as x. Before giving the detailed proof, we outline our strategy. Define the functions for Re s > 1.

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1 Chapter 7 The Prime number theorem for arithmetic progressions 7.1 The Prime number theorem Denote by π( the number of primes. We prove the Prime Number Theorem. Theorem 7.1. We have π( log as. Before giving the detailed proof, we outline our strategy. Define the functions θ( := log p, ψ( := log p = Λ(n, p n k,p: p k where Λ is the von Mangoldt function, given by Λ(n = log p if n = p k for some prime p and some k 1, and Λ(n = 0 otherwise. By Lemma 3.14 from Chapter 3 we have n=1 Λ(nn s = ζ (s ζ(s for Re s > 1. By applying Theorem 6.3 (the Tauberian theorem for Dirichlet series to the latter we obtain ψ( = 1 Λ(n 1 as. n 107

2 We prove that ψ( θ( is small. This gives θ(/ 1 as. Using partial summation, we deduce π( log / 1 as. We first verify the conditions of the Tauberian theorem. Lemma 7.. n=1 Λ(nn s can be etended to a function analytic on an open set containing {s C, Re s 1, s 1}, with a simple pole with residue 1 at s = 1. Proof. Let A := {s C, Re s 1, s 1}. By Theorem 5., ζ(s is analytic on an open set containing A with a simple pole at s = 1. Further, by Corollary 5.4 and Theorem 5.5, ζ(s 0 on A, and hence also ζ(s 0 on an open set containing A. So by Lemma.17, ζ (s/ζ(s is analytic on an open set containing A, with a simple pole with residue 1 at s = 1. This proves Lemma 7.. Lemma 7.3. (i θ( = O( as. (ii ψ( = θ( + O( as. (iii ψ( = O( as. Proof. (i By homework eercise 3a, we have p p 4 for. This implies θ( = p log p log 4 = O( as. (ii We have ψ( = p,k: p k log p = p log p + p = θ( + θ( 1/ + θ( 1/3 + log p + p 3 log p + Notice that θ(t = 0 if t <. So θ( 1/k = 0 if 1/k <, that is, if k > log / log. Hence ψ( θ( = [log / log ] k= θ( 1/k θ( + [log / log ] k=3 θ( 3 θ( ( log + log 3 θ( 3 = O ( + 3 log = O( as. (iii Combine (i and (ii. This follows also from homework eercise 6, but (ii will be needed anyhow. 108

3 Proof of Theorem 7.1. Lemmas 7. and 7.3 (iii imply that L Λ (s = n=1 Λ(nn s satisfies all conditions of Theorem 6.3, with α = 1. Hence ψ( From Lemma 7.3 (ii we infer θ( ψ( + O( = = 1 Λ(n 1 as. n = ψ( + O( 1/ 1 as. We now apply partial summation to obtain our result for π(. Thus, π( = p 1 = p log p = θ( log + θ(t t log t dt. 1 log p = θ( 1 log θ(t ( 1 dt log t By Lemma 7.3 (i there is a constant C > 0 such that θ(t Ct for all t. Together with homework eercise 1, this implies Hence θ(t t log dt C t π( log = θ( dt ( log t = O log ( log + O log = θ( + O ( 1 log as. 1 as. 7. The Prime number theorem for arithmetic progressions Let q, a be integers with q, gcd(a, q = 1. Define π(; q, a := number of primes p with p a (mod q. 109

4 Theorem 7.4. We have π(; q, a 1 ϕ(q log as. The proof is very similar to that of the Prime number theorem. quantities Let θ(; q, a := ψ(; q, a := p, p a (mod q log p, p,k, p k, p k a (mod q F (s := n=1, n a (mod q Let G(q be the group of characters modulo q. Lemma 7.5. For s C with Re s > 1 we have F (s = 1 ϕ(q χ G(q log p = Λ(nn s. n, n a (mod q χ(a L (s, χ L(s, χ. Λ(n. Define the Proof. By homework eercise 7a we have for χ G(q and for s C with Re s > 1, since χ G(q is a strongly multiplicative arithmetical function and L(s, χ converges absolutely, L (s, χ L(s, χ = χ(nλ(nn s. n=1 Using Theorem 4.9 (ii (one of the orthogonality relations for characters we obtain for s C with Re s > 1, χ G(q χ(a L (s, χ L(s, χ = = n=1 χ G(q χ G(q χ(a χ(nλ(nn s n=1 χ(aχ(n Λ(nn s = ϕ(q n=1, n a (mod q Λ(nn s. 110

5 Lemma 7.6. The function F (s can be continued to a function analytic on an open set containing {s C : Re s 1, s 1}, with a simple pole with residue ϕ(q 1 at s = 1. Proof. Let A := {s C : Re s 1, s 1}, B := {s C : Re s 1}. By Theorem 5.3 (iii, L(s, χ (q 0 is analytic on an open set containing A, with a simple pole at s = 1. Further, by Corollary 5.4 and Theorem 5.5, L(s, χ (q 0 0 for s A, and hence for s in an open set containing A. Therefore, L (s, χ (q 0 /L(s, χ (q 0 is analytic on an open set containing A. Further, by Lemma.17, it has a simple pole with residue 1 at s = 1. Let χ be a character mod q with χ χ (q 0. By Theorem 5.3 (ii, L(s, χ is analytic on an open set containing B, and by Corollary 5.4 and Theorem 5.5, it is non-zero on B, hence on an open set containing B. Therefore, L (s, χ/l(s, χ is analytic on an open set containing B. Now by Lemma 7.5, F (s is analytic on an open set containing A, with a simple pole with residue χ (q 0 (a/ϕ(q = ϕ(q 1 at s = 1. Lemma 7.7. (i θ(; q, a = O( as. (ii ψ(; q, a θ(; q, a = O( as. (iii ψ(; q, a = O( as. Proof. (i We have θ(; q, a θ( = O( as. (ii We have ψ(; q, a θ(; q, a = log p (iii Obvious. k,p, k,p k k,p, k,p k, p k a (mod q log p = ψ( θ( = O( as. Proof of Theorem 7.4. Notice that F (s satisfies the conditions of Theorem 6.3, with α = ϕ(q 1. Hence ψ(; q, a 1 ϕ(q 111 as,

6 and then by Lemma 7.7 (ii, θ(; q, a = ψ(; q, a + O( 1 ϕ(q as. By partial summation we have π(; q, a = p, p a (mod q log p 1 log p = θ(; q, a log + θ(t; q, a t log t dt. Now 0 θ(t; q, a t log t dt θ(t ( t log t dt = O log using the estimate from the proof of Theorem 7.1. So π(; q, a log This completes our proof. = θ(; q, a ( 1 + O log 1 ϕ(q as, as. In Chapter 1 we mentioned that there are sharper versions of the Prime Number Theorem, with an estimate for the error π( Li(. Similarly, there are refimenents of the Prime number theorem for arithmetic progressions with an estimate for the error π(; q, a Li(/ϕ(q. The simplest case is when we fi q and let, but for applications it is important to have also versions where q is allowed to move in some range when we let. By an absolute constant we mean a positive constant that does not depend on anything. The following result was proved by Walfisz in 1936, with important preliminary work by Landau and Siegel. Theorem 7.8. There are an absolute constant C 1, and for every real A > 0 there is a number C (A depending on A, such that the following holds. For every real 3, every integer q with q (log A and every integer a with gcd(q, a = 1, we have 1 π(; q, a ϕ(q Li( C 1e C (A log. 11

7 The constants C 1, C (A are ineffective, this means that by going through the proof of the theorem one cannot compute the constants, but only show that they eist. As we mentioned in Chapter 1, there is an intricate connection between the zero-free region of ζ(s and estimates for π( Li(, where Li( = dt/ log t. Similarly, there is a connection between zero-free regions of L-functions and estimates for π(; q, a Li(/ϕ(q. We recall the following, rather complicated, result of Landau (191 on the zero-free region of L-functions. Theorem 7.9. There is an absolute constant c > 0 such that for every integer q the following holds. Among all characters χ modulo q, there is at most one such that L(s, χ has a zero in the region R(q := { } c s C : Re s > 1. log(q(1 + Im s If such a character χ eists, it has no more than one zero in R(q and moreover, χ χ (q 0, χ is a real character and the zero is real. Any character χ modulo q having a zero in R(q is called an eceptional character mod q, and the zero of L(s, χ in R(q is called an eceptional zero. It is conjectured that eceptional characters do not eist. In order to obtain Theorem 7.8, one needs an estimate for the real part of a possible eceptional zero of an L-function. The following result was proved by Siegel (1935. Theorem For every ε > 0 there is a number c(ε > 0 such that for every integer q the following holds: if χ is an eceptional character modulo q and β an eceptional zero of L(s, χ, then Re β < 1 c(εq ε. Theorems 7.9 and 7.10 imply (after a lot of work Theorem 7.8. Proofs of Theorems may be found in H. Davenport, Multiplicative Number Theory, Graduate tets in mathematics 74, Springer Verlag, nd ed., Knowing that an arithmetic progression contains infinitely many primes, one would like to know when the first prime in such a progression occurs, i.e., the smallest such that π(; q, a > 0. The following estimate is due to Linnik (

8 Theorem Denote by P (q, a the smallest prime number p with p a (mod q. There are absolute constants c, L such that for every integer q and every integer a with gcd(a, q = 1 we have P (q, a cq L. The eponent L is known as Linnik s constant. Since the appearance of Linnik s paper, various people have tried to estimate it. The present record is L = 5.18, due to Xylouris (

Before giving the detailed proof, we outline our strategy. Define the functions. for Re s > 1.

Before giving the detailed proof, we outline our strategy. Define the functions. for Re s > 1. Chapter 7 The Prime number theorem for arithmetic progressions 7. The Prime number theorem We denote by π( the number of primes. We prove the Prime Number Theorem. Theorem 7.. We have π( as. log Before