Analytic number theory

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ANALYTIC NUMBER THEORY Fall 2014 Jan-Hendrik Evertse

e-mail: [email protected] office: 248 (Snellius) tel: 071-5277148

course website: http://pub.math.leidenuniv.nl/∼evertsejh/ant.shtml

Literature Below is a list of recommended additional literature. Much of the material of this course has been taken from the books of Davenport and Jameson. H. Davenport, Multiplicative Number Theory (2nd ed.), Springer Verlag, Graduate Texts in Mathematics 74, 1980. A.E. Ingham, The distribution of prime numbers, Cambridge University Press, 1932 (reissued in 1990). H. Iwaniec, E. Kowalski, Analytic Number Theory, American Mathematical Society Colloquium Publications 53, American Mathematical Society, 2004. G.J.O. Jameson, The Prime Number Theorem, London Mathematical Society, Student Texts 53, Cambridde University Press, 2003. S. Lang, Algebraic Number Theory, Addison-Wesley, 1970. S. Lang, Complex Analysis (4th. ed.), Springer Verlag, Graduate Texts in Mathematics 103, 1999. D.J. Newman, Analytic Number Theory, Springer Verlag, Graduate Texts in Mathematics 177, 1998. E.C. Titchmarsh, The theory of the Riemann zeta function (2nd. ed., revised by D.R. Heath-Brown), Oxford Science Publications, Clarendon Press Oxford, 1986.

i

Notation

• lim sup xn or limn→∞ xn n→∞


For a sequence of reals {xn } we define lim supn→∞ xn := limn→∞ supm>n xm . We have lim supn→∞ xn = ∞ if and only if the sequence {xn } is not bounded from above, i.e., if for every A > 0 there is n with xn > A. In case that the sequence {xn } is bounded from above, we have lim supn→∞ xn = α where α is the largest limit point (’limes superior’) of the sequence {xn }, in other words, for every ε > 0 there are infinitely many n such that xn > α − ε, while there are only finitely many n such that xn > α + ε.

• lim inf xn or limn→∞ xn n→∞


For a sequence of reals {xn } we define lim inf n→∞ xn := limn→∞ inf m>n xm . We have lim inf n→∞ xn = −∞ if the sequence {xn } is not bounded from below, and the smallest limit point (’limes inferior’) of the sequence {xn } otherwise.

• f (x) = g(x) + O(h(x)) as x → ∞ there are constants x0 , C such that |f (x) − g(x)| 6 Ch(x) for all x > x0

• f (x) = g(x) + o(h(x)) as x → ∞ f (x) − g(x) =0 x→∞ h(x) lim

• f (x) ∼ g(x) as x → ∞ lim

x→∞

f (x) =1 g(x)

ii

• f (x)  g(x), g(x)  f (x) (Vinogradov symbols; used only if f (x) > 0, g(x) > 0 for all x > x0 for some x0 > 0) f (x) = O(g(x)) as x → ∞, that is, there are constants x0 > 0, C > 0 such that f (x) 6 Cg(x) for all x > x0

• f (x)  g(x) f (x)  g(x) and f (x)  g(x), that is, there are constants x0 , C1 , C2 > 0 such that C1 f (x) 6 g(x) 6 C2 f (x) for x > x0

• f (x) = Ω(g(x)) as x → ∞ (defined only if g(x) > 0 for x > x0 for some x0 > 0) |f (x)| > 0, that is, there is a sequence {xn } with xn → ∞ as n → ∞ g(x) x→∞ |f (xn )| such that lim > 0 (possibly ∞) n→∞ g(xn )

lim sup

• f (x) = Ω± (g(x)) as x → ∞ (defined only if g(x) > 0 for x > x0 for some x0 > 0) f (x) f (x) > 0, lim inf < 0, that is, there are sequences {xn } and {yn } x→∞ g(x) x→∞ g(x) f (xn ) with xn → ∞, yn → ∞ as n → ∞ such that lim > 0 (possibly ∞) and n→∞ g(xn ) f (yn ) lim < 0 (possibly −∞) n→∞ g(yn ) lim sup

iii

• π(x) Number of primes 6 x

• π(x; q, a) Number of primes p with p ≡ a (mod q) and p 6 x

• Li(x) Z Li(x) = 2

x

dt ; this is a good approximation for π(x). log t

iv

Chapter 0 Prerequisites We have collected some facts from algebra and analysis which we will not discuss during our course, which will not be a subject of the oral examination, but to which we will have to refer quite often.

0.1

Groups

Literature: P. Stevenhagen: Collegedictaat Algebra 1 (Dutch), Universiteit Leiden. S. Lang: Algebra, 2nd ed., Addison-Wesley, 1984.

0.1.1

Definition

A group is a set G, together with an operation · : G×G → G satisfying the following axioms: • (g1 · g2 ) · g3 = g1 · (g2 · g3 ) for all g1 , g2 , g3 ∈ G; • there is eG ∈ G such that g · eG = eG · g = g for all g ∈ G; • for all g ∈ G there is h ∈ G with g · h = h · g = eG . 1

From these axioms it follows that the unit element eG is uniquely determined, and that the inverse h defined by the last axiom is uniquely determined; henceforth we write g −1 for this h. If moreover, g1 · g2 = g2 · g1 , we say that the group G is abelian or commutative. Remark. For n ∈ Z>0 , g ∈ G we write g n for g multiplied with itself n times. Further, g 0 := eG and g n := (g −1 )|n| for n ∈ Z 2, we have hgi = {eG = g 0 , g, g 2 , . . . , g n−1 },

g n = eG .

So g has order n. Example 1. µn = {ρ ∈ C∗ : ρn = 1}, that is the group of roots of unity of order n is a cyclic group of order n. For a generator of µn one may take any primitive root of unity of order n, i.e., e2πik/n with k ∈ Z, gcd(k, n) = 1. Example 2. Let p be a prime number, and (Z/pZ)∗ = {a mod p, gcd(a, p) = 1} the group of prime residue classes modulo p with multiplication. This is a cyclic group of order p − 1. Let G = hgi be a cyclic group and H a subgroup of G. Let k be the smallest positive integer such that g k ∈ H. Using, e.g., division with remainder, one shows that g r ∈ H if and only if r ≡ 0 (mod k). Hence H = hg k i and (G : H) = k.

0.1.7

Homomorphisms and isomorphisms

Let G1 , G2 be two groups. A homomorphism from G1 to G2 is a map f : G1 → G2 such that f (g1 g2 ) = f (g1 )f (g2 ) for all g1 , g2 ∈ G. This implies that f (eG1 ) = eG2 and f (g −1 ) = f (g)−1 for g ∈ G1 . Let f : G1 → G2 be a homomorphism. The kernel and image of f are given by Ker(f ) := {g ∈ G1 : f (g) = eG2 },

f (G1 ) = {f (g) : g ∈ G1 },

respectively. Notice that Ker(f ) is a normal subgroup of G1 . It is easy to check that f is injective if and only if Ker(f ) = {eG1 }. Let G be a group and H a normal subgroup of G. Then f : G → G/H : g 7→ gH 4

is a surjective homomorphism from G to G/H, the canonical homomorphism from G to G/H. Notice that the kernel of this homomorphism is H. Thus, every normal subgroup of G occurs as the kernel of some homomorphism. A homomorphism f : G1 → G2 which is bijective is called an isomorphism from G1 to G2 . In case that there is an isomorphism from G1 to G2 we say that G1 , G2 are isomorphic, notation G1 ∼ = G2 . Notice that a homomorphism f : G1 → G2 is an isomorphism if and only if Ker(f ) = {eG1 } and f (G1 ) = G2 . Further, in this case the inverse map f −1 : G2 → G1 is also an isomorphism. Let f : G1 → G2 be a homomorphism of groups and H = Ker(f ). This yields an isomorphism f : G1 /H → f (G1 ) : f (gH) = f (g). Proposition 0.1. Let C be a cyclic group. If C is infinite, then it is isomorphic to Z+ (the additive group of Z). If C has finite order n, then it is isomorphic to (Z/nZ)+ (the additive group of residue classes modulo n). Proof. Let C = hgi. Define f : Z+ → C by n 7→ g n . This is a surjective homomorphism; let H denote its kernel. Thus, Z+ /H ∼ = C. We have H = {0} if C is infinite, + and H = nZ if C has order n. This implies the proposition.

0.1.8

Direct products

Let G1 , . . . , Gr be groups. Denote by eGi the unit element of Gi . The direct product G1 × · · · × Gr is the set of tuples (g1 , . . . , gr ) with gi ∈ Gi for i = 1, . . . , r, endowed with the group operation (g1 , . . . , gr ) · (h1 , . . . , hr ) = (g1 h1 , . . . , gr hr ). This is obviously a group, with unit element (eG1 , . . . , eGr ) and inverse (g1 , . . . , gr )−1 = (g1−1 , . . . , gr−1 ). Proposition 0.2. Let G, G1 , . . . , Gr be groups. Then the following two assertions are equivalent: (i) G ∼ = G1 × · · · × Gr ; (ii) there are subgroups H1 , . . . , Hr of G satisfying the following properties: (a) Hi ∼ = Gi for i = 1, . . . , r; 5

(b) H1 , . . . , Hr commute, that is, Hi Hj = Hj Hi for i, j = 1, . . . , r; (c) G = H1 · · · Hr , i.e., every element of G can be expressed as g1 · · · gr with gi ∈ Hi for i = 1, . . . , r; (d) H1 , . . . , Hr are independent, i.e., if gi ∈ Hi (i = 1, . . . , r) are any elements such that g1 · · · gr = eG , then gi = eG for i = 1, . . . , r. Proof. (ii) ⇒ (i). Properties (b),(c),(d) imply that H1 × · · · × Hr → G : (g1 , . . . , gr ) 7→ g1 · · · gr is a group isomorphism. Together with (a) this implies (i). (i) ⇒ (ii). Let G0 := G1 × · · · × Gr and for i = 1, . . . , r, define the group G0i := {(eG1 , . . . , gi , . . . , eGr ) : gi ∈ Gi } where the i-th coordinate is gi and the other components are the unit elements of the respective groups. Clearly, G0i ∼ = Gi for i = 1, . . . , r, G01 , . . . , G0r commute, G0 = G01 · · · G0r and G01 , . . . , G0r are independent. Let f : G → G1 × · · · × Gr be an isomorphism and Hi := f −1 (G0i ) for i = 1, . . . , r. Then H1 , . . . , Hr satisfy (a)– (d). Notice that (b),(c),(d) imply that every element of G can be expressed uniquely as a product g1 · · · gr with gi ∈ Hi for i = 1, . . . , r. In what follows, if a group G has subgroups H1 , . . . , Hr satisfying (b),(c),(d), we say that G is the direct product of H1 , . . . , Hr , and denote this by G = H1 ×· · ·×Hr .

0.1.9

Abelian groups

The group operation of an abelian group is often denoted by +, but in this course we stick to the multiplicative notation. The unit element of an abelian group A is denoted by 1 or 1A . It is obvious that every subgroup of an abelian group is a normal subgroup. In Proposition 0.2, the condition that H1 , . . . , Hr commute holds automatically so it can be dropped. The following important theorem, which we state without proof, implies that the finite cyclic groups are the building blocks of the finite abelian groups. 6

Theorem 0.3. Every finite abelian group is a direct product of finite cyclic groups. Proof. See S. Lang, Algebra, 2nd ed. Addison-Wesley, 1984, Ch.1, §10. Let A be a finite, multiplicatively written abelian group of order > 2 with unit element 1. Theorem 0.3 implies that A is a direct product of cyclic subgroups, say C1 , . . . , Cr . Assume that Ci has order ni > 2; then Ci = hhi i, where hi ∈ A is an element of order ni . We call {h1 , . . . , hr } a basis for A. Every element of A can be expressed uniquely as g1 · · · gr , where gi ∈ Ci for i = 1, . . . , r. Further, every element of Ci can be expressed as a power hki , and hki = 1 if and only if k ≡ 0 (mod ni ). Together with Proposition 0.2 this implies the following characterization of a basis for A:  k1 kr   A = {h1 · · · hr : ki ∈ Z for i = 1, . . . , r}, (0.1) there are integers n1 , . . . , nr > 2 such that   k1 h1 · · · hkr r = 1 ⇐⇒ ki ≡ 0 (mod ni ) for i = 1, . . . , r.

0.2

Infinite products in analysis

The definition of an infinite product is somewhat subtle, since we want to guarantee that a convergent infinite product is non-zero as soon as all of its factors are non-zero. Q Q simply if limN →∞ N With this condition, defining ∞ n=1 An n=1 An to be convergent Q (1 − 1/n) = 0, while all the exists would not suffice. For instance, limN →∞ N n=2 factors in the product are non-zero. Q∞ Let {An }∞ n=1 be a sequence of complex numbers. We say that n=1 An converges, if the following holds: (i) at most finitely many of the numbers An are 0; Q (ii) if n0 is the smallest index such that An 6= 0 for all n > n0 , then limN →∞ N n=n0 An exists and is finite and 6= 0. If these two conditions are satisfied, we define ∞ Y n=1

An := lim

N →∞

N Y

An .

n=1

Thus, a convergent infinite product is non-zero if and only if all its factors are nonQ zero. Notice that we agree that ∞ n=1 An is divergent if infinitely many among the 7

An are 0. Q Q Further, we say that ∞ An converges absolutely if ∞ n=1 n=1 (1 + |An − 1|) converges, QN i.e., limN →∞ n=1 (1 + |An − 1|) < ∞. Note that here, all the factors are > 1, and the product can never converge to 0. Q QN Facts. 1) ∞ n=1 An converges if and only if limM,N →∞ n=M An = 1. Q∞ 2) If n=1 An converges absolutely, then it converges. More precisely, it converges unconditionally, that is, after any rearrangement of the terms An , the product Q∞ n=1 An remains convergent, and its value does not change. Q 3) ∞ A converges absolutely ⇐⇒ P∞ n=1 n P∞ n=1 log(1 + |An − 1|) converges ⇐⇒ n=1 |An − 1| converges (use the comparison criterion for convergent series, together with the fact that limz→0 log(1 + |z|)/|z| = 1).

0.3

Uniform convergence

Literature: W. Rudin, Principles of mathematical analysis, 3rd ed. McGraw-Hill, 1976, Chap. 7.

We consider functions f : D → C where D can be any set. We can express each such function as g + ih where g, h are functions from D to R. We write g = Re f and h = Im f . We recall that if D ⊂ Rn , then f is continuous if and only if Re f and Im f are continuous. In case that D ⊆ R, we say that f is differentiable if and only if Re f and Im f are differentiable, and in that case we define the derivative of f by f 0 = (Re f )0 +i(Im f )0 . Let D be any set and let {Fn }∞ n=1 be a sequence of functions from D to C. We say that {Fn } converges pointwise on D if F (z) := limn→∞ Fn (z) exists for all z ∈ D. We say that {Fn } converges uniformly on D if moreover, lim sup |Fn (z) − F (z)| = 0.

n→∞ z∈D

8

Fact. {Fn } converges uniformly on D if and only if lim sup |FM (z) − FN (z)| = 0.

M,N →∞ z∈D

Proposition 0.4. (i) Let D be a subset of Rn , and {Fn }∞ n=1 a sequence of continuous functions from D to C. Suppose that all Fn are continuous on D and that Fn → F uniformly on D. Then F is continuous on D. (ii) Let D be an open subset of R, and let {Fn } be a sequence of functions on D such that Fn → F pointwise on D. Assume that all Fn are differentiable and that Fn0 → G uniformly on D. Then Fn → F uniformly on D, and F 0 = G. Proof. See Rudin’s book, Chapter 7. Let again D be any set and {Fn : D → C}∞ n=1 a sequence of functions. We P∞ say that the series n=1 Fn converges pointwise/uniformly on D if the partial sums P∞ Pk on D. Further, we say that n=1 Fn is n=1 Fn converge pointwise/uniformly P ∞ pointwise absolutely convergent on D if n=1 |Fn (z)| converges for every z ∈ D. Proposition 0.5 (Weierstrass criterion for series). Let {Fn : D → C}∞ n=1 be a sequence of functions. Assume that there are finite real numbers Mn such that |Fn (z)| 6 Mn for z ∈ D, n > 1,

∞ X

Mn converges.

n=1

Then D.

P∞

n=1

Fn is both uniformly convergent, and pointwise absolutely convergent on

We need a similar result for infinite products of functions. Let again D be any Q set and {Fn : D → C}∞ a sequence of functions. We say that ∞ n=1 Fn converges Q∞ n=1 pointwise on D if n=1 Fn (z) converges for every z ∈ D. In that case we define the Q limit function ∞ n=1 Fn by ∞ Y n=1

Fn (z) := lim

N →∞

N Y

Fn (z) (z ∈ D).

n=1

Notice that the assumption that the infinite product converges pointwise includes that for all z ∈ D there are only finitely many n such that Fn (z) = 0. Further, z is 9

a zero of the infinite product if and only if z is a zero of at least one of the factors Fn . Q Q∞ We say that ∞ n=1 Fn is pointwise absolutely convergent if n=1 (1 + |Fn (z) − 1|) converges for every z ∈ D. Q QN Facts 1) ∞ n=1 Fn converges pointwise on D if and only if limM,N →∞ n=M Fn (z) = 1 for all z ∈ D. Q 2) If ∞ n=1 Fn is pointwise absolutely convergent on D then it is pointwise convergent on D and even unconditionally pointwise convergent on D, that is, the pointwise convergence nor the limit of the product are affected under any rearrangement of the functions Fn . Q P∞ 3) ∞ F is pointwise absolute convergent on D if and only if n n=1 n=1 |Fn (z) − 1| converges for every z ∈ D. Q Q on D if ∞ We say that ∞ n=1 Fn converges pointwise n=1 Fn converges uniformly Q∞ on D and if for the limit function F := n=1 Fn we have lim sup |F (z) −

N →∞ z∈D

N Y

Fn (z)| = 0.

n=1

Proposition 0.6 (Weierstrass criterion for infinite products). Let {Fn : D → C}∞ n=1 be a sequence of functions. Assume that there are finite real numbers Mn such that |Fn (z) − 1| 6 Mn for z ∈ D, n > 1,

∞ X

Mn converges.

n=1

Then

0.4

Q∞

n=1

Fn is pointwise absolutely convergent and uniformly convergent on D.

Integration

In this course, all integrals will be Lebesgue integrals of real or complex measurable functions on Rn (always with respect to the Lebesgue measure on Rn ). It is not really necessary to know what these are, and you will be perfectly able to follow the course without any knowledge of Lebesgue theory. But we will often have to deal with infinite integrals of infinite series of functions, and to handle these, Lebesgue theory is much more convenient than the theory of Riemann integrals. 10

It is important to mention here that Lebesgue integrals are equal to Riemann integrals whenever the latter are defined. However, Lebesgue integrals can be defined for a much larger class of functions. Further, in Lebesgue theory there are some very powerful convergence theorems for sequences of functions, theorems on interchanging multiple integrals, etc., which we will frequently apply. If you are willing to take for granted that all functions appearing in this course are measurable, there will be no problem to understand or apply these theorems. In this subsection we have collected a few useful facts, which are amply sufficient for our course.

0.4.1

Measurable sets

The length of a bounded interval I = [a, b], [a, b), (a, b] or (a, b), where a, b ∈ R, a < b, is given by l(I) := b − a. Let n ∈ Z>1 . An interval in Rn is a cartesian product of Q Q bounded intervals I = ni=1 Ii . We define the volume of I by l(I) := ni=1 l(Ii ). Let A be an arbitrary subset of Rn . We define the outer measure of A by λ∗ (A) := inf

∞ X

l(Ii ),

i=1

where the infimum is taken over all countable unions of intervals say that a set A is measurable if

S∞

i=1 Ii

⊃ A. We

λ∗ (S) = λ∗ (S ∩ A) + λ∗ (S ∩ Ac ) for every S ⊆ Rn , where Ac = Rn \ A is the complement of A. In this case we define the (Lebesgue) measure of A by λ(A) := λ∗ (A). This measure may be finite or infinite. It can be shown that intervals are measurable, and that λ(I) = l(I) for any interval I in Rn . Facts: S • A countable union ∞ i=1 Ai of measurable sets Ai is measurable. Further, the complement of a measurable set is measurable. Hence a countable intersection of measurable sets is measurable. • All open and closed subsets of Rn are measurable. 11

• Let A = ∪∞ Ai be a countable union of pairwise disjoint measurable sets. i=1P Then λ(A) = ∞ i=1 λ(Ai ), where we agree that λ(A) = 0 if λ(Ai ) = 0 for all i. • Under the assumption of the axiom of choice, one can construct non-measurable subsets of Rn . Let A be a measurable subset of Rn . We say that a particular condition holds for almost all x ∈ A, it if holds for all x ∈ A with the exception of a subset of Lebesgue measure 0. If the condition holds for almost all x ∈ Rn , we say that it holds almost everywhere. All sets occurring in this course will be measurable; we will never bother about the verification in individual cases.

0.4.2

Measurable functions

A function f : Rn → R is called measurable if for every a ∈ R, the set {x ∈ Rn : f (x) > a} is measurable. A function f : Rn → C is measurable if both Re f and Im f are measurable. Facts: • If A ⊂ Rn is measurable then its characteristic function, given by IA (x) = 1 if x ∈ A, IA (x) = 0 otherwise is measurable. • Every continuous function f : Rn → C is measurable. • If f, g : Rn → C are measurable then f + g and f g are measurable. Further, the function given by x 7→ f (x)/g(x) if g(x) 6= 0 and x 7→ 0 if g(x) = 0 is measurable. • If {fk : Rn → C} is a sequence of measurable functions and fk → f pointwise on Rn , then f is measurable. • If f, g : Rn → R are measurable, then so are max(f, g) and min(f, g). • If f : Rn → C is such that the set of discontinuities of f has Lebesgue measure 0, then f is measurable. 12

All functions occurring in our course can be proved to be measurable by combining the above facts. We will always omit such nasty verifications, and take the measurability of the functions for granted.

0.4.3

Lebesgue integrals

The Lebesgue integral is defined in various steps. P 1) An elementary function is a function of the type f = ri=1 ci IDi , where D1 , . . . , Dr are pairwise disjoint measurable subsets of Rn , and c1 , . . . , cr positive reals. Then R P we define f dx := ri=1 ci λ(Di ). R 2) Let f : Rn → R be measurable and f > 0 on Rn . Then we define f dx := R sup gdx where the supremum is taken over all elementary functions g 6 f . Thus, R f dx is defined and > 0 but it may be infinite. 3) Let f : Rn → R be an arbitrary measurable function. Then we define Z Z Z f dx := max(f, 0)dx − max(−f, 0)dx, provided that at least one of the integrals is finite. If both integrals are finite, we say that f is integrable or summable. 4) Let f : Rn → C be measurable. We say that f is integrable if Re f and Im f are integrable, and in that case we define Z Z Z f dx := (Re f )dx + i (Im f )dx. 5) Let D be a measurable subset of Rn . Let f be a complex function defined on a set containing D. We define f · ID by defining it to be equal to f on D and equal to 0 outside D. We say that f is measurable on D if f · ID is measurable. Further, we say that f is integrable over D if f · ID is integrable, and in that case we define R R f dx := f · ID dx. D Facts. • Let D be a measurable subset of Rn and f : D → C a measurable function. R Then f is integrable over D if and only if D |f |dx < ∞ and in that case, R R | D f dx| 6 D |f |dx. 13

• Let again D be a measurable subset of Rn and f : D → C, g : D → R>0 R measurable functions, such that D gdx < ∞ and |f | 6 g on D. Then f is R R integrable over D, and | f dx| 6 gdx. • Let D be a closed interval in Rn and f : D → C a bounded function which is Riemann integrable over D. Then f is Lebesgue integrable over D and the R R Lebesgue integral D f dx is equal to the Riemann integral D f dx. R∞ • Let f : [0, ∞) → C be such that the improper Riemann integral 0 |f (x)|dx R∞ converges. Then the improper Riemann integral 0 f (x)dx converges as well, R and it is equal to the Lebesgue integral [0,∞) f dx. However, an improper R∞ R∞ Riemann integral 0 f (x)dx which is convergent, but for which 0 |f (x)|dx = ∞ can not be interpreted as a Lebesgue integral. The same applies to the other Rb types of improper Riemann integrals, e.g., a f (x)dx where f is unbounded on (a, b). P • An absolutely convergent series of complex terms ∞ n=0 an may be interpreted as a Lebesgue integral. Define the function A by A(x) := an for x ∈ R with n 6 x < n + 1 and A(x) := 0 for x < 0. Then A is measurable and integrable, R P Adx. and ∞ n=0 an =

0.4.4

Important theorems

Theorem 0.7 (Dominated Convergence Theorem). Let D ⊆ Rn be a measurable set and {fk : D → C}k>0 a sequence of functions that are all integrable over D, and such that fk → f pointwise on D. Assume that there is an integrable function g : D → R>0 such that |fk (x)| 6 g(x) for all x ∈ D, k > 0. Then f is integrable R R over D, and D fk dx → D f dx. Corollary 0.8. let D ⊂ Rn be a measurable set of finite measure and {fk : D → C}k>0 a sequence of functions that are all integrable over D, and such that fk → f R R uniformly on D. Then f is integrable over D, and D fk dx → D f dx. Proof. Let ε > 0. There is k0 such that |f (x) − fk (x)| < ε for all x ∈ D, k > k0 . The constant function x 7→ ε is integrable over D since D has finite measure. Hence for k > k0 , f − fk is integrable over D, and so f is integrable over D. Consequently, 14

|f | is integrable over D. Now |fk | < ε + |f | for k > k0 . So by the Dominated R R Convergence Theorem, D fk dx → D f dx. In the theorem below, we write points of Rm+n as (x, y) with x ∈ Rm , y ∈ Rn . Further, dx, dy, d(x, y) denote the Lebesgue measures on Rm , Rn , Rm+n , respectively. Theorem 0.9 (Fubini-Tonelli). Let D1 , D2 be measurable subsets of Rm , Rn , respectively, and f : D1 × D2 → C a measurable function. Assume that at least one of the integrals   Z Z Z Z Z |f (x, y)|d(x, y), |f (x, y)|dy dx, |f (x, y)|dx dy D1 ×D2

D1

D2

D2

D1

is finite. Then they are all finite and equal. Further, f is integrable over D1 × D2 , x 7→ f (x, y) is integrable over D1 for almost all y ∈ D2 , y 7→ f (x, y) is integrable over D2 for almost all x ∈ D1 , and   Z Z Z Z Z f (x, y)d(x, y) = f (x, y)dy dx = f (x, y)dx dy. D1 ×D2

D1

D2

D2

D1

Corollary 0.10. Let D be a measurable subset of Rm and {fk : D → C}k>0 a P sequence of functions that are all integrable over D and such that ∞ k=0 |fk | converges pointwise on D. Assume that at least one of the quantities Z X ∞ ∞ Z  X |fk (x)| dx |fk (x)|dx, k=0

is finite. Then

P∞

k=0

D

D

k=0

fk is integrable over D and ∞ Z X k=0

fk (x)dx =

D

Z X ∞ D

 fk (x) dx.

k=0

Proof. Apply the Theorem of Fubini-Tonelli with n = 1, D1 = D, D2 = [0, ∞), F (x, y) = fk (x) where k is the integer with k 6 y < k + 1. Corollary 0.11. Let {akl }∞ k,l=0 be a double sequence of complex numbers such that at least one of ∞ X ∞ ∞ X ∞  X  X |akl | , |akl | k=0

l=0

l=0

15

k=0

converges. Then both ∞ X ∞ X k=0



akl ,

l=0

∞ X ∞ X l=0

akl



k=0

converge and are equal.

Proof. Apply the Theorem of Fubini-Tonelli with m = n = 1, D1 = D2 = [0, ∞), F (x, y) = akl where k, l are the integers with k 6 x < k + 1, l 6 y < l + 1.

0.5 0.5.1

Line integrals Paths in C

We consider continuous functions g : [a, b] → C, where a, b ∈ R and a < b. Two continuous functions g1 : [a, b] → C, g2 : [c, d] → C are called equivalent if there is a continuous monotone increasing function ϕ : [a, b] → [c, d] such that g1 = g2 ◦ ϕ. The equivalence classes of this relation are called paths (in C), and a function g : [a, b] → C representing a path is called a parametrization of the path. Roughly speaking, a path is a curve in C, together with a direction in which it is traversed. A (continuously) differentiable path is a path represented by a (continuously) differentiable function g : [a, b] → C. Let γ be a path. Choose a parametrization g : [a, b] → C of γ. We call g(a) the start point and g(b) the end point of γ. Further, g([a, b]) is called the support of γ. By saying that a function is continuous on γ, or that γ is contained in a particular set, etc., we mean the support of γ. The path γ is said to be closed if its end point is equal to its start point, i.e., if g(a) = g(b). The path γ is called a contour if it is closed, has no self-intersections, and is traversed counterclockwise (we will not give the cumbersome formal definition of this intuitively obvious notion). 16

Let γ1 , γ2 be paths, such that the end point of γ1 is equal to the start point of γ2 . We define γ1 + γ2 to be the path obtained by first traversing γ1 and then γ2 . For instance, if g1 : [a, b] → C is a parametrization of γ1 then we may choose a parametrization g2 : [b, c] → C of γ2 ; then g : [a, c] → C defined by g(t) := g1 (t) if a 6 t 6 b, g(t) := g2 (t) if b 6 t 6 c is a parametrization of γ1 + γ2 . Given a path γ, we define −γ to be the path traversed in the opposite direction, i.e., the start point of −γ is the end point of γ and conversely.

Let γ be a path and F : γ → C a continuous function on (the support of) γ. Then F (γ) is the path such that if g : [a, b] → C is a parametrization of γ then F ◦ g : [a, b] → C is a parametrization of F (γ). 17

0.5.2

Line integrals

All paths occurring in our course will be built up from circle segments and line segments. So for our purposes, it suffices to define integrals of continuous functions along piecewise continuously differentiable paths, these are paths of the shape γ1 + · · ·+γr , where γ1 , . . . , γr are continuously differentiable paths, and for i = 1, . . . , r−1, the end point of γi coincides with the start point of γi+1 . let γ be a continuously differentiable path, and f : γ → C a continuous function. Choose a continuously differentiable parametrization g : [a, b] → C of γ. Then we define Z Z b

f (g(t))g 0 (t)dt.

f (z)dz := a

γ

Further, we define the length of γ by b

Z

|g 0 (t)|dt.

L(γ) := a

These notions do not depend on the choice of g. If γ = γ1 + · · · + γr is a piecewise continuously differentiable path with continuously differentiable pieces γ1 , . . . , γr and f : γ → C is continuous, we define Z r Z X f (z)dz f (z)dz := γ

i=1

and

r X

L(γ) :=

γi

L(γi ).

i=1

H In case that γ is closed, we write γ f (z)dz. It can be shown that the value of this integral is independent of the choice of the common start point and end point of γ. R We mention here that line integrals γ f (z)dz can be defined also for paths γ that are not necessarily piecewise continuously differentiable. For piecewise continuously differentiable paths, this new definition coincides with the one given above. Let γ be any path and choose a parametrization g : [a, b] → C of γ. A partition of [a, b] is a tuple P = (t0 , . . . , ts ) where a = t0 < t1 < · · · < ts = b. We define the length of γ by s X L(γ) := sup |g(ti ) − g(ti−1 )|, P

i=1

18

where the supremum is taken over all partitions P of [a, b]. This does not depend on the choice of g. We call γ rectifiable if L(γ) < ∞ (in another language, this means that the function g is of bounded variation). Let γ be a rectifiable path, and g : [a, b] → C a parametrization of γ. Given a partition P = (t0 , . . . , ts ) of [a, b], we define the mesh of P by δ(P ) := max |ti − ti−1 |. 16i6s

A sequence of intermediate points of P is a tuple W = (w1 , . . . , ws ) such that t0 < w1 < t1 < w2 < t2 < · · · < ts . Let f : γ → C be a continuous function. For a partition P of [a, b] and a tuple of intermediate points W of P we define S(f, g, P, W ) :=

s X

f (g(wi ))(g(ti ) − g(ti−1 )).

i=1

R One can show that there is a finite number, denoted γ f (z)dz, such that for any choice of parametrization g : [a, b] → C of γ and any sequence (Pn , Wn )n>0 of partitions Pn of [a, b] and sequences of intermediate points Wn of Pn with δ(Pn ) → 0, Z f (z)dz = lim S(f, g, Pn , Wn ). n→∞

γ

In another language,

0.5.3

R γ

f (z)dz is equal to the Riemann-Stieltjes integral

Rb a

f (g(t))dg(t).

Properties of line integrals

Below (and in the remainder of the course), by a path we will mean a piecewise continuously differentiable path. In fact, all properties below hold for line integrals over rectifiable paths, but in textbooks on complex analysis, these properties are never proved in this generality. • Let γ be a path, and f : γ → C continuous. Then Z f (z)dz 6 L(γ) · sup |f (z)|. z∈γ γ

19

• Let γ1 , γ2 be two paths such that the end point of γ1 and the start point of γ2 coincide. Let f : γ1 + γ2 → C continuous. Then Z Z Z f (z)dz = f (z)dz + f (z)dz. γ1 +γ2

γ1

γ2

• Let γ be a path and f : γ → C continuous. Then Z Z f (z)dz = − f (z)dz. −γ

γ

• Let γ be a path and {fn : γ → C}∞ n=1 a sequence of continuous functions. Suppose that fn → f uniformly on γ, i.e., supz∈γ |fn (z)−f (z)| → 0 as n → ∞. R R Then f is continuous on γ, and γ fn (z)dz → γ f (z)dz as n → ∞. • Call a function F : U → C on an open subset U of C analytic if for every z ∈ U the limit F (z + h) − F (z) F 0 (z) = lim h∈C, h→0 h exists. Let γ be a path with start point z0 and end point z1 , and let F be an analytic function defined on an open set U ⊂ C that contains γ. Then Z F 0 (z)dz = F (z1 ) − F (z0 ). γ

• Let γ be a path and F an analytic function defined on some open set containing γ. Further, let f : F (γ) → C be continuous. Then Z Z f (w)dw = f (F (z))F 0 (z)dz. F (γ)

γ

Examples. 1. Let γa,r denote the circle with center a and radius r, traversed counterclockwise. For γa,r we may choose a parametrization t 7→ a + re2πit , t ∈ [0, 1]. Let n ∈ Z. Then I Z 1 n (z − a) dz = rn e2nπit · 2πi · re2πit dt γa,r

0

= 2πir

n+1

Z

1 2(n+1)πit

e 0

20

 dt =

2πi if n = −1; 0 if n 6= −1.

2. For z0 , z1 ∈ C, denote by [z0 , z1 ] the line segment from z0 to z1 . For [z0 , z1 ] we may choose a parametrization t 7→ z0 + t(z1 − z0 ), t ∈ [0, 1]. Let f : [z0 , z1 ] → C be continuous. Then Z 1 Z f (z0 + t(z1 − z0 ))(z1 − z0 )dt. f (z)d(z) = [z0 ,z1 ]

0.6

0

Topology

We recall some facts about the topology of C.

0.6.1

Basic facts

Let a ∈ C, r ∈ R>0 . We define the open disk and closed disk with center a and radius r, D(a, r) = {z ∈ C : |z − a| < r},

D(a, r) = {z ∈ C : |z − a| 6 r}.

Recall that a subset U of C is called open if either U = ∅, or for every a ∈ U there is δ > 0 with D(a, δ) ⊂ U . A subset U of C is closed if its complement U c = C \ U is open. It is easy to verify that the union of any possibly infinite collection of open subsets of C is open. Further, the intersection of finitely many open subsets is open. Consequently, the intersection of any possibly infinite collection of closed sets is closed, and the union of finitely many closed subsets is closed. By the Heine-Borel theorem, a subset of C is compact, i.e., every open cover of the subset has a finite subcover, if and only if it is closed and bounded. Let U be a non-empty subset of C. A point z0 ∈ C is called a limit point of U if there is a sequence {zn } in U such that all zn are distinct and zn → z0 as n → ∞. Recall that a non-empty subset U of C is closed if and only if each of its limit points belongs to U . Let U be a non-empty subset of C, and S ⊂ U . Then S is called discrete in U if it has no limit points in U . Recall that by the Bolzano-Weierstrass Theorem, every infinite subset of a compact set K has a limit point in K. This implies that S is discrete in U if and only if for every compact set K with K ⊂ U , the intersection K ∩ S is finite. 21

Let U be a non-empty, open subset of C. We say that U is connected if there are no non-empty open sets U1 , U2 with U = U1 ∪ U2 and U1 ∩ U2 = ∅. We say that U is pathwise connected if for every z0 , z1 ∈ U there is a path γ ⊂ U with start point z0 and end point z1 . A fact (typical for the topological space C) is that a non-empty open subset U of C is connected if and only if it is pathwise connected. Let U be any, non-empty open subset of C. We can express U as a disjoint union α∈I Uα , with I some index set, such that two points of U belong to the same set Uα if and only if they are connected by a path contained in U . The sets Uα are open, connected, and pairwise disjoint. We call these sets Uα the connected components of U .

S

0.6.2

Homotopy Let U ⊆ C and γ1 , γ2 two paths in U with start point z0 and end point z1 . Then γ1 , γ2 are homotopic in U if one can be continuously deformed into the other within U . More precisely this means the following: there are parametrizations f : [0, 1] → C of γ1 , g : [0, 1] → C of γ2 and a continuous map H : [0, 1] × [0, 1] → U with the following properties:

H(0, t) = f (t), H(1, t) = g(t) for 0 6 t 6 1; H(s, 0) = z0 , H(s, 1) = z1 for 0 6 s 6 1. 22

Let U ⊆ C be open and non-empty. We call U simply connected (’without holes’) if it is connected and if every closed path in U can be contracted to a point in U , that is, if z0 is any point in U and γ is any closed path in U containing z0 , then γ is homotopic in U to z0 .

A map f : D1 → D2 , where D1 , D2 are subsets of C, is called a homeomorphism if f is a bijection, and both f and f −1 are continuous. Homeomorphisms preserve topological properties of sets such as openness, closedness, boundedness, (simple) connectedness, etc. Theorem 0.12 (Schoenflies Theorem for curves). Let γ be a contour in C. Then there is a homeomorphism f : C → C such that f (γ0,1 ) = γ, where γ0,1 is the unit circle with center 0 and radius 1, traversed counterclockwise. Corollary 0.13 (Jordan Curve Theorem). Let γ be a contour in C. Then C \ γ has two connected components, U1 and U2 . The component U1 is bounded and simply connected, while U2 is unbounded.

The component U1 is called the interior of γ, notation int(γ), and U2 the exterior of γ, notation ext(γ).

23

Chapter 1 Introduction 1.1

The Prime Number Theorem

In this course, we focus on the theory of prime numbers. We use the following notation: we write f (x) ∼ g(x) as x → ∞ if limx→∞ f (x)/g(x) = 1, and denote by log x the natural logarithm. The central result is the Prime Number Theorem: Theorem 1.1 (Prime Number Theorem, Hadamard, de la Vall´ee Poussin, 1896). let π(x) denote the number of primes 6 x. Then π(x) ∼

x as x → ∞. log x

This result was conjectured by Legendre in 1798. In 1851/52, Chebyshev proved that if the limit limx→∞ π(x) log x/x exists, then it must be equal to 1, but he couldn’t prove the existence of the limit. However, Chebyshev came rather close, by showing that there is an x0 , such that for all x > x0 , 0.921

x x < π(x) < 1.056 . log x log x

¨ In 1859, Riemann published a very influential paper (B. Riemann, Uber die Anzahl der Primzahlen unter einer gegebenen Große, Monatshefte der Berliner Akademie der Wissenschaften 1859, 671–680; also in Gesammelte Werke, Leipzig 1892, 145–153) in which he related the distribution of prime numbers to properties 25

of the function in the complex variable s, ζ(s) =

∞ X

n−s

n=1

(nowadays called the Riemann zeta function). It is well-known that ζ(s) converges absolutely for all s ∈ C with Re s > 1, and that it diverges for s ∈ C with Re s 6 1. Moreover, ζ(s) defines an analytic (complex differentiable) function on {s ∈ C : P −s Re s > 1}. Riemann obtained another expression for ∞ that can be defined n=1 n everywhere on C \ {1} and defines an analytic function on this set; in fact it can be P −s shown that it is the only analytic function on C \ {1} that coincides with ∞ n=1 n on {s ∈ C : Re s > 1}. This analytic function is also denoted by ζ(s). Riemann proved the following properties of ζ(s): • ζ(s) has a pole of order 1 in s = 1 with residue 1, that is, lims→1 (s−1)ζ(s) = 1; • ζ(s) satisfies a functional equation that relates ζ(s) to ζ(1 − s); • ζ(s) has zeros in s = −2, −4, −6, . . . (the trivial zeros). The other zeros lie in the critical strip {s ∈ C : 0 < Re s < 1}. Riemann stated the following still unproved conjecture: Riemann Hypothesis (RH). All zeros of ζ(s) in the critical strip lie on the axis of symmetry of the functional equation, i.e., the line Re s = 21 .

Riemann made several other conjectures about the distribution of the zeros of ζ(s), and further, he stated without proof a formula that relates X θ(x) := log p (sum over all primes 6 x) p6x

26

to the zeros of ζ(s) in the critical strip. These other conjectures of Riemann were proved by Hadamard in 1893, and the said formula was proved by von Mangoldt in 1895. Finally, in 1896, Hadamard and de la Vall´ee Poussin independently proved the Prime Number Theorem. Their proofs used a fair amount of complex analysis. A crucial ingredient for their proofs is, that ζ(s) 6= 0 if Re s = 1 and s 6= 1. In 1899, de la Vall´ee Poussin obtained the following Prime Number Theorem with error term: Let Z x dt Li(x) := . 2 log t Then there is a constant c > 0 such that   √ π(x) = Li(x) + O xe−c log x as x → ∞.

(1.1)

Exercise 1.1. Prove that for every integer n > 1, Li(x) =

  x x x x + 1! 2 + · · · + (n − 1)! n + O as x → ∞, log x log x log x logn+1 x

where the constant implied by the O-symbol may depend on n. This shows that Li(x) is a much better approximation to π(x) than x/ log x. Indeed, since √ √ xe−c log x = en log log x−c log x → 0 as x → ∞ n x/ log x for any positive integer n, we have π(x) −

 x    √ x x −c log x = + O + O xe log x log2 x log3 x   x x  x 1  x = +O = · 1+ ∼ as x → ∞. 2 3 2 log x log x log x log x log2 x

Hence π(x) − Li(x) π(x) − Li(x) ∼ =O π(x) − x/ log x x/ log2 x 27

√

xe−c log x x/ log2 x

! → 0 as x → ∞.

In fact, in his proof of (1.1), de la Vall´ee Poussin used that for some constant c > 0, ζ(s) 6= 0 for all s with Re s > 1 −

c . log(|Im s| + 2)

A zero free region for ζ(s) is a subset S of the critical strip of which it is known that ζ(s) 6= 0 on S. In general, a larger provable zero free region for ζ(s) leads to a better estimate for π(x) − Li(x). In 1958, Korobov and Vinogradov independently showed that for every α > there is a constant c(α) > 0 such that ζ(s) 6= 0 for all s with Re s > 1 −

2 3

c(α) . (log(|Im s| + 2))α

From this, they deduced that for every β < 53 there is a constant c0 (β) > 0 with   0 β π(x) = Li(x) + O xe−c (β)(log x) as x → ∞. This has not been improved so far. On the other hand, in 1901 von Koch proved that the Riemann Hypothesis is equivalent to √  π(x) = Li(x) + O x log2 x as x → ∞, which is of course much better than the result of Korobov and Vinogradov. After Hadamard and de la Vall´ee Poussin, several other proofs of the Prime Number theorem were given, all based on complex analysis. In the 1930’s, Wiener and Ikehara proved a general so-called Tauberian theorem (from functional analysis) which implies the Prime Number Theorem in a very simple manner. In 1948, Erd˝os and Selberg independently found an elementary proof, “elementary” meaning that the proof avoids complex analysis or functional analysis, but definitely not that the proof is easy! In 1980, Newman gave a new, simple proof of the Prime Number 28

Theorem, based on complex analysis. Korevaar observed that Newman’s approach can be used to prove a simpler version of the Wiener-Ikehara Tauberian theorem with a not so difficult proof based on complex analysis alone and avoiding functional analysis. In this course, we prove the Tauberian theorem via Newman’s method, and deduce from this the Prime Number Theorem as well as the Prime Number Theorem for arithmetic progressions (see below).

1.2

Primes in arithmetic progressions

In 1839–1842 Dirichlet (the founder of analytic number theory) proved that every arithmetic progression contains infinitely many primes. Otherwise stated, he proved that for every integer q > 2 and every integer a with gcd(a, q) = 1, there are infinitely many primes p such that p ≡ a (mod q). His proof is based on so-called L-functions. To introduce these, we have to introduce the nottion of a Dirichlet character. A Dirichlet character modulo q is a function χ : Z → C with the following properties: • χ(1) = 1; • χ(b1 ) = χ(b2 ) for all b1 , b2 ∈ Z with b1 ≡ b2 (mod q); • χ(b1 b2 ) = χ(b1 )χ(b2 ) for all b1 , b2 ∈ Z; • χ(b) = 0 for all b ∈ Z with gcd(b, q) > 1. The principal character modulo q is given by (q)

(q)

χ0 (a) = 1 if gcd(a, q) = 1,

χ0 (a) = 0 if gcd(a, q) > 1.

Example. Let χ be a character modulo 5. Since 24 ≡ 1 (mod 5) we have χ(2)4 = 1. Hence χ(2) ∈ {1, i, −1, −i}. In fact, the Dirichlet characters modulo 5 are given by χj (j = 1, 2, 3, 4) where χ(b) = 1 χ(b) = i χ(b) = −1 χ(b) = −i χ(b) = 0

if if if if if

b ≡ 1 (mod 5), b ≡ 2 (mod 5), b ≡ 4 ≡ 22 (mod 5), b ≡ 3 ≡ 23 (mod 5), b ≡ 0 (mod 5). 29

In general, by the Euler-Fermat Theorem, if b, q are integers with q > 2 and gcd(b, q) = 1, then bϕ(q) ≡ 1 (mod q), where ϕ(q) is the number of positive integers a 6 q that are coprime with q. Hence χ(b)ϕ(q) = 1. The L-function associated with a Dirichlet character χ modulo q is given by L(s, χ) =

∞ X

χ(n)n−s .

n=1

Since |χ(n)| ∈ {0, 1} for all n, this series converges absolutely for all s ∈ C with Re s > 1. Further, many of the results for ζ(s) can be generalized to L-functions: • if χ is not a principal character, then then L(s, χ) has an analytic continuation (q) to C, while if χ = χ0 is the principal character modulo q it has an analytic Q (q) continuation to C \ {1}, with lims→1 (s − 1)L(s, χ0 ) = p|q (1 − p−1 ). • there is a functional equation, relating L(s, χ) to L(1 − s, χ), where χ is the complex conjugate character, defined by χ(b) := χ(b) for b ∈ Z. Furthermore, there is a generalization of the Riemann Hypothesis: Generalized Riemann Hypothesis (GRH): Let χ be a Dirichlet character modulo q for any integer q > 2. Then the zeros of L(s, χ) in the critical strip lie on the line Re s = 21 . De la Vall´ee Poussin managed to prove the following generalization of the Prime Number Theorem, using properties of L-functions instead of ζ(s): Theorem 1.2 (Prime Number Theorem for arithmetic progressions). let q, a be integers with q > 2 and gcd(a, q) = 1. Denote by π(x; q, a) the number of primes p with p 6 x and p ≡ a (mod q). Then π(x; q, a) ∼

x 1 · as x → ∞. ϕ(q) log x

Corollary 1.3. Let q be an integer > 2. Then for all integers a coprime with q we have π(x; q, a) 1 lim = . x→∞ π(x) ϕ(q) This shows that in some sense, the primes are evenly distributed over the prime residue classes modulo q. 30

1.3

An elementary result for prime numbers

We finish this introduction with an elementary proof of a weaker version of the Prime Number Theorem. Theorem 1.4. We have 1 2

x x 6 π(x) 6 2 for x > 3. log x log x

The proof is based on some simple lemmas. For an integer n 6= 0 and a prime number p, we denote by ordp (n) the largest integer k such that pk divides n. Further, we denote by [x] the largest integer 6 x. Lemma 1.5. Let n be an integer > 1 and p a prime number. Then  ∞  X n ordp (n!) = . pj j=1 Remark. This is a finite sum. Proof. We count the number of times that p divides n!. Each multiple of p that is 6 n contributes a factor p. Each multiple of p2 that is 6 n contributes another factor p, each multiple of p3 that is 6 n another factor p, and so on. Hence  ∞ ∞  X X n j ordp (n!) = (number of multiples of p 6 n) = . pj j=1 j=1

Lemma 1.6. Let a, b be integers with a > 2b > 0. Then   Y a p divides . b a−b+16p6a Proof. We have   a a(a − 1) · · · (a − b + 1) = , b 1 · 2···b

a − b + 1 > b.

Hence any prime with a − b + 1 6 p 6 a divides the numerator but not the denominator. 31

Lemma 1.7. Let a, b be integers with a > b > 0. Suppose that some prime power
pk divides ab . Then pk 6 a. Proof. Let p be a prime. By Lemma 1.5 we have     a a! ordp = ordp b b!(a − b)!      ∞  X a−b b a = − − j . j j p p p j=1 Each summand is either 0 or 1. Further, each summand with pj > a is 0. Hence 
ordp ab 6 α, where α is the largest j with pj 6 a. This proves our lemma. Lemma 1.8. Let n be an integer > 1. Then   2n n 6 6 2n−1 . n+1 [n/2] Proof.

n [n/2]

n







n




is the largest among the binomial coefficients 0 , . . . , n . Hence   n   X n n n 2 = 6 (n + 1) . j [n/2] j=0
n This establishes the lower bound for [n/2] . To prove the upper bound, we distinguish between the cases n = 2k + 1 odd and n = 2k even. First, let n = 2k + 1 be odd. Then         2k + 1 2k + 1 n 2k + 1 1 + = = 2 k k+1 [n/2] k   2k+1 1 X 2k + 1 6 = 22k = 2n−1 . 2 j=0 j
1 2k
2k Now, let n = 2k be even. Then since k−1 > 2 k for k > 1,           1 n 2k 2k 2k 2k = 6 + + [n/2] k 2 k−1 k k+1   2k 1 X 2k = 22k−1 = 2n−1 . 6 2 j=0 j

32

Proof of π(x) > 12 x/ log x. It is easy to check that π(x) > 21 x/ log x for 3 6 x 6 100. Assume that x > 100. Let n := [x]; then n 6 x < n + 1.
n Write [n/2] = pk11 · · · pkt t , where the pi are distinct primes, and the ki positive integers. By Lemma 1.7 we have pki i 6 n for i = 1, . . . , t. Then also pi 6 n for i = 1, . . . , t, hence t 6 π(n). It follows that   n 6 nπ(n) . [n/2] So by Lemma 1.8, nπ(n) > 2n /(n + 1). Consequently, n log 2 − log(n + 1) log n x (x − 1) log 2 − log(x + 1) > > 21 for x > 100. log x log x

π(x) = π(n) >

Proof of π(x) 6 2x/ log x. Let again n = [x]. Since t/ log t is an increasing function of t, it suffices to prove that π(n) 6 2 · n/ log n for all integers n > 3. We proceed by induction on n. It is straightforward to verify that π(n) 6 2 · n/ log n for 3 6 n 6 200. Let n > 200, and suppose that π(m) 6 2 · m/ log m for all integers m with 3 6 m < n. If n is even, then we can use π(n) = π(n − 1) and that t/ log t is increasing. Assume that n = 2k + 1 is odd. Then by Lemma 1.6, we have   Y 2k + 1 p > (k + 2)π(2k+1)−π(k+1) . > k k+26p62k+1 Using Lemma 1.8, this leads to (k + 2)π(2k+1)−π(k+1) 6 22k . Taking logarithms, we get π(2k + 1) log(k + 2) 6 2k log 2 + π(k + 1) log(k + 2), and applying the induction hypothesis to π(k + 1), using k + 2 > k + 1 > n/2, 2k log 2 2(k + 1) + log(k + 2) log(k + 1) (log 2 + 1)n + 1 n < 200. log(n/2) log n

π(n) = π(2k + 1) 6

33

34

Chapter 2 Complex analysis 2.1

Basics

Given z0 ∈ C, r > 0 we define D(z0 , r) := {z ∈ C : |z − z0 | < r} (open disk), D(z0 , r) := {z ∈ C : |z − z0 | 6 r} (closed disk). A subset U ⊆ C is called open if U = ∅ or if for every z0 ∈ C there is δ > 0 with D(z0 , δ) ⊂ U . A subset U ⊆ C is called closed if U c := C \ U is open. It is known that a subset of C is compact (i.e., every open cover of the set has a finite subcover) if and only if it is closed and bounded (see the Prerequisites). In what follows, U is a non-empty open subset of C and f : U → C a function. We say that f is holomorphic or analytic in z0 ∈ U if lim

z→z0

f (z) − f (z0 ) exists. z − z0

In that case, the limit is denoted by f 0 (z0 ). We say that f is analytic on U if f is analytic in every z ∈ U ; in that case, the derivative f 0 (z) is defined for every z ∈ U . We say that f is analytic around z0 if it is analytic on some open disk D(z0 , δ) for some δ > 0. Finally, given a not necessarily open subset A of C and a function f : A → C, we say that f is analytic on A if there is an open set U ⊇ A such that f is defined on U and analytic on U . An everywhere analytic function f : C → C is called entire. 35

For any two analytic functions f, g on some open set U ⊆ C,we have the usual rules for differentiation (f ±g)0 = f 0 ±g 0 , (f g)0 = f 0 g+f g 0 and (f /g)0 = (gf 0 −f g 0 )/g 2 (the latter is defined for any z with g(z) 6= 0). Further, given a non-empty set U ⊆ C, and analytic functions f : U → C, g : f (U ) → C, the composition g ◦ f is analytic on U and (g ◦ f )0 = (g 0 ◦ f ) · f 0 . Recall that a power series around z0 ∈ C is an infinite sum ∞ X f (z) = an (z − z0 )n n=0

with an ∈ C for all n ∈ Z>0 . The radius of convergence of this series is given by  −1 p n R = Rf = lim sup |an | . n→∞

We state without proof the following fact. P∞ n Theorem 2.1. Let z0 ∈ C and f (z) = n=0 an (z − z0 ) a power series around z0 ∈ C with radius of convergence R > 0. Then f defines a function on D(z0 , R) which is analytic infinitely often. For k > 0 the k-th derivative f (k) of f has a power series expansion with radius of convergence R given by ∞ X (k) n(n − 1) · · · (n − k + 1)an (z − z0 )n−k . f (z) = n=k

In each of the examples below, R denotes the radius of convergence of the given power series. ∞ X zn z , R = ∞, (ez )0 = ez . e = n! n=0 ∞ X z 2n iz −iz (−1)n cos z = (e + e )/2 = , R = ∞, cos 0 z = − sin z. (2n)! n=0 ∞

X z 2n+1 sin z = (eiz − e−iz )/2i = (−1)n , R = ∞, sin 0 z = cos z. (2n + 1)! n=0 ∞   X α (1 + z)α = zn, R = 1, ((1 + z)α )0 = α(1 + z)α−1 n n=0   α α(α − 1) · · · (α − n + 1) where α ∈ C, = . n n! ∞ X (−1)n−1 n log(1 + z) = ·z , R = 1, log 0 (1 + z) = (1 + z)−1 . n n=1 36

2.2

Cauchy’s Theorem and some applications

For the necessary definitions concerning paths, line integrals and topology we refer to the Prerequisites. In the remainder of this course, a path will always be a piecewise continuously differentiable path. Recall that for a piecewise continuously differentiable path γ, say γ = γ1 + · · · + γr where γ1 , . . . , γr are paths with continuously differentiable parametrizations gi : [ai , bi ] → C, and for a continuous function R P Rb f : γ → C we have γ f (z)dz = ri=1 aii f (gi (t))gi0 (t)dt. Theorem 2.2 (Cauchy). Let U ⊆ C be a non-empty open set and f : U → C an analytic function. Further, let γ1 , γ2 be two paths in U with the same start point and end point that are homotopic in U . Then Z

Z f (z)dz =

γ1

f (z)dz. γ2

Proof. Any textbook on complex analysis. Corollary 2.3. Let U ⊆ C be a non-empty, open, simply connected set, and f : U → C an analytic function. Then for any closed path γ in U , I f (z)dz = 0. γ

Proof. The path γ is homotopic in U to a point, and a line integral along a point is 0. Corollary 2.4. Let γ1 , γ2 be two contours (closed paths without self-intersections traversed counterclockwise), such that γ2 is contained in the interior of γ1 . Let U ⊂ C be an open set which contains γ1 , γ2 and the region between γ1 and γ2 . Further, let f : U → C be an analytic function. Then I

I f (z)dz =

γ1

f (z)dz. γ2

Proof. 37

Let z0 , z1 be points on γ1 , γ2 respectively, and let α be a path from z0 to z1 lying inside the region between γ1 and γ2 without self-intersections.

Then γ1 is homotopic in U to the path α + γ2 − α, which consists of first traversing α, then γ2 , and then α in the opposite direction. Hence Z

I f (z)dz = γ1

−

+ α

Z 

I

I f (z)dz =

γ2

α

f (z)dz. γ2

Corollary 2.5 (Cauchy’s Integral Formula). Let γ be a contour in C, U ⊂ C an open set containing γ and its interior, z0 a point in the interior of γ, and f : U → C an analytic function. Then I f (z) 1 · dz = f (z0 ). 2πi γ z − z0 Proof.

Let γz0 ,δ be the circle with center z0 and radius δ, traversed counterclockwise. Then by Corollary 2.4 we have for any sufficiently small δ > 0,

1 2πi

I γ

f (z) 1 · dz = z − z0 2πi 38

I γz0 ,δ

f (z) · dz. z − z0

Now, since f (z) is continuous, hence uniformly continuous on any sufficiently small compact set containing z0 , I I 1 1 f (z) f (z) · dz − f (z ) = · dz − f (z ) 0 0 2πi z − z 2πi z − z 0 0 γ γz0 ,δ Z 1 f (z0 + δe2πit ) 2πit = · δe dt − f (z ) 0 2πit δe 0 Z 1  2πit = f (z0 + δe ) − f (z0 ) dt 6 sup |f (z0 + δe2πit ) − f (z0 )| 06t61 0

→ 0 as δ ↓ 0. This completes our proof. We now show that every analytic function f on a simply connected set has an anti-derivative. We first prove a simple lemma. Lemma 2.6. Let U ⊆ C be a non-empty, open, connected set, and let f : U → C be an analytic function such that f 0 = 0 on U . Then f is constant on U . Proof. Fix a point z0 ∈ U and let z ∈ U be arbitrary. Take a path γz in U from z0 to z which exists since U is (pathwise) connected. Then Z f 0 (w)dw = 0. f (z) − f (z0 ) = γz

Corollary 2.7. Let U ⊂ C be a non-empty, open, simply connected set, and f : U → C an analytic function. Then there exists an analytic function F : U → C with F 0 = f . Further, F is determined uniquely up to addition with a constant. Proof (sketch). If F1 , F2 are any two analytic functions on U with F10 = F20 = f , then F10 − F20 is constant on U since U is connected. This shows that an anti-derivative of f is determined uniquely up to addition with a constant. It thus suffices to prove the existence of an analytic function F on U with F 0 = f . 39

Fix z0 ∈ U . Given z ∈ U , we define F (z) by Z f (w)dw, F (z) := γz

where γz is any path in U from z0 to z. This does not depend on the choice of γz . For let γ1 , γ2 be any two paths in U from z0 to z. Then γ1 − γ2 (the path consisting of first traversing γ1 and then γ2 in the opposite direction) is homotopic to z0 since U is simply connected, hence Z Z I f (z)dz − f (z)dz = f (z)dz = 0. γ1

γ1 −γ2

γ2

F (z+h)−F (z)

To prove that limh→0 = f (z), take a path γz from z0 to z and then the h line segment [z, z + h] from z to z + h. Then since f is uniformly continuous on any sufficiently small compact set around z, Z  Z Z f (w)dw = f (w)dw F (z + h) − F (z) = − Z =

γz +[z,z+h] 1

γz

[z,z+h]

  Z 1 f (z + th)hdt = h f (z) + (f (z + th) − f (z))dt .

0

0

So Z 1 F (z + h) − F (z) − f (z) = (f (z + th) − f (z))dt h 0 6 sup |f (z + th) − f (z)| → 0 as h → 0. 06t61

This completes our proof. Example. Let U ⊂ C be a non-empty, open, simply connected subset of C with 0 6∈ U . Then 1/z has an anti-derivative on U . For instance, if U = C \ {z ∈ C : Re z 6 0} we may take as anti-derivative of 1/z, Log z := log |z| + iArg z, 40

where Arg z is the argument of z in the interval (−π, π) (this is called the principal value of the logarithm). On {z ∈ C : |z − 1| < 1} we may take as anti-derivative of 1/z the power series ∞ X

(−1)n−1

n=1

2.3

(z − 1)n . n

Taylor series

Theorem 2.8. Let U ⊆ C be a non-empty, open set and f : U → C an analytic function. Further, let z0 ∈ U and R > 0 be such that D(z0 , R) ⊆ U . Then f has a Taylor series expansion f (z) =

∞ X

an (z − z0 )n converging for z ∈ D(z0 , R).

n=0

Further, we have for n ∈ Z>0 , I f (z) 1 · dz for any r with 0 < r < R. (2.1) an = 2πi γz0 ,r (z − z0 )n+1 Proof. We fix z ∈ D(z0 , R) and use w to indicate a complex variable. Choose r with |z − z0 | < r < R. By Cauchy’s integral formula, I f (w) 1 · dw. f (z) = 2πi γz0 ,r w − z We rewrite the integrand. We have  −1 f (w) f (w) f (w) z − z0 = = · 1− w−z (w − z0 ) − (z − z0 ) w − z0 w − z0 n X ∞  ∞ f (w) X z − z0 f (w) = · = · (z − z0 )n . n+1 w − z0 n=0 w − z0 (w − z ) 0 n=0 The latter series converges uniformly on γz0 ,r . For let M := supw∈γz0 ,r |f (w)|. Then   M |z − z0 | n f (w) n · (z − z0 ) 6 =: Mn sup n+1 r r w∈γz0 ,r (w − z0 ) 41

and

P∞

n=0

Mn converges since |z − z0 | < r. Consequently, I 1 f (w) f (z) = · dw 2πi γz0 ,r w − z  I ∞  X f (w) 1 n = · (z − z0 ) dw 2πi γz0 ,r n=0 (w − z0 )n+1 ( ) I ∞ X 1 f (w) = (z − z0 )n · dw . n+1 2πi (w − z ) 0 γ z ,r 0 n=0

Now Theorem 2.8 follows since by Corollary 2.4 the integral in (2.1) is independent of r. Corollary 2.9. Let U ⊆ C be a non-empty, open set, and f : U → C an analytic function. Then f is analytic on U infinitely often, that is, for every k > 0 the k-the derivative f (k) exists, and is analytic on U . Proof. Pick z ∈ U . Choose δ > 0 such that D(z, δ) ⊂ U . Then for w ∈ D(z, δ) we have I ∞ X 1 f (w) n an (w − z) with an = f (w) = · dw for 0 < r < δ. n+1 2πi (w − z) γ z,r n=0 Now for every k > 0, the k-th derivative f (k) (z) exists and is equal to k!ak . Corollary 2.10. Let γ be a contour in C, and U an open subset of C containing γ and its interior. Further, let f : U → C be an analytic function. Then for every z in the interior of γ and every k > 0 we have I k! f (w) (k) f (z) = · dw. 2πi γ (w − z)k+1 Proof. Choose δ > 0 such that γz,δ lies in the interior of γ. By Corollary 2.4, I I 1 f (w) 1 f (w) · dw = · dw. 2πi γ (w − z)k+1 2πi γz,δ (w − z)k+1 By the argument in Corollary 2.9, this is equal to f (k) (z)/k!. 42

We prove a generalization of Cauchy’s integral formula. Corollary 2.11. Let γ1 , γ2 be two contours such that γ1 is lying in the interior of γ2 . Let U ⊂ C be an open set which contains γ1 , γ2 and the region between γ1 , γ2 . Further, let f : U → C be an analytic function. Then for any z0 in the region between γ1 and γ2 we have I I 1 1 f (z) f (z) dz − dz. f (z0 ) = 2πi γ2 z − z0 2πi γ1 z − z0 Proof. We have seen that around z0 the function f has a Taylor expansion f (z) = P∞ n n=0 an (z − z0 ) . Define the function on U , g(z) :=

f (z) − a0 (z 6= z0 ); z − z0

g(z0 ) := a1 .

The function g is clearly analytic on U \ {z0 }. Further, ∞

g(z) − g(z0 ) X an (z − z0 )n−2 → a2 as z → z0 . = z − z0 n=2 Hence g is also analytic at z = z0 . In particular, g is analytic in the region between γ1 and γ2 . So by Corollary 2.4, I I g(z)dz = g(z)dz. γ1

γ2

Together with Corollaries 2.5, 2.4 this implies I 1 a0 f (z0 ) = a0 = · dz − 2πi γ2 z − z0 I 1 f (z) = · dz − 2πi γ2 z − z0

2.4

1 2πi

I

1 2πi

I

γ1

γ1

a0 · dz z − z0 f (z) · dz. z − z0

Isolated singularities, Laurent series, meromorphic functions

We define the punctured disk with center z0 ∈ C and radius r > 0 by D0 (z0 , r) = {z ∈ C : 0 < |z − z0 | < r}. 43

If f is an analytic function defined on D0 (z0 , r) for some r > 0, we call z0 an isolated singularity of f . In case that there exists an analytic function g on the non-punctured disk D(z0 , r) such that g(z) = f (z) for z ∈ D0 (z0 , r), we call z0 a removable singularity of f . Theorem 2.12. Let U ⊆ C be a non-empty, open set and f : U → C an analytic function. Further, let z0 ∈ U , and let R > 0 be such that D0 (z0 , R) ⊆ U . Then f has a Laurent series expansion f (z) =

∞ X

an (z − z0 )n converging for z ∈ D0 (z0 , R).

n=−∞

Further, we have for n ∈ Z, I 1 f (z) (2.2) an = · dz for any r with 0 < r < R. 2πi γz0 ,r (z − z0 )n+1 Proof. We fix z ∈ D0 (z0 , R) and use w to denote a complex variable. Choose r1 , r2 with 0 < r1 < |z − z0 | < r2 < R. By Corollary 2.11 we have I I 1 f (w) 1 f (w) (2.3) f (z) = · dw − · dw =: I1 − I2 , 2πi γz0 ,r2 w − z 2πi γz0 ,r1 w − z say. Completely similarly to Theorem 2.8, one shows that I1 =

∞ X

n

an (z − z0 )

n=0

1 with an = 2πi

I γz0 ,r2

f (w) · dw. (w − z0 )n+1

Notice that for w on the inner circle γz0 ,r1 we have  −1 f (w) f (w) f (w) w − z0 = =− · 1− w−z (w − z0 ) − (z − z0 ) z − z0 z − z0 ∞ X = − f (w)(z − z0 )−m−1 (w − z0 )m . m=0

Further, one easily shows that the latter series converges uniformly to f (w)/(w − z) 44

on γz0 ,r1 . After a substitution n = −m − 1, it follows that I2

−1 = 2πi

∞ X

I γz0 ,r2

−1 X

! f (w)(w − w0 )m (z − z0 )−m−1

· dw

m=0

1 = − an (z − z0 ) , with an = 2πi n=−∞ n

I γz0 ,r1

f (w) · dw. (w − z0 )n+1

By substituting the expressions for I1 , I2 obtained above into (2.3), we obtain ∞ X

f (z) = I1 − I2 =

an (z − z0 )n .

n=−∞

Further, by Corollary 2.4 we have 1 an = 2πi

I γz0 ,r

f (w) · dw (w − z0 )n+1

for any n ∈ Z and any r with 0 < r < R. This completes our proof. Let U ⊆ C be an open set, z0 ∈ U and f : U \ {z0 } → C an analytic function. Then z0 is an isolated singularity of f , and there is R > 0 such that f has a Laurent series expansion ∞ X f (z) = an (z − z0 )n n=−∞

converging for z ∈ D0 (z0 , R). Notice that z0 is a removable singularity of f if an = 0 for all n < 0. The point z0 is called • an essential singularity of f if there are infinitely many n < 0 with an 6= 0; • a pole of order k of f for some k > 0 if a−k 6= 0 and an = 0 for n < −k; a pole of order 1 is called simple; • a zero of order k of f for some k > 0 if ak 6= 0 and an = 0 for n < k; a zero of order 1 is called simple. 45

Notice that if f has a zero of order k at z0 then in particular, z0 is a removable singularity of f and so we may assume that f is defined and analytic at z0 . Moreover, z0 is a zero of order k of f if and only if f (j) (z0 ) = 0 for j = 0, . . . , k − 1, and f (k) (z0 ) 6= 0. For f, z0 as above, we define ordz0 (f ) := smallest k ∈ Z such that ak 6= 0. Thus, z0 essential singularity of f ⇐⇒ ordz0 (f ) = −∞; z0 pole of order k of f ⇐⇒ ordz0 (f ) = −k; z0 zero of order k of f ⇐⇒ ordz0 (f ) = k. Further, ordz0 (f ) = k if and only if there is a function g that is analytic around z0 such that f (z) = (z − z0 )k g(z) for z 6= z0 and g(z0 ) 6= 0. Lemma 2.13. Let R > 0 and let f, g : D0 (z0 , R) → C be two analytic functions. Assume that g 6= 0 on D0 (z0 , R), and that z0 is not an essential singularity of f or g. Then
ordz0 (f + g) > min ordz0 (f ), ordz0 (g) ; ordz0 (f g) = ordz0 (f ) + ordz0 (g); ordz0 (f /g) = ordz0 (f ) − ordz0 (g). Proof. Exercise. You may compare this with the function ordp for prime numbers p defined in Chapter 1. Both ordz0 and ordp are examples of discrete valuations. A discrete valuation on a field K is a surjective map v : K → Z ∪ {∞} such that v(0) = ∞; v(x) ∈ Z for x ∈ K, x 6= 0; v(xy) = v(x) + v(y) for x, y ∈ K; and v(x + y) > min(v(x), v(y)) for x, y ∈ K. Let U ⊂ C be an open set. A set S called discrete in U in U if S ⊂ U and for every compact set K with K ⊂ U , the intersection S ∩ K is finite. By the Bolzano-Weierstrass Theorem, S is discrete in U if and only if S ⊂ U and S has no limit point in U , that is, there are no z0 ∈ U and an infinite sequence {zn } of distinct elements of S such that limn→∞ zn = z0 . 46

A meromorphic function on U is a complex function f with the following properties: (i) there is a set S discrete in U such that f is defined and analytic on U \ S; (ii) all elements of S are poles of f . We say that a complex function f is meromorphic around z0 if f is analytic on D0 (z0 , r) for some r > 0, and z0 is a pole of f . It is easy to verify that if f, g are meromorphic functions on U then so are f + g and f · g. Later we will see that if U is connected, and g is a meromorphic, nonzero function on U , then the set of zeros of f is discrete in U . The zeros of g are poles of 1/g, and the poles of g are zeros of 1/g. Hence 1/g is meromorphic on U . Consequently, if U is an open, connected subset of C, then the functions meromorphic on U form a field.

2.5

Residues, logarithmic derivatives

Let z0 ∈ C, R > 0 and let f : D0 (z0 , R) → C be an analytic function. Then f has a Laurent series expansion converging on D0 (z0 , R): f (z) =

∞ X

an (z − z0 )n .

n=−∞

We define the residue of f at z0 by res(z0 , f ) := a−1 . In particular, if f is analytic at z0 then res(z0 , f ) = 0. By Theorem 2.12 we have I 1 res(z0 , f ) = f (z)dz 2πi γz0 ,r for any r with 0 < r < R. Theorem 2.14 (Residue Theorem). let γ be a contour in C. let z1 , . . . , zq be in the interior of γ. Let f be a complex function that is analytic on an open set containing γ and the interior of γ minus {z1 , . . . , zq }. Then I q X 1 f (z)dz = res(zi , f ). 2πi γ i=1 47

Proof. We proceed by induction on q. First let q = 1. Choose r > 0 such that γz1 ,r lies in the interior of γ. Then by Corollary 2.4, I I 1 1 f (z)dz = f (z)dz = res(z1 , f ). 2πi γ 2πi γz1 ,r Now let q > 1 and assume the Residue Theorem is true for fewer than q points. We cut γ into two pieces, the piece γ1 from a point w0 to w1 and the piece γ2 from w1 to w0 so that γ = γ1 + γ2 . Then we take a path γ3 from w1 to w0 inside the interior of γ without self-intersections; this gives two contours γ1 + γ3 and −γ3 + γ2 . We choose γ3 in such a way that it does not hit any of the points z1 , . . . , zq and both the interiors of these contours contain points from z1 , . . . , zq . Without loss of generality, we assume that the interior of γ1 +γ3 contains z1 , . . . , zm with 0 < m < q, while the interior of −γ3 + γ2 contains zm+1 , . . . , zq . Then by the induction hypothesis, I I I 1 1 1 f (z)dz = f (z)dz + f (z)dz 2πi γ 2πi γ1 2πi γ2 I I 1 1 f (z)dz + f (z)dz = 2πi γ1 +γ3 2πi −γ3 +γ2 q q m X X X = res(zi , f ) + res(zi , f ) = res(zi , f ), i=1

i=m+1

i=1

completing our proof. We have collected some useful facts about residues. Both f, g are analytic functions on D0 (z0 , r) for some r > 0. Lemma 2.15. (i) f has a simple pole or removable singularity at z0 with residue α α ⇐⇒ lim (z − z0 )f (z) = α ⇐⇒ f (z) − is analytic around z0 . z→z0 z − z0 (ii) Suppose f has a pole of order 1 at z0 and g is analytic and non-zero at z0 . Then res(z0 , f g) = g(z0 )res(z0 , f ). 48

(iii) Suppose that f is analytic and non-zero at z0 and g has a zero of order 1 at z0 . Then f /g has a pole of order 1 at z0 , and res(z0 , f /g) = f (z0 )/g 0 (z0 ). Proof. Exercise. Let U be a non-empty, open subset of C and f a meromorphic function on U which is not identically zero. We define the logarithmic derivative of f by f 0 /f. Suppose that U is simply connected and f is analytic and has no zeros on U . Then f 0 /f has an anti-derivative h : U → C. One easily verifies that (eh /f )0 = 0. Hence eh /f is constant on U . By adding a suitable constant to h we can achieve that eh = f . That is, we may view h as the logarithm of f , and f 0 /f as the derivative of this logarithm. But we will work with f 0 /f also if U is not simply connected and/or f has zeros or poles on U . In that case, we call f 0 /f also the logarithmic derivative of f , although it need not be the derivative of some function. The following facts are easy to prove: if f, g are two meromorphic functions on U that are not identically zero, then (f g)0 f0 g0 = + , fg f g

(f /g)0 f0 g0 = − . f /g f g

Further, if U is connected, then f0 g0 = ⇐⇒ f = cg for some constant c. f g Lemma 2.16. Let z0 ∈ C, r > 0 and let f : D0 (z0 , r) → C be analytic. Assume that z0 is either a removable singularity or a pole of f . Then z0 is a simple pole or (if z0 is neither a zero nor a pole of f ) a removable singularity of f 0 /f , and res(z0 , f 0 /f ) = ordz0 (f ). Proof. Let ordz0 (f ) = k. This means that f (z) = (z − z0 )k g(z) with g analytic around z0 and g(z0 ) 6= 0. Consequently, f0 (z − z0 )0 g 0 k g0 =k + = + . f z − z0 g z − z0 g 49

The function g 0 /g is analytic around z0 since g(z0 ) 6= 0. So by Lemma 2.15, res(z0 , f 0 /f ) = k. Corollary 2.17. Let γ be a contour in C, U an open subset of C containing γ and its interior, and f a meromorphic function on U . Assume that f has no zeros or poles on γ and let z1 , . . . , zq be the zeros and poles of f inside γ. Then 1 2πi

I γ

q

X f 0 (z) · dz = ordzi (f ) = Z − P, f (z) i=1

where Z, P denote the number of zeros and poles of f inside γ, counted with their multiplicities. Proof. By Theorem 2.14 and Lemma 2.16 we have 1 2πi

2.6

I γ

q q X X f 0 (z) 0 · dz = res(zi , f /f ) = ordzi (f ) = Z − P. f (z) i=1 i=1

Unicity of analytic functions

In this section we show that two analytic functions f, g defined on a connected open set U are equal on the whole set U , if they are equal on a sufficiently large subset of U . We start with the following result. Theorem 2.18. Let U be a non-empty, open, connected subset of C, and f : U → C an analytic function. Assume that f = 0 on an infinite subset of U having a limit point in U . Then f = 0 on U . Proof. Our assumption that U is connected means, that any non-empty subset S of U that is both open and closed in U , must be equal to U . Let Z be the set of z ∈ U with f (z) = 0. Let S be the set of z ∈ U such that z is a limit point of Z. By assumption, S is non-empty. Since f is continuous, we have S ⊆ Z. Any limit point in U of S is therefore a limit point of Z and so it belongs to 50

S. Hence S is closed in U . We show that S is also open; then it follows that S = U and we are done. Pick z0 ∈ S. We have to show that there is δ > 0 such that D(z0 , δ) ⊂ S. There is δ > 0 such that f has a Taylor expansion f (z) =

∞ X

an (z − z0 )n

n=0

converging on D(z0 , δ). Assume that f is not identically 0 on D(z0 , δ). Then not all coefficients an are 0. Assume that am 6= 0 and an = 0 for n < m, say. Then P n−m f (z) = (z − z0 )m h(z) with h(z) = ∞ . Since h(z0 ) = am 6= 0 and n=m an (z − z0 ) h is continuous, there is δ1 > 0 such that h(z) 6= 0 for all z ∈ D(z0 , δ1 ). But then f (z) 6= 0 for all z with 0 < |z − z0 | < δ1 , contradicting that z0 ∈ S. Hence f is identically 0 on D(z0 , δ). Clearly, every point of D(z0 , δ) is a limit point of D(z0 , δ), hence of Z. So D(z0 , δ) ⊂ S. This shows that indeed, S is open. Corollary 2.19. Let U be a non-empty, open, connected subset of C, and let f : U → C be an analytic function that is not identically 0 on U . Then the set of zeros of f in U is discrete in U , i.e., every compact subset of U contains only finitely many zeros of f . Proof. Suppose that some compact subset of U contains infinitely many zeros of f . Then by the Bolzano-Weierstrass Theorem, the set of these zeros would have a limit point in this compact set, implying that f = 0 on U . Corollary 2.20. Let U be a non-empty, open, connected subset of C, and f, g : U → C two analytic functions. Assume that f = g on an infinite subset of U having a limit point in U . Then f = g on U . Proof. Apply Theorem 2.18 to f − g. Let U, U 0 be open subsets of C with U ⊂ U 0 . Let f : U → C be an analytic function. An analytic continuation of f to U 0 is an analytic function g : U 0 → C such that g(z) = f (z) for z ∈ U . It is often a difficult problem to figure out whether such an analytic continuation exists. The next Corollary shows that if it exists, it must be unique. 51

Corollary 2.21. Let U, U 0 be non-empty, open subsets of C, such that U ⊂ U 0 and U 0 is connected. Let f : U → C be an analytic function. Then f has at most one analytic continuation to U 0 . Proof. . Let g1 , g2 be two analytic continuations of f to U 0 . Then g1 (z) = g2 (z) = f (z) for z ∈ U . Since U has a limit point in U 0 , it follows that g1 (z) = g2 (z) for z ∈ U 0. Another consequence of Theorem 2.18 is the so-called Schwarz’ reflection priciple, which states that analytic functions assuming real values on the real line have nice symmetric properties. Corollary 2.22 (Schwarz’ reflection principle). Let U be an open, connected subset of C, such that U ∩ R 6= ∅ and such that U is symmetric about R, i.e., z ∈ U for every z ∈ U . Further, let f : U → C be a nonidentically zero analytic function with the property that {z ∈ U ∩ R : f (z) ∈ R} has a limit point in U . Then f has the following properties: (i) f (z) ∈ R for z ∈ U ∩ R; (ii) f (z) = f (z) for z ∈ U ; (iii) If z0 and r > 0 are such that D0 (z0 , r) ⊂ U , then ordz0 (f ) = ordz0 (f ). Proof. We first show that the function z 7→ f (z) is analytic on U . Indeed, for z0 ∈ U , the limit   f (z) − f (z0 ) f (z) − f (z0 ) lim = lim = f 0 (z0 ) z→z0 z→z0 z − z0 z − z0 exists. Notice that for every z ∈ U ∩ R with f (z) ∈ R, we have f (z) = f (z). So by our assumption on f , the set of z ∈ U with f (z) = f (z) has a limit point in U . Now Corollary 2.20 implies that f (z) = f (z) for z ∈ U . This implies (i) and (ii). 52

We finish with proving (iii). Our assumption implies that f has a Laurent series expansion ∞ X f (z) = an (z − z0 )n n=−∞

converging on D (z0 , r). Then for z ∈ D0 (z0 , r) we have z ∈ D0 (z0 , r) and 0

f (z) = f (z) =

∞ X

! an (z − z0

)n

=

n=−∞

∞ X

an (z − z0 )n ,

n=−∞

which clearly implies (iii).

2.7

Analytic functions defined by integrals

In analytic number theory, quite often one has to deal with complex functions that are defined by infinite series, infinite products, infinite integrals, or even worse, infinite integrals of infinite series. In this section we have collected some useful results that allow us to verify in a not too difficult manner that such complicated functions are analytic. We start with a general theorem on analytic functions defined by an integral. Since we couldn’t find a good reference for a result of this type, we have included a proof. Theorem 2.23. Let D be a measurable subset of Rm , U an open subset of C and f : D × U → C a function with the following properties: (i) f is measurable on D × U (with U viewed as subset of R2 ); (ii) for every fixed x ∈ D, the function z 7→ f (x, z) is analytic on U ; (iii) for every compact subset K of U there is a measurable function MK : D → R such that Z |f (x, z)| 6 MK (x) for x ∈ D, z ∈ K, MK (x)dx < ∞. D

Then the function F given by Z F (z) :=

f (x, z)dx D

53

is analytic on U , and for every k > 1, F

(k)

Z

f (k) (x, z)dx,

(z) = D

(k)

where f (x, z) denotes the k-th derivative with respect to z of the analytic function z 7→ f (x, z). Proof. Fix z ∈ U . Choose r > 0 such that D(z, r) ⊂ U , and let 0 < δ < 21 r. We show that for w ∈ D(z, δ), F (w) can be expanded into a Taylor series around z; then it follows that F is analytic on D(z, δ) and so in particular in z. Let M (x) : D → R be a measurable function such that |f (x, w)| 6 M (x) for x ∈ D, w ∈ D(z, r) and R M (x)dx < ∞. D For w ∈ D(z, δ) we have by Corollary 2.5, ) Z ( Z I 1 f (x, ζ) f (x, w)dx = F (w) = · dζ dx. 2πi γz,2δ ζ − w D D By inserting f (x, ζ) f (x, ζ) f (x, ζ) = = ζ −w (ζ − z) − (w − z) ζ −z ∞ X

=

n=0

 −1 w−z 1− ζ −z

f (x, ζ) · (w − z)n n+1 (ζ − z)

we obtain Z ( F (w) = D

1 2πi

Z (Z = D

0

1

∞ X

! ) f (x, ζ) (w − z)n dζ dx n+1 (ζ − z) γz,2δ n=0 ! ) ∞ X f (x, z + 2δe2πit ) (w − z)n dt dx. 2πit )n (2δe n=0

I

We apply the Fubini-Tonelli Theorem (see the Prerequisites). Note that since |w − z| < δ, ! ) Z (Z 1 X ∞ f (x, z + 2δe2πit ) (w − z)n dt dx 2πit n (2δe ) D 0 n=0 ! ) Z (Z 1 X Z ∞ −n 6 M (x)2 dt dx 6 2M (x)dx < ∞. D

0

D

n=0

54

So the conditions of the Fubini-Tonelli Theorem are satisfied, and in the expression for F (w) derived above we can interchange the integrations and the summation. Performing this interchange and then applying Corollary 2.10, we obtain F (w) =

∞ X

(w − z)

n

n=0

=

∞ X

(w − z)n

n=0

=

∞ X

(w − z)

n

Z D

n=0

1

  f (x, z + 2δe2πit ) dt dx (2δe2πit )n D 0 ) ! I Z ( f (x, ζ) 1 · dζ dx 2πi γz,2δ (ζ − z)n+1 D

Z Z

 f (n) (x, z) · dx . n!

This shows that indeed, F (w) has a Taylor expansion around z converging on D(z, δ). So in particular, F is analytic in z. Further, F (k) (z) is equal to k! times R the coefficient of (w − z)k , that is, D f (k) (x, z)dx. This proves our Theorem. We deduce a result, which states that under certain conditions, the pointwise limit of a sequence of analytic functions is again analytic. Theorem 2.24. Let U ⊂ C be a non-empty open set, and {fn : U → C}∞ n=0 a sequence of analytic functions, converging pointwise to a function f on U . Assume that for every compact subset K of U there is a constant CK < ∞ such that |fn (z)| 6 CK for all z ∈ K, n > 0. (k)

Then f is analytic on U , and fn → f (k) pointwise on U for all k > 1. Proof. The set U can be covered by disks D(z0 , δ) with z0 ∈ U , δ > 0, such that D(z0 , 2δ) ⊂ U . We fix such a disk D(z0 , δ) and prove that f is analytic on D(z0 , δ) (k) and fn → f (k) pointwise on D(z0 , δ) for k > 1. This clearly suffices. Let z ∈ D(z0 , δ), k > 0. Then by Corollary 2.10, we have I k! fn (ζ) (k) fn (z) = · dζ 2πi γz0 ,2δ (ζ − z)k+1 Z 1 Z 1 fn (z0 + 2δe2πit )2δe2πit = k! · dt = k! gn,k (t, z)dt, (z0 + 2δe2πit − z)k+1 0 0 55

say. By assumption, there is C < ∞ such that |fn (w)| 6 C for w ∈ D(z0 , 2δ), n > 0. Further, for t ∈ [0, 1] we have |z0 + 2δe2πit − z| > δ. Hence |gn,k (t, z)| 6 C · 2δ/δ k+1 = 2Cδ −k for n, k > 0.

(2.4)

Notice that for k > 0, t ∈ [0, 1], z ∈ D(z0 , δ) we have gn,k (t, z) → gk (t, z) :=

f (z0 + 2δe2πit )2δe2πit . (z0 + 2δe2πit − z)k+1

Thanks to (2.4) we can apply the dominated convergence theorem, and obtain for every fixed z ∈ D(z0 , δ), k > 0, Z 1 (k) gk (t, z)dt. fn (z) → k! 0

Applying this with k = 0 and using fn → f pointwise, we obtain Z 1 f (z) = g0 (t, z)dt for z ∈ D(z0 , δ). 0

It follows from Theorem 2.23 that the right-hand side, and hence f , is analytic on D(z0 , δ), and moreover, Z 1 Z 1 (k) (k) gk (t, z)dt = lim fn(k) (z) for z ∈ D(z0 , δ), k > 1, g0 (t, z)dt = k! f (z) = 0

n→∞

0

(k)

where g0 is the k-th derivative of g0 with respect to z. Indeed, g0 (t, z) is measurable on [0, 1] × D(z0 , δ) and for every fixed t, the function z 7→ g0 (t, z) is analytic on D(z0 , δ). Further, by (2.4) and since gn,0 (t, z) → g0 (t, z), we have |g0 (t, z)| 6 2C for t ∈ [0, 1], z ∈ D(z0 , δ). So all conditions of Theorem 2.23 are satisfied. This completes our proof. Corollary 2.25. Let U ⊂ C be a non-empty open set, and {fn : U → C}∞ n=0 a sequence of analytic functions, converging to a function f pointwise on U , and uniformly on every compact subset of U . (k) Then f is analytic on U and fn → f (k) pointwise on U for every k > 1. Proof. Take a compact subset K of U . Let ε > 0. Then there is N such that |fn (z) − fm (z)| < ε for all z ∈ K, m, n > N . Choose m > N . Then there is C > 0 such that |fm (z)| 6 C for z ∈ K since fm is continuous. Hence |fn (z)| 6 C + ε for z ∈ K, n > N . Now our Corollary follows at once from Theorem 2.24. 56

Corollary 2.26. let U ⊂ C be a non-empty open set, and {fn : U → C}∞ n=0 a sequence of analytic functions, converging to a function f pointwise on U and uniformly on every compact subset of U . Then fn0 (z) f 0 (z) = n→∞ fn (z) f (z) lim

for all z ∈ U with f (z) 6= 0, where the limit is taken over those n for which fn (z) 6= 0. Proof. Obvious. Corollary 2.27. Let U ⊂ C be a non-empty open set and {fn : U → C}∞ n=0 a sequence of analytic functions. Assume that for every compact subset K of U there P are reals Mn,K such that |fn (z)| 6 Mn,K for z ∈ K and ∞ n=0 Mn,K converges. Then   (k) P P∞ P (k) = ∞ for k > 0, (i) ∞ n=0 fn n=0 fn n=0 fn is analytic on U , and Q∞ (ii) n=0 (1 + fn ) is analytic on U . P Proof. Our assumption on the functions fn implies that both the series ∞ n=0 fn and Q∞ the infinite product n=0 (1 + fn ) converge uniformly on every compact subset of U (see the Prerequisites). Now apply Corollary 2.25. Corollary 2.28. Let U , {fn }∞ in addition n=0 be as in Corollary 2.27 and assume Q∞ that fn 6= −1 on U for every n > 0. Then for the function F = n=0 (1 + fn ) we have ∞ F 0 X fn0 = . F 1 + f n n=0 Q Proof. Let Fm := m n=0 (1 + fn ). Then Fm → F uniformly on every compact subset of U . Hence by Corollary 2.26, m X F0 F0 fn0 = lim m = lim m→∞ Fm m→∞ F 1 + fn n=0

which clearly implies Corollary 2.28.

57

Chapter 3 Dirichlet series and arithmetic functions An arithmetic function is a function f : Z>0 → C. To such a function we associate its Dirichlet series ∞ X Lf (s) = f (n)n−s n=1

where s is a complex variable. It is common practice (although this doesn’t make sense) to write s = σ + it, where σ = Re s and t = Im s. It has shown very fruitful in number theory, to study an arithmetic function by means of its Dirichlet series. In this chapter, we prove some basic properties of Dirichlet series and arithmetic functions.

3.1

Dirichlet series

We want to develop a theory for Dirichlet series similar to that for power series. P n Every power series ∞ n=0 an z has a radius of convergence R such that the series converges if |z| < R and diverges if |z| > R. As we will see, a Dirichlet series Lf (s) = P∞ −s has an abscissa of convergence σ0 (f ) such that the series converges n=1 f (n)n for all s ∈ C with Re s > σ0 (f ) and diverges for all s ∈ C with Re s < σ0 (f ). For P −s instance, ζ(s) = ∞ has abscissa of convergence 1. n=1 n We start with an important summation result, which we shall use very frequently. 59

Theorem 3.1 (Partial summation, summation by parts). Let M, N be reals with M < N . Let x1 , . . . , xr be real numbers with M 6 x1 < · · · < xr 6 N , let P a(x1 ), . . . , a(xr ) be complex numbers, and put A(t) := xk 6t a(xk ) for t ∈ [M, N ]. Further, let g : [M, N ] → C be a differentiable function. Then r X

Z

N

A(t)g 0 (t)dt.

a(xk )g(xk ) = A(N )g(N ) − M

k=1

Proof. Let x0 < M and put A(x0 ) := 0. Then r X

a(xk )g(xk ) =

k=1

r X

(A(xk ) − A(xk−1 ))g(xk )

k=1

=

r X

A(xk )g(xk ) −

r−1 X

k=1

A(xk )g(xk+1 )

k=1

= A(xr )g(xr ) −

r−1 X

A(xk )(g(xk+1 ) − g(xk )).

k=1

Since A(t) = A(xk ) for xk 6 t < xk+1 we have Z

xk+1

A(xk )(g(xk+1 ) − g(xk )) =

A(t)g 0 (t)dt.

xk

Hence (3.1)

r X

a(xk )g(xk ) = A(xr )g(xr ) −

k=1

r−1 Z X

Zk=1xr = A(xr )g(xr ) −

xk+1

A(t)g 0 (t)dt

xk

A(t)g 0 (t)dt.

x1

In case that x1 = M ,xr = N we are done. if x1 > M , then A(t) = 0 for M 6 t < x1 Rx and thus, M1 A(t)g 0 (t)dt = 0. If xr < N , then A(t) = A(xr ) for xr 6 t 6 N , hence Z

N

A(t)g 0 (t)dt = A(N )g(N ) − A(xr )g(xr ).

xr

Together with (3.1) this implies our Theorem. 60

Theorem 3.2. Let f : Z>0 → C be an arithmetic function with the property that P there exists a constant C > 0 such that | N n=1 f (n)| 6 C for every N > 1. Then P∞ −s Lf (s) = n=1 f (n)n converges for every s ∈ C with Re s > 0. More precisely, on {s ∈ C : Re s > 0} the function Lf is analytic, and for its k-th derivative we have ∞ X (k) Lf (s) = f (n)(− log n)k n−s . n=1

Proof. Notice that on {s ∈ C : Re s > 0}, the partial sums Lf,N (s) :=

N X

f (n)n−s (N = 1, 2, . . .)

n=1

P (k) k −s are analytic, and Lf,N (s) = N for k > 0. We have to show n=1 f (n)(− log n) n that the partial sums converge on {s ∈ C : Re s > 0}, and that analyticity and the formula for the k-th derivative are maintained if we let N → ∞. Let s ∈ C, Re s > 0. We first rewrite Lf,N (s) using partial summation. Let P F (t) := 16n6t f (n). By Theorem 3.1 (with {x1 , . . . , xr } = {1, . . . , N } and g(t) = t−s ) we have Z N Z N Lf,N (s) = F (N )N −s − F (t)(−s)t−s−1 dt = F (N )N −s + s F (t)t−s−1 dt. 1

1

By assumption, there is C > 0 such that |F (t)| 6 C for every t > 1. Further |t−s−1 | = t−Re s−1 . Hence |F (t)t−s−1 | 6 Ct−Re s−1 . Since Re s > 0, the integral R ∞ −Re s−1 R∞ t dt converges, therefore, F (t)t−s−1 dt converges. Further, |F (N )N −s | 6 1 1 C · N −Re s → 0 as N → ∞. It follows that Lf (s) = limN →∞ Lf,N (s) converges if Re s > 0. We apply Theorem 2.24 to the sequence of partial sums {Lf,N (s)}. Let K be a compact subset of {s ∈ C : Re s > 0}. There are σ > 0, A > 0 such that Re s > σ, |s| 6 A for s ∈ K. Thus, for s ∈ K and N > 1, we have Z N −s |Lf,N (s)| 6 |F (N )N | + |s| |F (t)t−s−1 |dt 1

6 C · N −σ + A

Z

N

C · t−σ−1 dt = C · N −σ + AC · σ −1 (1 − N −σ )

1

6 C + AC · σ −1 , 61

which is an upper bound independent of s, N . Now Theorem 2.24 implies that for s ∈ C with Re s > 0, the series Lf (s) = limN →∞ Lf,N (s) is analytic and moreover, (k) Lf (s)

= lim

N →∞

(k) Lf,N (s)

=

∞ X

f (n)(− log n)k n−s .

n=1

Corollary 3.3. Let f : Z>0 → C be an arithmetic function and let s0 ∈ C be such P −s0 that ∞ converges. Then for s ∈ C with Re s > Re s0 the function Lf n=1 f (n)n converges and is analytic, and (k)

Lf (s) =

∞ X

f (n)(− log n)k n−s for k > 1.

n=1

Proof. Write s = s0 + s0 . Then Re s0 > 0 if Re s > Re s0 . There is C > 0 such that P P 0 −s0 −s0 )n−s . | 6 C for all N . Apply Theorem 3.2 to ∞ | N n=1 (f (n)n n=1 f (n)n Theorem 3.4. There exists a number σ0 (f ) with −∞ 6 σ0 (f ) 6 ∞ such that Lf (s) converges for all s ∈ C with Re s > σ0 (f ) and diverges for all s ∈ C with Re s < σ0 (f ). Moreover, if σ0 (f ) < ∞, then for s ∈ C with Re s > σ0 (f ) the function Lf is analytic, and (3.2)

(k) Lf (s)

=

∞ X

f (n)(− log n)k n−s for k > 1.

n=1

Proof. If there is no s ∈ C for which Lf (s) converges we have σ0 (f ) = ∞. Assume that Lf (s) converges for some s ∈ C and define  σ0 (f ) := inf σ : ∃s ∈ C such that Re s = σ, Lf (s) converges . Clearly, Lf (s) diverges if Re s < σ0 (f ). To prove that Lf (s) converges for Re s > σ0 (f ), take such s and choose s0 such that σ0 (f ) < Re s0 < Re s and Lf (s0 ) con(k) verges. By Corollary 3.3, Lf is convergent and analytic in s, and for Lf (s) we have expression (3.2). 62

The number σ0 (f ) is called the abscissa of convergence of Lf . There exists also a real number σa (f ), called the abscissa of absolute convergence of Lf such that Lf (s) converges absolutely if Re s > σa (f ), and does not converge absolutely if Re s < σa (f ). In fact, we have σa (f ) = σ0 (|f |), that is the abscissa of convergence of L|f | (s) = P∞ P∞ P∞ −s −s −σ |f (n)|n . For write σ = Re s. Then |f (n)n |= conn=1 n=1 n=1 |f (n)|n verges if σ > σ0 (|f |) and diverges if σ < σ0 (|f |). Theorem 3.5. For every arithmetic function f : Z>0 → C we have σ0 (f ) 6 σa (f ) 6 σ0 (f ) + 1. Proof. It is clear that σ0 (f ) 6 σa (f ). To prove σa (f ) 6 σ0 (f ) + 1, we have to show that Lf (s) converges absolutely if Re s > σ0 (f ) + 1. Take such s; then Re s = σ0 (f ) + 1 + ε with ε > 0. Put σ := σ0 (f ) + ε/2. The P −σ converges, hence there is a constant C such that |f (n)n−σ | 6 C series ∞ n=1 f (n)n for all n. Therefore, |f (n)n−s | = |f (n)| · n−Re s = |f (n)n−σ | · n−1−ε/2 6 Cn−1−ε/2 P P −s −1−ε/2 for n > 1. The series ∞ converges, hence ∞ n=1 |f (n)n | converges. n=1 n Exercise 3.1. Show that there exist arithmetic functions f such that σa (f ) = σ0 (f ) + 1. The next theorem implies that an arithmetic function is uniquely determined by its Dirichlet series. Theorem 3.6. Let f, g : Z>0 → C be two arithmetic functions for which there is σ ∈ R such that Lf (s), Lg (s) converge absolutely and Lf (s) = Lg (s) for all s ∈ C with Re s > σ. Then f = g. Proof. Let h := f − g. Our assumptions imply that Lh (s) converges absolutely, and Lh (s) = 0 for all s ∈ C with Re s > σ. We have to prove that h = 0. Assume that there are positive integers n with h(n) 6= 0, and let m be the smallest such n. Then for all s ∈ C with Re s > σ we have −s

h(m)m

=−

∞ X n=m+1

63

h(n)n−s .

Let σ1 > σ, and let s ∈ C with Re s > σ1 . Then |h(m)| 6

∞ X

|h(n)|(m/n)

Re s

=

n=m+1

6 mσ1

∞ X

|h(n)|(m/n)σ1 (m/n)Re s−σ1

n=m+1 ∞ X

! |h(n)| · n−σ1

· (m/(m + 1))Re s−σ1 .

n=m+1

The series between the parentheses is convergent, hence a finite number. So the right-hand side tends to 0 as Re s → ∞. This contradicts that h(m) 6= 0.

3.2

Arithmetic functions

A multiplicative function is an arithmetic function f such that f 6≡ 0 and f (mn) = f (m)f (n) for all positive integers m, n with gcd(m, n) = 1. A strongly multiplicative function is an arithmetic function f with the property that f 6≡ 0 and f (mn) = f (m)f (n) for all integers m, n. Notation. In expressions pk11 · · · pkt t it is always assumed that the pi are distinct prime numbers, and the ki positive integers. We start with some simple observations. Lemma 3.7. (i) Let f be a multiplicative function. Then f (1) = 1. Further, if n = pk11 · · · pkt t , then f (n) = f (pk11 ) · · · f (pkt t ). (ii) Let f, g be two multiplicative functions such that f (pk ) = g(pk ) for every prime p and k ∈ Z>1 . Then f = g. (iii) let f, g be two strongly multiplicative functions such that f (p) = g(p) for every prime p. Then f = g. Proof. Obvious. We define the convolution product f ∗ g of two arithmetic functions f, g by X (f ∗ g)(n) := f (n/d)g(d) for n ∈ Z>0 , d|n

where 0 d | n0 means that the sum is taken over all positive divisors of n. 64

Examples. Define the arithmetic functions e, E by e(1) = 1, e(n) = 0 for all n ∈ Z>1 , E(n) = 1 for all n ∈ Z>0 . Clearly, e is multiplicative, and E is strongly multiplicative. If f is any arithmetic function, then e ∗ f = f , while (E ∗ f )(n) =

X

f (d).

d|n

Lemma 3.8. (i) For any two arithmetic functions f, g we have f ∗ g = g ∗ f . (ii) For any three arithmetic functions f, g, h we have (f ∗ g) ∗ h = f ∗ (g ∗ h). Proof. Straightforward. Theorem 3.9. (i) Let A be the set of arithmetic functions f with f (1) 6= 0. Then A with ∗ is an abelian group with unit element e. (ii) Let M be the set of multiplicative functions. Then M with ∗ is a subgroup of A. Proof. (i) We know already that ∗ is commutative and associative and that e is the unit element of ∗. It remains to verify that every element of A has a (necessarily unique) inverse with respect to ∗. Let f ∈ A. Notice that for an arithmetic function g we have f ∗ g = e ⇐⇒ f (1)g(1) = 1,

X

f (n/d)g(d) = 0 for n > 1

d|n −1

⇐⇒ g(1) := f (1) ,

g(n) := −f (1)−1

X

f (n/d)g(d) for n > 1.

d|n, d0 . Then f (n) =

X

µ(n/d)F (d) for n ∈ Z>0 .

d|n

Proof. We have F = E ∗ f . Hence µ ∗ F = µ ∗ (E ∗ f ) = (µ ∗ E) ∗ f = e ∗ f = f.

Examples. 1) Define ϕ(n) := #{k ∈ Z : 1 6 k 6 n, gcd(k, n) = 1}. It is P well-known that d|n ϕ(d) = n for n ∈ Z>0 . This implies that ϕ(n) =

X

µ(n/d)d,

d|n

or ϕ = µ ∗ I1 , where we define Iα (n) = nα for n ∈ Z>0 , α ∈ C. As a consequence, ϕ is multiplicative, and for n = pk11 · · · pkt t we have ϕ(n) =

t Y

ϕ(pki i )

t Y = (pki i − piki −1 ).

i=1

i=1

P 2) Let α ∈ C and define σα (n) = d|n dα for n ∈ Z>0 . Then σα = E ∗ Iα , which implies that σα is multiplicative. Hence for n = pk11 · · · pkt t we have  t Y pα(ki +1)−1   i  if α 6= 0,  t  Y pαi − 1 ki i=1 σα (n) = σα (pi ) = t Y   i=1  (ki + 1) if α = 0.   i=1

67

3.3

Convolution product vs. Dirichlet series

We investigate the relation between the convolution product of two arithmetic functions and their associated Dirichlet series. Theorem 3.12. Let f, g be two arithmetic functions. Let s ∈ C be such that Lf (s) and Lg (s) converge absolutely. Then also Lf ∗g (s) converges absolutely, and Lf ∗g (s) = Lf (s)Lg (s). Proof. Since both Lf (s) and Lg (s) are absolutely convergent we can rearrange their product as a double series and then rearrange the terms: ∞ X

−s

f (m)m

∞  X

m=1

= =

−s

g(n)n



n=1 ∞ X ∞ X

−s

f (m)g(n)(mn)

=

m=1 n=1 ∞ X

∞  X X k=1

 f (m)g(n) k −s

mn=k

(f ∗ g)(k)k −s = Lf ∗g (s).

k=1

We now show that Lf ∗g (s) converges absolutely: ∞ X

−s

|(f ∗ g)(k)k | 6

k=1

=

∞  X X

 |f (m)| · |g(n)| · |k −s |

k=1

mn=k

∞ X

∞  X  |f (m)m | |g(n)n−s | < ∞ −s

m=1

n=1

by following the above reasoning in opposite direction and taking absolute values everywhere. This completes our proof. P P Q Q We define p (· · · ) = limN →∞ p6N (· · · ), and p (· · · ) = limN →∞ p6N (· · · ) where the sums and products are taken over the primes. Theorem 3.13. Let f be a multiplicative function. let s ∈ C be such that Lf (s) = P∞ −s converges absolutely. Then n=1 f (n)n (3.3)

Lf (s) =

∞ YX p

j=0

68

f (pj )p−js



and the right-hand side converges absolutely. P j −js Further, Lf (s) 6= 0 as soon as ∞ 6= 0 for every prime p. j=0 f (p )p Remark. The absolute convergence of the right-hand side means that for every prime p, the series ∞ X Ap (s) := |f (pj )p−js | j=0

converges, and that the infinite product

Q

p

Ap (s) converges.

Proof. We first show the absolute convergence of the right-hand side of (3.3). First notice that for every prime p, Ap (s) =

∞ X

j

|f (p )p

−js

|6

∞ X

|f (n)n−s | < ∞.

n=1

j=0

P Q Q Recall that p Ap (s) = p (1 + |Ap (s) − 1|) converges if p |Ap (s) − 1| converges (see the section on infinite products in the Prerequisites). But the latter holds, since X

|Ap (s) − 1| =

∞ XX p

p

|f (pj )p−js | 6

∞ X

|f (n)n−s | < ∞.

n=2

j=1

This proves the absolute convergence of the right-hand side of (3.3). P j −js . We have seen that the series Lp (s) all converge Put Lp (s) := ∞ j=0 f (p )p absolutely. Further, X

|Lp (s) − 1| 6

p

∞ XX p

|f (pj )p−js | < ∞.

j=1

Q Hence p Lp (s) converges, which implies that the product is 0 if and only if at least one of the factors Lp (s) is 0. Q It remains to prove that Lf (s) = p Lp (s). Let N > 1 and let p1 , . . . , pt be the prime numbers 6 N . Further, let SN be the set of integers composed of prime numbers 6 N and TN the set of remaining integers, i.e., divisible by at least one prime > N . Since the series Lp (s) (p prime) converge absolutely, we have X Y X t s Lp (s) = f (pj11 ) · · · f (pjt t )(p1−j1 · · · p−j ) = f (n)n−s . t p6N

j1 ,...,jt >0

n∈SN

69

Now clearly, ∞ X X X Y f (n)n−s − Lp (s) = f (n)n−s = f (n)n−s Lf (s) − n=1 n∈SN

p6N

∞ X

6

n∈TN

|f (n)n−s | → 0 as N → ∞.

n=N +1

This proves (3.3). Corollary 3.14. Let f be a strongly multiplicative function. Let s ∈ C be such that Lf (s) converges absolutely. Then Lf (s) =

Y p

1 . 1 − f (p)p−s

Further, Lf (s) 6= 0. Proof. Use that ∞ X

f (pj )p−js =

j=0

∞ X (f (p)p−s )j = j=0

1 . 1 − f (p)p−s

Further, all factors (1 − f (p)p−s )−1 are 6= 0, hence Lf (s) 6= 0. Examples. 1) For s ∈ C with Re s > 1 we have ζ(s) =

∞ X

n−s =

n=1

Y (1 − p−s )−1 (Euler) p

P −s converges absolutely, 2) For s ∈ C with Re s > 1, the series Lµ (s) = ∞ n=1 µ(n)n hence ∞ ∞ X X ζ(s)Lµ (s) = (E ∗ µ)(n)n−s = e(n)n−s = 1. n=1

P∞

−1

−s

That is, ζ(s) = n=1 µ(n)n prove this, is to observe that ζ(s)

−1

n=1

for s ∈ C with Re s > 1. An alternative way to

∞ ∞  X Y YX −s j −js = (1 − p ) = µ(p )p = µ(n)n−s . p

p

j=0

70

n=1

3) Recall that ϕ = µ ∗ I1 . The series LI1 (s) = absolutely for Re s > 2. Hence ∞ X

P∞

n=1

n/ns = ζ(s − 1) converges

ϕ(n)n−s = Lϕ(s) = Lµ (s)LI1 (s) = ζ(s − 1)/ζ(s)

n=1

and Lϕ (s) converges absolutely if Re s > 2. 4) The (very important) von Mangoldt function Λ is defined by  log p if n = pk for some prime p and some k > 1, Λ(n) = 0 otherwise. n Λ(n)

1 0

2 log 2

3 log 3

4 log 2

5 log 5

6 0

7 log 7

8 log 2

9 log 3

10 0

For n = pk11 · · · pkt t (unique prime factorization) we have X d|n

Λ(n) =

ki t X X

log pi =

i=1 j=1

t X

ki log pi = log n.

i=1

Hence E ∗ Λ = log, where log denotes the arithmetic function n 7→ log n. So Λ = µ ∗ log. P −s converges absoLemma 3.15. For s ∈ C with Re s > 1, the series ∞ n=1 Λ(n)n lutely, and ∞ X Λ(n)n−s = −ζ 0 (s)/ζ(s). n=1

Proof. We apply Theorem 3.12. First recall that Lµ (s) converges absolutely if Re s > P −s for Re s > 1. Hence 1. Further, by Theorem 3.4, we have ζ 0 (s) = ∞ n=1 (− log n)n ∞ X n=1

−s

| log(n)n | =

∞ X

(log n)n−Re s = −ζ 0 (Re s)

n=1

converges if Re s > 1. That is, Llog (s) converges absolutely if Re s > 1. It follows that LΛ (s) = Lµ (s)Llog (s) = −ζ(s)−1 ζ 0 (s) and LΛ (s) converges absolutely if Re s > 1. 71

Chapter 4 Characters and Gauss sums 4.1

Characters on finite abelian groups

In what follows, abelian groups are multiplicatively written, and the unit element of an abelian group A is denoted by 1. We denote the order (number of elements) of A by |A|. Let A be a finite abelian group. A character on A is a group homomorphism χ : A → C∗ (i.e., C \ {0} with multiplication). If |A| = n then an = 1, hence χ(a)n = 1 for each a ∈ A and each character χ on A. Therefore, a character on A maps A to the roots of unity. The product χ1 χ2 of two characters χ1 , χ2 on A is defined by (χ1 χ2 )(a) = χ1 (a)χ2 (a) for a ∈ A. With this product, the characters on A form an abelian b (or Hom(A, C∗ )). group, the so-called character group of A, which we denote by A (A) b is the trivial character χ0 that maps A to 1. Since any The unit element of A character on A maps A to the roots of unity, the inverse χ−1 : a 7→ χ(a)−1 of a character χ is equal to its complex conjugate χ : a 7→ χ(a). Let B be a subgroup of A and χ a character on B. By an extension of χ to A we mean a character χ0 on A such that χ0 |B = χ, i.e., χ0 (b) = χ(b) for b ∈ B. Lemma 4.1. Let A be a finite abelian group, B a proper subgroup of A, and χ a character on B. Then χ can be extended to A. 73

Proof. Choose g ∈ A \ B and define B 0 := Bhgi = {bg k : b ∈ B, k ∈ Z}. We first extend χ to B 0 . Let r be the smallest positive integer such that g r ∈ B, and let h := g r . Choose ρ such that ρr = χ(h) and define a character χ0 on B 0 by χ0 (bg k ) := χ(b)ρk . We have to show that this is well-defined, i.e., if b1 g k1 = b2 g k2 for certain b1 , b2 ∈ B, k1 , k2 ∈ Z, then χ(b1 )ρk1 = χ(b2 )ρk2 . Notice that g k1 −k2 = b2 b−1 ∈ B. By the 1 division with remainder algorithm we have k1 −k2 = tr+m with t, m ∈ Z, 0 6 m < r. k1 −k2 Then g m ∈ B, and so m = 0. Hence k1 − k2 = tr, implying b2 b−1 = ht . As 1 = g a consequence, χ(b1 )ρk1 = χ(b1 )χ(h)t ρk2 = χ(b1 ht )ρk2 = χ(b2 )ρk2 . This shows that χ0 is a well-defined map from B 0 to C∗ . It is clearly a character extending χ. Thus, we have extended χ to a character on a subgroup of A strictly larger than B. By repeating this process, after finitely many steps we obtain a character on A that extends χ. Lemma 4.2. Let A be a finite abelian group, and g ∈ A with g 6= 1. Then there is a character χ on A with χ(g) 6= 1. Proof. Assume g has order r. Let ρ ∈ C be a primitive r-th root of unity. It is easy to check that the map from hgi to C∗ given by g k 7→ ρk for k ∈ Z is a welldefined character on hgi. By extending this to A, we obtain a character χ on A with χ(g) 6= 1. Theorem 4.3. Let A be a finite abelian group. b = |A|. (i) |A| (ii) For any two characters χ1 , χ2 on A we have  X |A| if χ1 = χ2 , χ1 (a)χ2 (a) = 0 if χ1 = 6 χ2 . a∈A (iii) For any two elements a, b of A we have  X |A| χ(a)χ(b) = 0 b χ∈A

74

if a = b, if a = 6 b.

Remark. The identities (ii),(iii) are called the orthogonality relations for characters. −1 b and put S := P Proof. (ii). Let χ1 , χ2 ∈ A a∈A χ1 (a)χ2 (a). Let χ := χ1 χ2 = χ1 χ2 . P (A) Then S = a∈A χ(a). Clearly, if χ1 = χ2 then χ = χ0 , hence S = |A|. Let (A) χ1 6= χ2 . Then χ 6= χ0 , hence there is g ∈ A with χ(g) 6= 1. Further, X χ(g)S = χ(ga) = S, a∈A

since ga runs through the elements of A. Hence S = 0. b instead of |A|. Let a, b ∈ A and put T := (iii). We prove (iii) but with |A| P −1 b χ(c). If a = b, then c = 1, hence b χ(a)χ(b). Let c := ab . Then T = χ∈A χ∈A b b with T = |A|. Let a 6= b. Then c 6= 1, and so by Lemma 4.2 there is χ1 ∈ A χ1 (c) 6= 1. Hence X χ1 (c)T = (χ1 χ)(c) = T, P

b χ∈A

implying T = 0. (i). Define the standard inner product of two vectors x = (x1 , . . . , xn ), y = P (y1 , . . . , yn ) ∈ Cn by hx, yi := ni=1 xi yi . By (ii), the vectors vχ := (χ(a) : a ∈ A) b form an orthogonal system in C|A| with respect to the standard inner (χ ∈ A) product, hence they are linearly independent. A set of linearly independent vectors b 6 |A|. In a similar way we in C|A| cannot contain more than |A| elements, so |A| b deduce from (iii) that |A| 6 |A|. b b denote the character group of A. b Each element For a finite abelian group A, let A b given by b a ∈ A gives rise to a character b a on A, a(χ) := χ(a). Theorem 4.4 (Duality). Let A be a finite abelian group. Then the map a 7→ b a b b defines an isomorphism from A to A. b b Proof. The map ϕ : a 7→ b a obviously defines a group homomorphism from A to A. b We show that it is injective. Let a ∈ Ker(ϕ); then b a(χ) = 1 for all χ ∈ A, i.e., b χ(a) = 1 for all χ ∈ A, which by Lemma 4.2 implies that a = 1. So indeed, ϕ is b b = |A| b = |A|. injective. But then ϕ is surjective as well, since by Theorem 4.3 (i), |A| Hence ϕ is an isomorphism. 75

Below, we show that there is also an isomorphism from a finite abelian group A b but unlike the isomorphism in Theorem 4.4 this is not canonical, since it will to A, depend on a choice of generators for A. b is also a cyclic group of Lemma 4.5. Let A be a cyclic group of order n. Then A order n. Proof. Let A = hgi. Then A = {1, g, · · · g n−1 } and g n = 1. A character χ on A is determined by χ(g). Let ρ1 be a primitive n-th root of unity. It is easy to see that (A) there is a character χ1 on A with χ1 (g) = ρ1 , that χ0 , χ1 , . . . , χn−1 are distinct, 1 (A) n n and χ1 = χ0 . Further, if χ is any character on A, then χ(g) = 1, which implies b = hχ1 i is a cyclic group of order n. that χ is a power of χ1 . So A Lemma 4.6. Let A = A1 × · · · × Ar be the direct product of finite abelian groups b is isomorphic to A c1 × · · · × A cr . A1 , . . . , Ar . Then A Proof. It suffices to prove this for r = 2; then the proof of the lemma can be completed by induction on r. Denote by 1 the unit element of A. Let A = A1 ×A2 = {g1 g2 : g1 ∈ A1 , g2 ∈ A2 } where g1 g2 = 1 if and only if g1 = g2 = 1. Define a map c1 × A c2 → A b : (χ1 , χ2 ) 7→ χ1 χ2 , ϕ: A where χ1 χ2 (g1 g2 ) := χ1 (g1 )χ2 (g2 ) for g1 ∈ A1 , g2 ∈ A2 . It is easy to see that ϕ is a group homomorphism. Substituting g1 = 1, respectively g2 = 1, we see that χ2 , χ1 c1 × A c2 and A b have are uniquely determined by χ1 χ2 . Hence ϕ is injective. Since A the same cardinality, it follows also that ϕ is surjective. Proposition 4.7. Every finite abelian group is a direct product of cyclic groups. Proof. See S. Lang, Algebra, Chap.1, §10. Theorem 4.8. Let A be a finite abelian group. Then there exists an isomorphism b from A to A. Proof. By Proposition 4.7, A is a direct product C1 × · · · × Cr of finite cyclic groups. b is isomorphic to C c1 × · · · × C cr , where Cbi is a cyclic group By Lemmas 4.6, 4.5, A b can be of the same order as Ci , for i = 1, . . . , r. Now the isomorphism from A to A established by mapping a generator of Ci to one of Cbi , for i = 1, . . . , r. Remark. The isomorphism constructed above depends on choices of generators of Ci , Cbi , for i = 1, . . . , r. So it is not canonical. 76

4.2

Dirichlet characters

Let q ∈ Z>2 . Denote the residue class of a mod q by a. Recall that the prime residue classes mod q, (Z/qZ)∗ = {a : gcd(a, q) = 1} form a group of order ϕ(q) under multiplication of residue classes. We can lift any character χ e on (Z/qZ)∗ to a map χ : Z → C by setting  χ e(a) if gcd(a, q) = 1; χ(a) := 0 if gcd(a, q) > 1. Notice that χ has the following properties: (i) χ(1) = 1; (ii) χ(ab) = χ(a)χ(b) for a, b ∈ Z; (iii) χ(a) = χ(b) if a ≡ b (mod q); (iv) χ(a) = 0 if gcd(a, q) > 1. Any map χ : Z → C with properties (i)–(iv) is called a (Dirichlet) character modulo q. Conversely, from a character χ mod q one easily obtains a character χ e on (Z/qZ)∗ by setting χ e(a) := χ(a) for a ∈ Z with gcd(a, q) = 1. Let G(q) be the set of characters modulo q. We define the product χ1 χ2 of χ1 , χ2 ∈ G(q) by (χ1 χ2 )(a) = χ1 (a)χ2 (a) for a ∈ Z. With this operation, G(q) becomes a group, with unit element the principal character modulo q given by  1 if gcd(a, q) = 1; (q) χ0 (a) = 0 if gcd(a, q) > 1. The inverse of χ ∈ G(q) is its complex conjugate χ : a 7→ χ(a). It is clear, that this makes G(q) into a group that is isomorphic to the character group of (Z/qZ)∗ . One of the advantages of viewing characters as maps from Z to C is that this allows to multiply characters of different moduli: if χ1 is a character mod q1 and χ2 a character mod q2 , then their product χ1 χ2 is a character mod lcm(q1 , q2 ). We can easily translate the orthogonality relations for characters of (Z/qZ)∗ into orthogonality relations for Dirichlet characters modulo q. Recall that a complete 77

residue system modulo q is a set, consisting of precisely one integer from every residue class modulo q, e.g., {3, 5, 11, 22, 104} is a complete residue system modulo 5. Theorem 4.9. Let q ∈ Z>2 , and let Sq be a complete residue system modulo q. (i) Let χ1 , χ2 ∈ G(q). Then X

 χ1 (a)χ2 (a) =

a∈Sq

ϕ(q) if χ1 = χ2 ; 0 if χ1 6= χ2 .

(ii) Let a, b ∈ Z. Then   ϕ(q) if gcd(ab, q) = 1, a ≡ b (mod q); χ(a)χ(b) = 0 if gcd(ab, q) = 1, a ≡ 6 b (mod q);  χ∈G(q) 0 if gcd(ab, q) > 1. X

Proof. Exercise. Let χ be a character mod q and d a positive divisor of q. We say that q is induced by a character χ0 mod d if χ(a) = χ0 (a) for every a ∈ Z (1) with gcd(a, q) = 1. Here we define the principal character mod 1 by χ0 (a) = 1 (1) (q) for a ∈ Z. For instance, χ0 is induced by χ0 . Notice that if gcd(a, d) = 1 and gcd(a, q) > 1, then χ0 (a) 6= 0 but χ(a) = 0. The character χ is called primitive if there is no divisor d < q of q such that χ is induced by a character mod d. Theorem 4.10. Let q ∈ Z>2 and χ a character mod q. Then there are a unique divisor f of q, and a unique primitive character χ0 mod f , such that χ is induced by χ0 . The integer f from Theorem 4.10 is called the conductor of χ. To prove this, we need some lemmas. Lemma 4.11. Let a be an integer with gcd(a, d) = 1. Then there is b ∈ Z with a ≡ b (mod d), gcd(b, q) = 1. 78

Proof. Write q = q1 q2 , where q1 is composed of the primes occurring in the factorization of d, and where q2 is composed of primes not dividing d. By the Chinese Remainder Theorem, there is b ∈ Z with b ≡ a (mod d),

b ≡ 1 (mod q2 ).

This integer b is coprime with d, hence with q1 , and also coprime with q2 , so it is coprime with q. Lemma 4.12. Let d be a divisor of q. Then there is at most one character mod d that induces χ. Proof. Suppose that χ is induced by the character χ1 mod d. Let a ∈ Z with gcd(a, d) = 1 and choose b with a ≡ b (mod d) and gcd(b, q) = 1. Then χ1 (a) = χ1 (b) = χ(b). Hence χ1 is uniquely determined by χ. The next lemma gives a method to verify if a character χ is induced by a character mod d. Lemma 4.13. Let χ be a character mod q, and d a divisor of q. Then the following assertions are equivalent: (i) χ is induced by a character mod d; (ii) χ(a) = χ(b) for all a, b ∈ Z with a ≡ b (mod d) and gcd(ab, q) = 1; (iii) χ(a) = 1 for all a ∈ Z with a ≡ 1 (mod d) and gcd(a, q) = 1. Proof. The implications (i)⇒(ii)⇒(iii) are trivial. (iii) ⇒ (ii). Let a, b ∈ Z with a ≡ b (mod d) and gcd(ab, q) = 1. There is c ∈ Z with gcd(c, q) = 1 such that a ≡ bc (mod q). For this c we have c ≡ 1 (mod d). Now by (iii) we have χ(a) = χ(b)χ(c) = χ(b). (ii) ⇒ (i). We define a character χ0 mod d as follows. For a ∈ Z with gcd(a, d) > 1 put χ0 (a) := 0. For a ∈ Z with gcd(a, d) = 1, choose b ∈ Z such that a ≡ b (mod d) and gcd(b, q) = 1 (which is possible by Lemma 4.11), and put χ0 (a) := χ(b). By (ii) this gives a well-defined character mod d that clearly induces χ. Lemma 4.14. Let χ be a character mod q. Assume that χ is induced by characters χ1 mod d1 , χ2 mod d2 , where d1 , d2 are divisors of q. Then χ is induced by a character mod gcd(d1 , d2 ) which in turn induces χ1 , χ2 . 79

Proof. Let d = gcd(d1 , d2 ), d0 := lcm(d1 , d2 ). We first show that χ1 is induced by a character mod d. We apply criterion (iii) of the previous lemma. That is, we have to show that if a is an integer with gcd(a, d1 ) = 1 and a ≡ 1 (mod d), then χ1 (a) = 1. Take such a. Then a = 1 + td with t ∈ Z. There are x, y ∈ Z with xd1 + yd2 = d. Hence a = 1 + txd1 + tyd2 . The number c := 1 + tyd2 is coprime with d1 since a is coprime with d1 , and also coprime with d2 , hence it is coprime with d0 . By Lemma 4.11, there is b with b ≡ c (mod d0 ) and gcd(b, q) = 1. We have b ≡ a (mod d1 ), b ≡ 1 (mod d2 ), hence χ1 (a) = χ(b) = χ2 (1) = 1. It follows that χ1 is induced by a character, say χ3 mod d. Similarly, χ2 is induced by a character χ03 mod d. Both χ3 , χ03 induce χ. So by Lemma 4.12, χ3 = χ03 . Proof of Theorem 4.10. Let f be the smallest divisor of q such that χ is induced by a character mod f . This character, say χ0 , is necessarily primitive. Assume there is another primitive character χ00 mod f 0 that induces χ. By the previous lemma, χ is induced by a character χ000 mod gcd(f, f 0 ) that in turn induces χ0 and χ00 . But this is possible only if f = f 0 . By Lemma 4.12 it follows that also χ0 = χ00 .

4.3

Computation of G(q)

We give a method to compute the character group modulo q. We first make a reduction to prime powers. Theorem 4.15. Let q = pk11 · · · qtkt , where p1 , . . . , pt are distinct primes and k1 , . . . , kt positive integers. Then the map G(pk11 ) × · · · × G(pkt t ) → G(q) : (χ1 , . . . , χt ) 7→ χ1 · · · χt is a group isomorphism. Proof. Let f denote the map under consideration. Then f is a homomorphism. We (q) show that it is injective. Let χi ∈ G(pki i ) (i = 1, . . . , t) be such that χ1 · · · χt = χ0 . Let i ∈ {1, . . . , t} and choose a ∈ Z with gcd(a, pi ) = 1. By the Chinese Remainder Theorem, there is b ∈ Z such that b ≡ a (mod pki i ),

k

b ≡ 1 (mod pj j ) for j 6= i. 80

Then with this b we have χi (a) =

t Y

(q)

χj (b) = χ0 (b) = 1.

j=1 k

(p i )

Hence χi = χ0 i . This holds for i = 1, . . . , t, so f is injective. Now since G(pk11 ) × · · · × G(pkt t ) and G(q) have the same order ϕ(q), the map f is also surjective. To compute G(pk ) for a prime power pk , we need some information about the structure of (Z/pk Z)∗ . This is provided by the following theorem. Theorem 4.16. (i) Let p be a prime > 3. Then the group (Z/pk Z)∗ is cyclic of order pk−1 (p − 1). (ii) (Z/4Z)∗ is cyclic of order 2. Further, if k > 3 then (Z/2k Z)∗ =< −1 > × < 5 > is the direct product of a cyclic group of order 2 and a cyclic group of order 2k−2 . We skip the proof of k = 1 of (i), which belongs to a basic algebra course. For the proof of the remaining parts, we need a lemma. For a prime number p, and for a ∈ Z \ {0}, we denote by ordp (a) the largest integer k such that pk divides a. Lemma 4.17. Let p be a prime number and a an integer such that ordp (a − 1) > 1 if p > 3 and ordp (a − 1) > 2 if p = 2. Then k

ordp (ap − 1) = ordp (a − 1) + k. Proof. We prove the assertion only for k = 1; then the general statement follows easily by induction on k. Our assumption on a implies that a = 1 + pt b, where t > 1 if p > 3, t > 2 if p = 2, and where b is an integer not divisible by p. Now by the binomial formula, p   X p p a −1= (pt b)j = pt+1 bj + pt+2 (· · · ). j j=1
Here we have used that all binomial coefficients pj are divisible by p except the last. But the last term (pt b)p is divisible by ppt , and the exponent pt is larger than 81

t + 1 (the assumption t > 2 for p = 2 is needed to ensure this). This shows that ordp (ap − 1) = t + 1. Proof of Theorem 4.16. (i). We take for granted that (Z/pZ)∗ is cyclic of order p−1, and assume that k > 2. We construct a generator for (Z/pk Z)∗ . Let g be an integer such that g (mod p) is a generator of (Z/pZ)∗ . We show that we can choose g such that ordp (g p−1 − 1) = 1. Indeed, assume that ordp (g p−1 − 1) > 2 and take g + p. Then  p−1  X p − 1 p−1−j j p−1 g p = (p − 1)g p−2 p + p2 (· · · ) (g + p) −1 = j j=1 = −g p−2 p + p2 (· · · ) hence ordp ((g + p)p−1 − 1) = 1. So, replacing g by g + p if need be, we get an integer g such that g (mod p) generates (Z/pZ)∗ and ordp (g p−1 − 1) = 1. We show that g := g (mod pk ) generates (Z/pk Z)∗ . Let n be the order of g in (Z/pk Z)∗ ; that is, n is the smallest positive integer with g n ≡ 1 (mod pk ). On the one hand, g n ≡ 1 (mod p), hence p − 1 divides n. On the other hand, n divides the order of (Z/pk Z)∗ , that is, pk−1 (p − 1). So n = ps (p − 1) with s 6 k − 1. By Lemma 4.17 we have ordp (g n − 1) = ordp (g p−1 − 1) + s = s + 1. This has to be equal to k, so s = k − 1. Hence n = pk−1 (p − 1) is equal to the order of (Z/pk Z)∗ . It follows that (Z/pk Z)∗ = hgi. (ii). Assume that k > 3. Define the subgroup of index 2, H := {a ∈ (Z/2k Z)∗ : a ≡ 1 (mod 4)}. Then (Z/2k Z)∗ = H ∪ (−H) = {(−1)k a : k ∈ {0, 1}, a ∈ H} and (−1)k a = 1 if and only if k = 0 and a = 1. Hence (Z/2k Z)∗ =< −1 > ×H. Similarly as above, one shows that H is cyclic of order 2k−2 , and that H = h5i. Corollary 4.18. Let p be a prime and k > 1. (i) If p = 2, k = 1, 2 or p > 2 then G(pk ) is cyclic of order pk−1 (p − 1). (ii) If p = 2, k > 3, then G(pk ) is the direct product of a cyclic group of order 2 and a cyclic group of order 2k−2 . 82

Proof. Immediate consequence of Theorem 4.16 and Lemmas 4.5 and 4.6. Following the proofs of Lemmas 4.5, 4.6, we can give an explicit description for the groups G(pk ). (2)

(4)

Clearly, G(2) = {χ0 } and G(4) = {χ0 , χ4 }, where χ4 (a) = 1 if a ≡ 1 (mod 4), χ4 (a) = −1 if a ≡ 3 (mod 4), χ4 (a) = 0 if a is even. If p > 2, choose g ∈ Z such that g (mod pk ) generates (Z/pk Z)∗ , and choose a primitive pk−1 (p−1)-th root of unity ρ. Then G(pk ) = hχ1 i where χ1 is the Dirichlet character determined by χ1 (g) = ρ. As for 2k with k > 3, choose a primitive 2k−2 -th root of unity ρ. Then G(2k ) = hχ1 i × hχ2 i, where χ1 , χ2 are given by χ1 (−1) = −1, χ1 (5) = 1;

4.4

χ2 (−1) = 1, χ2 (5) = ρ.

Gauss sums

Let q ∈ Z>2 . For a character χ mod q and for b ∈ Z, we define the Gauss sum X χ(x)e2πibx/q , τ (b, χ) := x∈Sq

where Sq is a full system of representatives modulo q. This does not depend on the choice of Sq . The Gauss sum τ (1, χ) occurs for instance in the functional equation P −s (later). for the L-function L(s, χ) = ∞ n=1 χ(n)n We prove some basic properties of Gauss sums. Theorem 4.19. Let q ∈ Z>2 and let χ be a character mod q. Further, let b ∈ Z. (i) If gcd(b, q) = 1, then τ (b, χ) = χ(b) · τ (1, χ). (ii) If gcd(b, q) > 1 and χ is primitive, then τ (b, χ) = χ(b) · τ (1, χ) = 0. Proof. (i) Suppose gcd(b, q) = 1. If x runs through a complete residue system Sq mod q, then bx runs to another complete residue system Sq0 mod q. Write y = bx. Then χ(y) = χ(b)χ(x), hence χ(x) = χ(b)χ(y). Therefore, X X τ (b, χ) = χ(x)e2πibx/q = χ(b)χ(y)e2πiy/q y∈Sq0

x∈Sq

= χ(b)τ (1, χ). 83

(ii) We use the following observation: if q1 is any divisor of q with 1 6 q1 6 q, then there is c ∈ Z such that c ≡ 1 (mod q1 ), gcd(c, q) = 1, and χ(c) 6= 1. Indeed, this is obvious if q1 = q. If q1 < q, then Lemma 4.13 implies that if there is no such integer c then χ is induced by a character mod q1 , contrary to our assumption that χ is primitive. Now let d := gcd(b, q), put b1 := b/d, q1 := q/d, and choose c according to the observation. Then X χ(c)τ (b, χ) = χ(cx)e2πibx/q . x∈Sq

If x runs through a complete residue system Sq mod q, then y := cx runs through another complete residue system Sq0 mod q. Further, since c ≡ 1 (mod q1 ) we have e2πixb/q = e2πixb1 /q1 = e2πicxb1 /q1 = e2πiyb/q . Hence χ(c)τ (b, χ) =

X

χ(y)e2πiby/q = τ (b, χ).

y∈Sq

Since χ(c) 6= 1 this implies that τ (b, χ) = 0. Theorem 4.20. Let q ∈ Z>2 and let χ be a primitive character mod q. Then |τ (1, χ)| =

√ q.

Proof. We have by Theorem 4.19, |τ (1, χ)|2 = τ (1, χ) · τ (1, χ) =

q−1 X

χ(x)e−2πix/q τ (1, χ)

x=0

=

q−1 X

−2πix/q

e

τ (x, χ) =

x=0

=

=

q−1 X y=0

−2πix/q

e

x=0

q−1 q−1 X X x=0

q−1 X

q−1 X

! 2πixy/q

χ(y)e

y=0

! χ(y)e2πix(y−1)/q

y=0

χ(y)

q−1 X

! 2πix(y−1)/q

e

x=0

=

q−1 X y=0

84

χ(y)S(y), say.

If y = 1, then S(y) =

Pq−1

x=0

1 = q, while if y 6= 1, then S(y) =

e2πi(y−1) − 1 = 0. e2πi(y−1)/q − 1

Hence |τ (1, χ)|2 = χ(1)q = q. For later purposes we need the following variation on this result. A real character mod q is one which assumes only real values. This implies that χ(a) ∈ {±1} if gcd(a, q) = 1. Theorem 4.21. Let χ be a primitive real character mod q. Then τ (1, χ)2 = χ(−1)q. Proof. Similarly as in the proof of Theorem 4.20 we have 2

τ (1, χ) =

q−1 X

χ(x)e2πix/q τ (1, χ)

x=0

and by following the same reasoning, 2

τ (1, χ) =

q−1 X y=0

χ(y)

q−1 X

! e

2πix(y+1)/q

x=0

=

q−1 X

χ(y)T (y),

y=0

say. As is easily seen, T (q − 1) = q, while T (y) = 0 for y = 0, . . . , q − 2. This implies Theorem 4.21. √ Remark. Theorem 4.20 implies that εχ := τ (1, χ)/ q lies on the unit circle. Gauss gave an explicit expression for εχ in the case that εχ is a primitive real character mod q. There is no general method known to compute εχ for non-real characters χ modulo large values of q.

4.5

Quadratic reciprocity

For the interested reader, we give a proof of Gauss’ Quadratic Reciprocity Theorem using Gauss sums. This section requires a little bit more algebraic background. Let p > 2 be a prime number. An integer a is called a quadratic residue modulo p if x2 ≡ a (mod p) is solvable in x ∈ Z and p - a, and a quadratic non-residue modulo 85

p if x2 ≡ a (mod p) is not solvable in x ∈ Z. Further, a quadratic (non-)residue class modulo p is a residue class modulo p represented by a quadratic (non-)residue. We define the Legendre   1 a := −1  p 0

symbol if a is a quadratic residue modulo p; if a is a quadratic non-residue modulo p; if p|a.

Lemma   4.22. Let p be a prime > 2. (i)

· p

is a primitive character mod p.

(ii) There are precisely 21 (p − 1) quadratic residue classes, and precisely 12 (p − 1) quadratic  a  non-residue classes modulo p. (iii) ≡ a(p−1)/2 (mod p) for a ∈ Z. p Proof. (i) The group (Z/pZ)∗ is cyclic of order p − 1. Let g(mod p) be a generator of this group. Take a ∈ Z with gcd(a, p) = 1. Then there is t ∈ Z such that 2 a ≡ g t (mod p). in x ∈ Z if and only if t is  Now clearly, x ≡ a (mod p) issolvable  even. Hence

a p

= (−1)t . This shows that

· p

is a character mod p.

(ii) The group (Z/pZ)∗ consists of g t (mod p) (t = 0, . . . , p − 1). Clearly, the quadratic residue classes are those with t even, and the quadratic   non-residue classes those with t odd. This implies (ii). This shows also that

· p

is not the principal

character mod p, and so, since p is a prime, it must be primitive. (iii) The assertion is clearly true if p|a. Assume that p - a. Then there is t ∈ Z with a ≡ g t (mod p). Note that (g (p−1)/2 )2 ≡ 1 (mod p), hence g (p−1)/2 ≡ ±1 (mod p). But g (p−1)/2 6≡ 1 (mod p) since g (mod p) is a generator of (Z/pZ)∗ . Hence g (p−1)/2 ≡ −1 (mod p). As a consequence, a (p−1)/2 t a ≡ (−1) ≡ (mod p). p

The following is immediate: Corollary 4.23. Let p be a prime > 2. Then   −1  1 if p ≡ 1 (mod 4), (p−1)/2 = (−1) = −1 if p ≡ 3 (mod 4). p 86

Gauss’ Quadratic Reciprocity Theorem is as follows: Theorem 4.24. Let p, q be distinct primes > 2. Then   p  q  −1 if p ≡ q ≡ 3 (mod 4), (p−1)(q−1)/4 = (−1) = 1 otherwise. q p Furthermore, as a supplement we have: Theorem 4.25. Let p be a prime > 2. Then  2 1 if p ≡ ±1 (mod 8), (p2 −1)/8 = (−1) = −1 if p ≡ ±3 (mod 8). p Example. Check if x2 ≡ 33 (mod 97) is solvable.  33  97

 3   11   97   97  · = · 97 97 3 11  1   −2   1   −1   2  · = · · = 1 · (−1) · (−1) = 1. = 3 11 3 11 11 =

We prove only Theorem 4.24 and leave Theorem 4.25 as an exercise. We first make some preparations and then prove some lemmas. Let Q[X] denote the ring of polynomials with coefficients in Q. A number α ∈ C is called algebraic if there is a non-zero polynomial f ∈ Q[X] such that f (α) = 0. Among all non-zero polynomials from Q[X] having α as a zero, we choose one of minimal degree. By multiplying such a polynomial with a suitable constant, we obtain one which is monic, i.e., of which the coefficient of the highest power of X is 1. There is only one monic polynomial in Q[X] of minimal degree having α as a zero, for if there were two, their difference would give a non-zero polynomial in Q[X] of smaller degree having α as a zero. This unique monic polynomial in Q[X] of minimal degree having α as a zero is called the minimal polynomial of α, denoted by fα . We observe that fα must be irreducible in Q[X], that is, not a product of two non-constant polynomials from Q[X]. For otherwise, α would be a zero of one of these polynomials, which has degree smaller than that of fα . 87

Let q be a prime number > 2. We write ζq := e2πi/q . Define Rq := Z[ζq ] =

( r X

) ai ζqi : ai ∈ Z, r > 0 .

i=0

This set is closed under addition and multiplication, hence it is a ring. Lemma 4.26. Rq ∩ Q = Z. Proof. We use without proof, that the minimal polynomial of ζq is (X q −1)/(X−1) = P j X q−1 + · · · + X + 1. Hence ζqq−1 = − q−2 j=0 ζq . By repeatedly substituting this into Pr P j an expression i=0 ai ζqi with ai ∈ Z, we eventually get an expression q−2 j=0 bj ζq with bj ∈ Z for all j. Hence all elements of Rq can be expressed in this form. Now if α ∈ Rq ∩ Q, we get q−2 X α= bj ζqj j=0

with α ∈ Q and bj ∈ Z for all j. This implies that ζq is a zero of the polynomial bq−2 X q−2 + · · · + b0 − α. Since the minimal polynomial of ζq has degree q − 1, this is possible only if b0 = α and b1 = · · · = bq−2 = 0. Hence α ∈ Z. Given α, β ∈ Rq and n ∈ Z>0 , we write α ≡ β (mod n) in Rq if (α − β)/n ∈ Rq . Further, we write α ≡ β (mod n) in Z if (α − β)/n ∈ Z. By the Lemma we just proved, for α, β ∈ Z we have that α ≡ β (mod n) in Rq if and only if α ≡ β (mod n) in Z. Lemma 4.27. Let p be any prime number. Then for α1 , . . . , αr ∈ Rq we have (α1 + · · · + αr )p ≡ α1p + · · · + αrp (mod p) in Rq . Proof. By the multinomial theorem, (α1 + · · · + αr )p =

X i1 +···+ir

p! α1i1 · · · αrir . i ! · · · i ! r =p 1

All multinomial coefficients are divisible by p, except those where one index ij = p and the others are 0. 88

Proof of 4.24. We use Gauss sums. For notational convenience we write   Theorem · χq for q . We work in the ring Rq . Notice that by Theorem 4.21, and Corollary 4.23, τ (1, χq )2 = χq (−1)q = (−1)(q−1)/2 q.

(4.1)

Further, by Lemma 4.27 and Theorem 4.19, q−1 X

τ (1, χq )p ≡

χq (x)p ζqpx ≡

q−1 X

χq (x)ζqpx ≡ τ (p, χq ) ≡

x=0

x=0

p q

τ (1, χq ) (mod p) in Rq .

On multiplying with τ (1, χq ) and applying (4.1), we obtain p τ (1, χq )p+1 ≡ (−1)(q−1)/2 q · (mod p) in Rq . q On the other hand, by (4.1) and Lemma 4.26, τ (1, χq )p+1 = (−1)(q−1)(p+1)/4 q (p+1)/2 = (−1)(q−1)/2 q · (−1)(q−1)(p−1)/4 q (p−1)/2 q  ≡ (−1)(q−1)/2 q · (−1)(p−1)(q−1)/4 (mod p) in Rq . p As a consequence, (q−1)/2

(−1)

q·

p q

(q−1)/2

≡ (−1)

(q−1)(p−1)/4

q · (−1)

q  p

(mod p) in Z.

Since q is coprime with p, this gives p   (p−1)(q−1)/4 q ≡ (−1) (mod p) in Z. q p Since integers equal to ±1 can be congruent modulo p only if they are equal, this implies Theorem 4.24. Exercise 4.1. Prove Theorem 4.25. Hint. You have to follow the proof of Theorem 4.24, but instead of Rq , χq , you have to use the ring R8 = Z[ζ8 ] where ζ8 = e2πi/8 , and the character χ8 mod 8, given by   1 if a ≡ ±1 (mod 8), χ8 (a) = −1 if a ≡ ±3 (mod 8),  0 if a ≡ 0 (mod 2). Use that ζ8 has minimal polynomial X 4 + 1. 89

Chapter 5 The Riemann zeta function and L-functions 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 ∞ X L(s, χ) = χ(n)n−s . n=1

P∞

−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 ∈ Z. We first prove that ζ(s) has an analytic continuation to {s ∈ C : Re s > 0} \ {1}. 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 f (n) =

b

Z f (x)dx + f (a) +

a

n=a

b

(x − [x])f 0 (x)dx.

a

Remark. This result often occurs in the more symmetric form b X n=a

Z f (n) =

b

f (x)dx + a

1 (f (a) 2

Z + f (b)) + a

91

b

(x − [x] − 21 )f 0 (x)dx.

Proof. Let n ∈ {a, a + 1, . . . , b − 1}. Then Z n+1 Z n+1
0 x − [x] f (x)dx = (x − n)f 0 (x)dx n

n n+1

h in+1 Z = (x − n)f (x) − n

Z

n+1

f (x)dx = f (n + 1) −

n

f (x)dx. n

By summing over n we get Z

b

(x − [x])f 0 (x)dx =

a

b X

Z f (n) −

b

f (x)dx, a

n=a+1

which implies at once Lemma 5.1. Theorem 5.2. ζ(s) has a unique analytic continuation to the set {s ∈ C : Re s > 0, s 6= 1}, with a simple pole with residue 1 at s = 1. Proof. By Corollary 2.21 we know that an analytic continuation of ζ(s), if such exists, is unique. For the moment, let s ∈ C with Re s > 1. Then by Lemma 5.1, with f (x) = x−s , N X

n

−s

Z

N

1

1

=

(x − [x])(−sx−1−s )dx

x dx + 1 +

=

n=1

N

Z

−s

1 − N 1−s +1−s s−1

Z

N

(x − [x])x−1−s dx.

1

If we let N → ∞ then the left-hand side converges, and also the first term on the right-hand side, since |N −1−s | = N −1−Re s → 0. Hence the integral on the right-hand side must converge as well. Thus, letting N → ∞, we get for Re s > 1, Z ∞ 1 (5.1) ζ(s) = +1−s (x − [x])x−1−s dx. s−1 1 We now show that the integral on the right-hand side defines an analytic function on U := {s ∈ C : Re s > 0}, by means of Theorem 2.23. The function F (x, s) := (x − [x])x−1−s is measurable on [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. 92

Let K be a compact subset of U . Then there is σ > 0 such that Re s > σ for all s ∈ K. Now for x > 1 and s ∈ K we have |(x − [x])x−1−s | 6 x−1−σ R∞ and 1 x−1−σ dx < ∞. Hence all conditions of Theorem 2.23 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 {s ∈ C : Re s > 0, s 6= 1} 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 ∈ Z>2 , and let χ be a Dirichlet character mod q. Q (i) L(s, χ) = p (1 − χ(p)p−s )−1 for s ∈ C, Re s > 1. (q)

(ii) If χ 6= χ0 , then L(s, χ) converges, and is analytic on {s ∈ C : Re s > 0}. (q) (iii) L(s, χ0 ) can be continued to an analytic function on {s ∈ C : Re s > 0, s 6= 1}, and for s in this set we have Y (q) L(s, χ0 ) = ζ(s) · (1 − p−s ). p|q (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 P P Theorem 4.9), we have qm=1 χ(m) = 0, 2q m=q+1 χ(m) = 0, etc. Hence N X χ(n) = χ(tq + 1) + · · · + χ(tq + r) 6 r < q. n=1 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, Y Y (q) L(s, χ0 ) = (1 − p−s )−1 = ζ(s) (1 − p−s ). p-q

p|q

93

The right-hand side is defined and analytic on {s ∈ C : Re s > 0, s 6= 1}, 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 {s ∈ C : Re s > 1}. Proof. Use part (i) of the above theorem, together with Corollary 3.14. The remainder of this section is dedicated to the proof that ζ(s) 6= 0 if Re s = 1 and s 6= 1, and L(s, χ) 6= 0 for any s ∈ C with Re s = 1 and any non-principal character χ modulo an integer q > 2. We have to distinguish two cases, which are (1) treated quite differently. We interpret ζ(s) as L(s, χ0 ). Theorem 5.5. Let q ∈ Z>1 , χ a character mod q, and t a real. Assume that either (q) t 6= 0, or t = 0 but χ2 6= χ0 . Then L(1 + it, χ) 6= 0. Proof. We use a famous idea, due to Hadamard. It is based on the inequality 3 + 4 cos θ + cos 2θ = 2(1 + cos θ)2 > 0 for θ ∈ R.

(5.2)

Suppose that L(1 + it, χ) = 0. Consider the function (q)

F (s) := L(s, χ0 )3 · L(s + it, χ)4 · L(s + 2it, χ2 ). 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)
ords=1 (F ) = 3 · ords=1 L(s, χ0 ) + 4 · ords=1 L(s + it, χ) + ords=1 L(s + 2it, χ2 ) > −3 + 4 = 1. This shows that F is analytic around s = 1, and has a zero at s = 1. We now prove that |F (σ)| > 1 (or rather, log |F (σ)| > 0). This gives a contradiction since by continuity, lims↓1 |F (σ)| should be 0. So our assumption that L(1 + it, χ) = 0 must be false. From the definition of the function F we obtain   3 4 Y 1 1 1 ·   log |F (σ)| = log · (q) −σ−it 2 p−σ−2it −σ 1 − χ(p)p 1 − χ(p) 1 − χ (p)p 0 p  X 1 1 1 + 4 log + log = 3 log 1 − χ(p)p−σ−it 1 − χ(p)2 p−σ−2it . 1 − p−σ p-q

94

Note that if p - q then χ(p) is a root of unity. Hence |χ(p)p−it | = |χ(p)e−it log p | = 1. So we have χ(p)p−it = eiϕp with ϕp ∈ R. Hence  X 1 1 1 . log |F (σ)| = 3 log + 4 log + log −σ −σ iϕ −σ 2iϕ 1−p 1−p e p 1−p e p p-q

Recall that ∞

X 1 log = z n /n, 1−z n=1

1 = Re log 1 log 1 − z 1−z

Hence for r, ϕ ∈ R with 0 < r < 1,   1 1 = Re log log = Re 1 − reiϕ 1 − reiϕ =

∞ X rn n=1

n

inϕ

Re (e

)=

∞ X rn n=1

n

for z ∈ C with |z| < 1.

∞ X (reiϕ )n n=1

!

n

· cos nϕ.

This leads to log |F (σ)| =

X

3

p-q

=

∞ X p−nσ n=1

∞ XX p−nσ p-q n=1

n

n

+4

∞ X p−nσ n=1

n

· cos nϕp +

∞ X p−nσ n=1

n

! cos 2nϕp

(3 + 4 cos nϕp + cos 2nϕp ) > 0,

using (5.2). This shows that indeed, |F (σ)| > 1 for σ > 1, giving us the contradiction we want. (q)

It remains to prove that L(1, χ) 6= 0 for any character χ mod q such that χ 6= χ0 , (q) 2 χ = χ0 , i.e., for any real character χ not equal to the principal character. Dirichlet needed this fact already in his proof that for every pair of integers q, a with q > 3 and gcd(a, q) = 1 there are infinitely many primes p with p ≡ a (mod q). Dirichlet had a rather complicated proof that L(1, χ) 6= 0, based on Dirichlet series associated with quadratic forms (in modern language: Dedekind zeta functions for quadratic number fields) and class number formulas. Landau found a much more direct proof, which we give here, based on a simple result for Dirichlet series, which more or less asserts that a Dirichlet series with nonnegative real coefficients can not be continued analytically beyond the boundary of its half plane of convergence. 95

Lemma 5.6 (Landau). Let f : Z>0 → R be an arithmetical function with f (n) > 0 P −s for all n. Suppose that Lf (s) = ∞ has abscissa of convergence σ0 . Then n=1 f (n)n Lf (s) cannot be continued analytically to any open set containing {s ∈ C : Re s > σ0 } ∪ {σ0 }. Proof. Suppose Lf (s) can be continued to an analytic function g(s) on an open set containing {s ∈ C : Re s > σ0 } ∪ {σ0 }. Then there is δ > 0 such that g(s) is analytic on the open disk D(σ0 , δ) with center σ0 and radius δ. Let σ1 := σ0 + δ/3. Then D(σ1 , 2δ/3) ⊂ D(σ0 , δ), so g(s) is analytic and has a Taylor series expansion around σ1 converging on D(σ1 , 2δ/3). Now let σ0 − δ/3 < σ < σ0 , so that σ ∈ D(σ1 , 2δ/3). Using the Taylor series expansion of g(s) around σ1 , we get g(σ) =

∞ X g (k) (σ1 )

k!

k=0

· (σ − σ1 )k .

Since σ1 is larger than the abscissa of convergence σ0 of Lf (s), we have g

(k)

(σ1 ) =

(k) Lf (σ1 )

=

∞ X

f (n)(− log n)k n−σ1 for k > 0.

n=1

Hence ∞ X 1 g(σ) = k! k=0

∞ X

∞ X 1 = k! k=0

∞ X

! f (n)(− log n)k n−σ1

(σ − σ1 )k

n=1

! f (n)(log n)k n−σ1

n=1

96

(σ1 − σ)k .

Now all terms are non-negative, hence it is allowed to interchange the summations. Thus, ! ∞ ∞ X X 1 g(σ) = f (n)n−σ1 (log n)k (σ1 − σ)k k! n=1 k=0 =

∞ X

−σ (log n)(σ1 −σ)

f (n)n

e

=

∞ X

n=1

f (n)n

−σ1

σ1 −σ

n

=

∞ X

f (n)n−σ .

n=1

n=1

We see that Lf (s) converges for s = σ. But this is impossible, since σ is smaller than the abscissa of convergence σ0 of Lf (s). So our initial assumption that Lf (s) has an analytic continuation to an open set containing {s ∈ C : Re s > σ0 } ∪ {σ0 } must have been false. Remark. Lemma 5.6 becomes false if we drop the condition that f (n) > 0 for all n. P −s For instance, if χ is a non-principal character mod q, then L(s, χ) = ∞ n=1 χ(n)n diverges if Re s < 0, but one can show that L(s, χ) has an analytic continuation to the whole of C. (q)

Theorem 5.7. Let q ∈ Z>2 , and let χ be a character mod q with χ 6= χ0 (q) χ2 = χ0 . Then L(1, χ) 6= 0.

and

Proof. Assume that L(1, χ) = 0. Consider the function F (s) := L(s, χ)ζ(s). By Theorems 5.2, 5.3, this function is analytic at least on {s ∈ C : Re s > 0, s 6= 1}. But the simple pole of ζ(s) at s = 1 is cancelled by the zero of L(s, χ). Hence F (s) is analytic for all s with Re s > 0. We show that for s ∈ C with Re s > 1, F (s) is expressable as a Dirichlet series with non-negative coefficients. By Lemma 5.6, this Dirichlet series should have abscissa of convergence 6 0. But we show that the abscissa of convergence of this series is > 21 and derive a contradiction. P P −s −s The series ζ(s) = ∞ and ∞ converge absolutely if Re s > 1. n=1 n n=1 χ(n)n So by Theorem 3.12, F (s) = Lf (s) =

∞ X

f (n)n−s

n=1

97

for s ∈ C, Re s > 1,

where f = E ∗ χ, i.e., f (n) =

X

χ(d) for n ∈ Z>0 .

d|n

Hence f is a multiplicative function. We compute f in the prime powers. Since (q) χ2 = χ0 , we have χ(n) = ±1 for all n ∈ Z with gcd(n, q) = 1, while χ(n) = 0 if gcd(n, q) > 1. Hence, if p is a prime and k a non-negative integer, we have   1 if p|q,   k  X k + 1 if p - q, χ(p) = 1, f (pk ) = χ(p)j =  1 if p - q, χ(p) = −1, k even,  j=0   0 if p - q, χ(p) = −1, k odd. Therefore, f (pk ) > 0 for all prime powers pk . Since f is multiplicative, it follows that f (n) > 0 for all n ∈ Z>0 . The series Lf (s) has an analytic continuation to {s ∈ C : Re s > 0}, that is, F (s). So by Lemma 5.6, Lf (s) has abscissa of convergence σ0 (f ) 6 0. On the other hand, from the above table and from the fact that f is multiplicative, it follows that if n = m2 is a square, then f (n) > 1. Hence Lf (σ) =

∞ X n=1

−σ

f (n)n

>

∞ X

m−2σ = ∞ if σ 6 21 .

m=1

So σ0 (f ) > 12 . This gives a contradiction, and so our assumption that L(1, χ) = 0 has to be false.

98

Chapter 6 Tauberian theorems 6.1

Introduction

P n In 1826, Abel proved the following result for real power series. Let f (x) = ∞ n=0 an x be a power series with coefficients an ∈ R that converges on the real interval (−1, 1). P Assume that ∞ n=0 an converges. Then lim f (x) = x↑1

∞ X

an .

n=0

In general, the converse is not true, i.e., if limx↑1 f (x) exists one can not conclude P P n n For instance, if f (x) = (1 + x)−1 = ∞ that ∞ n=0 an converges. n=0 (−1) x , then P∞ 1 n limx↑1 f (x) = 2 , but n=0 (−1) diverges. In 1897, Tauber proved a converse to Abel’s Theorem, but under an additional P n hypothesis. Let again f (x) = ∞ n=0 an x be a power series with real coefficients converging on (−1, 1). Assume that (6.1)

lim f (x) =: α exists, x↑1

and moreover, (6.2)

lim nan = 0.

n→∞

Then (6.3)

∞ X

an converges and is equal to α.

n=0

99

Tauber’s result led to various other “Tauberian theorems,” which are are all of the following shape: - suppose one knows something about the behaviour of f (x) as x ↑ 1 (such as (6.1)); - further suppose one knows something about the growth of an as n → ∞ (such as (6.2)); P - then one can conclude something about the convergence of ∞ n=0 an (such as (6.3)). There is now a very general “Tauberian theory,” which is about Tauberian theorems for functions defined by integrals. These include as special cases Tauberian theorems for power series and Dirichlet series. We will prove a Tauberian theorem for Laplace transforms Z ∞ F (t)e−zt dt, G(z) := 0

where F : [0, ∞) → C is a ‘decent’ function and z is a complex variable. This Tauberian theorem has the following shape. - Assume that the integral converges for Re z > 0; - assume that one knows something about the limiting behaviour of G(z) as Re z ↓ 0; - assume that one knows something about the growth order of F ; R∞ - then one can conclude something about the convergence of 0 F (t)dt. With some modifications, we may view power series as special cases of Laplace P n transforms. Let g(x) = ∞ n=0 an x be a power series converging for |x| < 1. Define the function F (t) on [0, ∞) by F (t) := an if n 6 t < n + 1 (n ∈ Z>0 ). Then if Re z > 0, Z ∞ ∞ Z X −zt F (t)e dt = 0

=

n=0 ∞ X

Hence

an

dt =

∞ X n=0

Z

n+1

an n


1 −nz e − e−(n+1)z z

∞ 1 − e−z X an e−nz . z n=0

z g(e ) = 1 − e−z −z

F (t)e

−zt

n

n=0

=

n+1

Z

∞

F (t)e−zt dt if Re z > 0.

0

100

e−zt dt

Later, we show how a Dirichlet series can be expressed in terms of a Laplace transform. Around 1930, Wiener developed a general Tauberian theory, which is now part of functional analysis. From this, in 1931, Ikehara deduced a Tauberian theorem for Dirichlet series (now known as the Wiener-Ikehara Theorem), with which one can give simple proofs of the Prime Number Theorem and various generalizations thereof. In 1980, Newman published a new method to derive Tauberian theorems, based on a clever contour integration and avoiding any functional analysis. This was developed further by Korevaar. Using the ideas of Newman and Korevaar, we prove a Tauberian theorem for Laplace transforms, and deduce from this a weaker version of the Wiener-Ikehara theorem. This weaker version suffices for a proof of the Prime Number Theorem for arithmetic progressions. Literature: J. Korevaar, Tauberian Theory, A century of developments, Springer Verlag 2004, Grundlehren der mathematischen Wissenschaften, vol. 329. J. Korevaar, The Wiener-Ikehara Theorem by complex analysis, Proceedings of the American Mathematical Society, vol. 134, no. 4 (2005), 1107–1116.

6.2

A Tauberian theorem for Laplace transforms

Lemma 6.1. Let F : [0, ∞) → C be a measurable function. Further, assume there is a constant M such that |F (t)| 6 M for t > 1. Then

∞

Z

F (t)e−zt dt

G(z) := 0

converges, and defines an analytic function on {z ∈ C : Re z > 0}. Proof. We apply Theorem 2.23. We check that the conditions of that theorem are satisfied. Let U := {z ∈ C : Re z > 0}. First, F (t)e−zt is measurable on [0, ∞) × U . Second, for every fixed t ∈ [0, ∞), the function z 7→ F (t)e−tz is analytic on U . 101

Third, let K be a compact subset of U . Then there is δ > 0 such that Re z > δ for z ∈ K, and thus, |F (t)e−zt | 6 M e−δt for z ∈ K. R∞ The integral 0 M · e−δt dt converges. So indeed, all conditions of Theorem 2.23 are satisfied and thus, by that Theorem, G(z) is analytic on U . We are now ready to state our Tauberian theorem. Theorem 6.2. Let F : [0, ∞) → C be a function with the following properties: (i) F is measurable; (ii) there is M > 0 such that |F (t)| 6 M for all t > 0; (iii) there is a function G(z), which is analytic on an open set containing {z ∈ C : Re z > 0}, such that Z ∞ F (t)e−zt dt = G(z) for Re z > 0. 0

Then

R∞ 0

F (t)dt converges and is equal to G(0).

R∞ R∞ Remark. Theorem 6.2 states that limz→0, Re z>0 0 = 0 limz→0,Re z>0 . Although this seems plausible it is everything but trivial. Indeed, it will imply the Prime Number Theorem! Proof. The proof consists of several steps. Step 1. Reduction to the case G(0) = 0. We assume that Theorem 6.2 has been proved in the special case G(0) = 0 and deduce from this the general case. Assume that G(0) 6= 0. Define new functions Fe(t) := F (t) − G(0)e−t ,

G(0) e G(z) := G(z) − . z+1

e is analytic on an open set containing Then Fe satisfies (i),(ii), the function G e {z ∈ C : Re z > 0}, we have G(0) = 0, and for Re z > 0 we have Z ∞ Z ∞ Z ∞ G(0) −zt −zt e Fe(t)e dt = F (t)e dt − G(0) e−(z+1)t dt = G(z) − = G(z). z+1 0 0 0 102

R∞ e Hence Fe satisfies (iii). Now if we have proved that 0 Fe(t)dt = G(0) = 0, then it follows that Z ∞ Z ∞ e−t dt = G(0). F (t)dt = G(0) 0

0

Henceforth we assume, in addition to the conditions (i)–(iii), that G(0) = 0. Step 2. The function GT . For T > 0, define T

Z

F (t)e−zt dt.

GT (z) := 0

We show that GT is analytic on C. We apply again Theorem 2.23 and verify the conditions of that theorem. First, F (t)e−zt is measurable on [0, T ] × C. Second, for every fixed t ∈ [0, T ], z 7→ F (t)e−zt is analytic on C. To verify the third property, let K be a compact subset of C. Then for z ∈ K, there is A > 0 such that Re z > −A for z ∈ K. Hence |F (t)e−zt | 6 M eAt for 0 6 t 6 T, z ∈ K RT and clearly, 0 M · eAt dt < ∞ since we integrate over a bounded interval. So by Theorem 2.23, GT is indeed analytic on C. We clearly have Z

T

GT (0) =

F (t)dt. 0

So we have to prove: (6.4)

lim GT (0) = G(0) = 0.

T →∞

103

Step 3. An integral expression for GT (0).

We fix a parameter R > 0. It will be important in the proof that R can be chosen arbitrarily. Let: C + the semi-circle {z ∈ C : |z| = R, Re z > 0}, traversed counterclockwise; C − the semi-circle {z ∈ C : |z| = R, Re z 6 0}, traversed counterclockwise; L the line segment from −iR to iR, traversed upwards.

Define the auxiliary function (invented by Newman):

 z2  1 JR,T (z) := eT z 1 + 2 · . R z

The function GT (z) · JR,T (z) is analytic for z 6= 0, and at z = 0 it has a simple pole with residue GT (0) (or a removable singularity if GT (0) = 0). So by the residue theorem, Z 1 (A) GT (z)JR,T (z)dz = GT (0). 2πi C + +C − The function G(z) is analytic on an open set containing {Re z > 0}. Further, G(z)JR,T (z) is analytic on this open set. For it is clearly analytic if z 6= 0, and at z = 0 the simple pole of JR,T (z) is cancelled by the zero of G(z) at z = 0, thanks to our assumption G(0) = 0. So by Cauchy’s Theorem, Z 1 (B) G(z)JR,T (z)dz = 0. 2πi C + +(−L) 104

We subtract (B) from (A). This gives for GT (0) the expression Z Z 1 1 GT (z)JR,T (z)dz + GT (z)JR,T (z)dz GT (0) = 2πi C + 2πi C − Z Z 1 1 − G(z)JR,T (z)dz + G(z)JR,T (z)dz 2πi C + 2πi L Z Z 1 1 = (GT (z) − G(z))JR,T (z)dz + GT (z)JR,T (z)dz 2πi C + 2πi C − Z 1 + G(z)JR,T (z)dz 2πi L =: I1 + I2 + I3 , where I1 , I2 , I3 denote the three integrals. To R show that GT (0) → 0 as T → ∞, we have to estimate |I1 |, |I2 |, |I3 |. Here we use γ f (z)dz 6 length(γ) · supz∈γ |f (z)|. Step 4. Estimation of |I1 |. We first estimate |(GT (z) − G(z))JR,T (z)| for z ∈ C + . First assume that z ∈ C + , Re z > 0. Using the condition |F (t)| 6 M for t > 0, we obtain Z ∞ Z ∞ −zt |F (t)| · e−tRe z dt |GT (z) − G(z)| = F (t)e dt 6 T

T

Z

∞

6

M · e−tRe z dt =

T

M · e−T Re z . Re z

Further, for z ∈ C + , we have z · z = |z|2 = R2 . Hence  z 2  1 Re z T Re z z + z T Re z = e = 2eT Re z · 2 . |JR,T (z)| = e 1 + z·z z z·z R Hence for z ∈ C + , Re z > 0, |(GT (z) − G(z))JR,T (z)| 6

M Re z 2M · e−T Re z · 2eT Re z · 2 6 2 . Re z R R

By continuity, this is true also if Re z = 0. Hence |I1 | 6

1 1 2M length(C + ) · sup |(GT (z) − G(z))JR,T (z)| 6 · πR · 2 , 2π 2π R z∈C + 105

i.e., |I1 | 6

M . R

Step 5. Estimation of |I2 |. The argument is similar to the estimation of |I1 |. We start with estimating |GT (z)JR,T (z)| for z ∈ C − . First assume that z ∈ C − , Re z < 0. Then, using again |F (t)| 6 M for t > 0, Z T Z T −zt |F (t)| · e−tRe z dt F (t)e dt 6 |GT (z)| = 0

0

Z

T

M · e−tRe z dt =

6 0

while also T Re z

|JR,T (z)| = e


M M eT |Re z| − 1 6 · eT |Re z| |Re z| |Re z|

z + z = 2e−T |Re z| · |Re z| . · z·z R2

Hence for z ∈ C − with Re z < 0, |GT (z)JR,T (z)| 6

2M . R2

Again this holds true also if Re z = 0. So |I2 | 6

1 2M 1 length(C − ) · sup |GT (sz)JR,T (z)| 6 · πR · 2 , 2π 2π R z∈C −

leading to |I2 | 6

M . R

Step 6. Estimation of |I3 |. We choose for L the parametrization z = iy, −R 6 y 6 R. Thus, 1 I3 = 2πi where

Z

R

1 G(iy)JR,T (iy)d(iy) = 2π −R

Z

R

−R

 y2  1 HR (y) := G(iy) 1 − 2 . R iy 106

HR (y)eiT y dt,

Since by assumption, G(0) = 0, the function G(z)/z is analytic on an open set containing {z ∈ C : Re z > 0}. Hence HR (y) is continuously differentiable on [−R, R]. Since HR is independent of T , there is a constant A(R) independent of T such that |HR (y)| 6 A(R), |HR0 (y)| 6 A(R) for y ∈ [−R, R]. Using integration by parts, we get Z

R iT y

HR (y)e −R

Z R 1 dy = HR (y)deiT y iT −R   Z R 1 iT R −iT R 0 iT y = HR (R)e − HR (−R)e − HR (y)e dy . iT −R

Since |eiT y | = 1, we obtain Z

R iT y

HR (y)e

−R

  Z R 1 0 dy 6 A(R) + A(R) + |HR (y)|dy T −R 6

2A(R) + 2RA(R) . T

Hence C(R) , T where C(R) depends on R, but is independent of T . |I3 | 6

Step 7. Conclusion of the proof. We have to prove that limT →∞ GT (0) = G(0) = 0, in other words, for every ε > 0 there is T0 such that |GT (0)| < ε for all T > T0 . Combining steps 3–6, we get, for every choice of R, T , |GT (0)| 6 |I1 | + |I2 | + |I3 | 6

2M C(R) + . R T

Let ε > 0. Then choose R such that 2M/R < ε/2, and subsequently T0 with C(R)/T0 < ε/2. For these choices, it follows that for T > T0 , |GT (0)| < 21 ε + 12 ε = ε. This completes our proof. 107

6.3

A Tauberian theorem for Dirichlet series

Let Lf (s) =

P∞

n=1

f (n)n−s be a Dirichlet series. Put A(x) :=

X

f (n).

n6x

We prove the following Tauberian theorem. Theorem 6.3. Suppose Lf (s) satisfies the following conditions: (i) f (n) > 0 for all n; (ii) there are C > 0, σ > 0 such that |A(x)| 6 Cxσ for all x > 1; (iii) Lf (s) converges for s ∈ C with Re s > σ; (iv) Lf (s) can be continued to a function which is analytic on an open set containing {s ∈ C : Re s > σ} \ {σ} for which lims→σ (s − σ)Lf (s) = α. Then α A(x) = . x→∞ xσ σ lim

Remarks. 1) Condition (iii) follows from (ii) (see homework exercise 8a). Further, (iii) implies that Lf (s) is analytic for Re s > σ. 2) Condition (iv) means that Lf (s) has a simple pole with residue α at s = σ if α 6= 0, and a removable singularity at s = σ if α = 0. 3) The Wiener-Ikehara Theorem is the same as Theorem 6.3, except that only conditions (i),(iii),(iv) are required and (ii) can be dropped. We start with some preparations. Notice that condition (iv) of Theorem 6.3 implies that there is an analytic function g(s) on an open set containing {s ∈ C : Re s > σ} such that (6.5)

Lf (s) =

α + g(s) for s ∈ C with Re s > σ. s−σ

Further, we need some lemmas. Lemma 6.4. For s ∈ C with Re s > σ we have Z ∞ Lf (s) = s A(x)x−s−1 dx. 1

108

Proof. Let Re s > σ. Then by partial summation we have for every integer N > 1, N X

f (n)n

−s

= A(N )N

−s

Z +s

N

A(x)x−s−1 dx.

1

n=1

PN σ −s Since |A(N )| 6 6 CN σ · N −Re s → 0 as n=1 f (n) 6 CN , we have A(N )N N → ∞. By letting N → ∞, the lemma follows. Z ∞ A(x) − (α/σ)xσ Lemma 6.5. · dx = σ −1 g(σ) − σ −2 α converges. xσ+1 1 Proof. By substituting x = et , we see that the identity to be proved is equivalent to Z ∞
(6.6) e−σt A(et ) − α/σ dt = σ −1 g(σ) − σ −2 α. 0

We apply Theorem 6.2 to F (t) := e−σt A(et ) − α/σ. We check that this F satisfies conditions (i),(ii),(iii) of Theorem 6.2. First, F (t) is measurable (e.g., it has only countably many discontinuities). Second, by condition (ii) of Theorem 6.3, |F (t)| 6 C + |α/σ| for t > 0. Hence conditions (i),(ii) of Theorem 6.2 are satisfied. As for condition (iii), notice that by (6.6), (6.5) we have for Re z > 0, on substituting back x = et , Z ∞ Z ∞
−zt F (t)e dt = e−σt A(et ) − α/σ e−zt dt 0

0

Z =

∞

A(x)x

−z−σ−1

Z dx − (α/σ)

1

=

∞

x−z−1 dx

1

α

 α 1 α 1 Lf (z + σ) − = + g(z + σ) − , z+σ σz z+σ z σz

implying Z (6.7) 0

∞

F (t)e−zt dt =

1 g(z + σ) − α/σ) if Re z > 0. z+σ

The right-hand side is analytic on an open set containing {z ∈ C : Re z > 0}, hence (iii) is satisfied as well. So by Theorem 6.2, identity (6.7) extends to z = 0, and this gives precisely (6.6). 109

By condition (i), we have f (n) > 0 for all n. Hence the function A(t) is nondecreasing. Now Theorem 6.3 follows by combining Lemma 6.5 with the lemma below. Lemma 6.6. Let B : [1, ∞) → R be a non-decreasing function and let β ∈ R, σ > 0. Assume that Z ∞ B(x) − βxσ · dx converges. xσ+1 1 Then B(x) = β. x→∞ xσ lim

Proof. We may assume without loss of generality that β > 0. Indeed, assume that e β 6 0. Choose γ > 0 such that βe := β + γ > 0, and replace B(x) by B(x) := R∞ σ σ σ+1 e e is non-decreasing and e B(x) + γx . Then B (B(x) − βx )dx/x converges. If 1 σ σ e then limx→∞ B(x)/x = β follows. e we are able to prove that limx→∞ B(x)/x = β, So assume that β > 0. Assume that limx→∞ B(x)/xσ does not exist or is not equal to β. Then there are two possibilities: (a) there are ε > 0 and an increasing sequence {xn }∞ n=1 with xn → ∞ such that σ B(xn )/xn > β(1 + ε) for all n; (b) there are ε > 0 and an increasing sequence {xn }∞ n=1 with xn → ∞ such that σ B(xn )/xn 6 β(1 − ε) for all n. We consider only case (a); case (b) can be dealt with in the same manner. So R∞ assume (a). Then since 1 (B(x) − βxσ )dx/xσ+1 converges, we have Z (6.8)

y2

lim

y1 ,y2 →∞

y1

B(x) − βxσ · dx = lim y2 →∞ xσ+1

Z

y2

Z − lim

1

y1 →∞

y1

= 0. 1

We choose y1 , y2 appropriately and derive a contradiction. Notice that for x > xn we have, since B is non-decreasing, B(x) − βxσ B(xn ) − βxσ (1 + ε)xσn − xσ > > β · . xσ+1 xσ+1 xσ+1 This is > 0 for xn 6 x 6 (1 + ε)1/σ xn , so there is some hope that with the choice y1 = xn , y2 = (1 + ε)1/σ xn the integral in (6.8) becomes strictly positive and does 110

not converge to 0 as n → ∞. Indeed we have Z

(1+ε)1/σ xn

xn

B(x) − βxσ · dx > β xσ+1

Z

(1+ε)1/σ xn

xn

(1 + ε)xσn − xσ · dx xσ+1

(1+ε)1/σ

(1 + ε) − uσ · du (u = x/xn ) uσ+1 1 h i(1+ε)1/σ = β − (1 + ε)σ −1 u−σ − log u Z

= β

1

=


β ε − log(1 + ε) . σ

This last number is independent of n and strictly positive, since β > 0, σ > 0 and log(1 + ε) < ε. This contradicts (6.8). Hence case (a) is impossible.

111

Chapter 7 The Prime number theorem for arithmetic progressions 7.1

The Prime number theorem

Denote by π(x) the number of primes 6 x. We prove the Prime Number Theorem. x as x → ∞. Theorem 7.1. We have π(x) ∼ log x Before giving the detailed proof, we outline our strategy. Define the functions X X X Λ(n), log p, ψ(x) := log p = θ(x) := p6x

k,p: pk 6x

n6x

where Λ is the von Mangoldt function, given by Λ(n) = log p if n = pk for some prime p and some k > 1, and Λ(n) = 0 otherwise. By Lemma 3.14 from Chapter 3 we have ∞ X n=1

Λ(n)n−s = −

ζ 0 (s) for Re s > 1. ζ(s)

• By applying Theorem 6.3 (the Tauberian theorem for Dirichlet series) to the latter we obtain ψ(x) 1X = Λ(n) → 1 as x → ∞. x x n6x 113

• We prove that ψ(x) − θ(x) is small. This gives θ(x)/x → 1 as x → ∞. • Using partial summation, we deduce π(x) log x/x → 1 as x → ∞. We first verify the conditions of the Tauberian theorem. P −s Lemma 7.2. ∞ can be extended to a function analytic on an open set n=1 Λ(n)n containing {s ∈ C : Re s > 1, s 6= 1}, with a simple pole with residue 1 at s = 1. P −s Proof. Recall that ∞ = −ζ 0 (s)/ζ(s) for s ∈ C with Re s > 1 (see Lemma n=1 Λ(n)n 3.15). Let A := {s ∈ C : Re s > 1, s 6= 1}. By Theorem 5.2, 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) 6= 0 on A, and hence also ζ(s) 6= 0 on an open set containing A. So by Lemma 2.16, ζ 0 (s)/ζ(s) is analytic on an open set containing A, with a simple pole with residue −1 at s = 1. This proves Lemma 7.2. Lemma 7.3. (i) θ(x) = O(x) as x → ∞. √ (ii) ψ(x) = θ(x) + O( x) as x → ∞. (iii) ψ(x) = O(x) as x → ∞. Q Proof. (i) By homework exercise 3a, we have p6x p 6 4x for x > 2. This implies X θ(x) = log p 6 x log 4 = O(x) as x → ∞. p6x

(ii) We have ψ(x) =

X

log p =

X

log p +

p6x

p,k: pk 6x

= θ(x) + θ(x

1/2

X

log p +

p2 6x

X

log p + · · ·

p3 6x

) + θ(x1/3 ) + · · ·

Notice that θ(t) = 0 if t < 2. So θ(x1/k ) = 0 if x1/k < 2, that is, if k > log x/ log 2. Hence [log x/ log 2]

ψ(x) − θ(x) =

X

√ θ(x1/k ) 6 θ( x) +

k=2

[log x/ log 2]

X

√ θ( 3 x)

k=3

 log x  √
√ √ √ 6 θ( x) + − 3 θ( 3 x) = O x + 3 x · log x log 2 √ = O( x) as x → ∞. (iii) Combine (i) and (ii). This follows also from homework exercise 6, but (ii) will be needed anyhow. 114

Proof of Theorem 7.1. Lemmas 7.2 and 7.3 (iii) imply that LΛ (s) = satisfies all conditions of Theorem 6.3, with σ = 1, α = 1. Hence

P∞

n=1

Λ(n)n−s

1X ψ(x) = Λ(n) → 1 as x → ∞. x x n6x From Lemma 7.3 (ii) we infer √ θ(x) ψ(x) + O( x) ψ(x) = = + O(x−1/2 ) → 1 as x → ∞. x x x We now apply partial summation to obtain our result for π(x). Thus, Z x  1 0 X X 1 1 θ(t) · π(x) = 1= log p · = θ(x) − dt log p log x log t 2 p6x p6x Z x θ(t) θ(x) + = 2 dt. log x 2 t log t By Lemma 7.3 (i) there is a constant C > 0 such that θ(t) 6 Ct for all t > 2. Together with homework exercise 1, this implies Z x Z x  x  θ(t) dt · dt 6 C · = O as x → ∞. 2 2 log2 x 2 t log t 2 log t Hence  log x θ(x) x  π(x) log x = +O · x x x log2 x  1  θ(x) +O → 1 as x → ∞. = x log x

7.2

The Prime number theorem for arithmetic progressions

Let q, a be integers with q > 2, gcd(a, q) = 1. Define π(x; q, a) := number of primes p 6 x with p ≡ a (mod q). 115

Theorem 7.4. We have π(x; q, a) ∼

x 1 · as x → ∞. ϕ(q) log x

The proof is very similar to that of the Prime number theorem. Define the quantities X θ(x; q, a) := log p, p6x, p≡a (mod q)

X

ψ(x; q, a) :=

p,k, pk 6x, pk ≡a (mod q)

Let

∞ X

F (s) :=

X

log p =

Λ(n).

n6x, n≡a (mod q)

Λ(n)n−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) = −

X 1 L0 (s, χ) · . χ(a) · ϕ(q) L(s, χ) χ∈G(q)

Proof. By homework exercise 7a we have for χ ∈ G(q) and for s ∈ C with Re s > 1, since χ ∈ G(q) is a strongly multiplicative arithmetic function and L(s, χ) converges absolutely, ∞ X L0 (s, χ) =− χ(n)Λ(n)n−s . L(s, χ) n=1 Using Theorem 4.9 (ii) (one of the orthogonality relations for characters) we obtain for s ∈ C with Re s > 1, X χ∈G(q)

∞ X X L0 (s, χ) χ(a) · =− χ(a) χ(n)Λ(n)n−s L(s, χ) n=1

=−

χ∈G(q)

∞ X



 X

 n=1

χ(a)χ(n) Λ(n)n−s = −ϕ(q) ·

χ∈G(q)

∞ X n=1, n≡a (mod q)

116

Λ(n)n−s .

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 6= 1}, with a simple pole with residue ϕ(q)−1 at s = 1. Proof. Let A := {s ∈ C : Re s > 1, s 6= 1}, B := {s ∈ C : Re s > 1}. (q)

By Theorem 5.3 (iii), L(s, χ0 ) is analytic on an open set containing A, with a (q) simple pole at s = 1. Further, by Corollary 5.4 and Theorem 5.5, L(s, χ0 ) 6= 0 for (q) (q) s ∈ A, and hence for s in an open set containing A. Therefore, L0 (s, χ0 )/L(s, χ0 ) is analytic on an open set containing A. Further, by Lemma 2.16, it has a simple pole with residue 1 at s = 1. (q)

Let χ be a character mod q with χ 6= χ0 . By Theorem 5.3 (ii), L(s, χ) is analytic on an open set containing B, and by Corollary 5.4 and Theorems 5.5, 5.7, it is non-zero on B, hence on an open set containing B. Therefore, L0 (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 (q) pole with residue χ0 (a)/ϕ(q) = ϕ(q)−1 at s = 1. Lemma 7.7. (i) θ(x; q, a) = O(x) as x → ∞. √ (ii) ψ(x; q, a) − θ(x; q, a) = O( x) as x → ∞. (iii) ψ(x; q, a) = O(x) as x → ∞. Proof. (i) We have θ(x; q, a) 6 θ(x) = O(x) as x → ∞. (ii) We have X

ψ(x; q, a) − θ(x; q, a) =

log p

k,p, k>2,pk 6x, pk ≡a (mod q)

6

X

√ log p = ψ(x) − θ(x) = O( x) as x → ∞.

k,p, k>2,pk 6x

(iii) Obvious. Proof of Theorem 7.4. Notice that F (s) satisfies the conditions of Theorem 6.3, with α = ϕ(q)−1 . Hence ψ(x; q, a) 1 → as x → ∞, x ϕ(q) 117

and then by Lemma 7.7 (ii), √ ψ(x; q, a) + O( x) 1 θ(x; q, a) = → as x → ∞. x x ϕ(q) By partial summation we have π(x; q, a) =

X 26p6x, p≡a (mod q)

Now

Z 06 2

x

θ(t; q, a) · dt 6 t log2 t

1 θ(x; q, a) log p · = + log p log x

Z

x

2

Z 2

x

θ(t; q, a) · dt. t log2 t

 x  θ(t) · dt = O as x → ∞, t log2 t log2 x

using the estimate from the proof of Theorem 7.1. So  1  π(x; q, a) log x θ(x; q, a) 1 = +O → as x → ∞. x x log x ϕ(q) This completes our proof.

7.3

Related results

Riemann sketched a proof, and von Mangoldt gave the complete proof, of the following result, that relates the distribution of primes to the distribution of the zeros of the Riemann zeta function. Define  P ψ(x) = n6x Λ(n) if x is not a prime power, ψ0 (x) := ψ(x) − 21 Λ(x) if x is a prime power. Theorem 7.8. We have for x > 1, ψ0 (x) = x − lim

T →∞

X xρ − log 2π − 21 log(1 − x−2 ), ρ

|Im ρ| 0 there is a number C2 (A) depending on A, such that the following holds. For every real x > 3, every integer q with 2 6 q 6 (log x)A and every integer a with gcd(q, a) = 1, we have √ 1 π(x; q, a) − Li(x) 6 C1 xe−C2 (A) log x . ϕ(q) The constants C1 , C2 (A) are ineffective, this means that by going through the proof of the theorem one cannot compute the constants, but only show that they exist. As we mentioned in Chapter 1, there is an intricate connection between the Rx zero-free region of ζ(s) and estimates for |π(x) − Li(x)|, where Li(x) = 2 dt/ log t. Similarly, there is a connection between zero-free regions of L-functions and estimates for |π(x; q, a) − Li(x)/ϕ(q)|. We recall the following, rather complicated, result of Landau (1921) on the zero-free region of L-functions. Theorem 7.10. There is an absolute constant c > 0 such that for every integer q > 2 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   c R(q) := s ∈ C : Re s > 1 − . log(q(1 + |Im s|)) 119

If such a character χ exists, it has no more than one zero in R(q) and moreover, (q) χ 6= χ0 , χ is a real character and the zero is real. Any character χ modulo q having a zero in R(q) is called an exceptional character mod q, and the zero of L(s, χ) in R(q) is called an exceptional zero. It is conjectured that exceptional characters do not exist. In order to obtain Theorem 7.9, one needs an estimate for the real part of a possible exceptional zero of an L-function. The following result was proved by Siegel (1935). Theorem 7.11. For every ε > 0 there is a number c(ε) > 0 such that for every integer q > 2 the following holds: if χ is an exceptional character modulo q and β an exceptional zero of L(s, χ), then Re β < 1 − c(ε)q −ε . Theorems 7.10 and 7.11 imply (after a lot of work) Theorem 7.9. Proofs of Theorems 7.9–7.11 may be found in H. Davenport, Multiplicative Number Theory, Graduate texts in mathematics 74, Springer Verlag, 2nd ed., 1980. 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 x such that π(x; q, a) > 0. The following estimate is due to Linnik (1944). Theorem 7.12. 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 > 2 and every integer a with gcd(a, q) = 1 we have P (q, a) 6 cq L . The exponent 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 (2011).

120

Chapter 8 Euler’s Gamma function The Gamma function plays an important role in the functional equation for ζ(s) that we will derive in the next chapter. In the present chapter we have collected some properties of the Gamma function. For t ∈ R>0 , z ∈ C, define tz := ez log t , where log t is the ordinary real logarithm. Euler’s Gamma function is defined by the integral Z ∞ Γ(z) := e−t tz−1 dt (z ∈ C, Re z > 0). 0

Lemma 8.1. Γ(z) defines an analytic function on {z ∈ C : Re z > 0}. Proof. This is standard using Theorem 2.23. Let U := {z ∈ C : Re z > 0}. First, the function F (t, z) := e−t tz−1 is continuous, hence measurable on R>0 × U . Second, for each fixed t > 0, z 7→ e−t tz−1 is analytic on U . Third, let K be a compact subset of U . Then there exist δ, R > 0 such that δ 6 Re z 6 R for z ∈ K. This implies that for z ∈ K, t > 0,  δ−1 t for 0 6 t 6 1, −t z−1 |e t | 6 M (t) := −t R−1 −t/2 e t 6 Ce for t > 1, where C is some constant. Now we have Z ∞ Z 1 Z δ−1 M (t)dt = t dt + C 0

0

∞

e−t/2 dt = δ −1 + 2C < ∞.

1

Hence all conditions of Theorem 2.23 are satisfied, and thus, Γ(z) is analytic on U. 121

Using integration by parts, one easily shows that for z ∈ C with Re z > 0, Z ∞ −1 Γ(z) = z e−t dtz  0 Z ∞ −t z −t z t=∞ −1 e t dt = z −1 Γ(z + 1), [e t |t=0 + = z 0

that is, (8.1)

Γ(z + 1) = zΓ(z) if Re z > 0.

One easily shows that Γ(1) = 1 and then by induction, Γ(n) = (n − 1)! for n ∈ Z>0 . We now show that Γ has a meromorphic continuation to C. Theorem 8.2. There exists a unique meromorphic function Γ on C with the following properties: R∞ (i) Γ(z) = 0 e−t tz−1 dt for z ∈ C, Re z > 0; (ii) the function Γ is analytic on C \ {0, −1, −2, . . .}; (iii) Γ has a simple pole with residue (−1)n /n! at z = −n for n = 0, 1, 2, . . .; (iv) Γ(z + 1) = zΓ(z) for z ∈ C \ {0, −1, −2, . . .}. R∞ Proof. The function Γ has already been defined for Re z > 0 by 0 e−t tz−1 dt. By Corollary 2.21, Γ has at most one analytic continuation to any larger connected open set, hence there is at most one function Γ with properties (i)–(iv). We proceed to construct such a function. Let z ∈ C with Re z > 0. By repeatedly applying (8.1) we get (8.2)

Γ(z) =

1 · Γ(z + n) for Re z > 0, n = 1, 2, . . . . z(z + 1) · · · (z + n − 1)

We continue Γ to B := C \ {0, −1, −2, . . .} as follows. For z ∈ B, choose n ∈ Z>0 such that Re z + n > 0 and define Γ(z) by the right-hand side of (8.2). This does not depend on the choice of n. For if m, n are any two integers with m > n > −Re z, then by (8.2) with z + n, m − n instead of z, n we have Γ(z + n) =

1 · Γ(z + m), z + n) · · · (z + m − 1)

and so 1 1 · Γ(z + n) = · Γ(z + m). z(z + 1) · · · (z + n − 1) z(z + 1) · · · (z + m − 1) 122

Hence Γ is well-defined on B, and it is analytic on B since the right-hand side of (8.2) is analytic if Re z + n > 0. This proves (ii). We prove (iii). By (8.2) we have 1 Γ(z + n + 1) z→−n z(z + 1) · · · (z + n) (−1)n 1 Γ(1) = . = (−n)(−n + 1) · · · (−1) n!

lim (z + n)Γ(z) =

z→−n

lim (z + n)

Hence Γ has a simple pole at z = −n of residue (−1)n /n!. We prove (iv). Both functions Γ(z + 1) and zΓ(z) are analytic on B, and by (8.1), they are equal on the set {z ∈ C : Re z > 0} which has limit points in B. So by Corollary 2.20, Γ(z + 1) = zΓ(z) for z ∈ B. Theorem 8.3. We have Γ(z)Γ(1 − z) =

π for z ∈ C \ Z. sin πz

Proof. We prove that zΓ(z)Γ(1 − z) = πz/ sin πz, or equivalently, (8.3)

Γ(1 + z)Γ(1 − z) =

πz for z ∈ A := (C \ Z) ∪ {0}, sin πz

which implies Theorem 8.3. Notice that by Theorem 8.2 the left-hand side is analytic on A, while by limz→0 πz/ sin πz = 1 the right-hand side is also analytic on A. By Corollary 2.20, it suffices to prove that (8.3) holds for every z in an infinite subset 1 of A having a limit point in A. For this infinite set we take S := { 2n : n ∈ Z>0 }; this set has limit point 0 in A. Thus, (8.3), and hence Theorem 8.3, follows once we have proved that (8.4)

Γ(1 +

1 ) 2n

· Γ(1 −

1 ) 2n

=

π/2n sin π/2n

Z

∞

(n = 1, 2, . . .).

Notice that Γ(1 +

1 ) 2n

· Γ(1 −

1 ) 2n

=

−s 1/2n

e s Z0 ∞ Z ∞

= 0

123

0

Z ds ·

∞

e−t t−1/2n dt

0

e−s−t (s/t)1/2n dsdt.

Define new variables u = s + t, v = s/t. Then s = uv/(v + 1), t = u/(v + 1). The Jacobian of the substitution (s, t) 7→ (u, v) is u ∂s ∂s v 2 v + 1 (v + 1) ∂(s, t) = ∂u ∂v = 1 u ∂t ∂t ∂(u, v) v + 1 − (v + 1)2 ∂u ∂v −uv − u −u = = . 3 (v + 1) (v + 1)2 It follows that Γ(1 +

1 ) 2n

Z

· Γ(1 −

∞

Z

Z

e−u v 1/2n

0

∞

Z

∞

=

∞

= 0

1 ) 2n

0

0

u (v + 1)

∂(s, t) e v ∂(u, v) · dudv Z ∞ Z −u · dudv = e udu · 2 −u 1/2n

0

0

∞

v 1/2n dv. (v + 1)2

In the last product, the first integral is equal to 1, while for the second integral we have, by homework exercise 4, Z ∞ 1/2n Z ∞  1  v 1/2n v d dv = − (v + 1)2 v+1 0 0  1/2n ∞ Z ∞ Z ∞ v dw π/2n 1 1/2n · dv = = . =− + 2n v+1 0 v+1 w +1 sin π/2n 0 0 This implies (8.4), hence Theorem 8.3. √ Corollary 8.4. Γ( 21 ) = π. Proof. Substitute z =

1 2

in Theorem 8.3, and use Γ( 12 ) > 0.

Corollary 8.5. (i) Γ(z) 6= 0 for z ∈ C \ {0, −1, −2, . . .}. (ii) 1/Γ is analytic on C, and 1/Γ has simple zeros at z = 0, −1, −2, . . .. Proof. (i) Recall that Γ(n) = (n − 1)! 6= 0 for n = 1, 2, . . .. Further, by Theorem 8.3 we have Γ(z)Γ(1 − z) sin πz = π 6= 0 for z ∈ C \ Z. (ii) By (i), the function 1/Γ is analytic on C \ {0, −1, −2, . . .}. Further, at z = 0, −1, −2, . . ., Γ has a simple pole, hence 1/Γ is analytic and has a simple zero. 124

We give another expression for the Gamma function. Theorem 8.6. For z ∈ C \ {0, −1, −2, . . .} we have n! · nz . n→∞ z(z + 1) · · · (z + n)

Γ(z) = lim

Proof. Denote the limit by F (z). We prove by induction that for every non-negative integer n we have F (z) = Γ(z) for z ∈ C with Re z > −n and (if n > 0) z 6= 0, −1, . . . , 1 − n. For the moment, we assume that this assertion is true for n = 0, i.e., F (z) = Γ(z) for Re z > 0, and do the induction step. Assume our assertion holds for some integer n > 0. Let z ∈ C with Re z > −n − 1 and z 6= 0, . . . , n. We do not know a priori whether the limit F (z) exists, but at least we have n! · nz+1 (z + 1) · · · (z + n + 1) · n→∞ (z + 1) · · · (z + n + 1) (n + 1)!(n + 1)z  n z+1 = 1. = lim n→∞ n + 1

F (z + 1) = zF (z)

lim

By the induction hypothesis we know that F (z + 1) = Γ(z + 1), and so F (z) does exist, and F (z + 1) Γ(z + 1) F (z) = = = Γ(z). z z This completes the induction step. We now show that F (z) = Γ(z) for z ∈ C, Re z > 0. For this, we need some lemmas. Lemma 8.7. Let z ∈ C with Re z > 0. Then Z n t
n z−1 n! · nz = 1− t dt. z(z + 1) · · · (z + n) n 0 Proof. By substituting s = t/n, the integral becomes Z 1 z (1 − s)n sz−1 ds. n 0

The rest is left as an exercise. Lemma 8.8. For every integer n > 2 and every real t with 0 6 t 6 n we have −t

06e

2 t
n −t t − 1− 6 e · 2. n n

125

Proof. This is equivalent to t
n t2 6 1 (0 6 t 6 n, n > 2). 1 − 6 et 1 − n n Recall that if f, g are continuously differentiable, real functions with f (0) = g(0) and f 0 (x) 6 g 0 (x) for 0 6 x 6 A, say, then f (x) 6 g(x) for 0 6 x 6 A. From this observation, one easily deduces that 1 + x 6 ex , 1 − x 6 e−x , (1 − x)r > 1 − rx for 0 6 x 6 1, r > 0. This implies on the one hand, for n > 2, 0 6 t 6 n, t
n et 1 − 6 et (e−t/n )n 6 1, n on the other hand t
n t
n t2
n t2 t
n > 1+ · 1− = 1− 2 >1− . et 1 − n n n n n

Completion of the proof of Theorem 8.6. Let z ∈ C with Re z > 0. We prove that F (z) = Γ(z). By the integral expression for Γ(z) and by Lemma 8.7 we have Z n  Z n t
n z−1 −t z−1 e t dt − Γ(z) − F (z) = lim 1− t dt n→∞ n 0 0 Z n t
n  z−1 = lim e−t − 1 − t dt. n→∞ 0 n R R Now using | ...| 6 |...| and Lemma 8.8 we obtain Z n t2 e−t · |tz−1 |dt |Γ(z) − F (z)| 6 lim n→∞ 0 n Z ∞ 1 Γ(Re z + 2) 6 lim e−t tRe z+1 dt = lim = 0. n→∞ n 0 n→∞ n

We deduce some consequences. Recall that the Euler-Mascheroni constant γ is given by ! N X 1 γ = lim − log N. N →∞ n n=1 126

Corollary 8.9. We have Γ(z) = e−γz z −1

∞ Y

ez/n for z ∈ C \ {0, −1, −2, . . .}. 1 + z/n n=1

Proof. Let z ∈ C \ {0, −1, −2, . . .}. Then for N ∈ Z>0 we have ez log N Nz · N! = z −1 lim N →∞ (1 + z)(1 + z/2) · · · (1 + z/N ) N →∞ z(z + 1) · · · (z + N ) N Y 1 1 ez/n = z −1 lim e(log N −1− 2 −···− N )z N →∞ 1 + z/n n=1

Γ(z) =

lim

= e

−γz −1

z

∞ Y

ez/n . 1 + z/n n=1

As another consequence, we derive an infinite product expansion for sin πz. Corollary 8.10. We have  ∞  Y z2 1− 2 sin πz = πz · for z ∈ C. n n=1 Proof. For z ∈ C we have by Theorem 8.3, Corollary 8.5 and Corollary 8.9, π π = sin πz = Γ(z)Γ(1 − z) Γ(z)(−z)Γ(−z) ∞  ∞    Y Y z  z  −z/n −γz e 1+ · (−z)e e−z/n 1 − = π(−z) e z n n n=1 n=1 −1 γz

= πz

∞  Y n=1

 ∞  Y z  z z2 1− 1+ = πz · 1− 2 . n n n n=1

Recall that the Bernoulli numbers Bn (n > 0) are given by ∞

X Bn z = · z n (|z| < 2π). ez − 1 n=0 n! 127

Corollary 8.11. We have B0 = 1, B1 = − 12 , B3 = B5 = · · · = 0 and ζ(2n) = (−1)n−1 22n−1

B2n 2n π for n = 1, 2, . . . . (2n)!

Proof. Let z ∈ C with 0 < |z| < 1. Then sin πz 6= 0 and so, by taking the logarithmic derivative of sin πz, sin0 πz π cos πz π(eπiz + e−πiz )/2 = = πiz sin πz sin πz (e − e−πiz )/2i 1 2πiz · 2πiz z e −1 ∞ X Bn 1 · (2πi)n z n . = πi + z n=0 n! = πi +

(8.5)

We obtain another expression for the logarithmic derivative of sin πz by applying Corollary 2.28 to the product identity from Corollary 8.10. Note that for z ∈ C with P −2 converges. Hence the logarithmic |z| < 1 we have |z 2 /n2 | < n−2 and that ∞ n=1 n derivative of the infinite product is the infinite sum of the logarithmic derivatives of the factors, i.e., ∞

sin0 πz (πz)0 X (1 − z 2 /n2 )0 = + sin πz πz 1 − z 2 /n2 n=1 ∞ ∞ X 1 z 1 X −2z/n2 + = −2 = 2 2 z n=1 1 − z /n z n2 n=1

∞  2 X z k k=0

!

n2

∞ ∞ X X z 2k+1 1 = −2 (by absolute convergence) z n2k+2 k=0 n=1

(8.6)

∞ X 1 = −2 ζ(2k + 2)z 2k+1 . z k=0

Now Corollary 8.11 easily follows by comparing the coefficients of the Laurent series in (8.5) and (8.6). We finish with another important consequence of Theorem 8.6, the so-called duplication formula. 128

Corollary 8.12. We have 22z−1 Γ(2z) = √ · Γ(z)Γ(z + 21 ) for z ∈ C, z 6= 0, − 21 , −1, − 32 , −2, . . . . π Proof. Let A be the set of z indicated in the lemma. We show that the function F (z) := 22z Γ(z)Γ(z + 12 )/Γ(2z) is constant on A. Substituting z = 12 gives that the √ constant is 2 π, and then Corollary 8.12 follows. Let z ∈ A. To get nice cancellations in the numerator and denominator, we use the expressions Γ(z) = Γ(z + 21 ) =

2n+1 · n! · nz n! · nz = lim , n→∞ z(z + 1) · · · (z + n) n→∞ 2z(2z + 2) · · · (2z + 2n) lim

n! · nz+1/2 n→∞ (z + 1/2)(z + 3/2) · · · (z + n + 1/2) lim

2n+1 · n! · nz+1/2 , n→∞ (2z + 1)(2z + 3) · · · (2z + 2n + 1)

= lim Γ(2z) =

(2n + 1)! · (2n + 1)2z (limit over the odd integers). n→∞ 2z(2z + 1) · · · (2z + 2n + 1) lim

Thus, 22z Γ(z)Γ(z + 12 ) F (z) = Γ(2z)   2z(2z + 1) · · · (2z + 2n + 1) 22n+2 (n!)2 n2z+1/2 2z = 2 lim · n→∞ 2z(2z + 1) · · · (2z + 2n + 1) (2n + 1)! · (2n + 1)2z  2n+2 √  2 (n!)2 n = lim n→∞ (2n + 1)! since

22z · n2z = lim e2z log(2n/(2n+1)) = 1. n→∞ (2n + 1)2z n→∞ lim

This shows that indeed F (z) is constant. Remark. More generally, one can derive the multiplication formula of LegendreGauss, (2π)(n−1)/2 Γ(nz) = nnz−1/2 Γ(z)Γ(z + n1 ) · · · Γ(z + n−1 ) n 129

for every integer n > 2. The idea of the proof is similar to that of Corollary 8.12 (exercise).

130

Chapter 9 The functional equation for the Riemann zeta function We will eventually deduce a functional equation, relating ζ(s) to ζ(1 − s). There are various methods to derive this functional equation, see E.C. Titchmarsh, The theory of the Riemann zeta function. We give a proof based on a functional equation for P −πm2 z . We start with some preparations. the Jacobi theta function θ(z) = ∞ m=−∞ e

9.1

Poisson’s summation formula

We start with a simple result from Fourier analysis. Given a function f : [0, 1] → C, we define the Fourier coefficients of f by Z 1 cn (f ) := f (t)e−2πint dt for n ∈ Z. 0

Theorem 9.1. Let f be a complex analytic function, defined on an open subset of C containing the real interval [0, 1]. Then ( 1
N X f (0) + f (1) if x = 0 or x = 1, 2 lim cn (f )e2πinx = N →∞ f (x) if 0 < x < 1. n=−N Remarks 1. This version of Theorem 9.1 with the condition that f be analytic on an open subset containing [0, 1] is amply sufficient for our purposes. There are much 131

more general versions of this theorem, which are of course much more difficult to prove. For instance, Dirichlet proved the above theorem for functions f : [0, 1] → C that are differentiable and whose derivative is piecewise continuous. P P 2. It may be that a doubly infinite series ∞ an = limM,N →∞ N n=−∞ n=−M an diPN verges, while limN →∞ n=−N an converges. For instance, if a−n = −an for n ∈ P P∞ Z \ {0}, then limN →∞ N n=−N an = a0 , while n=−∞ an may be horribly divergent. Proof. We first assume that either 0 < x < 1, or that x ∈ {0, 1} and f (0) = f (1). We use the so-called Dirichlet kernel N X

DN (x) =

2πinx

e

−2πiN x

=e

2N X

e2πinx

n=0

n=−N

= e−2πiN x ·

2πi(2N +1)x

e

e2πix

−1 −1

eπi(2N +1)x − e−πi(2N +1)x sin(2N + 1)πx = . πix −πix e −e sin πx

= Further, we use

Z

1 2πint

e

 dt =

0

1 if n = 0, 0 if n 6= 0.

Using these facts, we obtain f (x) −

N X

2πinx

cn (f )e

= f (x) −

n=−N

n=−N

=

Z N X

Z N X

1

f (x)e

−2πint

f (t)e

−2πint

 dt e2πinx

0

 Z N X 2πinx dt e −

0

n=−N

1

n=−N

1

f (t)e

0

(the first integral is f (x) if n = 0 and 0 if n 6= 0) =

Z N X n=−N

Z

1


f (x) − f (t) · e−2πin(t−x) dt

0

1

f (x) − f (t)

= 0

Z = 0




N X

! e−2πin(t−x) dt

n=−N 1



sin (2N + 1)π(t − x) (f (x) − f (t) · · dt. sin π(t − x) 132

−2πint

 dt e2πinx

Fix x and define g(z) :=

f (x) − f (z) . sin π(z − x)

We show that g is analytic on an open set containing [0, 1]. First, suppose that 0 < x < 1. By assumption, f is analytic on an open set U ⊂ C containing [0, 1]. By shrinking U if needed, we may assume that U contains [0, 1] but not x + n for any non-zero integer n. Then sin π(z − x) has a simple zero at z = x but is otherwise non-zero on U . This shows that g(z) is analytic on U \ {x}. But g(z) is also analytic at z = x, since the simple zero of sin π(z − x) is cancelled by the zero of f (x) − f (z). In case that x ∈ {0, 1} and f (0) = f (1) one proceeds in the same manner. Using integration by parts, we obtain N X

f (x) −

cn (f )e2πinx =

=

1

g(t) sin{(2N + 1)π(t − x)}dt 0

n=−N

−1 = (2N + 1)π

Z

1

Z

g(t)d cos{(2N + 1)π(t − x)} 0

n −1 g(1) cos{(2N + 1)π(1 − x)} − g(0) cos{(2N + 1)πx} + (2N + 1)π Z 1 o 0 + g (t) cos{(2N + 1)π(t − x)}dt . 0

Since g is analytic, the functions g(t), g 0 (t) are continuous, hence their absolute values are bounded on [0, 1]. Further, the cosine terms have absolute values at most 1. It follows that the above expression converges to 0 as N → ∞. We are left with the case x ∈ {0, 1} and f (0) 6= f (1). Let fe(z) := f (z) + (f (0) − f (1))z. Then fe is analytic on U and fe(0) = fe(1) = f (0). It is easy to check that the function id : z 7→ z has Fourier coefficients c0 (id) = 21 , cn (id) = −1/2πin for n 6= 0. In particular, c−n (id) = −cn (id) for n 6= 0. Consequently, ! N N N X X
X cn (f ) = lim cn (fe) + f (1) − f (0) cn (id) lim N →∞

n=−N

N →∞

= f (0) +

n=−N 1 2


f (1) − f (0) = 133

n=−N 1 2


f (0) + f (1) .

This completes our proof. Theorem 9.2 (Poisson’s summation formula for finite sums). Let a, b be integers with a < b and let f be a complex analytic function, defined on an open set containing the interval [a, b]. Then b X

f (m) =

1 2

Z b N X
f (t)e−2πint dt f (a) + f (b) + lim N →∞

m=a

=

1 2

Z



f (a) + f (b) +

a

n=−N

b

f (t)dt + 2 a

∞ Z X n=1

b

f (t) cos 2πnt · dt.

a

Proof. Pick m ∈ {a, . . . , b − 1}. Then by Theorem 9.1, 1 2

Z N X




f (m) + f (m + 1) = lim

N →∞

Z

f (t)dt + lim

=

N →∞

m

Z

m+1

=

f (t)dt + 2 m

f (t)e−2πint dt

m

n=−N

m+1

m+1

N Z X

n=1


f (t) e2πint + e−2πint dt

m

n=1

∞ Z X

m+1

m+1

f (t) cos 2πnt · dt.

m

Now take the sum over m = a, a + 1, . . . , b − 1. We need a variation on Theorem 9.2, dealing with infinite sums

P∞

m=−∞

f (m).

Theorem 9.3. Let f be a complex function such that: (i) there is δ > 0 such that f (z) is analytic on U (δ) := {z ∈ C : |Im z| < δ}; (ii) there are C > 0, ε > 0 such that |f (z)| 6 C · (|z| + 1)−1−ε for z ∈ U (δ). Then

∞ X n=−∞

f (n) = lim

N →∞

Z N X n=−N

∞

f (t)e−2πint dt.

−∞

The idea is to apply Theorem 9.1 to the function F (z) := first prove some properties of this function. 134

P∞

m=−∞

f (z + m). We

P Lemma 9.4. (i) F (0) = F (1) = ∞ m=−∞ f (m). (ii) The function F (z) is analytic on an open set containing [0, 1]. R∞ R1 (iii) For every n ∈ Z we have 0 F (t)e−2πint dt = −∞ f (t)e−2πint dt. Proof. (i) Obvious. (ii) Let U := {z ∈ C : −δ < Re z < 1 + δ, |Im z| < δ}. Assuming that δ is sufficiently small, we have |f (z + m)| 6 C(|m| − δ)−1−ε =: Am for z ∈ U , P m ∈ Z \ {0}. All summands f (z + m) are analytic on U , and the series m6=0 Am converges. So by Corollary 2.27, the function F (z) is analytic on U . P (iii) Since |f (t + m)e−2πint | 6 Am for t ∈ [0, 1], m ∈ Z \ {0}, and m6=0 Am P −2πint converges, the series ∞ converges uniformly on [0, 1]. Therefore, m=∞ f (t + m)e we may interchange the integral and the infinite sum, and obtain Z

1

F (t)e

−2πint

Z

1

dt =

0

0

=

Z ∞  X −2πint f (t + m) e dt =

∞  X

m=−∞

m=−∞

∞ X

−2πin(t+m)

f (t + m)e

dt =

f (t + m)e−2πint dt

0

m+1

f (t)e−2πint dt

m

m=−∞

m=−∞

Z

Z ∞ X

1

∞

=

f (t)e−2πint dt.

−∞

R∞ In the last step we have used that the integral −∞ f (t)e−2πint dt converges, due to our assumption |f (z)| 6 C(|z| + 1)−1−ε for z ∈ U (δ). Proof of Theorem 9.3. By combining Theorem 9.1 with Lemma 9.4 we obtain ∞ X

f (m) =

1 2

Z N X
F (0) + F (1) = lim N →∞

m=−∞

=

lim

N →∞

Z N X n=−N

n=−N

∞

f (t)e−2πint dt.

−∞

135

0

1

F (t)e−2πint dt

9.2

A functional equation for the theta function

The Jacobi theta function is given by θ(z) :=

∞ X

e−πm

2z

(z ∈ C, Re z > 0).

m=−∞

Verify yourself that θ(z) converges and is analytic on {z ∈ C : Re z > 0}. √

Theorem 9.5. θ(z −1 ) = √ that | arg z| < π4 .

z · θ(z) for z ∈ C, Re z > 0, where

√

z is chosen such

Remark. Let A := {z ∈ C : Re z > 0}. We may choose the argument of z ∈ A √ √ such that | arg z| < π/2. Then indeed, we may choose z such that | arg z| < π/4. √ Proof. Both θ(z −1 ) and zθ(z) are analytic on A. Hence it suffices to prove the identity in Theorem 9.5 on a subset of A having a limit point in A. For this subset we take R>0 . Thus, it suffices to prove that ∞ X

−πm2 /x

e

=

√

x·

m=−∞

∞ X

2x

e−πm

for x > 0.

m=−∞ 2

We apply Theorem 9.3 to f (z) := e−πz /x with x > 0 fixed. Verify that f satisfies all conditions of that Theorem. Thus, for any x > 0, ∞ X m=−∞

e

−πm2 /x

= lim

N →∞

Z N X

∞

2 /x)−2πint

e−(πt

dt.

−∞

n=−N

√ We compute the integrals by substituting u = t x. Thus, Z ∞ Z ∞ √ √ 2 −(πt2 /x)−2πint e dt = x· e−πu −2πin x·u du −∞

−∞

=

√

Z

∞

x·

√

e−π(u+in

x)2 −πn2 x

du

−∞

=

√

xe

−πn2 x

Z

∞

−∞

136

e−π(u+in

√

x)2

du.

In the lemma below we prove that the last integral is equal to 1. Then it follows that N ∞ ∞ X X √ −πn2 x √ X 2 −πm2 /x xe = x e−πn x , e = lim N →∞

m=−∞

n=−∞

n=−N

since the last series converges. This proves our Theorem. Lemma 9.6. Let z ∈ C. Then

R∞ −∞

2

e−π(u+z) du = 1.

Proof. The following proof was suggested to me by Michiel Kosters. Let Z

∞

2

e−π(u+z) du.

F (z) := −∞

We show that this defines an analytic function on C. We apply Theorem 2.23. 2 First, (u, z) 7→ e−π(u+z) is continuous, hence measurable, on R × D(0, R). Second, 2 for every fixed u ∈ R, z 7→ e−π(u+z) is analytic on C. Third, let K be a compact subset of C, and choose R > 0 such that |z| 6 R for z ∈ K. Then for z ∈ K we have 2

2

2 +2πuRe z+πRe z 2 )

|e−π(u+z) | = e−Re π(u+z) = e−(πu 2 +2πRu+πR2

6 e−πu and

R∞ −∞

e−π(u−R)

2 +2πR2

2 +2πR2

= e−π(u−R)

,

du converges. So by Theorem 2.23, F is analytic on C.

Knowing that F is analytic on C, in order to prove that F (z) = 1 for z ∈ C it is sufficient to prove, for any set S ⊂ C with a limit point in C, that F (z) = 1 for z ∈ S. For the set S we take R. For z ∈ R we obtain, by substituting v = u + z, Z

∞

F (z) =

e

−π(u+z)2

Z

∞

du =

−πv 2

e

−∞

−∞

Z dv = 2

∞

0

Now a second substitution t = πv 2 yields F (z) = π

−1/2

Z 0

∞

e−t t−1/2 dt = π −1/2 Γ( 21 ) = 1.

137

2

e−πv dv.

9.3

The functional equation for the Riemann zeta function

Put ξ(s) := 12 s(s − 1)π −s/2 Γ( 12 s)ζ(s) = (s − 1)π −s/2 Γ( 21 s + 1)ζ(s), where we have used the identity 21 sΓ( 12 s) = Γ( 21 s + 1). Theorem 9.7. The function ξ has an analytic continuation to C. For this continuation we have ξ(1 − s) = ξ(s) for s ∈ C. Before proving this, we deduce some consequences. Corollary 9.8. The function ζ has an analytic continuation to C\{1} with a simple pole with residue 1 at s = 1. For this continuation we have ζ(1 − s) = 21−s π −s cos( 21 πs)Γ(s) · ζ(s) for s ∈ C \ {0, 1}. Proof. We define the analytic continuation of ζ by ζ(s) =

ξ(s)π s/2 · 1/Γ( 21 s + 1) . s−1

By Corollary 8.5, 1/Γ is analytic on C, and the other functions in the numerator are also analytic on C. Hence ζ is analytic on C \ {1}. The analytic continuation of ζ defined here coincides with the one defined in Theorem 5.2 on {s ∈ C : Re s > 0} \ {1} since analytic continuations to connected sets are uniquely determined. Hence ζ(s) has a simple pole with residue 1 at s = 1. We derive the functional equation. By Theorem 9.7 we have, for s ∈ C \ {0, 1}, ζ(1 − s) =

=

1 (1 2

ξ(1 − s) ξ(s) = 1 1 −(1−s)/2 −(1−s)/2 − s)(−s)π Γ( 2 (1 − s)) s(s − 1)π Γ( 21 (1 − s)) 2

− 1)π −s/2 Γ( 12 s) · ζ(s) = F (s)ζ(s), 1 s(s − 1)π −(1−s)/2 Γ( 21 (1 − s)) 2 1 s(s 2

138

say. Now we have Γ( 12 s)Γ( 21 s + 12 ) Γ( 21 − 12 s)Γ( 12 + 21 s) √ 21−s πΓ(s) (1/2)−s (by Corollary 8.12, Theorem 8.3) = π π/ sin(π( 12 − 12 s))

F (s) = π (1/2)−s ·

= π −s 21−s cos( 12 πs)Γ(s). This implies Corollary 9.8. Corollary 9.9. ζ has simple zeros at s = −2, −4, −6, . . .. ζ has no other zeros outside the critical strip {s ∈ C : 0 < Re s < 1}. Proof. We first show that ξ(s) 6= 0 if Re s > 1 or Re s 6 0. We use the second expression for ξ(s). By Corollary 5.4 and Theorem 4.5, we know that ζ(s) 6= 0 for s ∈ C with Re s > 1, s 6= 1. Further, lims→1 (s − 1)ζ(s) = 1, hence (s − 1)ζ(s) 6= 0 if Re s > 1. By Corollary 8.5, we know that Γ( 21 s + 1) 6= 0 if Re s > 1. hence ξ(s) 6= 0 if Re s > 1. But then by Theorem 9.7, ξ(s) 6= 0 if Re s 6 0. We consider ζ(s) for Re s 6 0. For s 6= −2, −4, −6, . . ., the function Γ( 12 s + 1) is analytic. Further, for these values of s, we have ξ(s) 6= 0, hence ζ(s) must be 6= 0 as well. The function Γ( 21 s) has simple poles at s = −2, −4, −6, . . .. To make ξ(s) analytic and non-zero for these values of s, the function ζ must have simple zeros at s = −2, −4, −6, . . .. Proof of Theorem 9.7 (Riemann). Let for the moment, s ∈ C, Re s > 1. Recall that Z ∞ 1 Γ( 2 s) = e−t t(s/2)−1 dt. 0

2

Substituting t = πn u gives Z ∞ Z −πn2 u 2 (s/2)−1 2 s/2 s 1 Γ( 2 s) = e (πn u) d(πn u) = π n 0

∞

0

Hence π

−s/2

Γ( 12 s)n−s

Z

∞

2

e−πn u u(s/2)−1 du,

= 0

and so, by summing over n, π

−s/2

Γ( 12 s)ζ(s)

=

∞ Z X n=1

0

139

∞

2u

e−πn

2

e−πn u u(s/2)−1 du.

· u(s/2)−1 du.

We justify that the infinite integral and infinite sum can be interchanged. We use the following special case of the Fubini-Tonelli theorem: if {fn : (0, ∞) → C}∞ n=1 R∞ P |f (u)|du converges, then is a sequence of measurable functions such that ∞ n n=1 P 0 R∞ all integrals 0 fn (u)du (n > 1) converge, the series ∞ n=1 fn (u) converges almost everywhere on (0, ∞) and moreover, ∞ Z X

∞

Z fn (u)du,

0

n=1

∞

0

∞ X

! fn (u) du

n=1

converge and are equal. In our situation we have that indeed (putting σ := Re s) ∞ Z X n=1

∞

|e

−πn2 u

(s/2)−1

·u

|du =

0

=

∞ Z X n=1

∞ X

∞

2

e−πn u u(σ/2)−1 du

0

π −σ/2 Γ( 21 σ)n−σ (reversing the above argument)

n=1

= π −σ/2 Γ( 21 σ)ζ(σ) converges. Thus, we conclude that for s ∈ C with Re s > 1, (9.1)

π −s/2 Γ( 12 s)ζ(s) =

Recall that θ(u) =

P∞

Z

∞

ω(u) · u(s/2)−1 du, where ω(u) =

0

2

e−πn u .

n=1 2u

n=−∞

∞ X

e−πn

= 1 + 2ω(u).

We want to replace the right-hand side of (9.1) by something that converges for every s ∈ C. Obviously, for s ∈ C with Re s < 0 there are problems if u ↓ 0. To R∞ R∞ R1 R1 overcome these, we split the integral 0 into 1 + 0 and then transform 0 into R∞ an integral 1 by means of a substitution v = u−1 . After this substitution, the integral contains a term ω(v −1 ). By Theorem 9.5, we have ω(v −1 ) = =

1 (θ(v −1 ) − 1) = 12 v 1/2 θ(v) − 21 2 1 1/2 v 2ω(v) + 1) − 12 = v 1/2 ω(v) 2

+ 21 v 1/2 − 12 .

We work out in detail the approach sketched above. We keep for the moment our 140

assumption Re s > 1. Thus, Z ∞ Z 1 − s (s/2)−1 1 π 2 Γ( 2 s)ζ(s) = ω(u)u du − 1

Z

ω(u)u(s/2)−1 du +

= 1

Z

∞

v 1/2 ω(v) + 21 v 1/2 −

1

1

Z

∞ 1 2

v

−(s+1)/2

ω(v −1 )v 1−s/2 dv −1

1

∞

=

∞

−v

−(s/2)−1


dv +

Z

1 2


1−s/2 −2 v v dv

∞


ω(v) v (s/2)−1 + v −(s+1)/2 dv

1

where we have combined the terms without ω into one integral, and the terms involving ω into another integral. Since we are still assuming Re s > 1, the first integral is equal to ∞  2 −s/2 1 2 −(s−1)/2 1 1 1 + v = v − = . − 2 s−1 s s − 1 s s(s − 1) 1 Hence π

−s/2

Γ( 12 s)ζ(s)

1 + = s(s − 1)

Z

∞


ω(v) v (s/2)−1 + v −(s+1)/2 dv.

1

For our function ξ(s) = 21 s(s − 1)π −s/2 Γ( 21 s)ζ(s) this gives Z ∞
1 1 (9.2) ξ(s) = 2 + 2 s(s − 1) ω(v) v (s/2)−1 + v −(s+1)/2 dv if Re s > 1. 1


R∞ Assume for the moment that F (s) := 1 ω(v) v (s/2)−1 + v −(s+1)/2 dv defines an analytic function on C. Then we can use the right-hand side of (9.2) to define the analytic continuation of ξ(s) to C. By substituting 1 − s for s in the right-hand side, we see that ξ(1 − s) = ξ(s). It remains to prove that F (s) defines an analytic function on C. We apply as
usual Theorem 2.23. We check that f (v, s) = ω(v) v (s/2)−1 + v −(s+1)/2 satisfies the conditions of that theorem. P −πn2 v a) f (v, s) is measurable on (1, ∞) × C. For ω(v) = ∞ is measurable, n=1 e being a pointwise convergent series of continuous, hence measurable functions, and also v (s/2)−1 + v −(s+1)/2 is measurable, since it is continuous.
b) s 7→ ω(v) v (s/2)−1 + v −(s+1)/2 is analytic on C for every fixed v. This is obvious. 141

c) Let K be a compact subset of C. Then there is a measurable function MK (v) R∞ on (1, ∞) such that |f (v, s)| 6 MK (v) for s ∈ K and 1 MK (v)dv < ∞. Indeed, choose A > 0 such that |Re s| 6 A for s ∈ K. we first have for v ∈ (1, ∞)
0 6 ω(v) 6 e−πv 1 + e−3πv + e−8πv 6 2e−πv and second, for v ∈ (1, ∞), s ∈ K, |v (s/2)−1 + v −(s+1)/2 | 6 v (A/2)−1 + v (−(A+1)/2 6 2v (A/2)−1 . Hence |f (v, s)| 6 4e−πv v (A/2)−1 =: MK (v). Further, Z

∞

Z MK (v)dv 6 4

1

0

∞

e−v v (A/2)−1) dv 6 4 · Γ( 21 A) < ∞.

So f (v, s) satisfies all conditions of Theorem 2.23, and it follows that F (s) = R∞ f (v, s)dv is analytic on C. 1

9.4

The functional equations for L-functions (q)

Let q be an integer > 2 and χ a Dirichlet character modulo q with χ 6= χ0 . We give, without proof, a functional equation for L(s, χ) in the case that χ is primitive, i.e., that it is not induced by a character modulo d for any proper divisor d of q. Notice that for any character χ modulo q we have χ(−1)2 = χ(1) = 1, hence χ(−1) ∈ {−1, 1}. A character χ is called even if χ(−1) = 1, and odd if χ(−1) = −1. There will be different functional equations for even and odd characters. In Chapter 4 we defined the Gauss sum related to a character χ mod q by τ (1, χ) =

q−1 X

χ(a)e2πia/q .

a=0

According to Theorem 4.19, if χ is primitive then |τ (1, χ)| = By χ we denote the complex conjugate of a character χ. 142

√ q.

Theorem 9.10. Let q be an integer with q > 2, and χ a primitive character mod q. Put √  q s/2 q 1 c(χ) := if χ is even, ξ(s, χ) := Γ( 2 s)L(s, χ), π τ (1, χ) √  q (s+1)/2
i q 1 ξ(s, χ) := Γ 2 (s + 1) L(s, χ), c(χ) := if χ is odd. π τ (1, χ) Then ξ(s, χ) has an analytic continuation to C, and ξ(1 − s, χ) = c(χ)ξ(s, χ) for s ∈ C. Remark. We know that |c(χ)| = 1. In general, it is a difficult problem to compute c(χ). The proof of Theorem 9.10 is similar to that of that of the functional equation for ζ(s), but with some additional technicalities, see H. Davenport, Multiplicative Number Theory, Chapter 9. In the next exercise we have collected some consequences. Exercise 9.1. Let q be an integer > 2 and χ a primitive character mod q. a) Prove that L(s, χ) has an analytic continuation to C. b) Prove the following: if χ is even, then L(s, χ) has simple zeros at s = 0, −2, −4, . . . and L(s, χ) 6= 0 if Re s < 0, s 6∈ {0, −2, −4, . . .}; if χ is odd, then L(s, χ) has simple zeros at s = −1, −3, −5, . . . and L(s, χ) 6= 0 if Re s < 0, s 6∈ {−1, −3, −5, . . .}. c) Prove a) and b) in the case that χ is non-principal, but not necessarily primitive.

143