1 log ζ(σ)= p−nσ. n p n=1 The double sum consists of positive terms. Hence we can change the order of summation to obtain X∞ −σ log ζ(σ)= cnn n=2 where ( 1 m m if n = p ,pprime cn = 0otherwise. The sum here continues to an analytic function in log ζ(s)= cnn , n=2 0 X∞ ζ (s) − −s = Λ(n)n , log n m ≤ Here we have used cn log n = m =logp if n = p .SinceΛ(n) log(n)we ≥ have uniform convergence in From (18) we have X∞ X∞ X∞ X −ζ0(s)= m−s n−sΛ(n)= n−s Λ(d). m=1 n=2 n=2 d|n Seminar Lecture Notes 18 Version: May 2 1996 B. E. Petersen Prime Number Theorem But X X Λ(d)= log p =logn. d|n pm|n It follows that X∞ 0 −s ζ (s)=− n log n, t Since log n = o(n ) for any t>0 the series converges absolutely in the Dirichlet series for the zeta function converges normally in Theorem 7.1. Z ∞ π(x) < log ζ(s)=s dx, e s>1. (20) s − 2 x (x 1) Proof. Since π(x) ≤ x the integral is absolutely convergent for defines an analytic function in the half-plane Hence from (16) we have for σ>1 X∞ log ζ(σ)=− π(n) − π(n − 1) log(1 − n−σ) n=2 XN XN = lim − π(n)log(1− n−σ)+ π(n − 1) log(1 − n−σ) N→∞ n=2 n=2 XN = lim − π(n) log(1 − n−σ) − log(1 − (n +1)−σ) N→∞ n=2 + π(1) log(1 − 2−σ)+π(N) log(1 − (N +1)−σ) Seminar Lecture Notes 19 Version: May 2 1996 B. E. Petersen Prime Number Theorem Now π(1) = 0 and ∞ X 1 π(N)log(1− (N +1)−σ) ≤ N (N +1)−kσ k k=1 ∞ N X ≤ (N +1)−k (N +1)σ k=0 1 = → 0 (N +1)σ−1 as N →∞since σ>1. It follows that X∞ log ζ(σ)=− π(n) log(1 − n−σ) − log(1 − (n +1)−σ) n=2 ∞ Z X n+1 d = π(x) log(1 − x−σ)dx n dx Zn=2 ∞ σπ(x) = dx. σ − 2 x(x 1) Theorem 7.2. 0 Z ∞ ζ (s) − −s−1 = s x ψ(x)dx, P − − Proof. Since ψ(x)= n≤x Λ(n)wehaveψ(n) ψ(n 1) = Λ(x). Then by (18) we have ζ0(s) XN − = lim (ψ(n) − ψ(n − 1))n−s (22) ζ(s) N→∞ n=2 XN = lim ψ(n)(n−s − (n +1)−s) (23) N→∞ n=2 (N +1)−sψ(N) − ψ(1) (24) X∞ = ψ(n)(n−s − (n +1)−s) (25) n=2 (26) Seminar Lecture Notes 20 Version: May 2 1996 B. E. Petersen Prime Number Theorem since ψ(1) = 0 and ψ(N) ≤ π(N)logN ≤ N log N.Thus ∞ Z Z ζ0(s) X n+1 d ∞ − = − ψ(x) x−sdx = s x−s−1ψ(x)dx ζ(s) dx n=2 n 0 where everything is justified for s 8 The Zeta Function on 6 i 6 Theorem 8.1. If t is real then ζ(1 + it) =0and ζ( t) =0. Proof. If s = σ + it, σ>1then X∞ −σ log | ζ(s) | = 3 4 i log ζ(σ) ζ(σ + it) ζ(σ + 2t) X∞ −σ = cnn (3 + 4 cos(t log n)+cos(2t log n)) n=2 ≥ 0 since 3+4cost +cos2t =2+4cost +2(cost)2 =2(1+cost)2 ≥ 0. Thus 4 3 ζ(σ + it) |≥ 1 ((σ − 1)ζ(σ)) | ζ(σ + i2t) (27) σ − 1 σ − 1 for σ>1andforallt.Sinceζ(s) has a simple pole at s = 1 with residue 1 we have lim (σ − 1)ζ(σ)=1. σ→1 Now suppose t =0and6 ζ(1 + it)=0.Then ζ(σ + it) 0 lim = ζ (1 + it) σ→1 σ − 1 and therefore (27) implies | ∞ lim | ζ(σ + i2t) = σ→1 6 which contradicts t =6 0. By the functional equation we now also have ζ(it) =0 if t is real and t =6 0. Finally the functional equation implies ζ(0) = −1/2. Seminar Lecture Notes 21 Version: May 2 1996 B. E. Petersen Prime Number Theorem Theorem 8.1 is due to Hadamard and de la Vall´ee Poussin. The proof given above is from Rademacher [31], page 94, who attributes this particular proof to Ingham [21], page 29. i We now proceed to estimate ζ(σ + it) and, for us more importantly, log ζ(σ + t) in σ ≥ 1and| t |≥2. The estimates we give are quite rough and unremittingly technical. The arguments here are adapted from [11], but they appear in a number of places. Much stronger (and more technical) results may be found in [22]. Lemma 8.2. Let C>0. Then there exists a constant C1 > 0 such that | ζ(s) |≤C1 log t − ≤ ≤ ≥ if s = σ + it, 1 C/log t σ 2 and t 2. Proof. Note √ | s | =(σ2 +t2)1/2 ≤ (4+t2)1/2 ≤ 2t ≤ 2t and | s − 1 | =((σ −1)2 +t2)1/2 ≥ t. Now Z Z ∞ − ∞ x [x] ≤ dx 1 s+1 dx σ+1 = σ . k x k x σk Consider a fixed t ≥ 2andchoosek such that k ≤ t Xk Xk 1 n−s ≤ eC ≤ ec(1 + log k) ≤ eC (1 + log t). n n=1 n=1 It follows by equation (15) if t ≥ 2andwechoosek with k ≤ t 2t k1−σ | ζ(s) |≤ + + eC(1 + log t) (28) σkσ t for 1−C/log t ≤ σ ≤ 2. Since t ≥ 2thenk ≥ 1 and therefore t/k ≤ 1+1/k ≤ 2. Since σ ≤ 2 it follows that we have (t/k)σ−1 ≤ 2. Thus k1−σ ≤ 2t−σ. t We have k>t− 1. In addition t ≥ 2 implies 1 − t−1 ≥ 1/2andsoσ ≤ 2 implies (1 − t−1)σ ≥ 1/4. Thus 2t 2t 2t1−σ 8t1−σ ≤ = ≤ . σkσ σ(t − 1)σ σ(1 − t−1)σ σ Seminar Lecture Notes 22 Version: May 2 1996 B. E. Petersen Prime Number Theorem From (28) we now have 8t1−σ | ζ(s) |≤ +2t−σ + eC (1 + log t) (29) σ for any t ≥ 2andσ with 1 − C/log t ≤ σ ≤ 2. Now choose t0 > 2 such that C 1 C 1 − > and 2 + 17e ≤ log t0. log t0 2 For 2 ≤ t ≤ t0,1− C/log t ≤ σ ≤ 2wehave C | ζ(s) |≤C ≤ 1 log t 1 log 2 for some constant C1 just by continuity. On the other hand t ≥ t0 implies 1 C < 1 − ≤ σ ≤ 2 2 log t and so 8t1−σ +2t−σ ≤ 16tC/log t +2t−1/2 ≤ 16eC +2. σ Thus | ζ(s) |≤17eC +2+eC log t C C ≤ log t0 + e log t ≤ (1 + e )logt. Since 1 − 1/log t ≤ 1ift ≥ 2wehave: Corollary 8.3. ζ(1 + it)=O(log t) Lemma 8.4. Let C>0. Then there exists a constant C1 > 0 such that 0 2 | ζ (s) |≤C1 (log t) − ≤ ≤ ≥ if s = σ + it, 1 C/log t σ 2 and t 2. Proof. By equation (15) we have Xk ζ0(s)=− n−s log n Z n=1 Z ∞ − ∞ − − x [x] (x [x]) log x s+1 dx + s s+1 dx k x k x k1−s k1−s log k −− − . (s − 1)2 s − 1 Seminar Lecture Notes 23 Version: May 2 1996 B. E. Petersen Prime Number Theorem Now we get much the same estimates as in the previous lemma but with an extra factor of log t. Corollary 8.5. 0 2 ζ (1 + it)=O((log t) ) Lemma 8.6. 0 ζ (1 + it) = O((log t)9) ζ(1 + it) Proof. From equation (27) we have 1 ≤| |3/4 1/4 | i | | ζ(σ) ζ(σ +2t) . (30) | ζ(σ + it) Since lims→1(s − 1)ζ(s) = 1 it follows that 1 (log t)1/4 ≤ C . | 1 3/4 | ζ(σ + it) | σ − 1 | Now for σ>1wehave Z σ 0 | i − i | ζ(1 + it) = ζ(σ + t) ζ (u + t) du 1Z σ 0 |− i ≥|ζ(σ + it) ζ (u + t) du 1 − 3/4 ≥ 1 (σ 1) − − 2 1/4 C2(σ 1)(log t) . C1 (log t) If A>0andB>0 then the function 4A g(u)= u3/4 − Bu, u ≥ 0, 3 has the maximum A4/(3B3)atthepointu = A4/B4. Therefore taking A = 1/4 2 4 4 3/(4C1(log t) ), B = C2(log t) and σ − 1=A /B we see that 3 |≥ 3 −7 | i ζ(1 + t) 4 4 3 (log t) . 4 C1 C2 0 | 2 Since | ζ (1 + it) = O((log t) ) the proof is complete. Lemma 8.7. There exist constants C and C1 suct that 0 ζ (σ + it) 9 ≤ C1(log t) ζ(σ + it) if t ≥ 2 and σ ≥ 1 − C/(log t)9. Seminar Lecture Notes 24 Version: May 2 1996 B. E. Petersen Prime Number Theorem Proof. Given any C>0 we have constants C2 and C3 such that Z σ 0 |≤| i |− i | ζ(σ + it) ζ(1 + t) σ (u + t) du 1 −7 2 ≥ C2(log t) − C3 | σ − 1 | (log t) if t ≥ 2andσ ≥ 1 − C/log t.Thusifσ ≥ C/(log t)9 ≥ 1 − C/log t then |≥ − −7 | ζ(σ + it) (C2 CC3)(log t) . Once we have found C2 and C3 for a given C>0thenmakingC smaller simply strengthens the hypothesis on σ. Hence we may choose C so that C4 = C2 − CC3 > 0. For this choice of C we know there is a constant C5 > 0such that 0 |≤ 2 | ζ (σ + it) C5(log t) for t ≥ 2andσ ≥ 1 − C/log t. We can now improve a bit on the theorem of Hadamard and de la Vall´ee Poussin. 6 ≥ Corollary 8.8. There exists a constant C>0 such that ζ(σ + it) =0if t 2 and σ ≥ 1 − C(log t)−9. A much stronger zero–free result, due to H.-E. Richert, is known (see Ivi´c’s book, [22]). Theorem 8.9. There exists a constant C>0 such that ζ(s) =06 for σ ≥ 1 − C(log t)−2/3(log log t)−1/3,t≥ 2. Remarkably enough no one has ever proved that there is a zero–free region of ≥ 1 1 are no zeros on the line We now obtain a uniform estimate for the logarithm of the zeta function. We will use this estimate in the proof of the prime number theorem. Theorem 8.10. There exists a constant C>0 such that |≤ 9 | log ζ(σ + it) C(log t) for t ≥ 2 and σ ≥ 1. Seminar Lecture Notes 25 Version: May 2 1996 B. E. Petersen Prime Number Theorem ≥ Proof. Let s = σ + it.Ifσ 2then X∞ −s |≤ ≤ −9 9 | log ζ(s) | = | log | ζ(s) | + i arg ζ(s) log ζ(2) + π/2=K K(log 2) (log t) if σ ≥ 2andt ≥ 2. If now 1 − C(log t)−9 ≤ σ ≤ 2andt ≥ 2then Z 2 0 ζ (u + it) |≤| i | | log ζ(σ + it) log ζ(2 + t) + du. σ ζ(u + it) The first term we estimated in the first part of the proof. Since 2 − σ ≤ 1the integral is bounded by 0 ζ (u + it) max ≤ K(log t)9. ≤ ≤ σ u 2 ζ(u + it) 9 Mellin Transforms If f is a function on (0, ∞) its Mellin transform F is formally defined by the integral Z ∞ F (s)= xs−1f(x)dx. (31) 0 A measurable function f on (0, ∞)issaidtobeoftype (α, β)ifxσ−1f(x)is Lebesgue integrable on (0, ∞)foreachσ with α<σ<β. This terminology, Seminar Lecture Notes 26 Version: May 2 1996 B. E. Petersen Prime Number Theorem from Patterson [30], though it overloads the word type,isconvenient.Iff has type (α, β) then the Mellin transform F (s) is analytic in the strip α< A classical example is the Gamma function Z ∞ Γ(s)= xs−1e−xdx. 0 Another example is the very first formula that Riemann derives for the zeta function in his 1859 paper, Z ∞ xs−1 Γ(s)ζ(s)= dx. x − 0 e 1 In (31) if we make the change of variable x = et we obtain Z ∞ F (s)= estf(et)dt. −∞ Thus the Mellin transform F of f is the Fourier–Laplace transform of the com- position f ◦ exp. In particular Z ∞ −iut σt t F (σ − iu)= e e f(e ) dt −∞ is the Fourier transform of eσtf(t). It follows that we have the inversion formula Z ∞ 1 σt t iut e f(e )= e F (σ − iu) du, 2π −∞ that is, Z ∞ 1 t −(σ+iu) f(e )= e F (σ + iu)du, 2π −∞ provided that Z ∞ | ∞ | F (σ + iu) du < . −∞ Indeed in this case f is continuous (after correction on a null set, if needed) and the inversion formula folds for each t. In the absence of the integrability hypothesis the inversion formula may continue to hold pointwise in a suitable principal–value sense, as well as in various non– pointwise senses, see Mellin [29], Hardy [15], [16], Titchmarsh [40], Patterson [30]. For example, by results of Dirichlet, Dini and Jordan, if f is locally of bounded variation then Z f(et +0)+f(et − 0) 1 T −(σ+iu) = lim e F (σ + iu)du. →∞ 2 T 2π −T Seminar Lecture Notes 27 Version: May 2 1996 B. E. Petersen Prime Number Theorem Theorem 9.1. If f is of type (α, β) on (0, ∞) then the Mellin transform Z ∞ F (s)= f(x) xs−1 dx 0 is analytic in the strip α< 2π i ∞ σ−i for each σ with α<σ<β for which Z ∞ | ∞ | F (σ + iu) du < . −∞ Let f be a nonnegative monotone nondecreasing function on [0, ∞) such that f(x)=0ifx<2andsuchthatf(x)=O(x). Let ∞ X 1 g(x)= f(x1/n). n n=1 Then log x f(x) ≤ g(x) ≤ f(x)+f(x1/2) log 2 and so g(x)=f(x)+O(x1/2 log x)=O(x). Since g(x)=0forx<2 we see that the Mellin transform Z ∞ G(−s)= g(x)x−s−1dx 0 defines a function analytic in The integration by parts is justified since g(x)x = o(1) for We also note Z ∞ ∞ X 1 G(−σ)= f(t1/n)t−σ−1dt n 0 n=1 ∞ Z X ∞ 1 = f(t1/n)t−σ−1dt n n=1 0 X∞ = f(x)x−ns−1dx, n=1 Seminar Lecture Notes 28 Version: May 2 1996 B. E. Petersen Prime Number Theorem where the interchange of the summation and integration is justified since all the terms are nonnegative. We now interchange the summation and integration again and note that we have a geometric series in x−s.Sincef(x)=0ifx<2 we can sum the series to obtain Z ∞ f(x) < G(−s)= dx, e s>1. s − 0 x(x 1) Indeed, we have this relation for s = σ>1, and both expressions are analytic in ∞ X 1 g(x)= f(x1/n) n n=1 Z ∞ Z ∞ −s−1 f(x) < G(−s)= g(x)x dx = dx, e s>1 s − 0 0 x(x 1) Z ∞ 1 σ+i g(x)= G(−s)xsds, σ > 1 2π i ∞ σ−i Z x g(t) g1(x)= dt 1 t Z ∞ σ+i − 1 G( s) s g1(x)= x ds, σ > 1 2π i ∞ s σ−i f(x)=g(x)+O(x1/2 log x). If we know G wemaybeabletoestimateg and then f.The integral for g1 in terms of G is better behaved at ∞, so perhaps we estimate g1 instead. In this case, we use the Tauberian lemma, lemma 2.1, to estimate g and then f as before. Mellin calisthenics 10 Sketches of the Proof of the PNT The proof goes by estimating an inverse Mellin transform in accord with the Mellin calesthenics above. In the sketches below we do not concern ourselves with integrability of the Mellin transform. In those cases were integrability holds the inversion formula is valid by theorem 9.1 above. In the other cases, where the integral in the inversion formula is taken in some principal value sense, results of Dirichlet, Dini and Jordan apply. Seminar Lecture Notes 29 Version: May 2 1996 B. E. Petersen Prime Number Theorem 10.1 Cebyˇˇ sev function method Since Z ∞ ζ0(s) − = ψ(x)x−s−1dx sζ(s) 0 we have Z ∞ 1 σ+i ζ0(s) ψ(x)=− xsds. 2π i ∞ sζ(s) σ−i We now evaluate the contour integral to obtain a formula for ψ(x), the von Mangoldt formula of 1895, [43] (see also Edwards [8]). We then use this formula to deduce Z x x2 ψ(t)dt ∼ . 1 2 This part uses the fact that there are no roots on 10.2 Modified Cebyˇˇ sev function method If Z x ψ(t) ψ1(x)= dt 1 t then we have Z ∞ σ+i 0 − 1 ζ (s) s ψ1(x)= 2 x ds. 2π i ∞ s ζ(s) σ−i If we estimate the contour integral to obtain ψ1(x) ∼ x then the Tauberian lemma, lemma 2.1, implies ψ(x)/x ∼ 1. 10.3 Still another Cebyˇˇ sev function method Since ζ(s) is meromorphic with a simple pole at s = 1, with residue 1, and no other poles, we have h(s) ζ(s)= s − 1 Seminar Lecture Notes 30 Version: May 2 1996 B. E. Petersen Prime Number Theorem where h(s)isentireandh(1) = 1. Thus ζ0(s) s h0(s) + = +1 ζ(s) s − 1 h(s) < ≥ is analytic in e s 1. Now 0 Z ∞ −1 ζ (s) s − −s−1 + − = (ψ(x) x) x dx s ζ(s) s 1 0 and so Z ∞ σ+i 0 ψ(x) − − 1 ζ (s) s 1 s−1 1= + − x ds x 2π i ∞ ζ(s) s 1 s σ−i for σ>1. Since the integrand is well-behaved on σ =1wecantrymovingthe contour of integration to σ = 1 to estimate 1 − ψ(x)/x. This is (roughly) the approach taken in Heins [19]. 10.4 Yet another Cebyˇˇ sev function method We estimate the integral Z ∞ 1 σ+i xs ζ0(s) Ψ(x)=− ds. 2π i ∞ s(s +1) ζ(s) σ−i It turns out that Z x X x2 ψ(t)dt = Λ(n)(x − n)=xΨ(x)= + o(x2) 2 1 n≤x and we use the Tauberian lemma as before. For this proof see Rademacher [31]. 10.5 Riemann’s method Riemann did not actually prove the prime number theorem, but he did have the ingredients described here. Since Z ∞ log ζ(s) π(x) = dx s − s 0 x(x 1) we have Z log ζ(s) ∞ = J(x)x−s−1dx s 0 Seminar Lecture Notes 31 Version: May 2 1996 B. E. Petersen Prime Number Theorem where ∞ X 1 J(x)= π(x1/n). n n=1 Then Z ∞ 1 σ+i log ζ(s) J(x)= xsds. 2π i ∞ s σ−i This integral is only conditionally convergent. We estimate it very carefully and deduce J(x) ∼ x/log x.Thenπ(x)=J(x)+O(x1/2 log x)=x/log x+o(x/log x). 10.6 Modified Riemann method If Z x J(t) J1(x)= dt 1 t then Z ∞ σ+i 1 log ζ(s) s J1(x)= 2 x ds. 2π i ∞ s σ−i We estimate the integral to show J1(x) ∼ x/log x. Then the Tauberian lemma, lemma 2.1, implies J(x)/x ∼ 1/log x. This is the method of proof used in Grosswald [11] and is the method we will use here. 10.7 Littlewood’s Method Littlewood [28] gives a proof by studying the inverse Mellin transform of ζ0(s) Γ(s) ζ(s) and relating it to the Cebyˇˇ sev ψ function X ψ(x)= Λ(n). n≤x 10.8 Ikehara Tauberian Theorem A very famous result concerning the asymptotic behavior of the inverse Mellin (or inverse Fourier) transform is the Tauberian theorem of Landau, Hardy, Lit- tlewood and Ikehara. It shows that a large part of the proof of the prime number theorem is valid in a more general context. See Landau [24], Hardy–Littlewood [17], Ikehara [20], Wiener [44] or Donoghue [7]. Seminar Lecture Notes 32 Version: May 2 1996 B. E. Petersen Prime Number Theorem Theorem 10.1. Ikehara. Let µ be a monotone nondecreasing function on (0, ∞) and let Z ∞ F (s)= x−sdµ(x). 1 If the integral converges absolutely for µ(x) ∼ Ax. If we apply the theorem to Z ∞ 1 ζ0(s) − = ψ(x)x−s−1dx s ζ(s) 1 with Z x ψ(t) µ(x)= dt 1 t then we obtain the prime number theorem. 11 ProofofthePrimeNumberTheorem Recall we have seen Z ∞ log ζ(s) π(x) < = dx, e s>1 s − s 0 x(x 1) and therefore if ∞ X 1 J(x)= π(x1/n) n n=1 then we have a Mellin transform Z ∞ log ζ(s) −s−1 = J(x)x dx, Seminar Lecture Notes 33 Version: May 2 1996 B. E. Petersen Prime Number Theorem (note J(t)=0ift<2) then an integration by parts yields the Mellin transform Z ∞ log ζ(s) −s−1 −2 9 | | This transform (for s = σ + it, σ>1 fixed) decays like t (log t) for large t and so is integrable along vertical lines. As we have seen it follows that the inverse transform formula is valid. Thus Z ∞ a+i 1 log ζ(s) s J1(x)= 2 x ds 2π i ∞ s a−i for any a>1. We will estimate this contour integral (essentially by moving the contour to a =1). Let h(s)=(s − 1)ζ(s)soh(s) is an entire analytic function and h(s) =0if6 ≥ log h(s) = log(σ − 1) + log ζ(s) where the principal branch of the logarithm is used. We know log ζ(s)isanalytic ≥ 6 in log h(s) = log(s − 1) + log ζ(s) ≥ 6 − for Z ∞ a+i 1 log h(s) s I1(x)= 2 x ds (32) 2π i ∞ s a−i Z ∞ a+i − 1 log(s 1) s H2(x)= 2 x ds. (33) 2π i ∞ s a−i (34) TheestimateofI1(x) will depend on ζ(s) =0when6 s =1+it. The estimate of H2(x), which is harder, has nothing to do with zeta functions. Lemma 11.1. Let Z ∞ a+i − 1 log(s 1) s Hm(x)= m x ds, 2π i ∞ s a−i a>1 and m>1.ThenHm(x) is independent of a and x x H (x)= + o . m log x log x Seminar Lecture Notes 34 Version: May 2 1996 B. E. Petersen Prime Number Theorem Proof. Consider the contour illustrated at the right. By Cauchy’s theorem we have Z σ+iT log(s − 1) xs ds = −IA − IB − IC − ID − IE sm σ−iT where IX indicates the integral over the segment labeled X in the illustration. Consider first the integral over the segment A: Z a+iT − − log(s 1) s IA = m x ds. A a + iT s 1+ iT Along the segment A the integrand is bounded in absolute log T value by C if T>2. The path of integration has B a T m length a − 1. We have the same estimate for IE.Since m>0weseeIA → 0andIE → 0asT →∞.The δ C integral over the semicircle of radius δ<1isgivenby Z D π/2 log(δe iθ) i 1+δ exp(iθ) θ E −IC = x iδe dθ. iθ m −π/2 (1 + δe ) - − i 1 − iT a T Thus Z π/2 log(δe iθ) | IC |≤ x1+δ δdθ − m −π/2 (1 δ) Z δ π/2 ≤ x1+δ (− log δ + π2/4) dθ − m (1 δ) −π/2 πx1+δ ≤ (−δ log δ + δπ2/4) (1 − δ)m → 0asδ → 0. Combining the integrals IB and ID we now have Z T 1 log(it) 1+ it Hm(x)=− lim lim where Z ∞ log(it) b ity Gm(y)= m e dt 0 (1 + it) is the Fourier transform of the integrable function log(it) Gm(t)= m Y (t) (1 + it) Seminar Lecture Notes 35 Version: May 2 1996 B. E. Petersen Prime Number Theorem where Y is Heaviside’s function. By the Riemann–Lebesgue lemma it follows d that Gm(y) → 0asy →∞, but we need more. In fact we need b − →∞ The traditional estimate of Fourier integral goes by integration by parts: Z T log(it) d b ity iyGm(y) = lim lim e dt → →∞ m δ 0 T δ (1 + it) dt i log(iT ) log( δ) i = lim lim eiTy e δy− → →∞ m m i δ 0 T (1 + iT ) (1 + δ) Z Z ! T ity T e mπ/2 − mi log t − ity m dt m+1 e dt . i δ t(1 + it) δ (1 + t) log T log t The first term is O( T m ). In the last integral the integrand is O( tm+1 )forlarge t and O(− log t) for small t. It follows the integrand is integrable, the limit of the last integral exists and this limit is the Fourier transform of an integrable function, and so has limit 0 at ∞ by the Riemann-Lebesgue lemma. We now have Z ∞ ity log(iδ) e b − iδy iyGm(y)= lim e + dt + o(1), (36) → m m+1 i δ 0 (1 + iδ) δ t(1 + t) as y →∞. Now let m+1 m 1 − 1 i t u(t)= m+1 + m+1 . i t(1 + it) t (1 + t) A quick calculation shows − (m +1)(m +2) 2 → u(t)=−(m +1)i t + O(t ),t 0 2 −2 (m +1)(m +2) −3 −4 →∞ u(t)=(m +1)it + t + O(t ),t . 2 It follows that u is integrable and therefore by the Riemann–Lebesgue lemma we now have Z T ity log(iδ) e b − iδy iyGm(y)=o(1) lim lim e + dt → →∞ m δ 0 T (1 + iδ) δ t Z ! T m ity m+1 t e − i m+1 dt δ (1 + it) Seminar Lecture Notes 36 Version: May 2 1996 B. E. Petersen Prime Number Theorem We integrate by parts again Z T tme ity iy m+1 dt (1 + it) Z δ T tm d ity = m+1 e dt (1 + it) dt δ Z m T T m−1 m t mt − it i ity − ty = m+1 e m+2 e dt. i (1 + it) δ δ (1 + t) It follows that Z ∞ Z ∞ − m ity m 1 m t e mt − it − ity iy m+1 dt = m+2 e dt. i 0 (1 + it) 0 (1 + t) m−1 m m+2 i Now (mt − it )/(1 + t) is integrable and so by the Riemann–Lebesgue lemma this last integral is o(1) as y →∞.Thus Z ∞ ity log(iδ) e b − iδy iyGm(y)=o(1) lim e + dt → m δ 0 (1 + iδ) δ t Z ∞ eity − iδy = o(1) lim log(iδ)e + dt . → δ 0 δ t i Since log(iδ)=logδ + π/2 if we take imaginary parts we obtain Z ∞ b − π sin ty Since the integral is π/2wehave b π 1 To prove the prime number theorem it remains to show if Z ∞ a+i 1 log h(s) s I1(x)= 2 x ds, a > 1, 2π i ∞ s a−i then x I (x)=o ,x→∞. 1 log x Seminar Lecture Notes 37 Version: May 2 1996 B. E. Petersen Prime Number Theorem Thecriticalissuehereisthatlogh(s), where h(s)=(s − 1)ζ(s), is analytic in ≥ By Cauchy’s theorem I1(x) = lim IA = lim −IB − IC − ID T →∞ T →∞ where IX indicates the contour integral over the segment B a + iT labelled X in the illustration in the right. To estimate the integrals we note xs s−2 log h(s) ≥ | | is analytic in and s = σ + it we have i σ+ t 6A s −2 x − x s log h(s) = 2 log ((s 1)ζ(s)) (σ + it) xσ D ≤ | | | | || - 2 log t +log ζ(s) − i t 1 − iT a T xσ ≤ C (log | t |)9. t2 It follows the integral over the horizontal segments goes to 0 as T →∞.Thus Z 1 T x1+ it I (x) = lim log h(1 + it) dt 1 →∞ 2 T 2π −T (1 + it) x = H(log x) 2π where Z T log h(1 + it) H(y) = lim e iyt dt. →∞ 2 T −T (1 + it) Integrating by parts we have Z T log h(1 + it) d iyt iyH(y) = lim e dt (37) →∞ 2 T −T (1 + it) dt 0 Z i T h (1+ t) − i (1 + it) 2logh(1 + t) h(1+it) iyt = − i lim e dt (38) →∞ 3 T −T (1 + it) (39) | | 9 since log h(1 + it)=O((log t ) ) implies the boundary terms have limit 0. ≥ We have already seen that log h(s)isanalyticin Seminar Lecture Notes 38 Version: May 2 1996 B. E. Petersen Prime Number Theorem at s = 1 it follows that ζ0(s)/ζ(s) has a simple pole with residue −1ats =1. Thus h(s)=(s − 1)ζ(s) implies h0(s) 1 ζ0(s) = + h(s) s − 1 ζ(s) is analytic in a neighborhood of s = 1 and also away from the zeros of ζ(s). ≥ Thus its analytic in 0 h (1 + it) ≤ C for | t |≤1. h(1 + it) From lemma 8.6 we have 0 0 i h (1 + it) ζ (1 + t) 1 = + i i h(1 + it) ζ(1 + t) t 9 ≤ C1 + c2 (log | t |) 9 ≤ C3(log | t |) for | t |≥1. It follows that the numerator in the last integrand in equation (37) is bounded by 9 9 C3 | t | (log | t |) + C4 (log | t |) . Hence the integrand is O t−2 (log | t |)9 , | t |→∞ and bounded for small | t |. Thus the integrand is integrable and so by the Riemann–Lebesgue lemma we have yH(y)=o(1), as | t |→∞. Then x x I (x)= H(log x)=o 1 2π log x which completes the proof of the prime number theorem. References [1] George E. Andrews, Number theory, W. B. Saunders Co., Philadelphia, 1971, (Reprinted by Dover, New York, 1994). [2] Bruce C. Berndt and Robert A. Rankin, Ramanujan. Letters and commen- tary, Amer. Math. Soc. and London Math. Soc., Providence, Rhode Island, 1995. Seminar Lecture Notes 39 Version: May 2 1996 B. E. Petersen Prime Number Theorem [3] P. L. Cebyˇˇ sev, La totalit´e des nombres premiers inf´erieurs a une limite donn´ee, J. Math. Pures Appl. 17 (1852), 341–365. [4] , Œuvres, Chelsea, New York, 1962, (Reprint). [5] Richard E. Crandall, Projects in scientific computation, Springer–Verlag, New York, Berlin, Heidelberg, 1994. [6] Charles-Jean de la Vall´ee Poussin, Recherches analytiques sur la th´eorie des nombres; Premi`ere partie: La fonction ζ(s) de Riemann et les nombres premiers en g´en´eral, Annales de la Soc. scientifiques de Bruxelles 20 (II) (1896), 183–256 (Second part 281–297 ??). [7] William F. Donoghue, Distributions and Fourier transforms,Academic Press, New York and London, 1969. [8] H. M. Edwards, Riemann’s zeta function, Academic Press, New York, Lon- don, 1974. 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Permission is granted to duplicate this document for non–profit educational purposes provided that no alterations are made and provided that this copyright notice is preserved on all copies. Bent E. Petersen 24 hour phone numbers Department of Mathematics office (541) 737-5163 Oregon State University home (541) 754-2315 Corvallis, OR 97331-4605 fax (541) 737-0517 [email protected] http://www.orst.edu/˜peterseb Seminar Lecture Notes 42 Version: May 2 1996