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Two Proofs of the Prime Number Theorem TWO PROOFS OF THE PRIME NUMBER THEOREM PO-LAM YUNG 1. Introduction Let π(x) be the number of primes ≤ x. The famous prime number theorem asserts the following: Theorem 1 (Prime number theorem). x (1) π(x) ∼ log x as x ! +1. (This means limx!+1(π(x) log x)=x = 1). 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 1=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 approximately 1= log x. Cheby- shev made some important progress in the 1850's. The landmark paper of Riemann [10] (and his only one on this subject) made clear the connection of the asymptotics of π(x) to the ζ function that now bears his name. Subsequently, in 1896, the first complete proof of the prime number theorem was given independently by Hadamard [5] and de la Vall´eePoussin [2]. About 50 years later, elementary approaches to the prime number theorem were also discovered, most notably by Erd¨os[3] and Selberg [8]. Our goal in this article is to elucidate a complex 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 function instead of '; see also the exposition in [12] for another account of Newman's proof.) Before we begin, we note here that Z x 1 Li(x) = dt 2 log t also satisfies Li(x) ∼ x= log x. So the prime number theorem can also be written as π(x) ∼ Li(x) as x ! +1. Indeed, Li(x) satisfies the following asymptotics: for any N > 0, one has x x x x x Li(x) = + + 2! + ··· + (N − 1)! + O : log x (log x)2 (log x)3 (log x)N (log x)N+1 Li(x) turns out to be a better approximation of π(x) than x= log x. Date: October 26, 2016. 1 2. Chebyshev's function The proofs of the prime number theorem we will give proceeds via Chebyshev's function: X log x (x) := log p; x > 0: log p p≤x The following proposition is well known:1 Proposition 2. The prime number theorem is equivalent to the assertion that (2) (x) ∼ x: Proof. Indeed, assume for the moment that (2) holds. Then since X (x) ≤ log x = π(x) log x; p≤x dividing both sides by x and letting x ! +1, we get π(x) log(x) 1 ≤ lim inf : x!1 x Also, for any α 2 (0; 1), we have X X (x) ≥ log p ≥ log p ≥ (π(x) − π(xα)) log(xα) ≥ α(π(x) − xα) log x: p≤x xα<p≤x Hence if (2) holds, then dividing the above inequality by x, and letting x ! 1, we get that (π(x) − xα) log x π(x) log x 1 ≥ α lim sup = α lim sup : x!+1 x x!+1 x Letting α ! 1−, we get π(x) log x 1 ≥ lim sup : x!+1 x Together we obtain (1), and the prime number theorem holds. The converse implication, namely that (1) implies (2), is not much harder. Since we do not need this direction of the implication, we leave this verification to the interested reader. Note that can be rewritten as X X X (x) = log p = Λ(n) p m2N: pm≤x n≤x where Λ is the von Mangoldt function, defined for n 2 N by ( log p; if n = pm for some prime p and some positive integer m Λ(n) = : 0; otherwise We are interested in the asymptotics of (x) as x ! +1. We will see, in the next section, that 1 one can study this by considering the Dirichlet series corresponding to fΛ(n)gn=1, namely 1 X Λ(n) : ns n=1 1See also Proposition 2.1 of Chapter 7 of [9]. 2 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; 1) ! C is a measurable function that vanishes on (0; 1). Suppose further that there exists some a 2 R, A > 0 such that jf(x)j ≤ Axa for all x 2 [1; 1). Let a0 be the infimum of all a 2 R, for which there exists A > 0 such that the above estimate holds. Then the Mellin transform of f is defined by Z 1 dx Mf(s) = f(x)x−s 0 x for all s 2 C with Re s > a0; 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 exists some a 2 R, A > 0 such that jF (t)j ≤ Aeat for all t 2 [0; 1). Let a0 be the infimum of all a 2 R, for which there exists A > 0 such that the above estimate holds. Then the Laplace transform of F is defined by Z 1 LF (s) = F (t)e−stdt −∞ for all s 2 C with Re s > a0; indeed the integral defining LF (s) converges absolutely there, and defines a holomorphic function of s in that half plane. If F (t) := f(et) where f is as in the above definition of the Mellin transform, then Mf(s) = LF (s) whenever they are defined. 1 Recall also the Fourier transform on R. If G 2 L (R), then its Fourier transform is defined by Z 1 Gb(τ) = G(t)e−itτ dt −∞ for all τ 2 R. If F is as in the above definition of the Laplace transform, then for all c > a0 and all −ct τ 2 R, we have LF (c + iτ) = Fcc(τ) where Fc(t) := F (t)e . P Our goal was to understand asymptotics of (x) = n≤x Λ(n) as x ! +1. The strategy we will follow is indeed fairly general, and the initial steps works perfectly well when the sequence 1 1 o(1) fΛ(n)gn=1 is replaced by any sequence of complex numbers fangn=1, as long as janj = n as n ! 1. We phrase it in the following proposition: 3 1 o(1) Proposition 3. Suppose fang is a sequence of complex numbers satisfying janj = n as n=1 P n ! 1. Let f : (0; 1) ! C be defined by f(x) = n≤x an. Then the Mellin transform of f is defined for all s 2 C with Re s > 1, and is given by 1 Mf(s) = D (s) s a for all such s, where 1 X an D (s) := a ns n=1 1 is the Dirichlet series corresponding to the sequence fangn=1 (also defined for Re s > 1). In partic- ular, the Mellin transform of is defined for all s 2 C with Re s > 1, and is given for all such s by 1 1 X Λ(n) M (s) = : s ns n=1 o(1) P Proof. Suppose janj = n as n ! 1, and f(x) = n≤x an. Then for any a > 1, there exists A > 0 such that f(x) ≤ Axa. Furthermore, we can rewrite f(x), as 1 X f(x) = anχ[n;1)(x) n=1 where χ[n;1) is the characteristic function of the interval [n; 1). Hence the Mellin transform of f is defined for all s 2 C with Re s > 1, and is given by 1 Z 1 1 X dx 1 X an 1 Mf(s) = a x−s = = D (s) n x s ns s a n=1 n n=1 for all such s. (The interchange of the sum with the integral can be justified using Fubini's theorem.) o(1) Since Λ(n) = n as n ! 1, applying the result to an = Λ(n) yields the desired conclusion for M (s). P The proposition suggests that in order to understand (x) = n≤x Λ(n) (or more generally P f(x) = n≤x an where fang is as in the proposition), it may be helpful to study the corresponding Dirichlet series DΛ(s) (or Da(s)); indeed, if we can invert the Mellin transform, then we can hope to convert information about the Dirichlet series DΛ(s) (or Da(s)) into information about (x) (or f(x)). 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 1 hypothesis on G. For instance, if both G and Gb are in L (R), then 1 Z 1 G(t) = Gb(τ)eitτ dτ 2π −∞ for all t 2 R. We will need a slightly different form of the Fourier inversion formula, when Gb is not necessarily integrable: 4 1 Proposition 4.
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