The Prime Number Theorem

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The Prime Number Theorem A Journey through The Prime Number Theorem Notation Along these notes we shall employ extensively Landau’s O- notation that we recall briefly here. The symbols O(g) and o(g) mean respectively a function f such that ¯ ¯ ¯ ¯ ¯f(x)¯ f(x) lim sup ¯ ¯ < 1 and lim = 0: x!1 g(x) x!1 g(x) Note that f = O(g) is only a short way of saying jf(x)j · Cjg(x)j for some positive constant C and x large enough. If f and g has the same asymptotic behavior, i.e. lim f=g = 1, we shall write f » g. Typically we shall consider the asymptotic behavior when x ! 1, otherwise it will be explicitly indicated. 1 Warming up The basic cornerstone in prime number distribution theory is the simple and beau- tiful Euler’s identity: µ ¶ X1 1 Y 1 ¡1 = 1 ¡ ns ps n=1 p where s > 1, to assure the convergence, and p runs over the prime numbers. This identity is equivalent to Fundamental Theorem of ArithmeticQ (unique factorization into primes), just noting that the right hand side is (1 + p¡s + p¡2s + p¡3s + ::: ). The importance of Euler’s identity stems from establishing a link between an analytic object, the Riemann zeta function X1 1 ³(s) = ; ns n=1 and an arithmetical object, prime numbers. For instance, Euler himself in 1737 realized that when s ! 1+ the divergence of harmonic series implies that there are infinitely many prime numbers. The giant step toward the understanding of the distribution of primes was given by Riemann who, in his celebrated memoir of 1859, considered ³ as a function of complex variable and proved that it can be extendend to a meromorphic function on the whole complex plane. There are quite elementary proofs of this fact. For instance, the simple identity µ ¶ 2 X1 (¡1)n¡1 1 ¡ ³(s) = 2s ns n=1 1 proves that ³ has a meromorphic continuation to f<s > 0g (note that jn¡s ¡ (n + 1)¡sj < C(s)jn¡s¡1j) with a simple pole at s = 1 with residue 1. More ¡s P m generally, Taylor expansion (1¡x) ¡1 = amx with am the generalized binomial ¡s+m¡1¢ coefficient m , implies X1 X1 X ¡s¡m ¡s ¡m 2 am³(s + m) = (2n) am(2n) m=1 n=1 m õ ¶ ! µ ¶ X1 1 ¡s 2 = (2n)¡s 1 ¡ ¡ 1 = 1 ¡ ³(s): 2n 2s n=1 Hence meromorphic continuation to f<s > kg implies meromorphic continuation to f<s > k ¡ 1g and ³ gets extended (of course uniquely) to a holomorphic function on C ¡ f1g with a simple pole at s = 1 with residue 1. Primes appear in an involved way in Euler’s identity. It would be desirable a relation between ³ and prime numbers counting function, i.e. X ¼(x) = 1 = jfp · x : p is a prime numbergj: p·x But it is technically simpler to establish this connection through the, in principle unnatural, function ( X log p if n = pk with p prime Ã(x) = Λ(n) with Λ(n) = n·x 0 otherwise Proposition 1.1 For <s > 1 Z ³0(s) X1 Λ(n) ³0(s) s 1 ¡ = and ¡ ¡ = s (Ã(x) ¡ x)x¡s¡1 dx: ³(s) ns ³(s) s ¡ 1 n=1 1 Proof: By logarithmic differentiation (Euler’s favorite trick) µ ¶ Y ³0(s) X log p X log p log p ³(s) = (1 ¡ p¡s)¡1 ) = ¡ p¡s = ¡ + + ::: : ³(s) 1 ¡ p¡s ps p2s p p p The convergence for <s > 1 is assured comparing with a geometric series and first formula follows. 2 Now it is not hard to obtain µ ¶ Z Z ³0(s) X1 1 1 X1 m+1 Ã(x) 1 Ã(x) ¡ = Ã(m) ¡ = s dx = s dx ³(s) ms (m + 1)s xs+1 xs+1 m=1 m=1 m 1 R 1 ¡s and the second formula is proved noting that s=(s ¡ 1) = s 1 x dx. 2 We know that ³ is analytic and ³(s) » (s ¡ 1)¡1 as s ! 1, hence ¡³0(s)=³(s) ¡ s=(s ¡ 1) ! 0, and the second identity of the proposition suggests that Ã(x) should be well approximated by x. It will be the content of the prime number theorem, but firstly we want to know what does it mean in terms of ¼(x). Proposition 1.2 Let E = E(x) be an increasing function such that Ã(x) = x + O(E(x)), then ¼(x) = li(x) + O(x1=2 + E(x)= log x) where li(x) is the integral loga- R x rithm 2 dt= log t. P 1=2 Proof: It is not difficult to prove that ¼(x) = 2·n·x Λ(n)= log n+O(x ) (use 1 1=2 1 1=3 1=n 1=n that the sum equals ¼(x)+ 2 ¼(x )+ 3 ¼(x )+::: and the bound ¼(x ) · x ). On the other hand, Z Z X Λ(n) Ã(x) X x dt Ã(x) x Ã(t) dt = + Λ(n) = + ; log n log x t log2 t log x t log2 t 2·n·x 2·n·x n 2 where the second equality is just partial summation in the form a2b2 + a3b3 + a4b4 + R x ¢ ¢ ¢ = a2(b2 ¡ b3) + a3(b3 ¡ b4) + ::: with an = Λ(n) and bn = n . This gives Z x Ã(x) ¡ x Ã(t) ¡ t 1=2 ¼(x) = li(x) + + 2 dt + O(x ) log x 2 t log t R R because integrating by parts dt= log t = x= log x + dt= log2 t. 2 According to these results, Ã(x) » x translate into ¼(x) » li(x) or equivalently into ¼(x) » x= log x (l’Hˆopitalrule proves li(x) » x= log x). Any of these asymptotic formulas is called Prime Number Theorem (abbreviated as PNT in the following). Usually one wants to go beyond when estimating the size of the error term (the function E(x) below). Theorem 1.3 (PNT with error term) It holds ¼(x) = li(x) + O(E(x)) for some function E(x) = o(li(x)). 3 p ¡ 1 log x In this notes we shall prove this theorem with E(x) = xe 6 . In fact, thanks to previous proposition we shall forget about ¼(x), and prove Ã(x) = x + O(E(x)). Even today it is not known a valid error term verifying E(x) = O(x®) for some ® < 1. As we shall see later, this is related to the so-called Riemann hypothesis. 2 PNT timelines ² 1849 Gauss conjetures that li(x) approximates ¼(x). ² 1851 Chebyshev proves C1x= log x < ¼(x) < C2x= log x with explicit Ci. ² 1859 Riemann writes his celebrated 8-paged memoir containing a proof of PNT with serious gaps, using complex analysis. ² 1896 Hadamard and de la Vall´eePoussin prove (independently) PNT. ² 1948 Erd˝osand Selberg find the first “elementary proof” of PNT. ² 1958 Vinogradov and Korobov find the best known error term. ² ????¸2003 Somebody proves Riemann Hypothesis. 3 A proof (?) of PNT for dreamers In this section we give a fake proof `ala Riemann that is non rigorous but contains all the ingredients of the real proof. At first sight it seems that the missing points are of technical nature and not difficult to fill, but probably subsequent pages will show a different truth. The starting point is the formula valid for <s > 1 ³0(s) X1 Λ(n) ¡ = : (3.1) ³(s) ns n=1 Let L be the line f<s = cg, for some 1 < c < 2, and RT , ST the “infinite rectangles” f=s · T; <s · cg, f=s · T; <s ¸ cg, respectively. For x > 2, x 62 Z, Z Z Z ³0(s) xs X ³x´s ds X ³x´s ds ¡ ds = Λ(n) + Λ(n) L ³(s) s L n s L n s n<x Z n>x Z X ³x´s ds X ³x´s ds = Λ(n) lim + Λ(n) lim T !1 n s T !1 n s n<x @RT n>x @ST 4 By Cauchy’s integral formula, the first integral equals 2¼i (there is a simple pole at s = 0) and the second integral equals 0 (no poles in ST ). Hence we have a neat analytic formula for our favorite arithmetical function: Z 1 ³0(s) Ã(x) = f(s) ds where f(s) = xs: 2¼i L s³(s) As ³ is meromorphic so is f(s). Residue theorem gives Z 1 ³0(s) xs X Ã(x) = ¡ lim ds = Res(f; s) T !1 2¼i ³(s) s @RT s2P where P is the set of poles of f. Recalling that s = 1 is the unique pole of ³, it follows P = f0; 1g [ Z where Z is the set of zeros of ³. Moreover Res(f; 0) = ¡³0(0)=³(0), Res(f; 1) = x (because ³(s) » 1=(s ¡ 1) as s ! 1); and if z 2 Z is a zero of multiplicity m, Res(f; z) = mxz=z. Therefore ³0(0) X xz Ã(x) = x ¡ ¡ (3.2) ³(0) z z2Z where each zero is repeated in the summation according to its multiplicity. After this amazing formula, one can claim that the answer to any question regarding to the distribution of prime numbers is embodied in the distribution of the zeros of Riemann’s zeta function. In particular, if <z < 1 for every z 2 Z, then jxzj = x<z = o(x) and, trusting on good convergence properties of the series, PNT follows in the form Ã(x) » x. Let us finish with an unbelievably ingenious proof due to Mertens of the missing point Z ½ f<s < 1g.
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