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The Zeta Function and the Prime Number Theorem

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The Zeta Function and the Prime Number Theorem The Zeta Function and the Prime Number Theorem Michael Taylor Introduction I was motivated to put together these notes while enjoying three books on prime numbers ([D], [J], and [S]) as 2003 Summer reading. The Prime Number Theorem, giving the asymptotic behavior as x ! +1 of ¼(x), the number of primes · x, has for its proof three ingredients: (I) Formulas for log ³(s) and ³0(s)=³(s) as Mellin transforms involving a func- tion J(x), closely related to ¼(x), and a more subtly related function, Ã(x). (II) Information that ³0(s)=³(s) has on fs : Re s ¸ 1g a singularity only at s = 1. (III) A Tauberian theorem that yields the asymptotic behavior of J(x) from the behavior of such a Mellin transform. Step (I) is achieved by taking the Euler product for ³(s), applying log, and following your nose, naturally ¯nding J(x). It is easy to show that J(x) and ¼(x) have similar behavior as x ! +1. Taking the s-derivative yields the relation involving ³0(s)=³(s), and naturally produces the function Ã(x). Step (II) is more subtle, though there is a clever and brief proof, largely due to de la Vall¶eePoussin, which is commonly presented (and which we will recall in Appendix A). Step (III) can be handled by a powerful result known as Ikehara's Tauberian Theorem, though it is often treated via a variant, slightly weaker and somewhat easier result, known as the Ingham-Newman theorem. One possibly novel aspect of our treatment (in xx4{5) it to produce another variant, relying more on real Fourier analysis and the fact that ³0(s)=³(s) is C1 on fs : Re s = 1g, except at s = 1. We came upon this approach while exploring similarities between arguments to implement step (III) and arguments used for \precise spectral asymptotics" of elliptic di®erential operators. In x1 we implement step (I), recast step (II) as a statement about the zeros of ³(s), state Ikehara's Tauberian Theorem, and use these results to prove the PNT. We give the Laplace transform version of Ikehara's theorem, and using it involves making a change of variable. In x2 we restate Ikehara's theorem in Mellin transform language, allowing one to avoid such a change of variable. Our calculations are done in terms of the Mellin transform of measures, often atomic measures. We avoid mention of Dirichlet series. This formalism allows us to avoid summation by parts arguments, and simply use integration by parts. We consider this an advantage, though some might regard it as merely a matter of taste. To be a little more explicit about matters hinted about above, in xx1{2 we go from X Ã(x) » x; Ã(x) = log p p;k;pk·x 1 2 to x X 1 J(x) » ;J(x) = ¼(x1=k) log x k k¸1 to x ¼(x) » : log x It turns out that formulas relating Ã(x) to J(x) bring in naturally Z x dy Li(x) = ; 2 log y and that Li(x) is a better approximation to J(x) and ¼(x) than is x= log x. In x3 we show how estimates on the error E(x) = Ã(x) ¡ x naturally lead to estimates on J(x) ¡ Li(x). While we do not produce estimates on E(x) here that imply the superiority of Li(x), these formulas do serve to make Li(x) look like the natural candidate. In x4 we use the results of steps (I) and (II), interpreted as information on the Fourier transform of a measure H derived from J(x), to produce complete asymptotic expansions of a class of molli¯cations of H. In x5 we apply fairly elementary methods of measure theory and Fourier analysis to derive from this the asymptotic behavior of J(x) and hence a proof of the PNT. There is an alternative approach to step (III), involving a much simpler Taube- rian theorem, at the cost of a more sophisticated Mellin transform and a ¯ner knowledge of ³0(s)=³(s) on fs : Re s ¸ 1g. We discuss this in x6, giving a treatment adapted from Chapter 13 of [A]. We have two appendices. Appendix A contains a proof of the result that ³(s) has no zeros on fs : Re s = 1g. Appendix B, not strongly related to the rest of these notes, gives a calculation of ³(k) when k is an even integer, as a result of k examining the Fourier series of fk(x) = x on [¡¼; ¼]. 1. The zeta function and the prime number theorem The zeta function, given for Re s > 1 by X1 1 (1.1) ³(s) = ; ns n=1 is connected to the study of primes via the Euler formula Y³ 1 ´¡1 (1.2) ³(s) = 1 ¡ : ps p 3 Here and below, p will run over the set of primes. The formula (1.2) can be rewritten X ³ 1 ´ log ³(s) = ¡ log 1 ¡ ps p (1.3) X X1 1 1 = : k pks p k=1 This can be rewritten as a Mellin transform, using the counting function (1.4) ¼(x) = #fp : p · xg; satisfying X 0 (1.5) ¼ (x) = ±p(x): p We see that if X 1 (1.6) J(x) = ¼(x1=k); k k¸1 then X X 0 1 (1.7) J (x) = ± k (x); k p k¸1 p and hence (1.3) is equivalent to Z 1 log ³(s) = J 0(x)x¡s dx 0 (1.8) Z 1 = s J(x)x¡s¡1 dx: 0 Thus there is a hope (vaguely realized) of obtaining J(x) via inversion of the Mellin transform. We make some comments on the close relationship of ¼(x) and J(x). Note that in (1.6) all terms vanish for k so large that x1=k < 2, i.e., the sum is restricted to k · log2 x. It readily follows that log x X2 1 ¼(x) < J(x) < ¼(x) + ¼(x1=2) k (1.9) k=2 1=2 < ¼(x) + ¼(x ) log(log2 x): 4 While we do not make use of it here, it is interesting to know that the relationship (1.6) can be inverted, to yield X ¹(n) (1.10) ¼(x) = J(x1=n); n n¸1 where ¹ is the MÄobiusfunction, given by ¹(n) = 0 if n has a repeated prime factor, (1.11) (¡1)#prime factors otherwise. At this point we recall some lore on ¼(x) and on ³(s). First, there is Chebyche®'s estimate: x x (1.12) C < ¼(x) < C ; 1 log x 2 log x which is more elementary than the Prime Number Theorem. (A direct proof of (1.12) will be contained in results we derive in x4.) Next, ³(s) is meromorphic on C, with just one pole, at s = 1. Also ³(s) has no zeros in fs : Re s > 1g, as a simple consequence of (1.2). In addition, ³(s) has no zeros on fs : Re s = 1g. This result is harder to prove, and lies at the heart of the proofs of Hadamard and de le Vall¶eePoussin of the Prime Number Theorem. (We will present what has become the standard proof of this fact in Appendix A.) Below we will show how the PNT follows from this fact, together with a certain Tauberian theorem, which we will also state below. Let us proceed to take s = 1 + it in (1.8) and rewrite it as a Fourier transform: Z log ³(1 + it) 1 (1.13) = J(ey)e¡y e¡ity dy = Fb(t); 1 + it ¡1 the Fourier transform of (1.14) F (y) = J(ey)e¡y: Note that F is supported on fy 2 R : y ¸ log 2g. By (1.9) and (1.12) we have C 0 < F (y) · 3 ; y which does not quite give F 2 L1(R). This is consistent with the fact that the left side of (1.13) has a logarithmic singularity at t = 0: log ³(1 + it) 1 (1.15) Fb(t) = » log ; t ! 0: 1 + it t 5 By comparison, if we set 1 G(y) = ; y ¸ 1; (1.16) y 0; y < 1; we have Gb 2 C1(R n 0) and 1 (1.17) Gb(t) » log ; t ! 0: t Hence the Fourier transform of F ¡ G is continuous on R. This leads one to believe that F (y) ¡ G(y) is relatively smaller as y ! 1, or otherwise put, that ey (1.18) J(ey) » ; y ! 1; y i.e., x x (1.19) J(x) » ; or ¼(x) » ; x ! 1: log x log x Of course, (1.19) is one version of the PNT. Unfortunately it is not so obvious how to deduce (1.18){(1.19) rigorously from (1.15){(1.17). One way to get asymptotics for J(x) from knowledge of ³(s) will be to exploit the following. Ikehara Tauberian Theorem. Let w % and consider Z 1 (1.20) f(s) = e¡sy dw(y): 0 Assume f is holomorphic on fs : Re s > 1g and that A (1.21) f(s) ¡ is continuous on fs : Re s ¸ 1g: s ¡ 1 Then (1.22) e¡yw(y) ! A as y ! +1: This result is applicable to what one gets upon applying d=ds to (1.8), i.e., Z ³0(s) 1 ¡ = J 0(x)(log x)x¡s dx ³(s) 0 (1.23) Z 1 = J 0(ey)y e¡syey dy: 0 6 This has the form (1.20) with Z y w(y) = xexJ 0(ex) dx 0 Z ey (1.24) = (log x)J 0(x) dx 0 X = log p: p;k;pk·ey In other words, X (1.25) w(y) = Ã(ey);Ã(x) = log p: p;k;pk·x Ikehara's theorem applies, with A = 1, to give: Ã(x) (1.26) ! 1; as x ! 1: x Note that Z x Ã(x) = (log y)J 0(y) dy (1.27) 0 Z x J(y) = (log x)J(x) ¡ dy: 0 y Now the estimate (1.12) implies Z J(y) C 1 x J(y) (1.28) · 3 ; hence lim dy = 0; y log y x!1 x 0 y so (1.26) is equivalent to log x log x (1.29) lim J(x) = 1; i.e., lim ¼(x) = 1; x!1 x x!1 x the PNT as stated in (1.19).
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