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TOWARDS the PRIME NUMBER THEOREM Contents 1. The TOWARDS THE PRIME NUMBER THEOREM ERIC ANTLEY Abstract. Originally proven by Jacques Salomon Hadamard and Charles Jean de la Valle-Poussin independently in 1896, the prime number theorem shows the number of primes less than or equal to a given number x approaches x log x as x tends towards infinity. In this paper, a proof using complex analysis will be given to prove this statement. Contents 1. The Chebyshev Functions and Their Properties 1 2. The Riemann Zeta Function 4 3. Some Integral Transforms 12 Acknowledgments 15 References 16 Let P be the set of primes. We define the function π(x) to be the prime counting function. Definition 0.1. Given a real number x, π(x) is the the cardinality of the set fp 2 Pjp ≤ xg. Theorem 0.2 (The Prime Number Theorem). We have the following limit behav- ior: x (0.3) π(x) ∼ : log x 1. The Chebyshev Functions and Their Properties We define the function ν(x) to be the first Chebyshev function. P Definition 1.1. Given a real number x, we define ν(x) to be the sum fp2Pjp≤xg log p. We define the function (x) to be the second Chebyshev function. Definition 1.2. Given a real number x, we define (x) to be the sum over primes P p and natural numbers k of fp2Pjpk≤xg log p. We define the function Λ(x) to be the von Mangoldt function. Definition 1.3. Given a real number x, we define Λ(x) to be log p if pk = x where p is some prime and 0 else, where k 2 Z. 1 2 ERIC ANTLEY P Clearly, (x) = n≤x Λ(n). The prime counting function π(x) is by definition what we are investigating, yet this function is very \unnatural." For now, the reader can understand this as simply meaning that it is a difficult function to work with, although easy to comprehend. On the other hand, the function (x) is a very \natural" function. We will see later that (x) is strongly related to a deceptively simple looking infinite sum known as the Riemann zeta function which is deeply rooted in number theory. Our first goal will be to examine (x) and related functions in order to show that properties of (x) imply the Prime Number Theorem, but first we will show that (x) is Θ(x). This means that for constants A and A0 with sufficiently large x, we have Ax ≤ (x) ≤ A0x. We will use this in section 3. From now on, we assume the symbol p denotes a prime number. P log x Lemma 1.4. (x) = p≤xb log p c log p. Proof. We now attempt to find an expression that assigns the correct \weight" to P each log p. By definition, (x) = p2P;pk≤x log p. Let p 2 P and k ≥ 0 be such that pk ≤ x and pk+1 > x. Note p contributes k log p to (x). Let x = pkα. k Then we see that logp x = logp (p α) = k logp p + logp α = k + logp α. Now, α is necessarily less than k, so bk + logy αc = k. Thus, we can express (x) as the P sum: p≤xblogp xc log p. Finally, we apply the change of log base formula to get P P log x p≤xblogp xc log p = (x) = p≤xb log p c log p. We now turn our attention to estimating the size of (x), but first we will require a few more definitions that will be used to describe the limiting behavior of functions. Definition 1.5. o(f(x)) Let f(x) and g(x) be functions. We say that g(x) = o(f(x)) if for all > 0 there exists an x0 such that for all x > x0 we have jg(x)j ≤ jf(x)j. Definition 1.6. O(f(x)) Let f(x) and g(x) be functions. We say that g(x) = O(f(x)) if there exists an M > 0 such that for all x > x0 we have jg(x)j ≤ Mjf(x)j. Definition 1.7. Θ(f(x)) Let f(x) and g(x) be functions. We say that g(x) = Θ(f(x)) if there exist A 0 0 and A such that for all x > x0 we have Ag(x) ≤ f(x) ≤ A g(x). 1 2 Lemma 1.8. (x) = ν(x) + O(x 2 (log x) ). 1 1 P m Proof. First, we note that (x) = m=1 ν(x ). This is a quick check which follows k 1 from the fact that p ≤ x is equivalent to p ≤ x k . Now, it is trivial to see that 1 1 1 for x > 1 we have ν(x) < x log x. Thus for m > 2, ν(x m ) < x m log x ≤ x 2 log x. 1 P m Now consider the sum 2≤m ν(x ). There are O(log x) terms of this sum because 1 1 ν(x m ) = 0 if m > log2 x, and each term is O(x 2 log x). Therefore: 1 X 1 X 1 1 (x) = ν(x m ) = ν(x) + ν(x m ) = ν(x) + O(x 2 log x)O(log x) (1.9) m=1 2≤m 1 2 = ν(x) + O(x 2 (log x) ): TOWARDS THE PRIME NUMBER THEOREM 3 Theorem 1.10. There exist constants A and A0 such that for large x 2 N we have Ax < (x) < A0x. Therefore (x) = Θ(x). 2m+1 (2m+1)! (2m+1)(2m):::(m+2) Proof. By the binomial theorem, m = m!(m+1)! = m! . Set M equal to this expression. Then since M occurs twice as a term of the expansion of (1 + 1)2m+1, 2M < 22m+1. Thus, M < 22m, and log M < 2m log 2. Now, it's obvious that if m + 1 < p ≤ 2m + 1, then p must be equal to only one of the terms in the numerator of M. Thus the product of all primes satisfying the previous Q inequality must divide the numerator of M, so m+1<p≤2m+1 p ≤ M. We have, X (1.11) ν(2m + 1) − ν(m + 1) = log p ≤ log M < 2m log 2: m+1<p≤2m+1 Now, for n = 1 or n = 2, it is clear that ν(n) < 2n log 2. To show this inequality holds for all n 2 N, we proceed by induction. Assume it holds for all n ≤ x − 1 for some natural number x > 2. If x is even, then x is not prime. Thus ν(x) = ν(x−1) and ν(x) < 2(x − 1) log 2 < 2x log 2. On the other hand, if x is odd, then for some n we have x = 2n + 1. By our previous work, we see that: (1.12) ν(x) = ν(2n + 1) − ν(n + 1) + ν(n + 1) < 2n log 2 + 2(n + 1) log 2 = 2x log 2: Note that we were able to bound ν(n + 1) because n + 1 < x, so our induction held up to n+1 by assumption. Since we have shown the inductive step holds for x odd or even, ν(n) < 2n log 2 = O(n), which in turn implies (n) = O(n). We will Q bx=pc+bx=p2c+bx=p3c+::: now show that the bound is tight. Recall that x! = p p . Let P1 b 2n c−2b n c 2n 2n! Q i=1 pi pi N = n = n!2 . We can write this as the product: p p . Notice that each term of the sum is either one or zero; this follows from simple properties of the floor function. Furthermore, every term of the sum is zero after pi exceeds 2n. Therefore: 1 blogp 2nc X 2n n X log 2n (1.13) b c − 2b c ≤ 1 = blog 2nc = b c: pi pi p log p i=1 i=1 We see that 2n ! X X 2n n X log 2n (1.14) log N = (b c − 2b c) log p ≤ b c log p = (2n): pi pi log p p≤2n i=1 p≤2n n+1 n+2 2n Finally, we see that N = 1 2 ::: n . Observe each term in this product is greater than or equal to 2 and there are n terms. Thus we have 2n ≤ N and 1 n log 2 ≤ log N ≤ (2n). If 1 ≤ n = b 2 xc we see that: 1 (1.15) x log 2 ≤ n log 2 ≤ (2n) ≤ (x): 4 0 Therefore, Ax < (x) < A x. (x) Corollary 1.16. x is bounded. 4 ERIC ANTLEY Proof. By Theorem 1:10, for natural numbers x we have 0 ≤ (x) = O(x). There- fore there exists an M such that for all x > x0 we have (x) ≤ Mx. Division by (x) nonzero x gives x ≤ M for x > x0. For the remaining number of finite natural (x) numbers n less than or equal to x0, (n) < 1. Therefore x is bounded. We now wish to restate the Prime Number Theorem in the language of a function whose properties are now familiar to us, (x). (x) Theorem 1.17. The Prime Number Theorem is equivalent to limx!1 x = 1. (x) π(x) Proof. We must show that limx!1 x = limx!1 x=(log(x)) . For that, we take the difference: (1.18) P P b log x c log p π(x) (x) p≤x log(x) p≤x log p 1 X − = − ≤ log(x) − log(p): x=(log(x)) x x x x p≤x Now, fix a > 0.
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