Chapter 5 The Riemann zeta function and L-functions 5.1 Basic facts We prove some results that will be used in the proof of the Prime Number Theorem (for arithmetic progressions). The L-function of a Dirichlet character χ modulo q is defined by 1 X L(s; χ) = χ(n)n−s: n=1 P1 −s We view ζ(s) = n=1 n as the L-function of the principal character modulo 1, (1) (1) more precisely, ζ(s) = L(s; χ0 ), where χ0 (n) = 1 for all n 2 Z. We first prove that ζ(s) has an analytic continuation to fs 2 C : Re s > 0gnf1g. We use an important summation formula, due to Euler. Lemma 5.1 (Euler's summation formula). Let a; b be integers with a < b and f :[a; b] ! C a continuously differentiable function. Then b X Z b Z b f(n) = f(x)dx + f(a) + (x − [x])f 0(x)dx: n=a a a 101 Remark. This result often occurs in the more symmetric form b Z b Z b X 1 1 0 f(n) = f(x)dx + 2 (f(a) + f(b)) + (x − [x] − 2 )f (x)dx: n=a a a Proof. Let n 2 fa; a + 1; : : : ; b − 1g. Then Z n+1 Z n+1 x − [x]f 0(x)dx = (x − n)f 0(x)dx n n h in+1 Z n+1 Z n+1 = (x − n)f(x) − f(x)dx = f(n + 1) − f(x)dx: n n n By summing over n we get b Z b X Z b (x − [x])f 0(x)dx = f(n) − f(x)dx; a n=a+1 a which implies at once Lemma 5.1. Theorem 5.2. ζ(s) has a unique analytic continuation to the set fs 2 C : Re s > 0; s 6= 1g, with a simple pole with residue 1 at s = 1. Proof. By Corollary 2.4 we know that an analytic continuation of ζ(s), if such exists, is unique. For the moment, let s 2 C with Re s > 1. Then by Lemma 5.1, with f(x) = x−s, N X Z N Z N n−s = x−sdx + 1 + (x − [x])(−sx−1−s)dx n=1 1 1 1 − N 1−s Z N = + 1 − s (x − [x])x−1−sdx: s − 1 1 If we let N ! 1 then the left-hand side converges, and also the first term on the right-hand side, since jN −1−sj = N −1−Re s ! 0. Hence the integral on the right-hand side must converge as well. Thus, letting N ! 1, we get for Re s > 1, 1 Z 1 (5.1) ζ(s) = + 1 − s (x − [x])x−1−sdx: s − 1 1 102 We now show that the integral on the right-hand side defines an analytic function on U := fs 2 C : Re s > 0g, by means of Theorem 2.6. The function F (x; s) := (x − [x])x−1−s is measurable on [1; 1) × U (by, e.g., the fact that its set of discontinuities has Lebesgue measure 0) and for every fixed x it is analytic in s. Let K be a compact subset of U. Then there is σ > 0 such that Re s > σ for all s 2 K. Now for x > 1 and s 2 K we have −1−s −1−σ j(x − [x])x j 6 x R 1 −1−σ and 1 x dx < 1. Hence all conditions of Theorem 2.6 are satisfied, and we may indeed conclude that the integral on the right-hand side of (5.1) defines an analytic function on U. Consequently, the right-hand of (5.1) is analytic on fs 2 C : Re s > 0; s 6= 1g and it has a simple pole at s = 1 with residue 1. We may take this as our analytic continuation of ζ(s). Theorem 5.3. Let q 2 Z>2, and let χ be a Dirichlet character mod q. Q −s −1 (i) L(s; χ) = p(1 − χ(p)p ) for s 2 C, Re s > 1. (q) (ii) If χ 6= χ0 , then L(s; χ) converges, and is analytic on fs 2 C : Re s > 0g. (q) (iii) L(s; χ0 ) can be continued to an analytic function on fs 2 C : Re s > 0; s 6= 1g, and for s in this set we have (q) Y −s L(s; χ0 ) = ζ(s) · (1 − p ): pjq (q) Hence L(s; χ0 ) has a simple pole at s = 1. Proof. (i) χ is a strongly multiplicative function, and L(s; χ) converges absolutely for Re s > 1. Apply Corollary 3.14. (ii) Let N be any positive integer. Then N = tq + r for certain integers t; r with t > 0 and 0 6 r < q. By one of the orthogonality relations for characters (see Pq P2q Theorem 4.11), we have m=1 χ(m) = 0, m=q+1 χ(m) = 0, etc. Hence N X χ(n) = χ(tq + 1) + ··· + χ(tq + r) 6 r < q: n=1 103 This last upper bound is independent of N. Now Theorem 3.2 implies that the L-series L(s; χ) converges and is analytic on Re s > 0. (iii) By (i) we have for Re s > 1, (q) Y −s −1 Y −s L(s; χ0 ) = (1 − p ) = ζ(s) (1 − p ): p-q pjq The right-hand side is defined and analytic on fs 2 C : Re s > 0; s 6= 1g, and so it (q) can be taken as an analytic continuation of L(s; χ0 ) on this set. Corollary 5.4. Both ζ(s) and L(s; χ) for any character χ modulo an integer q > 2 are 6= 0 on fs 2 C : Re s > 1g. Proof. Use part (i) of the above theorem, together with Corollary 3.14. 5.2 Non-vanishing on the line Re s = 1 We prove that ζ(s) 6= 0 if Re s = 1 and s 6= 1, and L(s; χ) 6= 0 for any s 2 C with Re s = 1 and any non-principal character χ modulo an integer q > 2. We have to distinguish two cases, which are treated quite differently. We interpret ζ(s) as (1) L(s; χ0 ). Theorem 5.5. Let q 2 Z>1, χ a character mod q, and t a real. Assume that either 2 (q) t 6= 0, or t = 0 but χ 6= χ0 . Then L(1 + it; χ) 6= 0. Proof. We use a famous idea, due to Hadamard. It is based on the inequality 2 (5.2) 3 + 4 cos θ + cos 2θ = 2(1 + cos θ) > 0 for θ 2 R: Suppose that L(1 + it; χ) = 0. Consider the function (q) 3 4 2 F (s) := L(s; χ0 ) · L(s + it; χ) · L(s + 2it; χ ): By our assumption on χ and t, L(s + 2it; χ2) is analytic around s = 1. Further, (q) L(s; χ0 ) has a simple pole at s = 1, while L(s + it; χ) has by assumption a zero at s = 1. Hence (q) 2 ords=1(F ) = 3 · ords=1 L(s; χ0 ) + 4 · ords=1 L(s + it; χ) + ords=1 L(s + 2it; χ ) > −3 + 4 = 1: 104 This shows that F is analytic around s = 1, and has a zero at s = 1. We now prove that jF (σ)j > 1 (or rather, log jF (σ)j > 0) for σ > 1. This gives a contradiction since by continuity, limσ#1 jF (σ)j should be 0. So our assumption that L(1 + it; χ) = 0 must be false. From the definition of the function F we obtain 0 1 3 4 Y 1 1 1 log jF (σ)j = log @ · · A (q) −σ 1 − χ(p)p−σ−it 1 − χ(p)2p−σ−2it p 1 − χ0 (p)p X 1 1 1 = 3 log + 4 log + log : 1 − p−σ 1 − χ(p)p−σ−it 1 − χ(p)2p−σ−2it p-q Note that if p - q then χ(p) is a root of unity. Hence jχ(p)p−itj = jχ(p)e−it log pj = 1. −it i'p So we have χ(p)p = e with 'p 2 R. Hence X 1 1 1 log jF (σ)j = 3 log + 4 log + log : 1 − p−σ 1 − p−σei'p 1 − p−σe2i'p p-q Recall that 1 1 X n 1 1 log = z =n; log = Re log for z 2 with jzj < 1. 1 − z 1 − z 1 − z C n=1 Hence for r; ' 2 R with 0 < r < 1, 1 i' n ! 1 1 X (re ) log = Re log = Re 1 − rei' 1 − rei' n n=1 1 1 X rn X rn = Re (ein') = · cos n': n n n=1 n=1 This leads to 1 1 1 ! X X p−nσ X p−nσ X p−nσ log jF (σ)j = 3 + 4 · cos n' + cos 2n' n n p n p p-q n=1 n=1 n=1 1 X X p−nσ = (3 + 4 cos n' + cos 2n' ) 0; n p p > p-q n=1 using (5.2). This shows that indeed, jF (σ)j > 1 for σ > 1, giving us the contradiction we want. 105 (q) It remains to prove that L(1; χ) 6= 0 for any character χ mod q such that χ 6= χ0 , 2 (q) χ = χ0 , i.e., for any real character χ not equal to the principal character.