Analytic Number Theory

Home , Dirichlet eta function, Euler product, Liouville function, Mertens function, Multiplication theorem, Prime omega function

Introduction to Analytic Number Theory

Selected Topics

Lecture Notes

Winter 2019/ 2020

Alois Pichler

Faculty of Mathematics

DRAFT Version as of April 23, 2021 2

Figure 1: Wizzard of Evergreen Terrace: Fermat’s last theorem wrong

rough draft: do not distribute Contents

1 Introduction 7

2 Deﬁnitions 9 2.1 Elementary properties...... 9 2.2 Fundamental theorem of arithmetic...... 12 2.3 Properties of primes...... 13 2.4 Chinese remainder theorem...... 14 2.5 Problems...... 16

3 Elementary number theory 19 3.1 Euler’s totient function...... 19 3.2 Euler’s theorem...... 21 3.3 Fermat Primality test...... 23 3.4 AKS Primality test...... 23 3.5 Problems...... 24

4 Continued fractions 25 4.1 Generalized continued fraction...... 25 4.2 Regular continued fraction...... 27 4.3 Elementary properties...... 29 4.4 Problems...... 29

5 Bernoulli numbers and polynomials 31 5.1 Deﬁnitions...... 31 5.2 Summation and multiplication theorem...... 32 5.3 Fourier series...... 33 5.4 Umbral calculus...... 34

6 Gamma function 35 6.1 Equivalent deﬁnitions...... 35 6.2 Euler’s reﬂection formula...... 36 6.3 Duplication formula...... 38

7 Euler–Maclaurin formula 39 7.1 Euler–Mascheroni constant...... 40 7.2 Stirling formula...... 40

8 Summability methods 43 8.1 Silverman–Toeplitz theorem...... 43 8.2 Cesàro summation...... 44 8.3 Euler’s series transformation...... 44 8.4 Summation by parts...... 45 8.4.1 Abel summation...... 46

3 4 CONTENTS

8.4.2 Lambert summation...... 46 8.5 Abel’s summation formula...... 47 8.6 Poisson summation formula...... 47 8.7 Borel summation...... 49 8.8 Abel–Plana formula...... 49 8.9 Mertens’ theorem...... 49

9 Multiplicative functions 51 9.1 Arithmetic functions...... 51 9.2 Examples of arithmetic functions...... 51 9.3 Dirichlet product and Möbius inversion...... 52 9.4 Möbius inversion...... 53 9.5 The sum function...... 54 9.6 Other sum functions...... 55 9.7 Problems...... 56

10 Euler’s product formula 57 10.1 Euler’s product formula...... 57 10.2 Problems...... 61

11 Dirichlet Series 63 11.1 Properties...... 63 11.2 Abscissa of convergence...... 63 11.3 Problems...... 64

12 Mellin transform 65 12.1 Inversion...... 65 12.2 Perron’s formula...... 67 12.3 Ramanujan’s master theorem...... 69 12.4 Convolutions...... 70

13 Riemann zeta function 73 13.1 Related Dirichlet series...... 73 13.2 Representations as integral...... 75 13.3 Globally convergent series...... 77 13.4 Power series expansions...... 79 13.5 Euler’s theorem...... 80 13.6 The functional equation...... 81 13.7 Riemann Hypothesis (RH)...... 85 13.8 Hadamard product...... 85

14 Further results and auxiliary relations 87 14.1 Occurrence of coprimes...... 87 14.2 Exponential integral...... 87 14.3 Logarithmic integral function...... 88 14.4 Analytic extensions...... 89

rough draft: do not distribute CONTENTS 5

15 Prime counting functions 91 15.1 Riemann prime counting functions and their relation...... 91 15.2 Chebyshev prime counting functions and their relation...... 93 15.3 Relation of prime counting functions...... 96

16 The prime number theorem 99 16.1 Zeta function on <(·) = 1 ...... 99 16.2 The prime number theorem...... 101 16.3 Consequences of the prime number theorem...... 102

17 Riemann’s approach by employing the zeta function 105 17.1 Mertens’ function...... 105 17.2 Chebyshev summatory function 휓 ...... 105 17.3 Riemann prime counting function...... 106 17.4 Prime counting function 휋 ...... 106

18 Further results 109 18.1 Results...... 109 18.2 Open problems...... 110

Bibliography 110

Version: April 23, 2021 6 CONTENTS

rough draft: do not distribute Introduction 1

Even before I had begun my more detailed investigations into higher arithmetic, one of my ﬁrst projects was to turn my attention to the decreasing frequency of primes, to which end I counted primes in several chiliads. I soon recognized that behind all of its ﬂuctuations, this frequency is on average inversely proportional to the logarithm.

Gauß, letter to Encke, Dec. 1849

For an introduction, see Zagier [21]. Hardy and Wright [10] and Davenport [5], as well as Apostol [2] are benchmarks for analytic number theory. Everything about the Riemann 휁 function can be found in Titchmarsh [18, 19] and Edwards [7]. Other useful references include Ivaniec and Kowalski [12] and Borwein et al. [4]. Some parts here follow the nice and recommended lecture notes Forster [8] or Sander [17]. This lecture note covers a complete proof of the prime number theorem (Section 16), which is based on a new, nice and short proof by Newman, cf. Newman [14], Zagier [22].

Please report mistakes, errors, violations of copyrights, improvements or necessary comple- tions. Updated version of these lecture notes: https://www.tu-chemnitz.de/mathematik/ﬁma/public/NumberTheory.pdf

Conjecture (Frank Morgan’s math chat). Suppose that there is a nice probability function 푃(푥) that a large integer 푥 is prime. As 푥 increases by Δ푥 = 1, the new potential divisor 푥 is prime with probability 푃(푥) and divides future numbers with probability 1/푥. Hence 푃 gets multiplied by (1 − 푃/푥), so that Δ푃 = (1 − 푃/푥)푃 − 푃 = −푃2/푥, or roughly

푃0 = −푃2/푥.

The general solution to this differential equation is

1 1 푃(푥) = ∼ . 푐 + log 푥 log 푥

7 8 INTRODUCTION

Figure 1.1: Gauss’ 1849 conjecture in his letter to the astronomer Johann Franz Encke, https://gauss.adw-goe.de/handle/gauss/199. Notably, all numbers 휋(·) are wrong in this letter: 휋(500 000) = 41 538, e.g.

rough draft: do not distribute Deﬁnitions 2

Before creation, God did just pure mathematics. Then he thought it would be a pleasant change to do some applied.

John Edensor Littlewood, 1885–1977

N = {1, 2, 3,... } the natural numbers P = {2, 3, 5,... } the prime numbers Z = {..., −2, −1, 0, 1, 2, 3 ... } the integers lcm: least common multiple, lcm(6, 15) = 30; sometimes also [푚, 푛] = lcm(푚, 푛) gcd: greatest common divisor, gcd(6, 15) = 3. We shall also write (6, 15) = 3. 푚 ⊥ 푛: the numbers 푚 and 푛 are co-prime, i.e., (푚, 푛) = 1 푓 (푥) 푓 ∼ 푔: the functions 푓 and 푔 satisfy lim푥 푔(푥) = 1.

푓 = O(푔): if there is 푀 > 0 such that | 푓 (푥)| ≤ 푀 · 푔(푥) for all 푥 > 푥0.

2.1 ELEMENTARY PROPERTIES

In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and that this product is unique, up to the order of the factors. Theorem 2.1 (Euclidean division1). Given two integers 푎, 푏 ∈ Z with 푏 ≠ 0, there exist unique integers 푞 and 푟 such that 푎 = 푏 푞 + 푟 and 0 ≤ 푟 < |푏|. Proof. Assume that 푏 > 0. Let 푟 > 0 be such that 푎 = 푏 푞 + 푟 (for example 푞 = 0, 푟 = 푎). If 푟 < 푏, then we are done. Otherwise, 푞2 B 푞1 + 1 and 푟2 B 푟1 − 푏 (with 푞1 B 푞, 푟1 B 푟) satisfy 푎 = 푏 푞2 + 푟2 and 0 ≤ 푟2 < 푟1. Repeating this process one gets eventually 푞 = 푞푘 and 푟 = 푟푘 such that 푎 = 푏 푞 + 푟 and 0 ≤ 푟 < 푏. If 푏 < 0, then set 푏0 B −푏 > 0 and 푎 = 푏0푞0 + 푟 with some 0 ≤ 푟 < 푏0 = |푏|. With 푞 B −푞0 it holds that 푎 = 푏0푞0 + 푟 = 푏푞 + 푟 and hence the result. Uniqueness: suppose that 푎 = 푏 푞 + 푟 = 푏 푞0 + 푟 0 with 0 ≤ 푟, 푟 0 < |푏|. Adding 0 ≤ 푟 < |푏| and −|푏| < −푟 0 ≤ 0 gives −|푏| < 푟 − 푟 0 ≤ |푏|, that is |푟 − 푟 0| ≤ |푏|. Subtracting the two equations yields 푏(푞0 − 푞) = 푟 − 푟 0. If |푟 − 푟 0| ≠ 0, then |푏| < |푟 − 푟 0|, a contradiction. Hence 푟 = 푟 0 and 푏(푞0 − 푞) = 0. As 푏 ≠ 0 it follows that 푞0 = 푞, proving uniqueness. 1Euclid, 300 bc

9 10 DEFINITIONS

Deﬁnition 2.2. If 푚 and 푛 are integers, and more generally, elements of an integral domain, it is said that 푚 divides 푛, 푚 is a divisor of 푛, or 푛 is a multiple of 푚, written as

푚 | 푛,

if there exists an integer 푘 ∈ Z such that 푚 푘 = 푛.

Remark 2.3. The following hold true.

(i) 푚 | 0 for all 푚 ∈ Z;

(ii) 0 | 푛 =⇒ 푛 = 0;

(iii) 1 | 푛 and −1 | 푛 for all 푛 ∈ Z;

(iv) if 푚 | 1, then 푚 = ±1;

(v) transitivity: if 푎 | 푏 and 푏 | 푐, then 푎 | 푐 and 푎 | 푏 푐 ;

(vi) if 푎 | 푏 and 푏 | 푎, then 푎 = 푏 or 푎 = −푏;

(vii) if 푎 | 푏 and 푎 | 푐, then 푎 | 푏 + 푐 and 푎 | 푏 − 푐.

Deﬁnition 2.4. A natural number 푛 ∈ N is composite if 푛 = 푘 · ℓ for 푘, ℓ ∈ N and 푘 > 1, ℓ > 1. A natural number 푛 > 1 which is not composite is called prime.

Composite numbers are {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24,... }, prime numbers are P = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 39, 41, 43, 47, 51,... }. Remark 2.5. Note that a number 푛 ∈ N is either composite, prime or the unit, 푛 = 1.

Deﬁnition 2.6. The greatest common divisor of the integers 푎1, . . . , 푎푛 is the largest natural number that dives all 푎푖, i.e.,

gcd(푎1, . . . , 푎푛) = max {푚 ≥ 1: 푚 | 푎1, . . . , 푚 | 푎푛} .

We shall also write (푎, 푏) B gcd(푎, 푏). Deﬁnition 2.7. The numbers 푎 and 푏 are coprime, relatively prime or mutually prime if (푎, 푏) = 1. We shall also write 푎 ⊥ 푏.

Lemma 2.8 (Bézout’s identity2). For every 푎, 푏 ∈ Z there exist integers 푥 and 푦 ∈ Z such that

푎 푥 + 푏 푦 = gcd(푎, 푏); (2.1)

more generally, {푎 푥 + 푏 푦 : 푥, 푦 ∈ Z} = {푧 · 푑 : 푧 ∈ Z}, where 푑 = gcd(푎, 푏).

Proof. Given 푎 ≠ 0 and 푏 ≠ 0, 푎, 푏 ∈ Z, deﬁne 푆 B {푎푥 + 푏푦 : 푥 ∈ Z, 푦 ∈ Z and 푎푥 + 푏푦 > 0}. The set is not empty, as 푎 ∈ 푆 or −푎 ∈ 푆 (indeed, choose 푥 = ±1 and 푦 = 0). By the well-ordering principle, there is a minimum element 푑 B min {푑0 : 푑0 ∈ S} = 푎 푠 + 푏 푡 ∈ 푆. We shall show that 푑 = gcd(푎, 푏). By Euclidean division we have that 푎 = 푑 푞 + 푟 with 0 ≤ 푟 < 푑. It holds that

푟 = 푎 − 푑 푞 = 푎 − (푎 푠 + 푏 푡)푞 = 푎(1 − 푠 푞) − 푏(푡 푞),

2Étienne Bézout, 1730–1783, proved the statement for polynomials.

rough draft: do not distribute 2.1 ELEMENTARY PROPERTIES 11 and thus 푟 ∈ 푆 ∪ {0}. As 푟 < 푑 = min S, it follows that 푟 = 0, i.e., 푑 | 푎; similarly, 푑 | 푏 and thus 푑 ≤ gcd(푎, 푏). Now let 푐 be any divisor of 푎 and 푏, that is, 푎 = 푐 푢 and 푏 = 푐 푣. Hence

푑 = 푎 푠 + 푏 푡 = 푐푢푠 + 푐푣푡 = 푐(푢푠 + 푣푡), that is 푐 | 푑 and therefore 푐 ≤ 푑, i.e., 푑 ≥ gcd(푎, 푏). Hence the result.

Corollary 2.9. There are integers 푥1, . . . , 푥푛 so that

gcd(푎1, . . . , 푎푛) = 푥1 푎1 + · · · + 푥푛 푎푛.

Corollary 2.10. It holds that 푎 and 푏 are coprime,

(푎, 푏) = 1 ⇐⇒ 푎 푠 + 푏 푡 = 1 (2.2) for some 푠, 푡 ∈ Z.

Proof. It remains to verify “ ⇐= ”: suppose that 푎 푠 + 푏 푡 = 1 and 푑 is a divisor of 푎 and 푏, i.e., 푎 = 푢 푑 and 푏 = 푣 푑. Then 1 = 푢푑푠 + 푣푑푡 = 푑(푢푠 + 푣푡), that is, 푑 | 1, thus 푑 = 1 and consequently (푎, 푏) = 1.

Lemma 2.11 (Euclid’s lemma). If 푝 ∈ P is prime and 푎 · 푝 a product of integers, then

푝 | (푎 · 푏) =⇒ 푝 | 푎 or 푝 | 푏.

Theorem 2.12 (Generalization of Euclid’s lemma). If

푛 | (푎 · 푏) and (푛, 푎) = 1, then 푛 | 푏.

Proof. We have from (2.2) that 푟 푛 + 푠 푎 = 1 for some 푟, 푠 ∈ Z. It follows that 푟 푛 푏 + 푠 푎 푏 = 푏. The ﬁrst term is divisible by 푛. By assumption, 푛 divides the second term as well, and hence 푛 | 푏.

The extended Euclidean algorithm (Algorithm1) computes the gcd of two numbers and the coefﬁcients in Bezout’s identity (2.1).

input :two integers 푎 and 푏 output :gcd(푎, 푏) and the coefﬁcients of Bézout’s identity (2.1) set (푑, 푠, 푡, 푑0, 푠0, 푡0) = (푎, 1, 0, 푏, 0, 1); initialize while 푑0 ≠ 0 do set 푞 = 푑 div 푑0 integer division set (푑, 푠, 푡, 푑0, 푠0, 푡0) = (푑0, 푠0, 푡0, 푑 − 푞 · 푑0, 푠 − 푞 · 푠0, 푡 − 푞 · 푡0) end return (푑, 푠, 푡) it holds that 푑 = gcd(푎, 푏) = 푎 푠 + 푏 푡 Algorithm 1: Extended Euclidean algorithm

Table 2.1 displays the results of the Euclidean algorithm for 푎 = 2490 and 푏 = 558: it holds that 13 ∗ 2490 − 58 ∗ 558 = 6 = gcd(푎, 푏).

Version: April 23, 2021 12 DEFINITIONS

푞 푑 푠 푡 푑0 푠0 푡0 2490 1 0 558 0 1 initialize 4 558 0 1 258 1 −4 2490/558 = 4 + 258/558 2 258 1 −4 42 −2 9 558/258 = 2 + 42/258 6 42 −2 9 6 13 −58 258/42 = 6 + 6/42 7 6 13 −58 0 −93 415 42/6 = 7 + 0

Table 2.1: Results of Algorithm1 for 푎 = 2490 and 푏 = 558

2.2 FUNDAMENTALTHEOREMOFARITHMETIC

The fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is a prime number itself or can be represented as the product of prime numbers and that, moreover, this representation is unique, up to the order of factors. Theorem 2.13 (Canonical representation of integers). Every integer 푛 > 1 has the unique representation 훼1 · · 훼휔 푛 = 푝1 ... 푝 휔 , (2.3)

where 푝1 < 푝2 < ··· < 푝 휔 and 훼푖 = 1, 2,... (훼푖 > 0). Proof. We need to show that every integer is prime or a product of primes. For the base case, 푛 = 2 is prime. Assume the assertion is true for all numbers < 푛. If 푛 is prime, there is nothing more to prove. Otherwise, 푛 is composite and there are integers so that 푛 = 푎푏 and 1 < 푎 ≤ 푏 < 푛. By induction hypothesis, 푎 = 푝1 ··· 푝 푗 and 푏 = 푞1 ··· 푞푘 , but then 푛 = 푝1 ··· 푝 푗 · 푞1 ··· 푞푘 is a product of primes. Uniqueness: suppose that 푛 is a product of primes which is not unique, i.e.,

푛 = 푝1 ··· 푝 푗

= 푞1 ··· 푞푘 .

Apparently, 푝1 | 푛. By Euclid’s lemma (Lemma 2.11) we have that 푝1 divides one of the 푞푖s, 푞1, say. But 푞1 is prime, thus 푞1 = 푝1. Now repeat the reasoning to

푛/푝1 = 푝2 ··· 푝 푗 and

= 푞2 ··· 푞푘 ,

so that 푞2 = 푝2, etc. Theorem 2.14 (Euclid’s theorem). There exist inﬁnity many primes.

Proof. Assume to the contrary that there are only ﬁnitely many primes, 푛, say, which are 푝1 = 2, 푝2 = 3, . . . , 푝푛. Consider the natural number

푞 B 푝1 · 푝2 · ... · 푝푛. If 푞 + 1 is prime, then the initial list of primes was not complete. If 푞 + 1 is not prime, then 푝 | (푞 + 1) for some prime number 푝. But 푝 | 푞, and hence, by Remark 2.3 (iv), it follows that 푝 | 1. But no prime number 푝 satisﬁes 푝 | 1 and hence 푞 + 1 is prime, but not in the list.

rough draft: do not distribute 2.3 PROPERTIESOFPRIMES 13

Theorem 2.15. Let 푝1 B 2, 푝2 B 3, etc. be the primes in increasing order. It holds that 2푛−3 푝푛 < 6 for 푛 ≥ 3.

Proof. The assertion is true for 푛 = 3. It follows from the proof of Theorem 2.14 that 푝푛+1 < 20 21 2푛−3 2푛−2 푝1 ··· 푝푛, thus 푝푛+1 < 2 · 3 · 6 · 6 ··· 6 = 6 , the result.

2.3 PROPERTIESOFPRIMES

Í푛−1 Deﬁnition 2.16 (Prime gap). The 푛-th prime gap is 푔푛 B 푝푛+1 − 푝푛. (I.e., 푝푛 = 2 + 푖=1 푔푖.) There exist prime gaps of arbitrary length.

Theorem 2.17 (Prime gap). For every 푔 ∈ N there exist 푛 ∈ N so that all 푛 + 1, . . . , 푛 + 푔 are not prime, i.e., there is an 푚 ∈ N so that 푝푚+1 ≥ 푝푚 + 푔.

Proof. Deﬁne 푛 B (푔 + 1)!, then 푑 | (푛 + 푑) for every 푑 ∈ {2, . . . , 푔 + 1}.

Remark 2.18. Alternatively, one may choose 푛 B lcm (1, 2, 3, . . . , 푔 + 1) in the preceding proof. Deﬁnition 2.19 (Prime gaps). Let 푝 ∈ P;

(i) if 푝 + 2 ∈ P, then they are called twin primes;

(ii) if 푝 + 4 ∈ P, they are called cousin primes;

(iii) if 푝 + 6 ∈ P, then they are called sexy primes.3

Note that every prime 푝 ∈ P is either 푝 = 2, or 푝 = 4푘 + 1 or 푝 = 4푘 + 3.

Theorem 2.20. There are inﬁnity many primes of the form 4푘 + 3.

Proof. Observe that

(4푘 + 1)·(4ℓ + 1) = 4 (4푘ℓ + 푘 + ℓ) + 1 and (2.4) (4푘 + 3)·(4ℓ + 3) = 4(4푘ℓ + 3푘 + 3ℓ + 2) + 1. (2.5)

By reductio ad absurdum, suppose that there are only ﬁnitely many primes in

P3 B {푝 ∈ P: 푝 = 4푘 + 3 for some 푘 ∈ N} = {3, 7, 11, 19,... } = {푝1, . . . , 푝푠 }.

2 2 Consider 푁 B 푝1 ··· 푝푠 + 2. By (2.5) and (2.4), 푁 = 4푘 + 3 and thus 2 - 푁. Let 푞1, . . . , 푞푟 = 푁 be the prime factors of 푁. As 2 - 푁, we have that 푞푖 = 4푘푖 + 1 or 푞푖 = 4푘푖 + 3. If all factors were of the form 푞푖 = 4푘푖 + 1, then 푁 = 4푘 + 1 by (2.4), but 푁 = 4푘 + 3. Hence there is some 푞푖 ∈ P3 for some 2 2 2 2 푖. But 푞푖 | 푁 and 푞푖 | 푝1 ··· 푝푠 (these are all primes in P3), thus 푞푖 | 푁 − 푝1 ··· 푝푠 , i.e., 푞푖 | 2 and this is a contradiction. The following theorem represents the beginning of rigorous analytic number theory.

Theorem 2.21 (Dirichlet’s theorem on arithmetic progressions, 1837). For any two positive coprime integers 푎, 푏 there are inﬁnity many primes of the form 푎 + 푛 푏, 푛 ∈ N.

Proposition 2.22. The following hold true:

3because 6 = sex in Latin

Version: April 23, 2021 14 DEFINITIONS

(i) if 2푛 − 1 ∈ P, then 푛 ∈ P;

(ii) if 2푛 + 1 ∈ P, then 푛 = 2푘 for some 푘 ∈ N.

ℓ ℓ Íℓ−1 푖 ℓ−푖−1 Proof. Suppose that 푛 ∉ P, thus 푛 = 푗·ℓ with 1 < 푗, ℓ < 푛. Recall that 푥 −푦 = (푥−푦)· 푖=0 푥 푦 , thus ℓ−1 ℓ ∑︁ 2푛 − 1 = 2 푗 − 1ℓ = 2 푗 − 1 · 2 푗·푖 푖=0 and thus 2 푗 − 1 | 2푛 − 1. But 2푛 − 1 ∈ P by assumption and so this contradicts the assertion (ii), as 1 < 푗 < 푛. As for (ii) suppose that 푛 ≠ 2푘 , then 푛 = 푗 · ℓ, with ℓ being odd and ℓ > 1. As above,

ℓ−1 ℓ ∑︁ 2푛 + 1 = 2 푗 − (−1)ℓ = 2 푗 − (−1) · (−1)ℓ−푖−1 2 푗·푖 푖=0

and hence 2 푗 +1 | 2푛 +1. The result follows by reductio ad absurdum, as 2푛 +1 ∈ P by assumption and 2 ≤ 푗 < 푛.

2푛 Deﬁnition 2.23 (Fermat4 prime). The Fermat numbers are 퐹푛 B 2 + 1. 퐹푛 is a Fermat prime, if 퐹푛 is prime.

As of 2019, the only known Fermat primes are 퐹0 = 3, 퐹1 = 5, 퐹2 = 17, 퐹3 = 257 and 퐹4 = 65 537. The factors of 퐹6, . . . , 퐹11 are known.

Theorem 2.24 (Gauss–Wantzel5 theorem). An 푛-sided regular polygon can be constructed with compass and straightedge if and only if 푛 is the product of a power of 2 and distinct Fermat 푘 primes: in other words, if and only if 푛 is of the form 푛 = 2 푝1 푝2 . . . 푝2, where 푘 is a nonnegative integer and the 푝푖 are distinct Fermat primes.

Deﬁnition 2.25 (Mersenne6 prime). Mersenne primes are prime numbers of the form 푀푛 B 2푛 − 1.

Some Mersenne primes include 푀2 = 3, 푀3 = 7, 푀5 = 31, 푀7 = 127, 푀13 = 8191. As of 2018, 51 Mersenne primes are known.

Proposition 2.26. It holds that

lcm(푎, 푏) ∗ gcd(푎, 푏) = 푎 ∗ 푏.

2.4 CHINESEREMAINDERTHEOREM

Deﬁnition 2.27. For 푚 ≠ 0 we shall say

푎 = 푏 mod 푚

if 푚 | (푎 − 푏).

4Pierre de Fermat, 1607–1665 5Pierre Wantzel, 1814–1848, French mathematician 6Marin Mersenne, 1588–1648, a French Minim friar

rough draft: do not distribute 2.4 CHINESEREMAINDERTHEOREM 15

Remark 2.28. By Definition 2.2 we have that 푎 = 푏 mod 푚 iff 푎 = 푏 + 푘 · 푚 for some 푘 ∈ Z. Proposition 2.29. It holds that (i) Reflexivity: 푎 = 푎 mod 푚 for all 푎 ∈ Z; (ii) Symmetry: 푎 = 푏 mod 푚, then 푏 = 푎 mod 푚 for all 푎, 푏 ∈ Z; (iii) Transitivity: 푎 = 푏 mod 푚 and 푏 = 푐 mod 푚, then 푎 = 푐 mod 푚; In what follows we assume that 푎 = 푎0 mod 푚 and 푏 = 푏0 mod 푚. (iv) 푎 ± 푏 = 푎0 ± 푏0 mod 푚; (v) 푎 푏 = 푎0 푏0 mod 푚; (vi) 푐 푎 = 푐 푎0 mod 푚 for all 푐 ∈ Z; (vii) (compatibility with exponentiation) 푎푘 = 푎0푘 mod 푚 for all 푘 ∈ N; (viii) 푝(푎) = 푝(푎0) mod 푚 for all polynomials 푝(·) with integer coefficients. Proposition 2.30. If gcd(푎, 푚) | 푐, then the problem 푎 푥 = 푐 mod 푚 has a solution (this is a possible converse to (vi)). 푠푐 푐 Proof. From (2.1) we have that 푑 B gcd(푎, 푚) = 푎푠+푚푡 and hence 푐 = 푎 푑 +푚푡 푑 . By assumption 푠푐 푐 we have that 푥 = 푑 is an integer and it follows that 푐 = 푎푥 + 푚 · 푡 푑 , that is, 푎푥 = 푐 mod 푚. Remark 2.31. Suppose that 0 < 푎 < 푝 for a prime 푝. Then 1 = gcd(푎, 푝) | 1 and hence the problem 푎푥 = 1 mod 푝 has a solution, irrespective of 푎. Algorithm1 provides numbers with 1 = 푎 푠 + 푝 푡, that is 푎−1 = 푠 mod 푝.

Theorem 2.32 (Chinese remainder theorem). Suppose that 푛1, . . . , 푛푘 are all pairwise coprime and 0 ≤ 푎푖 < 푛푖 for every 푖. Then there is exactly one integer with 푥 ≥ 0 and 푥 < 푛1 · ... · 푛푘 =: 푁 with

푥 = 푎1 mod 푛1, (2.6) . .

푥 = 푎푘 mod 푛푘 . (2.7)

Proof. The numbers 푛푖 and 푁푖 B 푁/푛푖 are coprime. Bézout’s identity provides integers 푀푖 and Í푛 푚푖 such that 푀푖 푁푖 + 푚푖 푛푖 = 1. Set 푥 B 푗=1 푎 푗 푀 푗 푁 푗 mod 푁. Recall that 푛푖 | 푁 푗 whenever 푖 ≠ 푗. Hence 푥 = 푎푖 푀푖 푁푖 = 푎푖 (1 − 푚푖 푛푖) = 푎푖 − 푎푖 푚푖 푛푖 mod 푛푖 and thus 푥 solves (2.6)–(2.7). Suppose there were two solutions 푥 and 푦 of (2.6)–(2.7). Then 푛푖 | 푥 − 푦 and, as the 푛푖 are all coprime, it follows that 푁 | 푥 − 푦. Hence 푥 = 푦, as it is required that 0 ≤ 푥, 푦 < 푁. Theorem 2.33 (Lucas’s theorem7). For integers 푛, 푘 and 푝 prime it holds that

ℓ 푛 Ö 푛 ≡ 푖 mod 푝, 푘 푘 푖=0 푖 ℓ ℓ−1 ℓ ℓ−1 where 푛 = 푛ℓ 푝 + 푛ℓ−1 푝 + · · · + 푛0 and 푘 = 푘ℓ 푝 + 푘ℓ−1 푝 + · · · + 푘0 are the base 푝 expansions with 0 ≤ 푛푖, 푘 < 푝. 7Édouard Lucas, 1842–1891, French mathematician

Version: April 23, 2021 16 DEFINITIONS

푝 푝·( 푝−1)···( 푝−푘+1) Proof. If 푝 is prime and 1 ≤ 푘 < 푝 an integer, then 푝 divides the numerator of 푘 = 푘·(푘−1)···1 , 푝 푝 푝 but not the denominator. Hence 푘 = 0 mod 푝. It follows that (1 + 푥) ≡ 1 + 푥 mod 푝. 푝푖 푖 푗 푗+1 Even more, 푝 | 푘 for every 0 < 푘 < 푝 . Indeed, for 푝 ≤ 푘 < 푝 it holds that

푝푖 푝푖 ··· 푝푖 − 푝 ··· 푝푖 − 푝2 ··· 푝푖 − 푝 푗 ··· 푝푖 − 푘 + 1 = . 푘 1 · 2 ··· 푝1 ··· 푝2 ··· 푝 푗 ··· 푘

The power of 푝 in the denominator are 1+2+· · ·+ 푗, the power of 푝 in the numerator is 푖+1+2+· · ·+ 푗 푝푖 푝푖 푝푖 and thus 푝 | 푘 . It follows that (1 + 푥) ≡ 1 + 푥 mod 푝. It follows that 푚 ℓ ∑︁ 푚 Ö 푖 푚푖 푥푘 = (1 + 푥)푚 = (1 + 푥) 푝 푘 푘=0 푖=0 ℓ ℓ 푚푖 Ö 푖 푚푖 Ö ∑︁ 푚 푖 ≡ 1 + 푥 푝 = 푖 푥푘푖 푝 푘푖 푖=0 푖=0 푘푖 =0 ℓ 푝−1 푚 ℓ Ö ∑︁ 푚 푖 ∑︁ Ö 푚푖 = 푖 푥푘푖 푝 = 푥푘 mod 푝 푘푖 푘푖 푖=0 푘푖 =0 푘=0 푖=0

where in the ﬁnal product, 푘푖 is the 푖th digit in the base 푝 representation of 푘. The statement of the theorem follows by comparing the coefﬁcients in the polynomial above. Proposition 2.34. It holds that

푛 푛 is prime ⇐⇒ ≡ 0 mod 푛 for all 푘 = 1, 2, . . . , 푛 − 1. 푘

푛 1 Proof. If 푛 = 푝 is prime, then 푘 = 푘 = 0 mod 푛 by Lucas’ theorem, as 푛 = 1 · 푝. 푛−1 Suppose that 푛 is composite and 푘 | 푛 is the smallest divisor. Then 푘−1 /푘 is not an integer. 푛 푛−1 Hence 푘 = 푛 · 푘−1 /푘 ≠ 0 mod 푛 and thus the equivalence.

2.5 PROBLEMS

Exercise 2.1. Verify the output −3 · 12 + 1 · 42 = 6 = gcd(12, 42) of Algorithm1.

Exercise 2.2. Show that 2−1 = 4, 3−1 = 5, 3−1 = 5, 4−1 = 2, 5−1 = 3 and 6−1 = 6 mod 7.

Exercise 2.3. Show that the set {3푘 + 2: 푘 ∈ Z} contains inﬁnity many primes.

Exercise 2.4. Show that {6푘 + 5: 푘 ∈ Z} contains inﬁnity many primes.

Exercise 2.5. If 푎푘 − 1 is prime (푘 > 1), then 푎 = 2.

Exercise 2.6. If 푎푘 + 1 is prime (푎, 푏 > 1), then 푎 is even and 푘 = 2푛 for some 푛 ≥ 1.

2 Exercise 2.7. Show that 퐹푛 = (퐹푛−1 − 1) + 1 and

2푛−1 (i) 퐹푛 = 퐹푛−1 + 2 퐹0 . . . 퐹푛−2,

2 2 (ii) 퐹푛 = 퐹푛−1 − 2(퐹푛−2 − 1) ,

rough draft: do not distribute 2.5 PROBLEMS 17

(iii) 퐹푛 = 퐹0 ··· 퐹푛−1 + 2. Exercise 2.8 (Goldbach’s theorem). No two Fermat numbers share a common integer factor greater than 1. (Hint: use (iii) in the preceding exercise.) Exercise 2.9. Solve this riddle, https://www.spiegel.de/karriere/das-rotierende-fuenfeck-raetsel- der-woche-a-1276765.html.

Version: April 23, 2021 18 DEFINITIONS

rough draft: do not distribute Elementary number theory 3

178212 + 184112 = 192212, 398712 + 436512 = 447212.

Fermat wrong

3.1 EULER’STOTIENTFUNCTION

Deﬁnition 3.1 (Euler’s totient function1). Euler’s totient function is ∑︁ 휑(푛) B 1 = {푚 ∈ {1, . . . , 푛} : (푚, 푛) = 1} . (3.1) 푚∈{1,...,푛} (푚,푛)=1 See Table 3.1 for some explicit values. Theorem 3.2 (휑 is multiplicative). It holds that 휑(푚 · 푛) = 휑(푚)· 휑(푛) if (푚, 푛) = 1.

Proof. Let 퐴푛 B {푖 ∈ N: 푖 < 푛 and (푖, 푛) = 1}. The Chinese remainder theorem (Theorem 2.32) provides a bijection 퐴푛 × 퐴푚 → 퐴푛·푚. Hence the result. Theorem 3.3. For 푝 prime it holds that 1 휑(푝푘 ) = 푝푘 − 푝푘−1 = 푝푘 1 − . 푝

Proof. The numbers 푚 ≤ 푝푘 which satisfy gcd 푝푘 , 푚 > 1 are precisely 푚 = 푝, 2 · 푝, . . . , 푝푘−1 · 푝, in total 푝푘−1 numbers. Hence 휑(푝푘 ) = 푝푘 − 푝푘−1, the result. Theorem 3.4 (Euler’s product formula). It holds that Ö 1 휑(푛) = 푛 · 1 − ; 푝 푝 |푛 in particular 휑(푝) = 푝 − 1. (3.2) 훼1 · 훼휔 Proof. The fundamental theorem of arithmetic (Theorem 2.13) states that 푛 = 푝1 . . . 푝 휔 . Now 휔 휔 휔 Ö Ö 1 Ö 1 휑(푛) = 휑 푝 훼푖 = 푝 훼푖 1 − = 푛 1 − . 푖 푖 푝 푝 푖=1 푖=1 푖 푖=1 푖 1Eulersche Phi-Funktion, eulersche Funktion

19 20 ELEMENTARYNUMBERTHEORY

푛 휏 휎 휑 휆 휇 휔 Ω 휆 휋 퐵푛 0 ———————— 0 1 1 1 1 1 1 1 1 0 0 1 0 − 2 1 2 2 3 1 1 −1 1 1 −1 1 6 3 2 4 2 2 −1 1 1 −1 2 0 1 4 3 7 2 2 0 1 2 1 2 − 30 5 2 6 4 4 −1 1 1 −1 3 0 1 6 4 12 2 2 1 2 2 1 3 42 7 2 8 6 6 −1 1 1 −1 4 0 1 8 4 15 4 2 0 1 3 −1 4 − 30 9 3 13 6 6 0 1 2 1 4 0 5 10 4 18 4 4 1 2 2 1 4 66 11 2 12 10 10 −1 1 1 −1 5 0 691 12 6 28 4 2 0 2 3 −1 5 − 2730 13 2 14 12 12 −1 1 1 −1 6 0 7 14 4 24 6 6 1 2 2 1 6 6 15 4 24 8 4 1 2 2 1 6 0 3617 16 5 31 8 4 0 1 4 1 6 − 510 17 2 18 16 16 −1 1 1 −1 7 0 43 867 18 6 39 6 6 0 2 3 −1 7 798 19 2 20 18 18 −1 1 1 −1 8 0 174 611 20 6 42 8 4 0 2 3 −1 8 − 330 21 4 32 12 6 1 2 2 1 8 0 854 513 22 4 36 10 10 1 2 2 1 8 138 23 2 24 22 22 −1 1 1 −1 9 0 236 364 091 24 8 60 8 2 0 2 4 1 9 − 2730 25 3 31 20 20 0 1 2 1 9 0 8 553 103 26 4 42 12 12 1 2 2 1 9 6 27 4 40 18 18 0 1 3 −1 9 0 23 749 461 029 28 6 56 12 6 0 2 3 −1 9 − 870 29 2 30 28 28 −1 1 1 −1 10 0 8 615 841 276 005 30 8 72 8 4 −1 3 3 −1 10 14 322 31 2 32 30 30 −1 1 1 −1 11 0

Table 3.1: Arithmetic functions

rough draft: do not distribute 3.2 EULER’STHEOREM 21

Theorem 3.5 (Gauss). It holds that ∑︁ 휑(푑) = 푛. 푑 |푛

1 1 3 1 1 3 7 2 9 1 11 3 13 7 3 4 17 9 19 1 Proof. Consider the fractions 20 , 10 , 20 , 5 , 4 , 10 , 20 , 5 , 20 , 2 , 20 , 5 , 20 , 10 , 4 , 5 , 20 , 10 , 20 , 1 . The fractions with denominator 푛 = 20 are precisely those with nominator coprime to 20, that is 휑(20) = 8 in total. There are 휑(10) = 4 fractions with denominator 10, and 휑(5) = 4 fractions with denominator 5, 휑(4) = 2, 휑(2) = 1 and 휑(1) = 1. In total, there are 푛 = 20 fractions, and thus Í 푛 = 푑 |푛 휑(푑).

3.2 EULER’STHEOREM

× The multiplicative group of modulo 푛 is (Z/푛Z) B {[푎]푛 : gcd(푎, 푛) = 1}. × Proposition 3.6. It holds that (Z/푛Z) = 휑(푛). 훼1 ··· 훼휔 Proof. Suppose that 푛 = 푝1 푝 휔 . By the Chinese remainder theorem (Theorem 2.32) we have that / / 훼1 × · · · × / 훼휔 Z 푛Z Z 푝1 Z Z 푝 휔 Z. Similarly, × × ( / )× / 훼1 × · · · × / 훼휔 Z 푛Z Z 푝1 Z Z 푝 휔 Z . × Z/ 훼푖 Z = 훼푖 − 훼푖 −1 = ( 훼푖 ) But 푝푖 푝푖 푝푖 휑 푝푖 and thus the result. Theorem 3.7 (Lagrange’s theorem). Let (퐺, ·) be a ﬁnite group. The order (number of elements) of every subgroup 퐻 of 퐺 divides the order of 퐺.

Proof. For 푎 ∈ 퐺 deﬁne the (left) cosets 푎퐻 B {푎ℎ : ℎ ∈ 퐻}. For the functions

푓푎,푏 : 푎퐻 → 푏퐻 푥 ↦→ 푏푎−1푥

−1 it holds that 푓푎,푏 = 푓푏,푎: the functions are invertible and bijective and thus the number of elements of 푎퐻 and 푏퐻 coincide for all 푎, 푏 ∈ 퐺. It follows that the order of 퐺 is the order of 퐻 × the number of distinct cosets. Theorem 3.8 (Euler). It holds that

푎 휑(푛) = 1 mod 푛 iff gcd(푎, 푛) = 1. (3.3)

Proof. If (푎, 푛) = 1 it holds that 푎, 푎2, . . . 푎푘 = 1 mod 푛 is a subgroup of (Z/푛Z)×. By Lagrange’s theorem we have that 푘 | 휑(푛), i.e., 푘 푀 = 휑(푛). It follows that

푎 휑(푛) = 푎푘 푀 = 푎푘 푀 = 1푀 = 1 mod 푛

and thus the result. Corollary 3.9. If 푝 is prime and 푝 - 푎, then

푎 푝−1 = 1 mod 푝. (3.4)

Proof. Apply (3.3) with (3.2).

Version: April 23, 2021 22 ELEMENTARYNUMBERTHEORY

Corollary 3.10 (Fermat’s little theorem). If 푝 is prime, then

푎 푝 = 푎 mod 푝.

Proof. If 푝 | 푎, then 푎 = 0 mod 푝 and the assertion is immediate; if 푝 - 푎, then the assertion follows from (3.4). Corollary 3.11 (Attributed to Euler). If 푝 is an odd prime, it holds that

푝−1 푎 2 = ±1 mod 푝, (3.5)

provided that (푎, 푝) = 1.

푝−1 푝−1 푝−1 푝−1 Proof. Indeed, it holds that 푎 2 − 1 푎 2 + 1 = 푎 − 1 = 0 mod 푝. Hence, 푝 | 푎 2 − 1 or 푝−1 푝 | 푎 2 + 1 and thus the result. Deﬁnition 3.12 (Legendre symbol). Let 푝 be an odd prime and 푎 an integer. The Legendre symbol is 푎 푝−1 푎 B 푎 2 mod 푝 and ∈ {−1, 0, +1}. 푝 푝 Deﬁnition 3.13 (Jacobi symbol). Let 푛 be any integer and 푎 an integer. The Jacobi symbol is

푎 훼1 푎 훼휔 · ... · , 푝1 푝 휔

훼1 · · 훼휔 where 푛 = 푝1 ... 푝 휔 as in (2.3). Deﬁnition 3.14 (Multiplicative order). The multiplicative order of 푎 ∈ Z modulo 푛 is the smallest positive integer 푘 = ord푛 (푎) with 푎푘 = 1 mod 푛.

1 2 3 Example 3.15. It holds that ord7 (4) = 3, as 4 = 4 mod 7, 4 = 2 mod 7, but 4 = 64 = 1 mod 7. Deﬁnition 3.16 (Carmichael function). The function 휆(푛), the smallest positive integer so that

푎휆(푛) = 1 for all 푎 with (푎, 푛) = 1

(cf. (3.3) and Table 3.1), is called Carmichael function.2

훼1 ··· 훼휔 Corollary 3.17 (Carmichael). For 푛 = 푝1 푝 휔 deﬁne ( ) ( − ) 훼1−1 ( − ) 훼휔 −1 휆 푛 B lcm 푝1 1 푝1 ,..., 푝 휔 1 푝 휔

then 푎휆(푛) = 1 mod 푛. 훼푖 훼 훼 휑( 푝푖 ) 푖 푖 Proof. By Euler’s theorem, 푎 = 1 mod 푝푖 . Note, that 휑(푝푖 ) | 휆(푛) by deﬁnition. It follows 휆(푛) 훼푖 휆(푛) that 푎 = 1 mod 푝푖 for all 푖 and thus 푎 = 1 mod 푛. Proposition 3.18. For all 푛 and 푎 it holds that

ord푛 (푎) | 휆(푛) and 휆(푛) | 휑(푛). 2Robert Daniel Carmichael, 1879–1967, American mathematician

rough draft: do not distribute 3.3 FERMAT PRIMALITY TEST 23

Theorem 3.19 (Wilson’s theorem3). It holds that

(푛 − 1)! = −1 mod 푛 iff 푛 is prime.

Proof. The statement is clear for 푛 = 2, 3 and 4. Suppose that 푛 is composite, 푛 = 푎 푏 with 푎 < 푏, then 푎 | (푛 − 1)! and 푏 | (푛 − 1)! and thus (푛 − 1)! = 0 mod 푛. If 푛 = 푞2, then 푞 | 1 · 2 ··· 푞 ··· 2푞 ···(푛 − 1) and thus (푛 − 1)! = 0 mod 푛. Suppose now that 푛 =: 푝 is prime. Note that (푝 − 1)−1 = 푝 − 1 = −1 mod 푝 (indeed, (푝 − 1)2 = 푝2 − 2푝 + 1 = 1 mod 푝). Further, the product 2 ···(푝 − 2) has an even number of factors and for each 푟 ∈ {2, . . . , 푝 − 2} it holds that 푟−1 ∈ {2, . . . , 푝 − 2}. Thus Wilson’s theorem.

3.3 FERMAT PRIMALITY TEST

By (3.5) we have that 푎 푝−1 = 1 mod 푝 if (푎, 푝) = 1. Note as well that the equation 푎푛−1 = 1 mod 푛 trivially holds true for 푎 = 1 and it is also trivial for 푎 = 푛 − 1 = −1 mod 푛 and 푛 odd.

Deﬁnition 3.20 (Fermat pseudoprime). Suppose that 푎 ∈ {2, 3, . . . , 푝 − 2}. If 푎푛−1 = 1 mod 푛 when 푛 is composite, then 푎 is known as Fermat liar. In this case 푛 is called a Fermat pseudoprime to base 푎.

Deﬁnition 3.21 (Carmichael number). A composite number 푛 is a Carmichael number if 푎푛−1 = 1 mod 푛 for all integers 푎.

Example 3.22. The ﬁrst Carmichael number was given by Carmichael in 1910. The ﬁrst are 561 = 3 ∗ 11 ∗ 17, 1105 = 5 ∗ 13 ∗ 17, 1729 = 7 ∗ 13 ∗ 19, 2465 = 5 ∗ 17 ∗ 29, 2821 = 7 ∗ 13 ∗ 31, 6601 = 7 ∗ 23 ∗ 41, 8911 = 7 ∗ 19 ∗ 67.

Theorem 3.23. There are inﬁnitely many Camichael numbers.

Theorem 3.24 (Korselt’s criterion,4 1899). A positive composite integer 푛 is a Carmichael number iff 푛 is square–free and for all prime divisors 푝 of 푛, it is true that 푝 − 1 | 푛 − 1.

3.4 AKSPRIMALITYTEST

A starting point for the AKS primality test5 (Algorithm2) is the following theorem.

Theorem 3.25 (AKS). It holds that

푛 is prime ⇐⇒ (푥 + 푎)푛 ≡ 푥푛 + 푎 mod 푛

for all (푎, 푛) = 1.

Proof. Proposition 2.34 together with Fermat’s little theorem.

Remark 3.26 (Primes is P). The AKS algorithm was the ﬁrst to determine whether any given number is prime or composite within polynomial time.

3John Wilson, 1741–1793, British mathematician 4Alwin Reinhold Korselt, 1864 (Mittelherwigsdorf)–1947 (Plauen) 5Also known as Agrawal–Kayal–Saxena primality test; published by Manindra Agrawal, Neeraj Kayal and Nitin Sax- ena in 2002.

Version: April 23, 2021 24 ELEMENTARYNUMBERTHEORY

input :an integer 푛 output :푛 is prime or 푛 is composite if 푛 = 푎푘 for some 푎 ∈ N and 푘 > 1 then return 푛 is composite end 2 ﬁnd smalles 푟 such that ord푟 (푛) > log(푛) . if 1 < (푎, 푛) < 푛 for some 푎 < 푟 then return 푛 is composite end if 푎 < 푛 then return 푛 is prime end for 푎 = 1 to √︁휑(푟) log 푛 do if (푥 + 푎)푛 ≠ 푥푛 + 푎 mod (푥푟 − 1, 푛) then return 푛 is composite end end return 푛 is prime Algorithm 2: AKS primality test

3.5 PROBLEMS

Exercise 3.1. Show that the function 휔 is additive (in the sense of Deﬁnition 9.2 below), while Ω is completely additive.

Exercise 3.2. Discuss additivity/ multiplicativity for the other arithmetic functions introduced in this section. Í푛 Í푛 푛 Í푛 Í푛 푛 Exercise 3.3. Show that 푘=1 휏(푘) = 푘=1 푘 and 푘=1 휎(푘) = 푘=1 푘 푘 .

rough draft: do not distribute Continued fractions 4

80 435 758 145 817 5153 − 80 538 738 812 075 9743 + 12 602 123 297 335 6313 = 42.

The Hitchhiker’s Guide to the Galaxy

For the standard theory see Wall [20] or Duverney [6]; for the relation to orthogonal polynomials see Khrushchev [13].

4.1 GENERALIZEDCONTINUEDFRACTION

Deﬁnition 4.1 (Continued fraction). Different notations for the (generalized) continued fraction

푎1 푟푛 B 푏0 + (4.1) 푎2 푏1 + . . . . . 푏2 + . 푎푛 푏푛 + 푏푛

include

푛 푎푖 푎1 | 푎2 | 푎푛 | 푎1 푎2 푎푛 푟푛 = 푏0 + K = 푏0 + + + · · · + = 푏0 + ... . 푖=1 푏푖 | 푏1 | 푏2 | 푏푛 푏1 + 푏2 + 푏푛

Theorem 4.2. For every 푛 ≥ 0 we have that 푟 = 푝푛 , where 푛 푞푛

푝 = 1, 푝 = 푏 , 푝 = 푏 푝 + 푎 푝 , −1 0 0 푛 푛 푛−1 푛 푛−2 (4.2) 푞−1 = 0, 푞0 = 1, 푞푛 = 푏푛푞푛−1 + 푎푛푞푛−2.

Proof. Observe ﬁrst that 푟 = 푏 = 푝0 and 푟 = 푏 + 푎1 = 푏1푏0+푎1 = 푝1 . For 푛 = 2 we obtain 0 0 푞0 1 0 푏1 푏1 푞1 푎1 푏2 (푏1푏0+푎1)+푎2푏0 푝2 푟2 = 푏0 + 푎2 = = . We continue by induction. It holds that 푏1+ 푏2푏1+푎21 푞2 푏2

0 푎1 푝푛 푟푛 = 푏0 + =: 0 . 푎푛−1 푞푛 푏1 + ... 푎푛 푏푛−1 + 푎푛+1 푏푛 + 푏푛+1

25 26 CONTINUEDFRACTIONS

푎푛+1 Replacing 푏푛 in (4.1) by 푏푛 + we obtain by the induction hypothesis that 푏푛+1 ! 0 푎푛+1 푝푛 = 푏푛 + 푝푛−1 + 푎푛 푝푛−2 and 푏푛+1 ! 0 푎푛+1 푞푛 = 푏푛 + 푞푛−1 + 푎푛푞푛−2, 푏푛+1

which implies

0 푏푛+1 푝푛 = 푏푛+1 (푏푛 푝푛−1 + 푎푛 푝푛−2) + 푎푛+1 푝푛−1 and 0 푏푛+1푞푛 = 푏푛+1 (푏푛푞푛−1 + 푎푛푞푛−2) + 푎푛+1푞푛−1,

and, again by the induction hypothesis,

0 푏푛+1 푝푛 = 푏푛+1 푝푛 + 푎푛+1 푝푛−1 and 0 푏푛+1푞푛 = 푏푛+1푞푛 + 푎푛+1푞푛−1.

0 0 푝푛+1 Now set 푝 + 푏 + 푝 and 푞 + 푏 + 푞 to get 푟 + = . 푛 1 B 푛 1 푛 푛 1 B 푛 1 푛 푛 1 푞푛+1

Theorem 4.3 (Determinant formula). For every 푛 ≥ 1 it holds that

푛 푝푛−1 푞푛 − 푝푛 푞푛−1 = (−1) 푎1푎2 . . . 푎푛 and (4.3) 푛−1 푎1푎2 . . . 푎푛 푟푛 − 푟푛−1 = (−1) if 푞푛푞푛−1 ≠ 0; 푞푛−1 푞푛

further it holds that

푛−1 푝푛−2 푞푛 − 푝푛 푞푛−2 = (−1) 푎1푎2 . . . 푎푛−1 · 푏푛 and

푛 푎1푎2 . . . 푎푛 · 푏푛 푟푛 − 푟푛−2 = (−1) . 푞푛 푞푛−2

Proof. For 푛 = 0, the statement reads 1 · 1 − 푏0 · 0 = 1, which is the assertion. By induction and (4.2),

푝푛−1푞푛 − 푝푛푞푛−1 = 푝푛−1 (푏푛푞푛−1 + 푎푛푞푛−2) − (푏푛 푝푛−1 + 푎푛 푝푛−2)푞푛−1

= −푎푛 (푝푛−2푞푛−1 − 푝푛−1푞푛−2) .

Further we have that

푝푛−2푞푛 − 푝푛푞푛−2 = 푝푛−2 (푏푛푞푛−1 + 푎푛푞푛−2) − (푏푛 푝푛−1 + 푎푛 푝푛−2) 푞푛−2

= 푏푛 (푝푛−2푞푛−1 − 푝푛−1푞푛−2) 푛−1 = (−1) 푎1푎2 . . . 푎푛−1 · 푏푛

by (4.3). The remaining assertions follow by dividing accordantly.

Corollary 4.4. It holds that 푟 = 푏 − Í푛 (−1)푘 푎1 푎2...푎푘 . 푛 0 푘=1 푞푘−1 푞푘

rough draft: do not distribute 4.2 REGULARCONTINUEDFRACTION 27

4.2 REGULARCONTINUEDFRACTION

Remark 4.5 (Equivalence transformation). For any sequence 푐푖 with 푐푖 ≠ 0 it holds that

푎1 푐1푎1 푟푛 = 푏0 + = 푏0 + . 푎2 푐1푐2푎2 푏1 + 푐1푏1 + 푎3 푐2푐3푎3 푏2 + 푐2푏2 + 푎4 푐3푐4푎4 푏3 + 푐3푏3 + 푏4 + ... 푐4푏4 + ...

푛 1 푎1 푎2 1 푎푖 If we choose 푐1 B , 푐2 B , 푐3 = and generally 푐푛+1 B we get that 푟푛 = 푏0 + K = 푎1 푎2 푎1 푎3 푐푛 푎푛+1 푖=1 푏푖 푛 1 푏0 + K . 푖=1 푐푖 푏푖

Deﬁnition 4.6 (Regular continued fraction). The continued fraction with 푎푖 = 1 is called regular. The convergents of a regular continued fraction are denoted 푝푛 =: [푏 ; 푏 , . . . , 푏 ]. 푞푛 0 1 푛

푏 푏 (푏 푏 +1+푏 +(푏 푏 +1) Remark 4.7. The ﬁrst convergents are 푏 , 푏1푏0+1 , 푏2 (푏1푏0+1)+푏0 , 3 2 1 0 0 1 0 , etc. 0 푏1 푏2푏1+1 푏3 (푏2푏1+1)+푏1 Remark 4.8. Note, that [푏0; 푏1, . . . , 푏푛−1, 푏푛, 1] = [푏0; 푏1, . . . , 푏푛−1, 푏푛 + 1] and the continued fraction of a rational thus is not unique.

Note that the denominators of a regular continued fracttion satisfy 푝푛−1푞푛 − 푝푛푞푛−1 = ±1 by (4.3) and thus they are relatively prime, gcd(푞푛, 푞푛−1) = 1, by Bézout’s identity (2.1) and Corollary 2.10. By the same equality it follows that 푝 and 푞 are coprime as well, so 푟 = 푝푛 is 푛 푛 푛 푞푛 free of common factors.

Lemma 4.9 (Reciprocals). It holds that [0; 푏 , . . . 푏 ] = 1 . 1 푛 [푏1; 푏2,...푏푛 ] Consider a real number 푟. Let 푖 B b푟c be the integer part of 푟 and 푓 B 푟 − 푖 be the fractional part of 푟. Then the continued fraction representation of 푟 is [푖; 푎1, 푎2,...], where 1/ 푓 = [푎1; 푎2,...] is the continued fraction representation of 1/ 푓 . Example 4.10 (Euclid’s algorithm). Table 2.1 displays the extended Euclidean algorithm (Algo- 0 rithm1) for 푎 = 2490 and 푏 = 558. The successive fractions of −푡 /푠0 are

4, 1 4 + = 4.5, 2 1 58 4 + = = 4.4615 ..., 1 13 2 + 6 1 415 2490 4 + = = = 4.4623 ..., 1 2 + 1 93 558 6+ 7

푎 which are improved approximations of 푏 . The sequence of integer quotients is 푏0 = 4, 푏1 = 2, 푏2 = 6 and 푏3 = 7, cf. Table 2.1.

Remark 4.11. Euclid’s algorithm (Algorithm1) produces for 푎/푏 and for 푚푎/푚푏 the same sequence 푞 of integer quotients (푚 ∈ Z\{0}). For the reduced fraction 푎/푏 we have that 푠 푎 + 푡 푏 =

Version: April 23, 2021 28 CONTINUEDFRACTIONS

gcd(푎, 푏) = 1, the last coefficient 푏푛+1 in the continued fraction of 푎/푏 thus is 푏푛+1 = 1, i.e., 푎/푏 = [푏0; 푏1, . . . , 푏푛−1, 푏푛, 1]. Every infinite continued fraction is irrational, and every irrational number can be represented in precisely one way as an infinite continued fraction.

Example 4.12. Applying the Euclidean algorithm to 푎 = 휋 and 푏 = 1 gives the successive 22 333 355 approximations 3, 7 = 3.142 ... , 106 = 3.141 50 ... and 113 = 3.141 592 9 ... Theorem 4.13 (Legendre’s best approximation I). The convergents are best approximations, 푝푛 i.e., for 훼 lim →∞ it holds that B 푛 푞푛

|푞 훼 − 푝| < |푞푛 훼 − 푝푛 | =⇒ 푞 > 푞푛,

where 푝 ∈ Z and 푞 ∈ N.

Proof. Assume, by contraposition, that

|푞 훼 − 푝| < |푞푛 훼 − 푝푛 | and 푞 ≤ 푞푛. (4.4)

푝 푝 푥 푝 푥 푞 −푝 푝 The equations 푛 푛+1 = have integral solution = ± 푛+1 푛+1 ∈ Z2, 푞푛 푞푛+1 푦 푞 푦 −푞푛 푝푛 푞 as the determinant is 푝푛푞푛+1 − 푝푛+1푞푛 = ±1 by (4.3).

⊲ If 푦 = 0, then 푥 ≠ 0 and 푝 = 푝푛푥 and 푞 = 푞푛푥. It follows that |푞 훼 − 푝| = |푥| · |푞푛 훼 − 푝푛 | ≥ |푞푛 훼 − 푝푛 |, which contradicts the assumption (4.4).

⊲ If 푥 = 0, then 푦 ≠ 0 and 푞 = 푞푛+1푦, but from the assumption (4.4) we have that 푞 ≤ 푞푛 < 푞푛+1. So we conclude that 푥 ≠ 0 and 푦 ≠ 0.

⊲ If 푥 < 0 and 푦 < 0, then 0 < 푞 = 푞푛푥 + 푞푛+1푦, which cannot hold true.

⊲ If 푥 > 0 and 푦 > 0, then 푞푛푥 + 푞푛+1푦 = 푞 ≤ 푞푛 < 푞푛+1, which cannot hold true.

Hence 푥 and 푦 have opposite signs. Recall from (4.3) that 푞푛훼 − 푝푛 and 푞푛+1훼 − 푝푛+1 have opposite signs as well. We further have that

푞훼 − 푝 = 푥 (푞푛훼 − 푝푛) + 푦 (푞푛+1훼 − 푝푛+1) ,

where the products 푥 (푞푛훼 − 푝푛) and 푦 (푞푛+1훼 − 푝푛+1) have the same sign by the above reasoning. It follows that

|푞훼 − 푝| = |푥 (푞푛훼 − 푝푛)| + |푦 (푞푛+1훼 − 푝푛+1)|

≥ |푥 (푞푛훼 − 푝푛)|

≥ |푞푛훼 − 푝푛 | ,

again a contradiction to (4.4). It follows that 푞 > 푞푛.

Corollary 4.14 (Legendre’s best approximation II). The convergents are best approximations, 푝푛 i.e., with 훼 lim →∞ it holds that B 푛 푞푛

푝 푝푛 훼 − < 훼 − =⇒ 푞 > 푞푛. (4.5) 푞 푞푛

rough draft: do not distribute 4.3 ELEMENTARY PROPERTIES 29

Proof. Assume that 푞 ≤ 푞푛. Then we may multiply with (4.5) to obtain |푞 훼 − 푝| < |푞푛 훼 − 푝푛 |, but the preceding theorem implies 푞 > 푞푛. This contradicts the assumption and hence the assertion.

푝푛 Theorem 4.15. For 훼 lim →∞ it holds that B 푛 푞푛

푏 푝 1 1 푛+2 < 훼 − 푛 < < . 2 푞푛 푞푛+2 푞푛 푞푛 푞푛+1 푞푛 Further, for 훼 irrational we have that 푝0 < 푝2 < ··· < 훼 < ··· < 푝3 < 푝1 . 푞0 푞2 푞3 푞1

푝푛 Proof. The assertion follows from (4.3), as the convergents 푟 = , 푟 + and 푟 + oscillate 푛 푞푛 푛 1 푛 2 around 훼.

4.3 ELEMENTARY PROPERTIES

Theorem 4.16 (Gauss’s continued fraction). Let 푓0, 푓1, 푓2,... be a sequence of functions so that 푓푖−1 (푧) − 푓푖 (푧) = 푘푖 푧 푓푖+1 (푧), then

푓 (푧) 1 1 = . 푓0 (푧) 푘1푧 1 + 푘2푧 1 + 푘3푧 1 + 1 + ...

4.4 PROBLEMS

√ 1+ 5 Exercise 4.1 (Golden ratio). Show that 휙 = 2 = [1; 1, 1, 1,... ]. √ Exercise 4.2. Show that 2 = [1; 2, 2, 2,... ]. √ Exercise 4.3. Show that 5 = [2; 4, 4, 4,... ].

Version: April 23, 2021 30 CONTINUEDFRACTIONS

rough draft: do not distribute Bernoulli numbers and polynomials 5

1729 = 13 + 123 = 93 + 103.

Taxicab

5.1 DEFINITIONS

Definition 5.1. Bernoulli numbers1 퐵푘 , 푘 = 0, 1,... , are defined by 푧 1 = 푒푧 − 1 푧 푧2 1 + 2! + 3! + ... ∑︁ 퐵 = 푘 푧푘 (5.1) 푘! 푘=0 푧 푧2 푧4 푧6 푧8 푧10 = 1 − + − + − + ∓ ... ; 2 12 720 30 240 1 209 600 47 900 160 Bernoulli polynomials are ∞ 푧 푒푧푥 ∑︁ 퐵 (푥) = 푘 푧푘 . (5.2) 푒푧 − 1 푘! 푘=0 Table 3.1 presents some explicit Bernoulli numbers and Table 5.1 below lists the first Bernoulli polynomials. Remark 5.2. It is evident by comparing (5.1) and (5.2) that

퐵푘 = 퐵푘 (0). (5.3)

− = cosh 푧 = 푒푧 +푒 푧 = 푒2푧 −1+2 = 2 + Remark 5.3. Note that coth 푧 sinh 푧 푒푧 −푒−푧 푒2푧 −1 푒2푧 −1 1 is an odd function, thus

∞ 푧 푧 푧 푧 ∑︁ 퐵 coth = + = 2푘 푧2푘 2 2 푒푧 − 1 2 (2푘)! 푘=0 and it follows that 퐵2푘+1 = 0 (5.4) 1 for 푘 ≥ 1 (note, however, that 퐵1 = − 2 ). Proposition 5.4 (Explicit formula). Bernoulli polynomials (and thus Bernoulli numbers) are given explicitly by 푘 푛 ∑︁ 1 ∑︁ 푛 퐵 (푥) = (−1)ℓ (ℓ + 푥)푘 . (5.5) 푘 푛 + 1 ℓ 푛=0 ℓ=0 1Jacob I Bernoulli, 1654–1705

31 32 BERNOULLINUMBERSANDPOLYNOMIALS

+ 푧 Í (1−푒푧 )푛 1 Proof. It holds that 푧 = log 1 − (1 − 푒 ) = − 푛=0 푛+1 and thus

∞ 푛 ∞ 푛 푧 푒푧 푥 ∑︁ 1 − 푒푧 ∑︁ 1 ∑︁ 푛 = 푒푧푥 = (−1)ℓ 푒푧 (푥+ℓ) 푒푧 − 1 푛 + 1 푛 + 1 ℓ 푛=0 푛=0 ℓ=0 ∞ ∞ 푛 ∑︁ 푧푘 ∑︁ 1 ∑︁ 푛 = · (−1)ℓ (ℓ + 푥)푘 . 푘! 푛 + 1 ℓ 푘=0 푛=0 ℓ=0

The sum over 푛 in the latter display terminates at 푘, as 푥 ↦→ (ℓ + 푥)푘 is a polynomial of degree 푘 Í푛 ℓ 푛 푘 and thus ℓ=0 (−1) ℓ (ℓ +푥) = 0 for 푛 ≥ 푘 . The assertion follows by comparing coefﬁcients. Remark 5.5 (Translation). We have that

∞ ∞ ∞ 푘 푧 푒푧 (푥+푦) ∑︁ (푧 푦)푘 ∑︁ 퐵 (푥) ∑︁ 푧푘 ∑︁ 푘 = · ℓ 푧ℓ = 퐵 (푥)푦푘−ℓ 푒푧 − 1 푘! ℓ! 푘! ℓ ℓ 푘=0 ℓ=0 푘=0 ℓ=0 so that 푘 ∑︁ 푘 퐵 (푥 + 푦) = 퐵 (푥) 푦푘−ℓ 푘 ℓ ℓ ℓ=0 by comparing with (5.2). 푒(1−푥) 푧 푒−푥푧 Remark 5.6 (Symmetry). Comparing the coefficients in the identity 푧 푒푧 −1 = −푧 푒−푧 −1 reveals that 푘 퐵푘 (1 − 푥) = (−1) 퐵푘 (푥). (5.6) 푘 In particular we find that 퐵푘 (1) = (−1) 퐵푘 . Remark 5.7. Differentiating (5.2) with respect to 푥 and comparing the coefficients at 푧푘 reveals that 0 퐵푘 (푥) = 푘 · 퐵푘−1 (푥). (5.7)

5.2 SUMMATION AND MULTIPLICATION THEOREM

Theorem 5.8 (Faulhaber’s formula2). For 푝 = 0, 1, 2,... , 푥 ∈ R and 푛 ∈ N (actually 푛 ∈ Z, if we Í푏 Í푎−1 set 푘=푎 B − 푘=푏+1 whenever 푏 < 푎) it holds that

푛−1 ∑︁ 퐵푝+1 (푛 + 푥) − 퐵푝+1 (푥) (푘 + 푥) 푝 = . 푝 + 1 푘=0

Í푛−1 푝 Proof. Set 푆 푝 (푛) B 푘=0 (푘 + 푥) , then the generating function is

∞ 푛−1 ∞ 푝 푛−1 (푛+푥)푧 푥푧 ∑︁ 푆 푝 (푛) ∑︁ ∑︁ (푘 + 푥) ∑︁ 푒 푒 푧 푝 = 푧 푝 = 푒(푘+푥)푧 = − . 푝! 푝! 푒푧 − 1 푒푧 − 1 푝=0 푘=0 푝=0 푘=0 ∑︁ 퐵 (푛 + 푥) − 퐵 (푥) = 푧푘−1 푘 푘 , 푘! 푘=0

from which the assertion follows by comparing the coefﬁcients (푘 = 푝 + 1). 2Johann Faulhaber, 1580–1635

rough draft: do not distribute 5.3 FOURIERSERIES 33

Proposition 5.9 (Multiplication theorem). For 푚 ∈ N holds that

푚−1 ∑︁ ℓ 퐵 (푚푥) = 푚푘−1 퐵 푥 + . 푘 푘 푚 ℓ=0 Proof. Indeed, the result follows by comparing the coefﬁcients in the identity

푚−1 ∞ 푘 푚−1 (푥+ ℓ )푧 푥푧 푧 푚 푥푧 ∑︁ ∑︁ ℓ 푧 ∑︁ 푒 푚 푧푒 푒 푚 − 1 푒 퐵푘 푥 + = 푧 = = 푧 . 푚 푘! 푒푧 − 1 푒푧 − 1 푒푧/푚 − 1 푒푧/푚 − 1 ℓ=0 푘=0 ℓ=0

5.3 FOURIERSERIES

In what follows we deﬁne the periodic function (with period 1)

훽푘 (푥) B 퐵푘 (푥 − b푥c), 푥 ∈ R. (5.8)

(푘−2) It follows from (5.4) and (5.6) that 훽푘 is continuous for 푘 ≥ 2. Even more, by (5.7), 훽푘 ∈ 퐶 (R). Theorem 5.10. The Bernoulli polynomials are given, for 푘 ≥ 1, by the Fourier series3

푘 휋 휋푖푛푥 ∞ cos 2휋푛푥 − 푘! ∑︁ 푒2 ∑︁ 2 퐵 (푥) = − = −2 · 푘! , 푥 ∈ (0, 1). (5.9) 푘 (2휋푖)푘 푛푘 (2휋푛)푘 푛∈Z, 푛=1 푛≠0

푧 푒푧푥 Proof. Consider the function 푥 ↦→ 푒푧 −1 . Its Fourier coefﬁcients, for 푛 ∈ Z, are

1 ∫ 1 푧 푒푧푥 푧 ∫ 1 푧 푒푥 (푧−2휋푖푛) −2휋푖푛푥 = 푥 (푧−2휋푖푛) = 푒 푧 d푥 푧 푒 d푥 푧 0 푒 − 1 푒 − 1 0 푒 − 1 푧 − 2휋푖푛 푥=0 푧 푒푧 − 1 푧 = = . 푒푧 − 1 푧 − 2휋푖푛 푧 − 2휋푖푛 The Fourier series thus is ∑︁ 푧푘 푧 푒푧푥 ∑︁ 푧 퐵 (푥) = = 푒2휋푖푛푥 푘 푘! 푒푧 − 1 푧 − 2휋푖푛 푘=0 푛∈Z ∑︁ 푧 1 = 1 − 푒2휋푖푛푥 2휋푖푛 1 − 푧 푛≠0 2휋푖푛 ∞ ∑︁ ∑︁ 푧 푘 = 1 − 푒2휋푖푛푥 2휋푖푛 푛≠0 푘=1 ∞ ∑︁ 푧푘 ∑︁ 푒2휋푖푛푥 = 1 − · 푘! . 푘! (2휋푖푛)푘 푘=1 푛≠0

The result follows by comparing the coefﬁcients.

3 휋 휋 Recall that sin 푥 + 2 = cos 푥 and cos 푥 + 2 = − sin 푥.

Version: April 23, 2021 34 BERNOULLINUMBERSANDPOLYNOMIALS

Polynomial 퐵푘 (푥) Fourier series 훽푘 (푥)

퐵0 (푥) = 1 1 Í∞ sin 2휋푛푥 퐵1 (푥) = 푥 − 2 = −2 · 푛=1 2휋푛 sawtooth wave (5.10) ( ) = 2 − + 1 = · Í∞ cos 2휋푛푥 퐵2 푥 푥 푥 6 4 푛=1 (2휋푛)2 ( ) = 3 − 3 2 + 1 = · Í∞ sin 2휋푛푥 퐵3 푥 푥 2 푥 2 푥 12 푛=1 (2휋푛)3 ( ) = 4 − 3 + 2 − 1 = − · Í∞ cos 2휋푛푥 퐵4 푥 푥 2푥 푥 30 48 푛=1 (2휋푛)4 ( ) = 5 − 5 4 + 5 3 − 푥 = − · Í∞ sin 2휋푛푥 퐵5 푥 푥 2 푥 3 푥 6 240 푛=1 (2휋푛)5

Table 5.1: Bernoulli polynomials

5.4 UMBRALCALCULUS

푘 퐵푥 푥 (퐵+푛) 푥 푥 푒푛푥 As a formal power series and interpreting 퐵 as 퐵푘 we have 푒 = 푒푥 −1 and thus 푒 = 푒푥 −1 and also 푒푛푥 − 1 푒(퐵+푛) 푥 − 푒퐵푥 = 푥 = 푥 푒0푥 + 푒푥 + · · · + 푒(푛−1) 푥 , 푒푥 − 1 + + 푥푘+1 (퐵+푛) 푘 1−퐵푘 1 푘 푘 푘 and by comparing the coefﬁcients of 푘! thus 푘+1 = 0 + 1 + · · · + (푛 − 1) , i.e., 1 푘 + 1 푘 + 1 푘 + 1 0푘 + 2푘 + · · · + (푛 − 1)푘 = 푛푘+1 + 퐵 푛푘 + 퐵 푛푘−1 + · · · + 퐵 푛 . 푘 + 1 1 1 1 2 푘 푘

rough draft: do not distribute Gamma function 6

6.1 EQUIVALENT DEFINITIONS

Deﬁnition 6.1. Euler’s integral of the second kind, aka. Gamma function, is (the analytic extension) of ∫ ∞ Γ(푠) B 푥푠−1푒−푥 d푥, <(푠) > 0. 0 Remark 6.2. By integration by parts it holds that Γ(푠 + 1) = 푠 Γ(푠), so the analytic extension to C\{0, −1, −2,... } is apparent. The derivatives are ∫ ∞ Γ(푘) (푠) = 푥푠−1 (log 푥)푘 푒−푥 d푥, <(푠) > 0. (6.1) 0 Proposition 6.3 (Gauß’ deﬁnition of the Γ-function). It holds that 푛! 푛푠 Γ(푠) = lim , 푠 ∉ {0, −1, −2,... }. (6.2) 푛→∞ 푠(푠 + 1)···(푠 + 푛) 푥 푛 −푥 Proof. Recall that 1 − 푛 → 푒 as 푛 → ∞ for every 푥 ∈ C. Then ∫ 푛 푛 ∫ ∞ ∫ ∞ 푥 푠−1 푠−1 −푥 푠−1 1 − 푥 d푥 = 푓푛 (푥)푥 d푥 −−−−→ 푒 푥 d푥 0 푛 0 푛→∞ 0 푥 푛 1 for 푓푛 (푥) = 1 − 푛 · (0,푛) by Lebesgue’s dominated convergence theorem. By integration by parts we obtain

∫ 푛 푛 푛 푠 푛 ∫ 푛 푛−1 푠 푥 푠−1 푥 푥 푥 푥 1 − 푥 d푥 = 1 − + 1 − d푥 0+ 푛 푛 푠 푥=0+ 0+ 푛 푠 1 ∫ 푛 푥 푛−1 = · 1 − 푥푠 d푥. 푠 0+ 푛 Repeating the argument 푛 − 1 times gives ∫ 푛 푥 푛 푛 − 1 1 ∫ 푛 푥 푛−2 1 − 푥푠−1 d푥 = · 1 − 푥푠+1 d푥 0+ 푛 푛 푠(푠 + 1) 0+ 푛 (푛 − 1)(푛 − 2)··· 1 1 ∫ 푛 = 푥푠+푛−1 d푥 푛−1 푛 푠(푠 + 1)···(푠 + 푛 − 1) 0 푛! 푛푠+푛 푛! 푛푠 = = 푛푛 푠(푠 + 1)···(푠 + 푛) 푠(푠 + 1)···(푠 + 푛) and thus the result. Corollary 6.4. It holds that 푠 1 + 1 1 Ö 푛 Γ(푠) = . (6.3) 푠 1 + 푠 푛=1 푛

35 36 GAMMAFUNCTION

1 1 1 2 3 푛 Proof. Note that 1 + 1 1 + 2 ... 1 + 푛−1 = 1 · 2 · ... · 푛−1 = 푛, so the result follows from (6.2).

Theorem 6.5 (Schlömilch formula, Weierstrass’ deﬁnition). It holds that

푒−훾푠 Ö 푒푠/푛 Γ(푠) = , (6.4) 푠 1 + 푠 푛=1 푛

where 훾 is the Euler–Mascheroni constant.1 Proof. Use that Í푛 1 − log 푛 −−−−→ 훾 and thus exp −훾푠 + Í푛 푠 ∼ 푛푠. The result follows 푗=1 푗 푛→∞ 푗=1 푗 with (6.3).

Theorem 6.6. The Taylor series expansion is given by

∑︁ 휁 (푘) log Γ(1 + 푠) = −훾푠 + (−푠)푘 , |푠| < 1. 푘 푘=2

Proof. With (6.4) we get

∑︁ 푠 푠 log 푠 · Γ(푠) = −훾푠 + − log 1 + 푛 푛 푛=1 ! ∑︁ 푠 ∑︁ 1 푠 푘 = −훾푠 + + (−1)푘 푛 푘 푛 푛=1 푘=1 ∑︁ 1 ∑︁ 푠 푘 = −훾푠 + (−1)푘 푘 푛 푘=2 푛=1 ∑︁ 휁 (푘) = −훾푠 + (−푠)푘 , (6.5) 푘 푘=2

the result.

1 푠 2 휋2 2 푠 2 휋2 Corollary 6.7. It holds that Γ(푠) = 푠 − 훾 + 2 훾 + 6 + O 푠 and Γ(푠 + 1) = 1 − 훾푠 + 2 훾 + 6 + O 푠2. The residues at 푠 = −푛 (푛 = 0, 1,... ) are given by

(−1)푛 Γ(푠) = + O(1). (6.6) (푠 + 푛)· 푛!

6.2 EULER’S REFLECTION FORMULA

Proposition 6.8. It holds that

∞ 1 ∑︁ 2푥 휋 coth 휋푥 = + , 푥 ∉6 Z. (6.7) 푥 푥2 + 푛2 푛=1 1Lorenzo Mascheroni, 1750–1800

rough draft: do not distribute 6.2 EULER’S REFLECTION FORMULA 37

Proof. Consider the function 푡 ↦→ cosh(푥 푡) for 푥 > 0 on (−휋, 휋). The function is even, thus the 푎0 Í Taylor series expansion is cosh 푥푡 = 2 + 푛=1 푎푛 cos 푛푡 with 1 ∫ 휋 2 푎0 = cosh 푥푡 d푡 = sinh 푥휋 휋 −휋 푥 휋 and 1 ∫ 휋 1 ∫ 휋 푒푖푛푡 + 푒−푖푛푡 푒푥푡 + 푒−푥푡 푎푛 = cos 푛푡 cosh 푥푡 d푡 = d푡 휋 −휋 휋 −휋 2 2 (푖푛+푥)푡 (푖푛−푥)푡 (−푖푛+푥)푡 (−푖푛−푥)푡 휋 1 푒 푒 푒 푒 = + + + 4휋 푖푛 + 푥 푖푛 − 푥 −푖푛 + 푥 −푖푛 − 푥 푡=−휋 (−1)푛 푒푥 휋 − 푒−푥 휋 푒−푥 휋 − 푒푥 휋 푒푥 휋 − 푒−푥 휋 푒−푥 휋 − 푒푥 휋 = + + + 4휋 푖푛 + 푥 푖푛 − 푥 −푖푛 + 푥 −푖푛 − 푥 (−1)푛 4푥 = sinh 휋푥 2휋 푛2 + 푥2 and thus sinh 휋푥 sinh 휋푥 ∑︁ 2푥 cos 푛푡 cosh 푥푡 = + (−1)푛 . (6.8) 휋푥 휋 푛2 + 푥2 푛=1 The result follows for 푡 = 휋. Proposition 6.9 (Euler’s inﬁnite product for the sine function). It holds that

∞ Ö 푥2 sin 휋푥 = 휋푥 1 − . (6.9) 푘2 푘=1

0 2 ( ) 푓 (푥) = ( ) Î∞ − 푥 Proof. Set 푓 푥 B sin 휋푥 thus 푓 (푥) 휋 cot 휋푥; set 푔 푥 B 휋푥 푘=1 1 푘2 and it follows that

푔0(푥) 1 ∑︁ 2푥/푘2 1 ∑︁ 2푥 = − = − 푔(푥) 푥 1 − 푥2/푘2 푥 푘2 − 푥2 푘=1 푘=1

푔0 (푖푥) 푓 0 (푥) and using (6.7) thus 푖 푔(푖푥) = 푓 (푥) . It follows that 푓 (푥) = 푐푔(푖푥) for some constant 푐 ∈ C. The constant is clear by letting 푥 → 0 in (6.9). Corollary 6.10 (Wallis’2 product for 휋). It holds that

휋 Ö 4푘2 2 2 4 4 6 6 8 8 = = ..., 2 4푘2 − 1 1 3 3 5 5 7 7 9 푛=1 or equivalently, √ 휋푘 2푘 → 1, (6.10) 22푘 푘 as 푘 → ∞. = 1 − 1 = 2푛−1 2푛+1 Proof. Choose 푥 2 in (6.9) and observe that 1 4푛2 2푛 2푛 . 1 2푘 = 1·3···(2푘−1) The second formula follows from 22푘 푘 2·4···(2푘) . 2John Wallis, 1616–1703

Version: April 23, 2021 38 GAMMAFUNCTION

Proposition 6.11 (Reﬂection formula, functional equation). It holds that 휋 Γ(푠) Γ(1 − 푠) = . sin(휋푠)

Proof. We use (6.2) to see that

푛! 푛푠 푛! 푛1−푠 Γ(푠) Γ(1 − 푠) ← · 푠(1 + 푠)···(푛 + 푠) (1 − 푠)(1 − 푠 + 1)···(1 − 푠 + 푛) 1 푛!2 푛 = · 푠 (1 − 푠2)(22 − 푠2) ... (푛2 − 푠2) 1 − 푠 + 푛 1 1 푛 = 휋 · 휋푠 − 푠2 − 푠2 − 푠2 1 − 푠 + 푛 1 12 1 22 ... 1 푛2 휋 → sin 휋푠 as 푛 → ∞ by (6.9) and thus the result.

1 √ Corollary 6.12. It holds that Γ 2 = 휋.

6.3 DUPLICATION FORMULA

Proposition 6.13 (Legendre duplication formula). It holds that

1 √ Γ(푠) Γ 푠 + = 21−2푠 휋 Γ(2푠); 2

more generally, for 푚 ∈ {2, 3, 4,... }, the multiplication theorem

푚−1 Ö 푘 푚−1 1 −푚푠 Γ 푠 + = (2휋) 2 푚 2 Γ(푚푠) 푚 푘=0 holds true. Proof. Indeed, with (6.2) and (6.10),

1 푠 푠+ Γ( ) Γ + 1 푛! 푛 · 푛! 푛 2 푠 푠 2 푠(1+푠)···(푛+푠) (푠+ 1 )(1+푠+ 1 )···(푛+푠+ 1 ) ←−−−− 2 2 2 Γ(2푠) 푛→∞ (2푛)! (2푛)2푠 2푠(1+2푠)···(2푛+2푠) 푛1/2 · 2−2푠 2푠(2푠 + 2) ... (2푠 + 2푛) (2푠 + 1) ... (2푠 + 2푛 − 1) = · · 2푛 푠(푠 + 1)···(푠 + 푛) ( + 1 )( + 3 )···( + − 1 )·( + + 1 ) 푛 푠 2 푠 2 푠 푛 2 푠 푛 2 2−2푠 푛 √ = √ 2푛+12푛 · −−−−→ 21−2푠 휋 2푛 + 푛+1 푛→∞ 푛 푛 푠 2 and thus the result; the remaining statement follows similarly.

rough draft: do not distribute Euler–Maclaurin formula 7

+ Í푛 ( ) = ∫ 푛 ( ) b c By Riemann–Stieltjes integration it follows that 푘=1 푓 푘 0+ 푓 푥 d 푥 . We thus have

푛 ∑︁ ∫ 푛 ∫ 푛+ 푓 (푘) − 푓 (푥) d푥 = 푓 (푥) d(b푥c − 푥). 푘=1 0 0+ By integration by parts thus,

푛 ∑︁ ∫ 푛 ∫ 푛 푓 (푘) − 푓 (푥) d푥 = (푥 − b푥c) 푓 0(푥) d푥, 푘=1 0 0 or 푛 ∑︁ ∫ 푛 푓 (푛) − 푓 (0) ∫ 푛 1 푓 (푘) = 푓 (푥) d푥 + + 푥 − b푥c − 푓 0(푥) d푥. (7.1) 2 2 푘=1 0 0

Now recall from (5.8) the functions 훽푘 (푥) = 퐵푘 (푥 − b푥c) and from (5.10) the function 퐵1 (푥), thus 푛−1 ∑︁ ∫ 푛 푓 (푛) − 푓 (0) ∫ 푛 푓 (푘) = 푓 (푥) d푥 − + 훽 (푥) 푓 0(푥) d푥. (7.2) 2 1 푘=0 0 0 0 ( ) 훽푘+1 푥 Recall from (5.7) that 훽푘 (푥) = 푘+1 and from (5.3) that 훽푘 (푛) = 훽푘 (0) = 퐵푘 . Integrating by parts thus gives

푛−1 ∑︁ ∫ 푛 퐵 ∫ 푛 훽 (푥) 푓 (푘) = 푓 (푥) d푥 + 퐵 푓 (푛) − 푓 (0) + 2 · ( 푓 0(푛) − 푓 0(0)) − 2 푓 (2) (푥) d푥. 1 2 2 푘=0 0 0

Repeating the procedure and noting that 퐵3 = 0,

푛−1 ∑︁ ∫ 푛 퐵 퐵 ∫ 푛 훽 (푥) 푓 (푘) = 푓 (푥) d푥 + 1 푓 (푛) − 푓 (0) + 2 · ( 푓 0(푛) − 푓 0(0)) + 3 푓 (3) (푥) d푥. 1 2 6 푘=0 0 0

Repeating the procedure (in total 푝 times) gives the Euler–Maclaurin1 summation formula.

Theorem 7.1. For 푝 ∈ N and 푓 ∈ 퐶 푝 ([0, 푛]) it holds that

− 푝 푛 1 ∫ 푛 푛 푝 ∫ 푛 ∑︁ ∑︁ 퐵ℓ (ℓ−1) (−1) ( 푝) 푓 (푘) = 푓 (푥) d푥 + 푓 (푥) − 훽푝 (푥) 푓 (푥) d푥, ℓ! 푥=0 푝! 푘=0 0 ℓ=1 0 or 푝 푛 ∫ 푛 푛 푝 ∫ 푛 ∑︁ ∑︁ ℓ 퐵ℓ (ℓ−1) (−1) ( 푝) 푓 (푘) = 푓 (푥) d푥 + (−1) 푓 (푥) − 훽푝 (푥) 푓 (푥) d푥. ℓ! 푥=0 푝! 푘=1 0 ℓ=1 0 1Colin Maclaurin, 1698–1746, Scottish

39 40 EULER–MACLAURINFORMULA

7.1 EULER–MASCHERONI CONSTANT

Example 7.2. The Euler–Mascheroni constant is

푛−1 ∑︁ 1 훾 B lim − log 푛 = 0.577 215 664 9 .... (7.3) 푛→∞ 푘 푘=1

1 Set 푓 (푥) B 푥+1 , then, by (7.2),

푛 푛−1 ∑︁ 1 ∑︁ ∫ 푛 1 − 1 ∫ 푛 훽 (푥) − log(푛 + 1) = 푓 (푘) − 푓 (푥) d푥 = − 푛+1 − 1 d푥. 푗 2 (푥 + 1)2 푗=1 푘=0 0 0

∞ → ∞ = 1 − ∫ 훽1 (푥) Letting 푛 gives the convergent integral 훾 2 1 푥2 d푥.

7.2 STIRLINGFORMULA

Choose 푓 (푥) B log 푥 in (7.1), then

푛 ∑︁ ∫ 푛 1 ∫ 푛 훽 (푥) log 푛! = log 푘 = log 푥 d푥 + log 푛 + 1 d푥 2 푥 푘=2 1 1

푛 − + 1 + = + ∫ 훽1 (푥) → ∞ and thus log 푛! 푛 2 log 푛 푛 1 1 푥 d푥 exists for 푛 . 푛 ( ) 2푛 2푛 (2푛) 푛!푒√ 1 ← 푏2푛 = 2푛 !푒√ · 푛 푛 = √︁ 푛 푛 → √1 With 푏푛 B 푛 and Corollary 6.10 we ﬁnd that 푏 2 2푛 2 2푛 2 2푛 푛 푛 √ 푏푛 (2푛) 2푛 푛! 푒 2 2휋 푛 푛 and thus, asymptotically, 푛! ∼ 2휋푛 푒 . A more thorough analysis (cf. Abramowitz and Stegun [1, 6.1.42]) gives the asymptotic expansion 푛 1 √ ∑︁ 퐵 log Γ(푧) = 푧 − log 푧 − 푧 + log 2휋 + 2푚 . 2 2푚(2푚 − 1)푧2푚−1 푚=1

Alternative proof, following an idea of P. Billingsley. Let 푋푖 be independent Poisson variables with parameter 1. The random variable 푋1 + · · · + 푋푛 follows a Poisson distribution with parameter 푛. Set ℎ(푥) B max(0, 푥) (the ramp function), then

∞ 푋 + · · · + 푋 − 푛 ∑︁ 푒−푛푛푘 푘 − 푛 푒−푛 ∑︁ 푛푘 푛푘+1 푒−푛 푛푛+1 E ℎ 1 √ 푛 = √ = √ − = √ . 푘! (푘 − 1)! 푘! 푛! 푛 푘=푛+1 푛 푛 푘=푛+1 푛

By the central limit theorem 푋1+···+√ 푋푛−푛 converges in distribution to a random variable 푍 ∼ 푛 2 2 ∞ ∫ ∞ −푥 /2 −푥 /2 N(0, 1) for which E ℎ(푍) = 푥 푒√ d푥 = − 푒√ = √1 . Thus the result. 0 2휋 2휋 2휋 푥=0 Corollary 7.3. It holds that

푧 (−1)푘 푘 + 푧 푘 푧 ≈ and ≈ 푘 Γ(−푧)푘 푧+1 푘 Γ(푧 + 1)

as 푘 → ∞.

rough draft: do not distribute 7.2 STIRLINGFORMULA 41

Proof. Indeed, with (6.3) we have that

−푧−1 ∞ 1 + 1 푧 −푧 + 푘 − 1 1 Ö 푗 (−1)푘 = = 푘 푘 Γ(−푧)(푘 + 1)푧+1 푠+1 푗=푘+1 1 − 푗 and thus the result.

Version: April 23, 2021 42 EULER–MACLAURINFORMULA

rough draft: do not distribute Summability methods 8

8.1 SILVERMAN–TOEPLITZTHEOREM

∞ Deﬁnition 8.1. We shall say that the summability method 푡 = (푡푖푘 )푖,푘=0 conﬁnes a regular summability method if the following hold true for every convergent sequence with 푎푘 −−−−→ 훼: 푘→∞ Í S1. 푘=0 푡푖푘 푎푘 converges for every 푖 = 0, 1,... and Í S2. 푘=0 푡푖푘 푎푘 −−−−→ 훼. 푖→∞

Theorem 8.2 (Silverman–Toeplitz). The matrix 푡 = (푡푖 푗 ) with 푡푖푘 ∈ C conﬁnes a regular summability method if and only if it satisﬁes the following properties: Í T1. 푘=0 |푡푖푘 | ≤ 푀 < ∞ for all 푖 = 0, 1,... ,

T2. 푡푖푘 −−−−→ 0 for all 푘 = 0, 1,... and 푖→∞ Í∞ T3. 푡푖푘 −−−−→ 1. 푘=0 푖→∞

∞ Proof. For 푎 = (푎푘 )푘=0 ∈ 푐 (the Banach space of sequences with a limit) deﬁne 푡푖 (푎) B Í ∗ Í 푘=0 푡푖푘 푎푘 , a linear functional on 푐. Recall that 푡푖 ∈ 푐 is continuous with norm k푡푖 k = 푘=0 |푡푖푘 | (i.e., (푡푖푘 )푘 ∈ ℓ1 for all 푖) which converges by S1 for every point 푎 = (푎푘 ) ∈ 푐. By the uniform bounded principle (Banach–Steinhaus theorem) there is constant 푀 such that sup푖 k푡푖 k ≤ 푀 < ∞, hence T1. Choose 푎 = 푒푘 ∈ 푐0, then 푡푖푘 = 푡푖 (푒푘 ) −−−−→ 0 by S2, hence T2. Finally choose 푎 = (1, 1,... ) ∈ 푖→∞ Í 푐 with limit 푎푘 → 훼 = 1 to get 푘=0 푡푖푘 푎푘 −−−−→ 1 by S2, hence T3. 푖→∞ Í As for the converse observe that for 푎 = 훼 ·(1, 1,... ) + 푘=0 (푎푘 − 훼) · 푒푘 and hence 푡푖 (푥) = Í Í Í 훼 · 푘=0 푡푖푘 + 푘 (푎푘 − 훼) 푡푖푘 , where 푘=0 푡푖푘 −−−−→ 1 by T3. For the remaining term we have 푖→∞ Í Í | 푘=0 (푎푘 − 훼) 푡푖푘 | ≤ 푘 |푎푘 − 훼| |푡푖푘 | + 푀 sup푘>푟 |푎푘 − 훼| by T2 and hence, for 푟 large enough, 푓푖 (푎푘 ) −−−−→ 훼, i.e., S2 and S1. 푖→∞

Remark 8.3. For the summability methods below we investigate the sequences of partial sums Í푘 푠푘 B 푗=0 푎 푗 . Note, that ∞ ∑︁ ∑︁ ∑︁ 푡푖푘 푠푘 = 푎 푗 · 푡푖푘 푘=0 푗=0 푘= 푗 and ∑︁ ∑︁ 푇푖푘 − 푇푖,푘+1 푠푘 = 푎 푗 · 푇푖 푗 , 푘=0 푗=0 Í where 푇푖 푗 = 푘= 푗 푡푖푘 and 푡푖 푗 = 푇푖 푗 − 푇푖, 푗+1.

43 44 SUMMABILITYMETHODS

8.2 CESÀRO SUMMATION

Í푛 Í∞ Consider the 푛th partial sum of the series 푠푛 B 푗=0 푎 푗 → 푠 B 푗=0 푎 푗 . Then

푛 + 1 − 0 푛 + 1 − 1 푛 + 1 − 푛 푠 + 푠 + · · · + 푠 푎 + 푎 + · · · + 푎 = 0 1 푛 → 푠, (8.1) 푛 + 1 0 푛 + 1 1 푛 푛 푛 + 1

( 1 푘+1 if 푘 ≤ 푖 the limit does not change and Theorem 8.2 applies with 푡푖푘 = The procedure may 0 else. be repeated, which gives rise for the following, more general method.

Deﬁnition 8.4. For 훼 ∈ C we shall call the limit

푛 푛 ∑︁ 푗 ∑︁ 푛+훼 푎 푗 =: 푎 푗 푗=0 푗 푗=0

the Cesàro mean.1

푛 ( 푗) 푛+1− 푗 Note that, for 훼 = 1, 푛+1 = 푛+1 and thus (8.1). ( 푗 ) Í 푘 1 Remark 8.5. The Cesàro limit of Grandi’s series is 푘=0 (−1) = 2 , cf. Figure 13.3.

8.3 EULER’S SERIES TRANSFORMATION

Theorem 8.6 (Euler transform). It holds that

푖 ∑︁ ∑︁ 1 ∑︁ 푖 푎 = 푦 푗+1푎 , (8.2) 푗 (1 + 푦)푖+1 푗 푗 푗=0 푖=0 푗=0

where 푦 ∈ C is a parameter.

Proof. We have that ∞ + ∑︁ 푖 1 푖 1 = 푦− 푗−1. (8.3) 푗 1 + 푦 푖= 푗

Indeed, for 푗 = 0 the assertion follows from the usual geometric series. Differentiating (8.3) reveals the claim for 푗 ← 푗 + 1 after obvious rearrangements. By rearranging the series (8.2) we ﬁnd that

푖 ∞ ∞ ∞ ∑︁ 1 ∑︁ 푖 ∑︁ ∑︁ 푖 1 ∑︁ 푦 푗+1푎 = 푎 푦 푗+1 = 푎 , (1 + 푦)푖+1 푗 푗 푗 푗 (1 + 푦)푖+1 푗 푖=0 푗=0 푗=0 푖= 푗 푗=0

thus the assertion.

1Ernesto Cesàro, 1859–1906

rough draft: do not distribute 8.4 SUMMATION BY PARTS 45

8.4 SUMMATION BY PARTS

The rearrangement

푛 푛 ∑︁ ∑︁ 푓푘 (푔푘+1 − 푔푘 ) = [ 푓푛푔푛+1 − 푓푚푔푚] − 푔푘 ( 푓푘 − 푓푘−1). 푘=푚 푘=푚+1 is called summation by parts, or Abel2 transformation. Repeating the procedure 푀 times gives the assertion of the following statement. Proposition 8.7. For 푀 = 0, 1,... it holds that

푛 푀−1 푛−푀 ∑︁ ∑︁ ∑︁ (푖) (푖+1) + (푀) (푀) 푓푘 푔푘 = 푓0 퐺푖 푓 푗 퐺 푗+푀 푘=0 푖=0 푗=0 푀−1 푛−푀 ∑︁ 푖 (푖) ˜ (푖+1) 푀 ∑︁ (푀) ˜ (푀) = (−1) 푓푛−푖퐺푛−푖 + (−1) 푓 푗 퐺 푗 , 푖=0 푗=0 where 푀 (푀) ∑︁ 푀−푘 푀 푓 B (−1) 푓 푗+푘 푗 푘 푘=0 and 푛 (푀) ∑︁ 푘 − 푗 + 푀 − 1 퐺 B 푔푘 , 푗 푀 − 1 푘= 푗 푗 (푀) ∑︁ 푗 − 푘 + 푀 − 1 퐺˜ B 푔푘 . 푗 푀 − 1 푘=0 ∞ Proposition 8.8 (Abel’s test). Suppose the sequence (푏푘 )푘=0 is of bounded variation (i.e., Í Í∞ Í 푗=0 푏 푗+1 − 푏 푗 < ∞) and 푘=0 푎푘 converges. Then 푘=0 푎푘 푏푘 converges too.

Proof. Note ﬁrst that 푏푘 is uniformly bounded, as

푘−1 ∞ ∑︁ ∑︁ |푏푘 | = 푏0 + 푏 푗+1 − 푏 푗 ≤ |푏0| + 푏 푗+1 − 푏 푗 .

푗=0 푗=0 By summation by parts it holds that

푁 푁 푁 −1 푁 ∑︁ ∑︁ ∑︁ ∑︁ 푎푘 푏푘 = 푏푀 푎푘 + (푏 푗+1 − 푏 푗 ) 푎푘 . (8.4) 푘=푀 푘=푀 푗=푀 푘= 푗+1 Í푁 For 휀 > 0 ﬁnd 푛0 ∈ N such that 푘=푀 푎푘 < 휀 for every 푁, 푀 > 푛0. It follows that

푁 푁 −1 ∑︁ ∑︁ ∑︁ ≤ | | + − ≤ ©| | + − ª 푎푘 푏푘 푏푀 휀 푏 푗+1 푏 푗 휀 휀 푏0 2 푏 푗+1 푏 푗 ® . 푘=푀 푗=푀 푗=0 « ¬ The assertion follows, as 휀 > 0 is arbitrary. 2Niels Henrik Abel, 1802–1829, Norwegian

Version: April 23, 2021 46 SUMMABILITYMETHODS

Proposition 8.9 (Continuity). Suppose that

(i) |1 − 푏푘 (푟)| −−−−→ 0 for all 푘 ∈ N (pointwise convergence) and 푟→1− Í (ii) 푘=0 |푏푘 (푟) − 푏푘+1 (푟)| < 퐶 for some 퐶 < ∞ and 푟 < 1 (uniform bounded variation), then ∞ ∑︁ ∑︁ lim 푏푘 (푟)· 푎푘 = 푎푘 , 푟→1− 푘=0 푘=0 provided that the latter sum exists. Í∞ Proof. Let 푀 ∈ N be large enough so that 푘=ℓ 푎ℓ < 휀 for all ℓ > 푀 and 푟0 < 1 large enough 휀 such that |1 − 푏푘 (푟)| < for all 푘 = 0, 1, . . . , 푀 − 1 and |1 − 푏푀 (푟)| < 2 for all 푟 ∈ (푟0, 1). 푀 ·supℓ∈N |푎ℓ | Similarly to (8.4) it follows for 푟 > 푟0 that

∞ ∞ 푀−1 ∞ ∑︁ ∑︁ ∑︁ ∑︁ 푎푘 − 푎푘 푏푘 (푟) = 푎푘 1 − 푏푘 (푟) + 푎푘 1 − 푏푘 (푟) 푘=0 푘=0 푘=0 푘=푀 푀−1 ∞ ∞ ∞ ∑︁ ∑︁ ∑︁ ∑︁ = 푎푘 1 − 푏푘 (푟) + 1 − 푏푀 (푟) 푎푘 + 푏 푗 (푟) − 푏 푗+1 (푟) 푎푘 . 푘=0 푘=푀 푗=푀 푘= 푗+1

Hence ∞ ∞ ∞ ∑︁ ∑︁ 휀 ∑︁ 푎푘 − 푎푘 푏푘 (푟) ≤ 푀 + 2휀 + 휀 푏 푗 (푟) − 푏 푗+1 (푟) < (3 + 퐶)휀 푀 푘=0 푘=0 푗=푀 and thus the assertion.

8.4.1 Abel summation Theorem 8.10. It holds that ∞ ∞ ∑︁ 푘 ∑︁ lim 푟 · 푎푘 = 푎푘 , 푟→1− 푘=0 푘=0 provided that the latter sum exists.

푘 푘 Í Í 푘 푘+1 Proof. Choose 푏푘 (푟) B 푟 , then 푏푘 (푟) = 푟 −−−−→ 1 and 푘=0 |푏푘 (푟) − 푏푘+1 (푟)| = 푘=0 푟 − 푟 = 푟→1− 1 for 푟 ∈ (0, 1). The assertion follows with Proposition 8.9.

8.4.2 Lambert summation Theorem 8.11 (Lambert3 summation). It holds that

∞ ∞ ∑︁ 푘 푟 푘 ∑︁ lim(1 − 푟)· · 푎푘 = 푎푘 , (8.5) 푟 ↑1 1 − 푟 푘 푘=1 푘=1

provided that the latter sum exists.

3Johann Heinrich Lambert, 1728–1777, Swiss polymath

rough draft: do not distribute 8.5 ABEL’S SUMMATION FORMULA 47

푘 ( ) (1−푟)푘 푟 Proof. Choose 푏푘 푟 B 1−푟 푘 . By de L’Hôpital’s rule, (1 − 푟)푘 푟 푘 −푘 푟 푘 + (1 − 푟)푘2 푟 푘−1 lim 푏푘 (푟) = lim = lim = 1. 푟 ↑1 푟 ↑1 1 − 푟 푘 푟 ↑1 −푘푟 푘−1

The assertion follows with Proposition 8.9 by monotonicity, as 푏푘 (푟) > 푏푘+1 (푟) > 0 for 푟 ∈ (0, 1).

8.5 ABEL’S SUMMATION FORMULA

Theorem 8.12 (Abel’s summation formula). For 휙 : [1, ∞) → C differentiable it holds that ∫ 푥 ∑︁ 0 푎푛 휙(푛) = 퐴(푥) 휙(푥) − 퐴(푢) 휙 (푢) d푢, (8.6) 1≤푛≤푥 1 Í where 퐴(푥) B 1≤푛≤푥 푎푛 is the summatory function. Proof. Choose 휀 ∈ (0, 1). By Riemann-Stieltjes integration by parts we have that ∫ 푥 ∫ 푥 푥 휙(푢) d퐴(푢) + 퐴(푢) d휙(푢) = 휙(푢) 퐴(푢)|푢=1−휀 = 퐴(푥) 휙(푥) 1−휀 1−휀 and thus the result. Alternative proof. Note, that 퐴(푥) = 0 for 푥 ≤ 1 and 퐴(푢) = 퐴(푛) for 푢 ∈ [푛, 푛 + 1), 푛 ∈ N. Thus, for integer 푥, 푥 ∑︁ ∑︁ 푎푛 휙(푛) = 퐴(푛) − 퐴(푛 − 1) 휙(푛) 1≤푛≤푥 푛=1 푥−1 푥−1 ∑︁ ∑︁ = 퐴(푛)휙(푛) + 퐴(푥)휙(푥) − 퐴(0)휙(1) − 퐴(푛)휙(푛 + 1) 푛=1 푛=1 푥−1 ∑︁ = 퐴(푥)휙(푥) − 퐴(푛) 휙(푛 + 1) − 휙(푛) 푛=1 푥−1 ∑︁ ∫ 푛+1 = 퐴(푥)휙(푥) − 퐴(푛) 휙0(푢) d푢 푛=1 푛 푥−1 ∑︁ ∫ 푛+1 = 퐴(푥)휙(푥) − 퐴(푢)휙0(푢) d푢 푛=1 푛 ∫ 푥 = 퐴(푥)휙(푥) − 퐴(푢)휙0(푢) d푢. 1 ∫ 푥 ( ) 0( ) = ( ) ( ) − (b c) (b c) Finally note that b푥c 퐴 푢 휙 푢 d푢 퐴 푥 휙 푥 퐴 푥 휙 푥 and thus the assertion.

8.6 POISSONSUMMATIONFORMULA

Deﬁnition 8.13. The Fourier transform of a function 푓 : R → C is ∫ ∞ 푓ˆ(푘) B 푓 (푡) 푒−2휋푖푘푡 d푡, 푘 ∈ R. −∞

Version: April 23, 2021 48 SUMMABILITYMETHODS

Theorem 8.14 (Poisson summation formula). It holds that ∑︁ ∑︁ 푓 (푘) = 푓ˆ(푘). (8.7) 푘 ∈Z 푘 ∈Z Í Proof. Set 푔(푡) B 푛∈Z 푓 (푡 + 푛) and note that 푔 has period 1, i.e., 푔(푡 + 1) = 푔(푡). Its Fourier Í 2휋푖푘푡 series is 푔(푡) = 푘 ∈Z 푐푘 · 푒 , where

∫ 1 −2휋푖푘푡 푐푘 = 푒 푔(푡) d푡 0 ∫ 1 ∑︁ = 푒−2휋푖푘푡 푓 (푡 + 푛) d푡 0 푛∈Z ∑︁ ∫ 푛+1 = 푒−2휋푖푘 (푡−푛) 푓 (푡) d푡 푛∈Z 푛 ∑︁ ∫ 푛+1 = 푒−2휋푖푘푡 푓 (푡) d푡 푛∈Z 푛 ∫ ∞ = 푒−2휋푖푘푡 푓 (푡) d푡 −∞ = 푓ˆ(푘).

It follows that ∑︁ ∑︁ 2휋푖푘푡 ∑︁ 2휋푖푘푡 푓 (푡 + 푘) = 푔(푡) = 푒 푐푘 = 푒 푓ˆ(푘) 푘 ∈Z 푘 ∈Z 푘 ∈Z and the assertion by choosing 푡 = 0.

Example 8.15 (Jacobi4 theta function). For 푥 > 0 deﬁne the function

∑︁ 2 ∑︁ 2 휗(푥) B 푒−푘 휋푥 = 1 + 2 푒−푘 휋푥, (8.8) 푘 ∈Z 푘=1

then 1 1 휗(푥) = √ 휗 . (8.9) 푥 푥 Proof. We have that

2 2 ∞ − 푘 휋 ∞ 1 푖푘 2 − 푘 휋 ∫ 2 푥 ∫ − (푡+ ) 푥 −2휋푖푘푡 −푡 휋푥 푒 1 2 1 푥 푒 푒 · 푒 d푡 = √ · 푒 2휋푥 d푡 = √ , −∞ 푥 −∞ √︃ 1 푥 2휋 · 2휋푥 | {z } 푖푘 1 pdf of N (− 푥 , 2휋푥 )

2 2 푘2 휋 푖 푘 where we have used that 2휋 푖 푘 푡 + 푡 휋푥 = 푥 + 휋푥 푡 + 푥 . Thus the Fourier transform of 2 −푡2 휋푥 1 − 푘 휋 푓 (푡) 푒 is 푓ˆ(푘) = √ 푒 푥 . Applying the Poisson summation formula (8.7) reveals the B 푥 result.

4Carl Gustav Jacob Jacobi, 1804–1851

rough draft: do not distribute 8.7 BOREL SUMMATION 49

8.7 BOREL SUMMATION

Borel summation allows summing divergent series.

∞ Deﬁnition 8.16. The Borel summation of the sequence (푎푘 )푘=0 is

∞ ∫ ∞ ∫ ∞ ∑︁ 푘 −푡 1 −푡/푧 푎푘 푧 B B푎(푡 푧) 푒 d푡 = B푎(푡) 푒 d푡, 푧 푘=0 0 0

Í∞ 푎푘 푘 where B푎(푡) B 푘=0 푘! 푡 . Í∞ 푘 Proposition 8.17. If 푘=0 푎푘 푧 converges, then the sum coincides with its Borel summation. Proof. Indeed, one may interchange the summation with integration and it holds that

∞ ∞ ∑︁ ∑︁ 푎 ∫ ∞ 푎 푧푘 = 푘 푧푘 · 푒−푡 푡푘 d푡 푘 푘! 푘=0 푘=0 0 ∞ ∑︁ ∫ ∞ 푎 = 푒−푡 · 푘 (푡 푧)푘 d푡 푘! 푘=0 0 ∫ ∞ = 푒−푡 ·B푎(푡 푧) d푡 0

and thus the assertion.

Í∞ 푘 Í∞ 푘 1 Example 8.18. The sum 푎(푧) B 푘=0 (−푧) 푘! does not converge. But B푎(푡) = 푘=0 (−푡) = 1+푡 and the Borel summation is ∫ ∞ d푡 푎(푧) = 푒−푡 0 1 + 푧푡 ∫ ∞ 1 = 푒1/푧 푒−푡 d푡 1 푡←푡− /푧 1/푧 푧푡 1 1 = 푒1/푧 Γ 0, , 푧 푧 Í∞ 푘 1 1/푧 1 that is, 푘=0 (−푧) 푘! = 푧 푒 Γ 0, 푧 . Clearly, the series is not summable in the classical sense.

8.8 ABEL–PLANAFORMULA

8.9 MERTENS’ THEOREM

Í∞ Í∞ Theorem 8.19. Suppose that 푖=0 푎푖 =: 퐴 converges absolutely and 푖=0 푏푖 =: 퐵 converges, Í∞ Í푘 then the Cauchy product 푖=0 푐푖 = 퐴 · 퐵 converges as well, where 푐푘 B ℓ=0 푎ℓ 푏푘−ℓ . Í푛 Í푛 Í푛 Proof. Deﬁne the partial sums 퐴푛 B 푖=0 푎푖, 퐵푛 B 푖=0 푏푖 and 퐶푛 B 푖=0 푐푖. Then 퐶푛 = Í푛 Í푛 푖=0 푎푛−푖 퐵푖 = 푖=0 푎푛−푖 (퐵푖 − 퐵) + 퐴푛 퐵 by rearrangement.

Version: April 23, 2021 50 SUMMABILITYMETHODS

There exists an integer 푁 so that 휀 |퐵푛 − 퐵| < Í∞ (8.10) 1 + 푘=0 |푎푘 | for all 푛 > 푁. Further, there is 푀 > 0 so that 휀 |푎푛 | < (8.11) 푁 1 + sup푖∈{0,1,...,푁 −1} |퐵푖 − 퐵|

for all 푛 > 푀. And there is 퐿 > 0 so that for all 푛 > 퐿 also 휀 |퐴 − 퐴| < . (8.12) 푛 |퐵| + 1

Now let 푛 > max {푀 + 푁, 퐿}. It follows that

푛 ∑︁ | − | ( − ) + ( − ) 퐶푛 퐴퐵 = 푎푛−푖 퐵푖 퐵 퐴푛 퐴 퐵 푖=0 푁 −1 푛 ∑︁ ∑︁ ≤ 푎 푛 − 푖 · |퐵푖 − 퐵| + |푎푛−푖 | · |퐵푖 − 퐵| + |퐴푛 − 퐴| · |퐵| 푖=0 |{z} 푖=푁 | {z } ≥푀 | {z } <휀 by (8.12) | {z } <휀 by (8.10) <휀/푁 by (8.11)

≤ 3휀

and hence the assertion.

rough draft: do not distribute Multiplicative functions 9

Young men ought to be conceited, but they oughtn’t to be imbecile.

G. H. Hardy, 1877–1947

9.1 ARITHMETICFUNCTIONS

Deﬁnition 9.1. A function 푓 : N → C is said to be arithmetic; the class of all arithmetic functions is A. Deﬁnition 9.2. An arithmetic function 푓 : N → C is1 additive, if 푓 (푚 · 푛) = 푓 (푚) + 푓 (푛) provided that (푚, 푛) = 1, totally (completely) additive, if 푓 (푚 · 푛) = 푓 (푚) + 푓 (푛) for all 푚, 푛 ∈ N. An arithmetic function is multiplikative, if 푓 (푚 · 푛) = 푓 (푚)· 푓 (푛) provided that (푚, 푛) = 1, totally (completely) multiplicative, if 푓 (푚 · 푛) = 푓 (푚)· 푓 (푛) for all 푚, 푛 ∈ N. The class of all multiplicative functions is M.

Remark 9.3. If 푓 (·) is multiplicative, then 푓 (1) = 1 or 푓 (·) = 0. Further, if 푓1, . . . , 푓푘 are multiplicative then 푛 ↦→ 푓1 (푛)··· 푓푘 (푛) is multiplicative.

9.2 EXAMPLESOFARITHMETICFUNCTIONS

Deﬁnition 9.4. The big Omega function 휔 ∑︁ ∑︁ Ω(푛) = 훼푖 = 1 푖=1 푝∈P: 푝 훼 |푛

훼1 훼휔 counts the total number of prime factors in the canonical representation (2.3), 푛 = 푝1 . . . 푝 휔 , while the prime omega function ∑︁ 휔(푛) B 휔 = 1 (9.1) 푝∈P: 푝 |푛 counts the distinct number of prime factors of 푛. The Liouville function2 is (the second 휆 in Table 3.1)

휆(푛) B (−1)Ω(푛) . (9.2) 1streng additiv, dt. 2Joseph Liouville, 1809–1882, French

51 52 MULTIPLICATIVE FUNCTIONS

Proposition 9.5. It holds that (i) The function big Omega Ω is completely additive, (ii) Liouville’s 휆 is completely multiplicative, (iii) the prime omega function 휔 is additive and (iv) Euler’s totient function 휑 is multiplicative (cf. Theorem 3.2). Deﬁnition 9.6. The number-of-divisors function ∑︁ 휏(푛) B 1 (9.3) 푑 |푛

(for the German Teiler = divisors) counts the number of divisors; often, 휎0 = 푑 = 휈 B 휏. Deﬁnition 9.7. The sigma function, or sum-of-divisors function is ∑︁ 휎(푛) B 푑. 푑 |푛 Deﬁnition 9.8. More generally, the divisor function is ∑︁ 푘 휎푘 (푛) B 푑 . (9.4) 푑 |푛

Proposition 9.9. The divisor function 휎푘 are multiplicative, but not completely multiplicative.

9.3 DIRICHLETPRODUCTANDMÖBIUSINVERSION

Recall the definition of arithmetic (Definition 9.1), additive and multiplicative (Definition 9.2) functions. Definition 9.10 (Dirichlet convolution). The Dirichlet3 product of arithmetic functions 푓 and 푔 is the arithmetic function 푓 ∗ 푔, where ∑︁ ( 푓 ∗ 푔)(푛) B 푓 (푑1)· 푔(푑2). (9.5) 푑1 ·푑2=푛 The product is commutative and associative: ∑︁ ( 푓 ∗ 푔) ∗ ℎ (푛) = 푓 ∗ (푔 ∗ ℎ) (푛) = 푓 ∗ 푔 ∗ ℎ (푛) = 푓 (푑1) 푔(푑2) ℎ(푑3). 푑1 푑2 푑3=푛 Definition 9.11. Define the arithmetic functions ( 1 if 푛 = 1, 훿1 (푛) B 0 else and 1(푛) B 1 for 푛 ∈ N.

Remark 9.12. It holds that 훿1 ∗ 푓 = 푓 ∗ 훿1 = 푓 (훿1 is the neutral element with respect to the Dirichlet product). −1 −1 Deﬁnition 9.13 (Dirichlet inverse). The function 푓 with 푓 ∗ 푓 = 훿1 is the Dirichlet inverse of 푓 with respect to ∗. 3Johann Peter Gustav Lejeune Dirichlet, 1805–1859

rough draft: do not distribute 9.4 MÖBIUSINVERSION 53

9.4 MÖBIUSINVERSION

Deﬁnition 9.14 (Möbius function). The Möbius4 function 휇 : N → {−1, 0, 1} (cf. Table 3.1) is

1 if 푛 = 1,   푘 휇(푛) B (−1) if 푛 = 푝1 ··· 푝푘 with distinct primes 푝1 < . . . < 푝푘 ,  0 if 푛 has a square (prime) factor.  Remark 9.15. The function 푛 ↦→ |휇(푛)| is the indicator function for the square-free integers.

Proposition 9.16. The function 휇(·) is multiplicative (not totally multiplicative, though).

Theorem 9.17 (Dirichlet inverse of the Möbius function). It holds that ( ∑︁ 1 if 푛 = 1, 휇(푑) = (9.6) 0 else. 푑 |푛

I.e., the Möbius function is the Dirichlet inverse of 1, 1−1 = 휇 or

휇 ∗ 1 = 훿1; (9.7)

훼1 ··· 훼휔 Proof. For 푛 = 푝1 푝 휔 > 1 (with 휔 distinct prime factors) we have that

휔 ∑︁ ∑︁ ∑︁ ∑︁ 휇(푑) = 휇(푑) = 휇(1) + 휇 푝푖1 ··· 푝푖푟 , 푑 |푛 푑 |푛, 푟=1 1≤푖1<...푖푟 ≤휔 푑 square-free

where the latter sum runs over all square-free divisors of 푛. For combinatorial reasons,

휔 ∑︁ ∑︁ 휔 휇(푑) = (−1)푟 = (1 − 1) 휔 = 0 푟 푑 |푛 푟=0

and thus the result.

1 −1 휇(푛) Example 9.18. The Dirichlet inverse of the arithmetic function 훼 : 푛 ↦→ 푛푠 is 훼 : 푛 ↦→ 푛푠 . 푠 Í 푛 −1 Í 푑 휇(푑) 1 Í 1 1 1 Proof. Indeed, 푑 |푛 훼 푑 훼 (푑) = 푑 |푛 푛 · 푑푠 = 푛푠 푑 |푛 휇(푑) = 푛푠 휇 ∗ (푛) = 푛푠 훿1 (푛) = 훿1 (푛) by (9.7).

Deﬁnition 9.19. The sum function is ∑︁ 푆 푓 (푛) B 푓 (푑). (9.8) 푑 |푛

Remark 9.20. We have that 푆 푓 = 푓 ∗ 1 = 1 ∗ 푓 . (However, note that 1 · 푓 = 푓 · 1 = 푓 for the usual, pointwise product.)

4August Ferdinand Möbius, 1790–1868

Version: April 23, 2021 54 MULTIPLICATIVE FUNCTIONS

Theorem 9.21 (Möbius inversion formula; Dedekind5). Any arithmetic function 푓 can be expressed in terms of its sum function 푆 푓 as ∑︁ 푓 (푛) = 휇(푑1)· 푆 푓 (푑2) (9.9) 푑1 푑2=푛

for every integer 푛 ≥ 1, where 푆 푓 is the sum function (9.8).

Proof. We have that 푓 = 훿1 ∗ 푓 = (휇 ∗ 1) ∗ 푓 = 휇 ∗ (1 ∗ 푓 ) = 휇 ∗ 푆 푓 , this is the result. For a more explicit proof, ∑︁ 푛 ∑︁ 푛 ∑︁ 푛 ∑︁ 휇(푑) 푆 = 휇 푆 (푑) = 휇 푓 (푑 ) 푓 푑 푑 푓 푑 1 푑 |푛 푑 |푛 푑 |푛 푑1 |푑 ∑︁ ∑︁ 푛 ∑︁ ∑︁ 푚 = 푓 (푑1) 휇 = 푓 (푑1) 휇 , 푑 푑2 푑1 |푛 푑1 |푑 |푛 푑1 |푛 푑2 |푚

where 푚 = 푛/푑1, 푑2 = 푑/푑1. By (9.6), the second sum is non-zero when 푚 = 1, i.e., 푛 = 푑1, and hence the whole expression equals 푓 (푛). Theorem 9.22. The set M of multiplicative functions is closed under the Dirichlet product, i.e., 푓 ∗ 푔 ∈ M whenever 푓 , 푔 ∈ M. Proof. For (푎, 푏) = 1 note that

푑1 | 푎 and 푑2 | 푏 =⇒ (푑1, 푑2) = 1 and {푑 : 푑 | 푎 · 푏} = {푑1 · 푑2 : 푑1 | 푎 and 푑2 | 푏} . Set ℎ B 푓 ∗ 푔, then ∑︁ 푎 ∑︁ 푏 ℎ(푎) ℎ(푏) = ( 푓 ∗ 푔)(푎)·( 푓 ∗ 푔)(푏) = 푓 (푑1) 푔 · 푓 (푑2) 푔 푑1 푑2 푑1 |푎 푑2 |푏 ∑︁ 푎 푏 ∑︁ 푎푏 = 푓 (푑1) 푓 (푑2)· 푔 푔 = 푓 (푑1푑2)· 푔 푑1 푑2 푑1푑2 푑1 |푎, 푑2 |푏 푑1 |푎, 푑2 |푏 ∑︁ 푎 푏 = 푓 (푑) 푔 = ℎ(푎 푏). 푑 푑 |푎푏

Thus, ℎ is also multiplicative so that 푓 ∗ 푔 ∈ M.

9.5 THESUMFUNCTION

Theorem 9.23. The function 푓 (·) is multiplicative iff its sum-function 푆 푓 is multiplicative.

Proof. Assuming that 푓 is multiplicative we have that 푆 푓 = 푓 ∗1 is multiplicative by Theorem 9.22, as 1 is multiplicative. As for the converse it holds that 푓 = 휇 ∗ 푆 푓 by the Möbius inversion formula (9.9). Assum- ing that 푆 푓 is multiplicative, it follows again with Theorem 9.22 that 푓 is multiplicative, as 휇 is multiplicative by Proposition 9.16. 5Richard Dedekind, 1831–1916, German

rough draft: do not distribute 9.6 OTHERSUMFUNCTIONS 55

Corollary 9.24. The sum function of a multiplicative function 푓 is given by

휔 Ö 2 훼푖 푆 푓 (푛) = 1 + 푓 (푝푖) + 푓 (푝푖 ) + · · · + 푓 (푝푖 ) . 푖=1

If 푓 is totally multiplicative, then it is sufﬁcient to know the restriction 푓 |P, as

휔 Ö 푓 (푝 ) 훼푖 +1 − 1 푆 (푛) = 푖 . 푓 푓 (푝 ) − 1 푖=1 푖

9.6 OTHERSUMFUNCTIONS

Corollary 9.25 (Corollary to Theorem 9.17; Möbius). For a function 푓 : [1, ∞) → C deﬁne the summatory function ∑︁ 푥 퐹(푥) B 푓 , (9.10) 푘 푘 ≤푥, 푘 ∈N then ∑︁ 푥 푓 (푥) = 휇(푘) 퐹 ; (9.11) 푘 푘 ≤푥, 푘 ∈N conversely, (9.11) implies (9.10). Proof. For an arithmetic function 훼 : N → C and 푓 : [0, ∞) → C deﬁne the operation ∑︁ 푥 (훼 • 푓 )(푥) B 훼(푘) 푓 . 푘 푘 ≤푥, 푘 ∈N The following hold true: (i) 훼 •( 푓 + 푔) = 훼 • 푓 + 훼 • 푔, (ii) (훼 + 훽)• 푓 = 훼 • 푓 + 훽 • 푓 , (iii) 훼 •(훽 • 푓 ) = (훼 ∗ 훽)• 푓 and

(iv) 훿1 • 푓 = 푓 . (i) and (ii) are clear. As for (iii), ∑︁ 푥 ∑︁ ∑︁ 푥 훼 •(훽 • 푓 )(푥) = 훼(푘)(훽 • 푓 ) = 훼(푘) 훽(ℓ) 푓 푘 푘ℓ ≤ ≤ 푥 푘 푥 푘 푥 ℓ ≤ 푘 ∑︁ 푥 = 훼(푘)훽(ℓ) 푓 푘ℓ 푘ℓ ≤푥 ∑︁ ∑︁ 푥 = 훼(푘)훽(ℓ) 푓 푛 푛≤푥 푘ℓ=푛 ∑︁ 푥 = (훼 ∗ 훽)(푛) 푓 = (훼 ∗ 훽)• 푓 )(푥). 푛 푛≤푥 Í 푥 (iv) ﬁnally follows trivially from (훿1 • 푓 )(푥) = 푘 ≤푥 훿1 (푘) 푓 푘 = 푓 (푥).

Version: April 23, 2021 56 MULTIPLICATIVE FUNCTIONS

Now note that 퐹 = 1 • 푓 . Hence

휇 • 퐹 = 휇 •(1 • 푓 ) = (휇 ∗ 1)• 푓 = 훿1 • 푓 = 푓

by (10.3), which is (9.11). Finally

1 • 푓 = 1 •(휇 • 퐹) = (1 ∗휇)• 퐹 = 훿1 • 퐹 = 퐹,

which is (9.10).

Remark 9.26. Set 푓 (푥) B 1, then 퐹(푥) = b푥c. It follows that ∑︁ j 푥 k 1 = 휇(푘)· . 푘 푘 ≤푥

For example (푥 = 5)

1 = 5휇(1) + 2휇(2) + 휇(3) + 휇(4) + 휇(5) or 1 = 6휇(1) + 3휇(2) + 2휇(3) + 휇(4) + 휇(5) + 휇(6)

(푥 = 6).

9.7 PROBLEMS

Exercise 9.1. Show the Lambert series for the Möbius function, ∑︁ 푟푛 휇(푛) = 푟, |푟| < 1 1 − 푟푛 푛=1 or more generally, ∑︁ ∑︁ 푥푛 푎 푥푛 = (휇 ∗ 푎)(푛) . 푛 1 − 푥푛 푛=1 Exercise 9.2. Show that ∑︁ 휇(푛) = 0, 푛 푛=1 provided that the sum exists (hint: Lambert summation 8.5).

Exercise 9.3. Show that ∑︁ 푎 ∑︁ (푎 ∗ 휇)(푛) 푛 = 휁 (푠)· . 푛푠 푛푠 푛=1 푛=1 Exercise 9.4. Show that 푛 ∑︁ 2 ∑︁ 푥 푥푛 = 휆 (푛) , 1 − 푥푛 푛=1 푛=1 where 휆 is the Liouville function.

rough draft: do not distribute Euler’s product formula 10

Lisez Euler, lisez Euler, c’est notre maître à tous.

Pierre-Simon Laplace, 1749–1827

10.1 EULER’SPRODUCTFORMULA

Deﬁnition 10.1. Riemann’s 휁-function is the analytic continuation of the Dirichlet series

∞ ∑︁ 1 휁 (푠) B , (10.1) 푛푠 푛=1

which converges for <(푠) > 1.

Proposition 10.2 (Euler’s product formula). It holds that

Ö 1 휁 (푠) = , <(푠) > 1. (10.2) − 1 푝 prime 1 푝푠

Proof. The assertion follows with 푓 (푛) = 1 from the following, more general statement.

Í 푓 (푛) Theorem 10.3. Suppose that 푓 is multiplicative and 푛=1 푛푠 converges absolutely for some 푠 ∈ C, then ∑︁ 푓 (푛) Ö ∑︁ 푓 (푝푘 ) = . 푛푠 푝푘푠 푛=1 푝 prime 푘=0 If 푓 is totally multiplicative, then

∑︁ 푓 (푛) Ö 1 = . 푛푠 푓 ( 푝) 푛=1 푝 prime 1 − 푝푠

Proof. Choose large enough so that Í 푓 (푛) . Set P { ∈ P ≤ }, then 푁 푛>푁 푛푠 < 휀 푁 B 푝 : 푝 푁

∞ Ö ∑︁ 푓 (푝푘 ) ∑︁ 푓 (푛) = 푝푘푠 푛푠 푝∈P푁 푘=0 푛∈N, 푝 |푛⇒푝∈P푁

푘 and hence Í∞ 푓 (푛) − Î Í∞ 푓 ( 푝 ) ≤ Í 푓 (푛) ≤ Í 푓 (푛) < 휀 from which the 푛=1 푛푠 푝∈P푁 푘=0 푝푘푠 푛>푁 푛푠 푛>푁 푛푠 assertion follows.

57 58 EULER’SPRODUCTFORMULA

Proposition 10.4 (Liouville function). It holds that

− 휁 (2푠) Ö 1 1 ∑︁ 휆(푛) = 1 + = , <(푠) > 1, 휁 (푠) 푝푠 푛푠 푝 prime 푛=1

where 휆(푛) is the Liouville function, cf. (9.2).

1 1− · Proof. Indeed, 휁 (푠) = Î 푝푠 2 = Î 1 + 1 , the ﬁrst assertion. The remaining assertion 휁 (2푠) 푝∈P − 1 푝∈P 푝푠 1 푝푠 푘 follows with 1 = Í (−1) . + 1 푘=0 푝푘푠 1 푝푠

Î 1 Remark 10.5. It follows from 휁 (푠) = 휁 (2푠) × 푝∈P 1 + 푝푠 that every number is the product of simple primes (square-free) × the square of a natural number.

Theorem 10.6. Let 푄푘 (푛) be the number of 푘th-power free integers up to 푛, then 푛 푄 (푛) ∼ 푘 휁 (푘) as 푛 → ∞. = 푄2 (푛) ∼ 6 ≈ Square-free numbers, in particular, have for 푘 2 the asymptotic density 푛 휋2 0.6079. − 1 푘 − 1 Sketch of the proof. For large 푛, a fraction 1 2푘 of numbers is not divisible by 2 , 1 3푘 are not divisible by 3푘 , etc. These ratios are multiplicative (cf. the Chinese remainder theorem, Theorem 2.32) and we thus obtain the approximation

Ö 1 푛 푛 푄 (푛) ∼ 푛 · 1 − = = , 푘 푝푘 Î 1 휁 (푘) 푝 prime 푝 prime 1− 1 푝푘

the assertion. Theorem 10.7. It holds that ∑︁ 푎 ∑︁ 푏 ∑︁ 푐 푛 · 푛 = 푛 , (10.3) 푛푠 푛푠 푛푠 푛=1 푛=1 푛=1 Í Í where 푐푛 = 푑 |푛 푎푑 · 푏푛/푑 = 푗·푘=푛 푎 푗 · 푏푘 .

Í 푎푛 Í 푏푚 Í 푎푛 ·푏푚 Proof. Indeed, 푛=1 푛푠 · 푚=1 푚푠 = 푛,푚=1 (푛·푚)푠 . The result follows by comparing terms. Proposition 10.8. It holds that

1 ∑︁ 휇(푛) = , <(푠) > 1. (10.4) 휁 (푠) 푛푠 푛=1

Proof. The proof is immediate by (10.3) and (9.6). Corollary 10.9 (Corollary to Proposition 10.4). It holds that

∞ 휁 (푠) ∑︁ |휇(푛)| ∑︁ 1 = = . 휁 (2푠) 푛푠 푛푠 푛=1 푛=1, 푛 square-free

rough draft: do not distribute 10.1 EULER’SPRODUCTFORMULA 59

Corollary 10.10 (Landau1). It holds that

∑︁ 휇(푛) ∑︁ 휇(푛) log 푛 = 0 and = −1. (10.5) 푛 푛 푛=1 푛=1

Proof. Consider (10.4) and its derivative for 푠 → 1. Í Proposition 10.11. The number of divisors of 푛 (including 1 and 푛) is 휏(푛) B 푑 |푛 1. It holds that ∑︁ 휏(푛) 휁 (푠)2 = , <(푠) > 1. 푛푠 푛=1 Proof. The proof is immediate by (10.3). Proposition 10.12. It holds that

휁 (푠 − 1) ∑︁ 휑(푛) = , <(푠) > 1, 휁 (푠) 푛푠 푛=1 where 휑(·) is Euler’s totient function (3.1).

Proof. The proof is immediate by (10.3), as

∑︁ 휑(푛) ∑︁ 1 ∑︁ ∑︁ 푛 휁 (푠)· = 1 · 휑(푑) = = 휁 (푠 − 1) 푛푠 푛푠 푛푠 푛=1 푛=1 푑 |푛 푛=1 by Gauss’ theorem, Theorem 3.5. Proposition 10.13. The following hold true (cf. Titchmarsh [19]):

휁 (푠)2 Í 2휔(푛) ⊲ 휁 (2푠) = 푛=1 푛푠 , where 휔 is the number of different prime factors, cf. (9.1);

Í 휎푎 (푛) ⊲휁 (푠) 휁 (푠 − 푎) = 푛=1 푛푠 , where 휎푎 is the divisor function, cf. (9.4); 휁 (푠)3 Í 휏 푛2 ⊲ 휁 (2푠) = 푛=1 푛푠 , where 휏 is the number of divisors, cf. (9.3);

휁 (푠)4 Í 휏 (푛)2 ⊲ 휁 (2푠) = 푛=1 푛푠 ;

푘 Í 푓푘 (푛) ⊲ 휁 (푠) − 1 = 푛=2 푛푠 , where 푓푘 (푛) is the number of representations of 푛 as a product of 푘 factors, each greater than unity when 푛 > 1, the order of the factors being essential;

1 Í 푓 (푛) ⊲ 2−휁 (푠) = 푛=1 푛푠 , where 푓 (푛) is the number of representations of 푛 as a product of factors greater than unity, representations with factors in a different order being considered as distinct, and 푓 (1) = 1.

Deﬁnition 10.14. The von Mangoldt2 function is ( log 푝 if 푛 = 푝푘 (푝 prime, 푘 integer), Λ(푛) B (10.6) 0 else. 1Edmund Landau, 1877–1938, German 2Hans Carl Friedrich von Mangoldt, 1854–1925, German

Version: April 23, 2021 60 EULER’SPRODUCTFORMULA

Proposition 10.15. We have that ∑︁ ∑︁ log 푛 = Λ(푑) and Λ(푛) = − 휇(푑) log 푑. 푑 |푛 푑 |푛

Proof. Use Möbius inversion formula (9.9). Proposition 10.16. It holds that ∑︁ Λ(푛) log 휁 (푠) = . 푛푠 log 푛 푛=1 Proof. From (10.2) we deduce that

∞ ∑︁ 1 ∑︁ ∑︁ 1 log 휁 (푠) = − log 1 − = . 푝푠 푘 푝푘 푠 푝 prime 푝 prime 푘=1

푘 Λ( 푝 ) = 1 = 푘 Now note that log 푝푘 푘 for 푛 푝 and 0 else, thus the result. Theorem 10.17. For <(푠) > 1 it holds that 휁 0(푠) ∑︁ Λ(푛) − = (10.7) 휁 (푠) 푛푠 푛=1 ∑︁ log 푝 ∑︁ log 푝 ∑︁ log 푝 = = + ; (10.8) 푝푠 − 1 푝푠 푝푠 (푝푠 − 1) 푝 prime 푝 prime 푝 prime

1 the latter sum converges for <(푠) > 2 . Í 1 Proof. From (10.2) we deduce that log 휁 (푠) = − 푝∈P log 1 − 푝푠 . Differentiating with respect to 0 1 log 1 푠 thus 휁 (푠) = Í 푝푠 푝 = − Í log 푝 ; the second identity is immediate. 휁 (푠) 푝∈P − 1 푝∈P 푝푠−1 1 푝푠 Convergence follows from the integral test for convergence or the inequality ∑︁ log 푝 log 2 ∑︁ log 푛 log 2 ∫ ∞ log 푥 ≤ + ≤ + d푥 < ∞. 푝푠 (푝푠 − 1) 2푠 (2푠 − 1) 푛푠 (푛푠 − 1) 2푠 (2푠 − 1) 푥푠 (푥푠 − 1) 푝∈P 푛=3 2 Deﬁnition 10.18. The prime zeta function is 1 1 1 ∑︁ 1 푃(푠) B + + + · · · = . (10.9) 2푠 3푠 5푠 푝푠 푝 prime Theorem 10.19. For <(푠) > 1 it holds that ∑︁ 1 log 휁 (푠) = 푃(ℓ 푠) (10.10) ℓ ℓ=1 and conversely,

∞ ∑︁ 1 ∑︁ 휇(푘) 푃(푠) = = log 휁 (푘 푠). 푝푠 푘 푝 prime 푘=1

rough draft: do not distribute 10.2 PROBLEMS 61

Proof. From Euler’s product (10.2) it follows that

∑︁ 1 ∑︁ ∑︁ 1 1 log 휁 (푠) = − log 1 − = (10.11) 푝푠 ℓ 푝ℓ푠 푝 prime 푝 prime ℓ=1 ∑︁ 1 ∑︁ 1 ∑︁ 1 = = 푃(ℓ 푠). ℓ 푝ℓ 푠 ℓ ℓ=1 푝 prime ℓ=1

Hence,

∑︁ 휇(푘) ∑︁ 휇(푘) 푃(ℓ 푘 푠) ∑︁ ∑︁ 푃(푛 푠) ∑︁ 푃(푛 푠) ∑︁ log 휁 (푘푠) = = 휇(푘) = 휇(푘) = 푃(푠), 푘 푘 ℓ 푛 푛 푘=1 푘,ℓ=1 푛=1 푘 ℓ=푛 푛=1 푘 ℓ=푛 | {z } 훿1 (푛) by (9.6)

the result. 1 Proposition 10.20. For 푘 ≥ 1 and <(푠) > 1 it holds that |푃(푘푠)| ≤ 푘−1 . Proof. Indeed,

∑︁ 1 ∑︁ 1 ∑︁ ∫ 푛 d푥 ∫ ∞ d푥 1 |푃(푘푠)| ≤ 푃(푘) = ≤ ≤ = = . 푝푘 푛푘 푥푘 푥푘 푘 − 1 푝 prime 푛=2 푛=2 푛−1 1

10.2 PROBLEMS

Exercise 10.1. Verify the properties in Proposition 10.13.

Version: April 23, 2021 62 EULER’SPRODUCTFORMULA

rough draft: do not distribute Dirichlet Series 11

푓 (푛) If 푛 ↦→ 푓 (푛) is multiplicative, then so is 푛 ↦→ 푛푠 for every 푠 ∈ C. Deﬁnition 11.1. The Dirichlet series associated with an arithmetic function 푎 is ∞ ∑︁ 푎푛 D푎(푠) B . 푛푠 푛=1

11.1 PROPERTIES

Lemma 11.2. If 푎1 + · · · + 푎푛 is bounded, then D푎(푠) is holomorphic in {푠 : <(푠) > 0}. Theorem 11.3. It holds that ∑︁ 푎 ∑︁ 푏 ∑︁ (푎 ∗ 푏)(푛) 푛 · 푛 = , 푛푠 푛푠 푛푠 푛=1 푛=1 푛=1 Í Í where ’∗’ is the Dirichlet convolution (9.5), 푐푛 = 푑 |푛 푎푑 푏푛/푑 = 푗·푘=푛 푎 푗 푏푘 . Proof. See (10.3).

1 Í 푏푛 Í 푛 Theorem 11.4. It holds that Í 푎푛 = 푛=1 푛푠 , where 푏푛 B 푑 |푛 휇 푑 푎푑 . 푛=1 푛푠

푎푛 푏푚 푐푛 0 Í Í Í Í Í 0 Proof. It holds that 푛=1 푛푠 · 푚=1 푚푠 = 푛=1 푛푠 with 푐푛 = 푑 |푛 푎푑 푑0 |푛/푑 휇(푑 )푏푛/푑푑 .

11.2 ABSCISSAOFCONVERGENCE

Í∞ 푎푛 Theorem 11.5 (Cf. Hardy and Riesz [9, Theorem 2]). Suppose that 푛=1 푛푠0 converges and 휋 Í∞ 푎푛 휗 < 2 . Then the Dirichlet series D푎(푠) = 푛=1 푛푠 converges uniformly on

퐺 B {푠 ∈ C: arg(푠 − 푠0) ≤ 휗}

and D푎(푠) is holomorphic on {푠 : <(푠) > <(푠0)}. Í 푎푛 Proof. The series converges at 푠0, hence, for 휀 > 0, there is 푀 > 0 so that 푀 ≤푛<푡 푛푠0 < 휀 <( ) <( ) ( ) 1 for all 푡 > 푀. Set 휎 B 푠 and 휎0 B 푠0 and 휙 푛 B 푛푠−푠0 , then with Abel’s summation formula (8.6), ∑︁ 푎 ∑︁ 푎 1 ∑︁ 푎 1 ∫ 푁 ∑︁ 푎 푠 − 푠 푛 = 푛 · = 푛 · + 푛 · 0 d푢. 푛푠 푛푠0 푛푠−푠0 푛푠0 푁 푠−푠0 푛푠0 푢푠−푠0+1 푀 ≤푛<푁 푀 ≤푛<푁 푀 ≤푛<푁 푀 푀 ≤푛<푢

As 휎0 < 휎 it follows that

∑︁ 푎 휀 ∫ 푁 푠 − 푠 푛 < + 휀 0 d푢 푛푠 푁 휎−휎0 푢푠−푠0+1 푀 ≤푛<푁 푀 1 |푠 − 푠 | 1 1 |푠 − 푠 | = + 0 − ≤ + 0 휀 휎−휎 휎−휎 휎−휎 휀 1 2 . 푁 0 휎0 − 휎 푁 0 푀 0 |휎 − 휎0|

63 64 DIRICHLETSERIES

Deﬁne the angle sin 훼 | 휎−휎0 | . Assuming that 푠 ∈ 퐺 it holds that 휋 − 훼 < 휗, thus | 휎−휎0 | = B |푠−푠0 | 2 |푠−푠0 | sin 훼 = cos 휋 − 훼 > cos 휗, i.e., |푠−푠0 | < 1 . It follows that 2 | 휎−휎0 | cos 휗

∑︁ 푎 2 푛 < 휀 1 + 푛푠 cos 휗 푀 ≤푛<푁

uniformly in 퐺 and hence the assertion.

Í 푎푛 Í 푏푛 Theorem 11.6 (Uniqueness). Suppose that 퐹(푠) B 푛=1 푛푠 and 퐺(푠) B 푛=1 푛푠 converge for 푠 = 푠0 and 퐹(휎) = 퐺(휎) for all 휎 ∈ R sufﬁciently large. Then 푎푛 = 푏푛 for all 푛 ∈ N.

Proof. Convergence is uniformly on [<(푠0) + 1, ∞), thus ∑︁ 1 lim 퐹(휎) = 푎푛 · lim = 푎1. 휎→∞ 휎→∞ 푛휎 푛=1

The same argument applied to 퐺 gives that 푎1 = 푏1. 푠 푠 Í 2 푠 Consider next the functions 퐹2 (푠) B 2 (퐹(푠) − 푎1) = 푛=2 푎푛 푛 and 퐺2 (푠) B 2 (퐺(푠) − 푏1). As 푎1 = 푏1 it holds that 퐹2 = 퐺2. As above we ﬁnd that 푎2 = 푏2 and the assertion follows by repeating the arguments successively. Theorem 11.7 (Abscissa of convergence). Let

Í푁 log 푛=1 푎푛 휎0 B lim sup , 푛→∞ log 푁

Í∞ 푎푛 then the Dirichlet series D푎(푠) = 푛=1 푛푠 converges for all 푠 ∈ C with <(푠) > 휎0 and diverges for all <(푠) < 휎0.

11.3 PROBLEMS

Í 훼푛 Exercise 11.1. Show that 푛=1 푛푠 converges for all 푠 ∈ C, if |훼| < 1, but diverges for all 푠 ∈ C if |훼| > 1.

rough draft: do not distribute Mellin transform 12

Deﬁnition 12.1. The Mellin1 transform of a function 푓 : (0, ∞) → C is the function M 푓 : C → C, where ∫ ∞ M 푓 (푠) B 푥푠−1 푓 (푥) d푥, 푠 ∈ C. 0

Example 12.2. Bateman [3, pp. 303] provides a comprehensive list of Mellin transforms. Ta- ble 12.1 provides few examples.

12.1 INVERSION

Proposition 12.3 (Inversion of the Mellin transform). It holds that

1 ∫ 푐+푖∞ M−1퐹(푥) = 퐹(푠) 푥−푠 d푠 = 푓 (푥), (12.1) 2휋푖 푐−푖∞

∞ ( ) M ( ) = ∫ ( ) 푠−1 ∈ { ∈ C <( ) } where 퐹 푠 B 푓 푠 0 푓 푥 푥 d푥 and 푐 is within the strip of analyticity, 푐 푠 : 푎 < 푠 < 푏 .

Proof. We shall derive the result from Fourier transforms. With 푐 fixed, define ℎ(푥) B 푥푐 · 푓 (푥) so that 0 is in the strip of analyticity of the function ℎ. Note first that ∫ ∞ ∫ ∞ Mℎ(−푖푡) = 푥−푖푡−1 ℎ(푥) d푥 = 푒−푖푡 푦 ℎ푒푦 d푦 = F ℎ ◦ exp (푡), 푦 0 푥←푒 −∞

where ∫ ∞ F(푔)(푡) B 푒−푖푡 푦 푔(푦) d푦 −∞ is the usual Fourier transform. Employing the inverse Fourier transform gives

1 ∫ ∞ 1 ∫ +푖∞ ℎ푒푦 = 푒푖푡 푦 Mℎ(−푖푡) d푡 = 푒−푠푦 Mℎ(푠) d푠. (12.3) 2휋 −∞ 푡←−푠/푖 2휋푖 −푖∞

1Hjelmar Mellin, 1854–1933, Finnish 2The function 0 if 푥 < 0,  ( )  1 퐻 푥 B 2 if 푥 = 0, (12.2)  1 if 푥 > 1  is the Heaviside step function. 3aka. Cahen–Mellin integral 4Hermite polynomials

65 66 MELLINTRANSFORM

function Mellin transform remark ∞ ( )M ( ) = ∫ 푠−1 ( ) 푓 푥 푓 푠 0 푥 푓 푥 d푥 + ∞ 1 ∫ 푐 푖 ( ) −푠 ( ) 2휋푖 푐−푖∞ 퐹 푠 푥 d푠 퐹 푠 cf. (12.1) 훿(푥 − 푏) 푏푠−1 푎 2 푏푎+푠 푥 퐻(푏 − 푥) 푎+푠 푎 푏푎+푠 푥 퐻(푥 − 푏) − 푎+푠 Lemma 12.8 푠+휆 휆 푏 1 − 훽 푠+휆 푥 푓 푎푥 훽 푎 M 푓 훽 −푏푥 3 Γ(푠) 푒 푏푠 1 2 √ 2 − log 푥 푛 푠 /2 푒 2 · 퐻푛 (log 푥) 4 2휋(−푖) 푒 · 퐻푛 (푖 푠) Γ(푠) 휋푠 sin 훼푥 훼푠 sin 2 Example 12.15 Γ(푠) 휋푠 cos 훼푥 훼푠 cos 2 −푥 sin(푠 arctan 훼) 푒 sin 훼푥 Γ(푠) 푠/2 (1+훼2) −푥 cos(푠 arctan 훼) 푒 cos 훼푥 Γ(푠) 푠/2 (1+훼2) 2 푠(푠−1) Γ(푠/2) 2 휓(푥 ) 휉(푠) = 푠 휁 (푠) cf. (13.20) 2 휋 /2 Í 휙(푘) 푘 푘=0 푘! (−푥) Γ(푠)· 휙(−푠) cf. (12.6) Dirichlet series Í −푛푥 푛=1 푎푛푒 Γ(푠)·D푎(푠) Theorem 12.9 Í 푛 푛=1 푎푛 푓 ( 푥 )D푎(−푠)·M 푓 (−푠) cf. (12.5) Derivatives (푛) 푛 Γ(푠) 푓 (푥) (−1) Γ(푠−푛) M 푓 (푠 − 푛) 푛 d 푛 푛 푥 d푥 푓 (푥) (−1) 푠 M 푓 (푠) Convolutions ∞ 훼 ∫ 훽 푥 ( ) d푦 M ( + )·M ( + + ) 푥 0 푦 푓 푦 푔 푦 푦 푓 푠 훼 푔 푠 훼 훽 cf. (12.8) ∞ 훼 ∫ 훽 ( ) ( ) d푦 M ( + )·M ( − − ) 푥 0 푦 푓 푥푦 푔 푦 푦 푓 푠 훼 푔 훽 푠 훼 cf. (12.9) + ∞ ( ) ( ) 1 ∫ 푐 푖 M ( )·M ( − ) 푓 푥 푔 푥 2휋푖 푐−푖∞ 푓 푧 푔 푠 푧 d푧 Proposition 12.13

Table 12.1: Mellin transforms

rough draft: do not distribute 12.2 PERRON’SFORMULA 67

Now note that Mℎ(푠) = M 푓 (푠 + 푐) and after replacing 푦 ← log 푥 in (12.3) thus

푥−푐 ∫ +푖∞ 푓 (푥) = 푥−푐 · ℎ푥 = 푥−푠 ·Mℎ(푠) d푠 2휋푖 −푖∞ 1 ∫ +푖∞ = 푥−(푠+푐) ·M 푓 (푠 + 푐) d푠 2휋푖 −푖∞ 1 ∫ 푐+푖∞ = 푥−푠 ·M 푓 (푠) d푠, 푠←푠−푐 2휋푖 푐−푖∞ the assertion.

12.2 PERRON’SFORMULA

Theorem 12.4 (Perron’s formula, see also Proposition 12.6 below5). Suppose that D푎(푠) B Í 푎푛 푛=1 푛푠 converges uniformly on {푠 : <(푠) > 푐 − 휀} for some 휀 > 0 and 푐 > 0. Then it holds that

∫ 푐+푖∞ 푠 ∗ ∑︁∗ 1 푥 퐴 (푥) B 푎푛 = D푎(푠) d푠 for all (sic!) 푥 ∈ R, (12.4) 휋푖 푠 푛≤푥 2 푐−푖∞

where ! ( ∑︁∗ 1 ∑︁ ∑︁ Í 푎 + 1 푎 if 푥 is an integer + 푛<푥 푛 2 푥 푎푛 B 푎푛 푎푛 = Í 푛≤푥 2 푛<푥 푛≤푥 푛<푥 푎푛 else is the summatory function. Í Remark 12.5. Notice that Perron’s formula allows computing 푛≤푥 푎푛, even if D푎 does not converge at 푠 = 0. Proof. By Abel’s summation formula (8.6) it follows that

∑︁ 푎 ∫ ∞ 1 ∫ ∞ 퐴(푥) D푎(푠) = 푛 = d퐴(푥) = 푠 d푥 = 푠 ·M 퐴(−푠). 푛푠 푥푠 푥푠+1 푛=1 0 0 From Mellin inversion (12.1) we deduce that

1 ∫ 푐+푖∞ 퐴(푥) = M 퐴(푠)푥−푠 d푠 2휋푖 푐−푖∞ 1 ∫ 푐+푖∞ D푎(−푠) = · 푥−푠 d푠 2휋푖 푐−푖∞ −푠 1 ∫ 푐+푖∞ 푥푠 = D푎(푠)· d푠, 2휋푖 푐−푖∞ 푠 the assertion. Proposition 12.6 (Discrete convolution). It holds that

∞ ∑︁ 푛 1 ∫ 푐+푖∞ 푎 · 푓 = D푎(푠)M 푓 (푠) 푥푠 d푠. (12.5) 푛 푥 2휋푖 푛=1 푐−푖∞ 5Oskar Perron, 1880–1975, German mathematician

Version: April 23, 2021 68 MELLINTRANSFORM

Remark 12.7. The preceding result generalizes Perron’s formula. Indeed, choose 푓 (푥) B 퐻(푥 − 1 1), then M 푓 (푠) = 푠 (cf. Table 12.1) and this gives Perron’s formula (12.4).

Proof. We have that ! ∑︁ 푛 ∫ ∞ ∑︁ 푛 M 푎 · 푓 (푠) = 푥푠−1 푎 · 푓 d푥 푛 푥 푛 푥 푛=1 0 푛=1 ∫ ∞ ∑︁ 푠 1−푠−2 = 푎푛푛 · 푥 푓 (푥) d푥 푥← 푛 푥 푛=1 0 = D푎(−푠)·M 푓 (−푠).

By inverting the formula with (12.1) we obtain

∑︁ 푛 1 ∫ 푐+푖∞ 푎 · 푓 = D푎(−푠)M 푓 (−푠) 푥−푠 d푠 푛 푥 2휋푖 푛=1 푐−푖∞ 1 ∫ 푐+푖∞ = D푎(푠)M 푓 (푠) 푥푠 d푠, 2휋푖 푐−푖∞

the result.

However, the latter identity is only almost everywhere and 퐴(푥) is not continuous at integers. To verify Perron’s formula rigorously (i.e., for all 푥 ∈ R) we shall employ the following auxiliary lemma.

Lemma 12.8. Let 푐 > 0, 푦 > 0 and 푇 > 0. Further, set

푐+푖푇 푠 0 if 푦 < 0, 1 ∫ 푦  퐼(푦, 푇) B d푠 and 퐻(푦) B 1/2 if 푦 = 0, 2휋푖 푐−푖푇 푠  1 if 푦 > 0  (Heaviside step function, cf. (12.2)). It holds that

( 푦푐 min 1, 1 if 푦 ≠ 1, |퐼(푦, 푇) − 퐻(푦 − 1)| < 푇 |log 푦 | 푐 휋푇 if 푦 = 1

(cf. Table 12.1 with 푏 = 1 and 푎 = 0).

Proof. For 푦 = 1,

1 ∫ 푇 d푡 1 ∫ 푇 푐 − 푖푡 1 ∫ 푇 푐 퐼(1, 푇) = = d푡 = d푡 2 2 2 2 2휋 −푇 푐 + 푖푡 2휋 −푇 푐 + 푡 휋 0 푐 + 푡 1 ∫ 푇 /푐 1 1 ∫ ∞ 1 1 ∫ ∞ 1 1 1 ∫ ∞ 1 = d푡 = d푡 − d푡 = − d푡 2 2 2 2 휋 0 1 + 푡 휋 0 1 + 푡 휋 푇 /푐 1 + 푡 2 휋 푇 /푐 1 + 푡

∞ 0 = 1 ( ) − 1 1 ∫ d푡 = 푐 (recall that arctan 푧 1+푧2 ) and hence 퐼 1, 푇 2 < 휋 푇 /푐 푡2 휋푇 .

rough draft: do not distribute 12.3 RAMANUJAN’SMASTERTHEOREM 69

푡 푡 푐 + 푖푇 푟 + 푖푇 −푟 + 푖푇 푐 + 푖푇

0 푐 푟 −푟 0 푐

푐 − 푖푇 푟 − 푖푇 −푟 − 푖푇 푐 − 푖푇

(a) Contour for 푦 < 1 (b) Contour for 푦 > 1

Figure 12.1: Contours

∫ 푦푠 If 0 < 푦 < 1, then 0 = 푠 d푠 for the contour according Figure 12.1a and hence 1 ∫ 푟−푖푇 ∫ 푟+푖푇 ∫ 푟+푖푇 푦푠 퐼(푦, 푇) = + − d푠 2휋푖 푐−푖푇 푟−푖푇 푐+푖푇 푠 1 ∫ ∞−푖푇 ∫ ∞+푖푇 푦푠 −−−−→ − d푠 푟→∞ 2휋푖 푐−푖푇 푐+푖푇 푠 1 ∫ ∞ 푦푟−푖푇 ∫ ∞ 푦푟+푖푇 = d푟 − d푟 2휋푖 푐 푟 − 푖푇 푐 푟 + 푖푇 ∞ 푟 푐 | ( )| 1 ∫ 푦 ≤ 푦 and thus 퐼 푦, 푇 < 2 2휋 푐 푇 d푟 푇 휋 |log 푦 | and thus the result. 푦푠 For 푦 > 1, the function 푠 ↦→ 푠 has a pole with residue 1 at 푠 = 0. For the integral along the contour in Figure 12.1b thus,

1 ∫ −푟+푖푇 ∫ −푟−푖푇 ∫ 푐푟−푖푇 푦푠 퐼(푦, 푇) = 1 − + + d푠. 2휋푖 푐+푖푇 −푟+푖푇 −푟−푖푇 푠 This integral can be treated as above and thus the result. Proof of Perron’s formula (Theorem 12.4). It holds that

∞ ∑︁∗ ∑︁ 푥 퐴(푥) = 푎 = 푎 퐻 − 1 푛 푛 푛 푛≤푥 푛=1 ∞ ∞ ! ∑︁ 1 ∫ 푐+푖푇 푥 푠 d푠 1 ∫ 푐+푖∞ ∑︁ 푎 푥푠 = 푎 = 푛 d푠, 푛 2휋푖 푛 푠 2휋푖 푛푠 푠 푛=1 푐−푖푇 푐−푖∞ 푛=1 the assertion

12.3 RAMANUJAN’SMASTERTHEOREM

Í 푎푛 Í −푛푥 Theorem 12.9. Let D푎(푠) B 푛=1 푛푠 and 푓푎 (푥) B 푛=1 푎푛푒 , then

Γ(푠)·D푎(푠) = M 푓푎 (푠),

Version: April 23, 2021 70 MELLINTRANSFORM

cf. Table 12.1. Proof. Indeed, ∫ ∞ ∑︁ 푠−1 −푛푥 M 푓푎 (푠) = 푎푛 · 푥 푒 d푥 푛=1 0 ∑︁ 푎 ∫ ∞ = 푛 · 푥푠−1푒−푥 d푥 푛푠 푛=1 0 = D푎(푠)· Γ(푠), the assertion. Perron’s formula6 describes the inverse Mellin transform applied to a Dirichlet series. 7 | ( )Γ( )| 1 Theorem 12.10 (Ramanujan’s master theorem ). Suppose that 휙 is entire and 휙 푠 푠 < 푅1+휀 in {푠 : =(푠) < 푐, |푠| ≥ 푅} for increasing 푅. Then the Mellin transform of ∑︁ 휙(푘) 푓 (푥) B (−푥)푘 푘! 푘=0 is ∫ ∞ M 푓 (푠) = 푥푠−1 푓 (푥) d푥 = Γ(푠)· 휙(−푠). (12.6) 0 The identity is often employed as formal identity. Proof. The Gamma function 푠 ↦→ Γ(푠) has poles at 푠 ∈ {−푘 : 푘 = 0, 1, 2,... } and the residues at (−1) 푘 푠 = −푘 are given by Γ(푠) = 푘!(푠+푘) + O(1), cf. (6.6). By Cauchy’s residue theorem, applied to the contour Figure 12.1b displays, 1 ∫ 푐+푖∞ ∑︁ (−1)푘 ∑︁ 휙(푘) Γ(푠)휙(−푠)푥−푠 d푠 = 휙(푘)푥푘 = (−푥)푘 = 푓 (푥), 2휋푖 푘! 푘! 푐−푖∞ 푘=0 푘=0 and the assertion follows by Mellin inversion (12.1).

12.4 CONVOLUTIONS

Deﬁnition 12.11 (Multiplicative convolution). The convolution of two functions 푓 , 푔 : (0, ∞) → C is ∫ ∞ 푥 d푦 ( 푓 ∗ 푔)(푥) = 푓 푔(푦) . 0 푦 푦 Proposition 12.12 (Convolution). It holds that M( 푓 ∗ 푔) = M 푓 ·M푔; (12.7) and further (cf. Table 12.1), ∫ ∞ 푥 d푦 M 푥 훼 푦훽 푓 푔(푦) (푠) = M 푓 (푠 + 훼)·M푔(푠 + 훼 + 훽), (12.8) 0 푦 푦 ∫ ∞ d푦 M 푥 훼 푦훽 푓 (푥푦)푔(푦) (푠) = M 푓 (푠 + 훼)·M푔(훽 − 푠 − 훼). (12.9) 0 푦 6Oskar Perron, 1880–1975, German 7Srinivasa Ramanujan, 1887–1920, Indian

rough draft: do not distribute 12.4 CONVOLUTIONS 71

Proof. With Fubini,

∫ ∞ 푥 d푦 ∫ ∞ ∫ ∞ 푥 M 푥 훼 푦훽 푓 푔(푦) (푠) = 푥푠+훼−1 푦훽−1 푓 푔(푦) d푦d푥 0 푦 푦 0 0 푦 ∫ ∞ ∫ ∞ = 푦훽+푠+훼−1푔(푦) d푦 · 푥푠+훼−1 푓 (푥) d푥 푥←푥푦 0 0 = M 푓 (푠 + 훼)·M푔(훽 + 푠 + 훼) and thus (12.8) and (12.7). Further,

∫ ∞ d푦 ∫ ∞ ∫ ∞ M 푥 훼 푦훽 푓 (푥푦)푔(푦) (푠) = 푥푠+훼−1 푦훽−1 푓 (푥푦) 푔(푦) d푦d푥 0 푦 0 0 ∫ ∞ ∫ ∞ = 푥푠+훼−1 푓 (푥) d푥 · 푦훽−푠−훼−1푔(푦) d푦 푥←푥/푦 0 0 = M 푓 (푠 + 훼)·M푔(훽 − 훼 − 푠) and thus (12.9).

Proposition 12.13. We have that

1 ∫ 푐+푖∞ M( 푓 · 푔)(푠) = M 푓 (푧)·M푔(푠 − 푧) d푧. 2휋푖 푐−푖∞ Proof. Recall from Table 12.1 that M−11(·) = 훿(· − 1), thus

1 ∫ 푐+푖∞ 1 ∫ 푐+푖∞ ∫ ∞ ∫ ∞ M 푓 (푧)·M푔(푠 − 푧) d푧 = 푥푧−1 푓 (푥) d푥 · 푦푠−푧−1푔(푦) d푦d푧 2휋푖 푐−푖∞ 2휋푖 푐−푖∞ 0 0 ∫ ∞ ∫ ∞ 1 ∫ 푐+푖∞ 푦 −푧 푓 (푥) = d푧 d푥 푔(푦)푦푠−1 d푦 0 0 2휋푖 푐−푖∞ 푥 푥 | {z } M−11(푦/푥) ∫ ∞ ∫ ∞ 푦 푓 (푥) 푠−1 = 훿 − 1 d 푥 푦 푔(푦) d푦 (12.1) 0 0 푥 푥 ∫ ∞ ∫ ∞ = 훿 (푦 − 1) 푔(푥푦)푦푠−1 d푦 푥푠−1 푓 (푥) d푥 푦←푥푦 0 0 ∫ ∞ = 푔(푥)푥푠−1 푓 (푥) d푥 0 = M( 푓 · 푔)(푠) and thus the assertion.

Example 12.14. Let 휙(푘) = 1, then 푓 (푥) = 푒−푥 and M 푓 (푠) = Γ(푠).

푘 푘 휋 Example 12.15. Let 휙(푘) B −훼 sin 2 so that

∑︁ 휙(푘) ∑︁ 훼2푛+1 푓 (푥) = (−푥)푘 = (−1)푛 푥2푛+1 = sin 훼푥. 푘! (2푛 + 1)! 푘=0 푛=0

Γ(푠) 푠 휋 It follows that M(sin 훼푥)(푠) = 훼푠 sin 2 (cf. Talbe 12.1).

Version: April 23, 2021 72 MELLINTRANSFORM

(a) Godfrey Harold Hardy, 1877–1947 (b) Srinivasa Ramanujan, 1887–1920: the man who knew inﬁnity

Figure 12.2: I remember once going to see him (Ramanujan) when he was lying ill at Putney. I had ridden in taxi-cab No. 1729, and remarked that the number seemed to be rather a dull one, and that I hoped it was not an unfavourable omen. ‘No’, he replied, ‘it is a very interesting number; it is the smallest number expressible as the sum of two [positive] cubes in two different ways.’

푘 푘 휋 Example 12.16. Let 휙(푘) B 훼 cos 2 so that

∑︁ 휙(푘) ∑︁ 훼2푛 푓 (푥) = (−푥)푘 = (−1)푛 (−푥)2푛 = cos 훼푥 푘! (2푛)! 푘=0 푛=0

Γ(푠) 푠 휋 and thus M(cos 훼푥)(푠) = 훼푠 cos 2 .

rough draft: do not distribute Riemann zeta function 13

Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate.

Leonhard Euler, 1707–1783

Riemann1 (cf. Riemann [15]) was the ﬁrst to consider the function 휁 on the complex plane. 푘 (푘) 푘 Í∞ log 푛 Remark 13.1 (Derivatives). The derivatives of the zeta function are 휁 (푠) = (−1) 푛=2 푛푠 , 푘 ≥ 1.

Proposition 13.2 (Explicit values). For 푛 = 1, 2,... it holds that

퐵 휁 (2푛) = (−1)푛−1 2푛 (2휋)2푛. 2 (2푛)!

Proof. Choose 푥 = 0 in the Fourier series (5.9) and note that 훽푘 is continuous for 푘 ≥ 2 with 훽푘 (0) = 퐵푘 .

Example 13.3 (Basel problem). In particular we ﬁnd the relations

∑︁ 1 휋2 ∑︁ 1 휋4 ∑︁ 1 휋6 휁 (2) = = , 휁 (4) = = , 휁 (6) = = , (13.1) 푘2 6 푘4 90 푘6 945 푘=1 푘=1 푘=1

etc.

1 1 Note that 푛푠 = 푛휎 for 푠 = 휎 + 푖푡 so that (10.1) converges indeed for <(푠) > 1. The following provides an extension to <(푠) > 0.

13.1 RELATEDDIRICHLETSERIES

Lemma 13.4. It holds that

1 ∑︁ (−1)푛−1 휁 (푠) = , <(푠) > 0. (13.2) 1 − 21−푠 푛푠 푛=1

Í (−1)푛−1 Remark 13.5. Note, that the abscissa of convergence (cf. Theorem 11.7) of the series 푛=1 푛푠 − 1−푠 Í (−1)푛 1 is 휎0 = 0, although 1 − 2 휁 (푠) = 푛=1 푛푠 does not have a singularity at <(푠) = 푠0. This is a notable difference to power series.

1Bernhard Riemann, 1826–1866, German

73 74 RIEMANN ZETA FUNCTION

Proof. Indeed,2

2 1 1 1 1 1 1 1 휂(푠) B 1 − 휁 (푠) = 1 + + + + + + + + ... 2푠 2푠 3푠 4푠 5푠 6푠 7푠 8푠 2 2 2 2 − − − − 2푠 4푠 6푠 8푠 1 1 1 1 1 1 = 1 − + − + − + ∓ ... 2푠 3푠 4푠 5푠 6푠 7푠

and thus the result.

Proposition 13.6. It holds that

∞ 1 ∑︁ 푛 푛 − 푠 휁 (푠) = − , (13.3) 푠 − 1 (푛 + 1)푠 푛푠 푛=1

the series converges for <(푠) > 0.

Proof. Indeed,

∑︁ 푛 푛 − 푠 ∑︁ 푛 − 1 ∑︁ 푛 − 푠 ∑︁ 푛 − 1 − 푛 + 푠 − = − = = (푠 − 1)휁 (푠), (푛 + 1)푠 푛푠 푛푠 푛푠 푛푠 푛=1 푛=1 푛=1 푛=1

푠 Í 푠 푘 hence the identity. As for convergence recall that (1 + 푥) = 푘=0 푘 푥 . Thus

푛 푛 − 푠 1 1 −푠 − = 푛 1 + − 푛 + 푠 (푛 + 1)푠 푛푠 푛푠 푛 1 푠 푠(푠 + 1) 1 = 푛 − 푛 + 푛 + O − 푛 + 푠 푛푠 푛 2푛2 푛2 푠(푠 + 1) 1 = + O 2푛푠+1 푛푠+2

and thus the result.

Í 푛 푛−1 Í 1 1 Remark 13.7. The series converges for 푠 = 1 to 푛=1 푛+1 − 푛 = 푛=1 − 푛+1 + 푛 = 1 so that 휁 has a pole at 푠 = 1 with residue 1, i.e.,

1 휁 (푠) = + O(1). 푠 − 1

Proposition 13.8. The series

∞ 1 ∑︁ 푛(푛 + 1) 2푛 + 3 + 푠 2푛 − 1 − 푠 휁 (푠) = − 푠 − 1 2 (푛 + 1)푠+2 푛푠+2 푛=1

converges for <(푠) > −1.

2The function 휂 is also known as Dirichlet eta function.

rough draft: do not distribute 13.2 REPRESENTATIONS AS INTEGRAL 75

Proof. As above,

∑︁ 푛(푛 + 1) 2푛 + 3 + 푠 2푛 − 1 − 푠 − 2 (푛 + 1)푠+2 푛푠+2 푛=1 ∑︁ (푛 − 1)푛 2(푛 − 1) + 3 + 푠 푛(푛 + 1) 2푛 − 1 − 푠 = · − · 2 푛푠+2 2 푛푠+2 푛=1 ∑︁ 2푛2 (푠 − 1) = = (푠 − 1) 휁 (푠), 2푛푠+2 푛=1

after simpliﬁcations. Convergence can be established as above, it holds that

푛(푛 + 1) 2푛 + 3 + 푠 2푛 − 1 − 푠 1 푠 + 3 − = + O 푛−푠−3 2 (푛 + 1)푠+2 푛푠+2 2푛푠+2 3

and thus the assertion. Remark 13.9. Note that3 ∞ ∞ 1 ∑︁ 2푛 + 3 2푛 − 1 1 ∑︁ 1 1 1 휁 (0) = − 푛 − (푛 + 1) = − − = − . 2 푛 + 1 푛 2 푛 푛 + 1 2 푛=1 푛=1

13.2 REPRESENTATIONS AS INTEGRAL

Proposition 13.10 (Mellin transform). It holds that

∫ ∞ 푥푠−1 ( ) Γ( ) = (<( ) ) 휁 푠 푠 푥 d푥 푠 > 1 . (13.4) 0 푒 − 1

Í −푛푥 1 Proof. Apply Theorem 12.9 with 푓 (푥) = 푛=1 푒 = 푒푥 −1 . Theorem 13.11. It holds that ∫ ∞ b푢c 휁 (푠) = 푠 d푢, <(푠) > 1. (13.5) 푠+1 1 푢

1 Proof. Choose 푎푛 B 1 so that 퐴(푥) = b푥c and 휙(푥) = 푥푠 . Then apply Abel’s summation formula (8.6). Theorem 13.12. It holds that ∫ ∞ 1 휁 (푠) = −푠 frac 푡푠−1 d푡, 0 < <(푠) < 1 0 푡

where frac(푥) B 푥 − b푥c is the fractional part of the number 푥 ∈ R. Proof. From (13.5) we infer that

∫ 1 1 휁 (푠) = 푠 푢푠−1 d푢, <(푠) > 1. 0 푢 3 1 I.e., 1 + 1 + 1 + · · · = − 2 ,cf. Figure 13.3

Version: April 23, 2021 76 RIEMANN ZETA FUNCTION

− 1 ∫ 1 1 푠−1 ∫ 1 푠−2 푢푠 1 1 It holds that 푢 d푢 = 푢 d푢 = − = − for <(푠) > 1 and thus 0 푢 0 푠 1 푢=0 푠 1

푠 ∫ 1 1 1 휁 (푠) − = 푠 − 푢푠−1 d푢. (13.6) 푠 − 1 0 푢 푢

− ∞ ∫ ∞ 1 푠−1 ∫ ∞ 푠−2 푢푠 1 1 Now note that 푢 d푢 = 푢 d푢 = − = − − for <(푠) < 1 and thus 1 푢 1 푠 1 푢=1 푠 1 푠 ∫ ∞ 1 1 ∫ ∞ 1 휁 (푠) − = 푠 − 푢푠−1 d푢 + 푠 푢푠−1 d푢. 푠 − 1 0 푢 푢 1 푢 ∫ ∞ 1 1 푠 = 푠 − 푢푠−1 d푢 − 0 푢 푢 푠 − 1

and thus the result.

Remark 13.13. The equation (13.6) holds for <(푠) > 0 and thus proves that 휁 has a single pole at 푠 = 1.

Deﬁnition 13.14 (Mertens function). The Mertens4 function is the summatory function of the Möbius function, ∑︁ 푀(푥) B 휇(푛). (13.7) 푛≤푥 Theorem 13.15. It holds that 1 ∫ ∞ 푀(푢) = 푠 d푢, <(푠) > 1. 푠+1 휁 (푠) 1 푢

1 Proof. In Abel’s summation formula (8.6) choose 푎푛 B 휇(푛) so that 퐴(푥) = 푀(푥) and 휙(푥) = 푥푠 , then the result derives from (10.4).

Theorem 13.16 (Riemann, cf. Edwards [7]). It holds that (recall that (−푥)푠 = exp 푠 · log(−푥), which is not deﬁned for 푥 ∈ R≥0)

∮ (−푥)푠−1 ( ) Γ( ) ( ) = 2 sin 휋푠 푠 휁 푠 푖 푥 d푥, (13.8) 퐻 푒 − 1 where 퐻 is the Hankel5 contour.

Remark 13.17. The Hankel contour (aka. keyhole contour, cf. Figure 13.1) is a path in the complex plane which extends from [+∞, 훿], around the origin counter clockwise and back to [+∞, −훿], where 훿 is an arbitrarily small positive number.

Proof. We have that

∮ (−푥)푠−1 ∫ 훿 ∫ ∫ ∞ (−푥)푠 d푥 = − + + 푥 d푥 푥 , (13.9) 퐻 푒 − 1 +∞ |푥 |=훿 훿 (푒 − 1)푥

although, strictly speaking, the path of integration must be taken to be slightly above the real axis as it descends from +∞ to 0 and slightly below the real axis as it goes from 0 back to +∞.

4Franz Mertens, 1840–1927 5Hermann Hankel, 1839–1873, German mathematician

rough draft: do not distribute 13.3 GLOBALLYCONVERGENTSERIES 77

Figure 13.1: Hankel contour

3 0.10

0.05 1

-15 -10 -5 5 -14 -12 -10 -8 -6 -4 -2 -1

-2 -0.05

-3

-4 -0.10 (a) reals (b) negative reals

Figure 13.2: 휁 on the real line

Now note that ∞ ∫ (−푥)푠 d푥 ∫ ∑︁ (−푥)푘 = 푥푠−2 퐵 d푥 (푒푥 − 1)푥 푘 푘! |푥 |=훿 |푥 |=훿 푘=0 ∫ 2휋 ∞ 푘 푖푘푡 푠−1 푖푡 (푠−1) ∑︁ 푘 훿 푒 = 훿 · 푒 (−1) 퐵푘 d푡 −−−−→ 0 푘! 훿→0 0 푘=0 for 푠 > 1 by (5.1). The remaining terms in the integral (13.9) are

∫ 훿 exp(푠 log 푥 − 푖휋) ∫ +∞ exp(푠 log 푥 + 푖휋) ∫ ∞ 푥푠−1 + = 푖 휋푠 − −푖 휋푠 lim 푥 d푥 푥 d푥 푒 푒 푥 d푥. 훿→0 +∞ (푒 − 1)푥 훿 (푒 − 1)푥 0 푒 − 1

푠−1 ∮ (−푥) = − ( )Γ( ) ( ) With (13.4) we thus have 퐻 푒푥 −1 d푥 2푖 sin 휋푠 푠 휁 푠 and thus the assertion.

13.3 GLOBALLYCONVERGENTSERIES

Theorem 13.18. It holds that 푛 1 ∑︁ 1 ∑︁ 푛 1 휁 (푠) = (−1)푘 , 푠 ∈ C, 1 − 21−푠 2푛+1 푘 (푘 + 1)푠 푛=0 푘=0

Version: April 23, 2021 78 RIEMANN ZETA FUNCTION

1 Figure 13.3: Ramanujan’s proof of 휁 (−1) = 1 + 2 + 3 + · · · = − 12 in his letter to Hardy

the series converges for all 푠 ∈ C. Proof. Apply Euler transform (cf. (8.2)) to (13.2). Theorem 13.19 (Cf. Hasse [11]). It holds that

푛 1 ∑︁ 1 ∑︁ 푛 (−1)푘 휁 (푠) = ; (13.10) 푠 − 1 푛 + 1 푘 (푘 + 1)푠−1 푛=0 푘=0 the series converges for all 푠 ∈ C. ∞ 1 = 1 ∫ −푘푡 푠−1 Proof. Note ﬁrst that 푘푠 Γ(푠) 0 푒 푡 d푡 and

푛 푛 ∑︁ 푛 (−1)푘 1 ∫ ∞ ∑︁ 푛 = (−1)푘 푒−(푘+1)푡 푡푠−1 d푡 푘 (푘 + 1)푠 Γ(푠) 푘 푘=0 0 푘=0 1 ∫ ∞ = 1 − 푒−푡 푛 푒−푡 푡푠−1 d푡. Γ(푠) 0

Í 푥푛+1 Finally, as log(1 − 푥) = − 푛=0 푛+1 , 푛 ∞ ∑︁ 1 ∑︁ 푛 (−1)푘 1 ∫ ∞ ∑︁ (1 − 푒−푡 )푛 = 푒−푡 푡푠−1 d푡 푛 + 1 푘 (푘 + 1)푠 Γ(푠) 푛 + 1 푛=0 푘=0 0 푛=0 1 ∫ ∞ 푡 = −푡 푠−1 −푡 푒 푡 d푡 Γ(푠) 0 1 − 푒 1 ∫ ∞ 푡푠 = 푡 d푡 Γ(푠) 0 푒 − 1 1 = 휁 (푠 + 1) Γ(푠 + 1) Γ(푠) by (13.4) and thus the result.

1 Corollary 13.20 (Explicit values, cf. Remark 13.9). It holds that 휁 (0) = − 2 and 퐵 휁 (−푛) = − 푛+1 , 푛 = 1, 2, 3,.... 푛 + 1

퐵푛+1 (1) 퐵푛+1 Proof. By comparing (5.5) and (13.10) it follows with (5.6) that 휁 (−푛) = − 푛+1 = − 푛+1 .

rough draft: do not distribute 13.4 POWER SERIES EXPANSIONS 79

Alternative proof. The assertion follows as well from the residue theorem with (5.1) and (13.8). 1 Remark 13.21 (Cf. Figure 13.3). This result is often stated as 1 + 1 + 1 + · · · = 휁 (0) = − 2 or 1 1 + 2 + 3 + · · · = 휁 (−1) = − 12 . Corollary 13.22. It holds (cf. Remark 5.3 and Figure 13.2) that 휁 (푠) = 0, 푠 = −2, −4, −6,... These are called the trivial zeros of the Riemann zeta function.

13.4 POWER SERIES EXPANSIONS

푛 푛−푠 1 Í∞ Deﬁne the function 푓푛 (푠) B (푛+1)푠 − 푛푠 and recall from (13.3) that 휁 (푠) = 푠−1 푛=1 푓푛 (푠). The series converges for <(푠) > 0. From Taylor we obtain that 1 ∑︁ (푠 − 1)2 휁 (푠) = 푓 (1) + (푠 − 1) 푓 0(1) + 푓 00(1) + ... 푠 − 1 푛 푛 2 푛 푛=1 Í 푓 (1) ∑︁ 푠 − 1 ∑︁ = 푛=1 푛 + 푓 0(1) + 푓 00(1) + ... 푠 − 1 푛 2 푛 푛=1 푛=1 Now note, for 푘 ≥ 1, that log푘−1 푛 (푛 − 푠) log푘 푛 푛 log푘 (푛 + 1) (−1)푘 푓 (푘) (푠) = −푘 − + 푛 푛푠 푛푠 (푛 + 1)푠 and (−1)푘 푓 (푘+1) (1) log푘 푛 1 푛 푛 − 1 푛 = − log푘+1 (푛 + 1) − log푘+1 푛 . (13.11) 푘 + 1 푛 푘 + 1 푛 + 1 푛 Deﬁnition 13.23. The numbers 푚 ∑︁ log푘 푛 log푘+1 푚 훾푘 B lim − (13.12) 푚→∞ 푛 푘 + 1 푛=1 are Stieltjes constants.6

Remark 13.24. The constant 훾0 = 훾 = 0.577 215 ... is Euler’s constant or Euler–Mascheroni constant, cf. (7.3). The following constants are 훾1 = −0.072 815 ... , 훾2 = −0.009 690 ... , etc. 푘 푘+1 ∫ 푚 log 푛 = log 푚 Remark 13.25. Note in (13.12) that 1 푛 d푛 푘+1 . Theorem (Global Laurent series expansion). It holds that ∞ 1 ∑︁ (−1)푘 훾 휁 (푠) = + 푘 (푠 − 1)푘 (13.13) 푠 − 1 푘! 푘=0 1 훾 = + 훾 − 훾 (푠 − 1) + 2 (푠 − 1)2 ∓ ..., 푠 − 1 1 2

for every 푠 ∈ C\{0}, where 훾푘 are Stieltjes constants. 휁 has a simple pole at 푠 = 1 with residue Res1 (휁) = 1. 6Thomas Jean Stieltjes, 1856–1894, Dutch

Version: April 23, 2021 80 RIEMANN ZETA FUNCTION

Proof. Note ﬁrst that ∑︁ ∑︁ 1 푓 (1) = = 1 푛 푛(푛 + 1) 푛=1 푛=1 It follows with (13.11) that

푚 푘 푘+1 푚 푘 (푘+1) ∑︁ log 푛 푚 log (푚 + 1) ∑︁ (−1) 푓푛 (1) 훾푘 = lim − = lim , 푚→∞ 푛 푚 + 1 푘 + 1 푚→∞ 푘 + 1 푛=1 푛=1 as the sum is telescoping. Note further that

푚 푚 1 푘+1 log푘+1 (푚 + 1) − log푘+1 푚 = log 푚 + log 1 + − log푘+1 푚 푚 + 1 푚 + 1 푚 푚 1 푘+1 log푘 푚 = log 푚 + − log푘+1 푚 + O 푚 + 1 푚 푚 1 log푘 푚 = − log푘+1 푚 + O −−−−−→ 0 푚 + 1 푚 푚→∞

and thus the assertion.

13.5 EULER’STHEOREM

Corollary 13.26 (Euler’s theorem). The series of reciprocal primes diverges, ∑︁ 1 = ∞. 푝 푝 prime

Remark 13.27 (Euler’s informal proof). We provide Euler’s informal proof ﬁrst. Note, that

∑︁ 1 Ö 1 ∑︁ 1 log = log = − log 1 − 푛 1 − 푝−1 푝 푛=1 푝 prime 푝 prime ∑︁ 1 1 1 = + + + ... 푝 2푝2 3푝3 푝 prime ∑︁ 1 1 ∑︁ 1 1 ∑︁ 1 = + + + ... 푝 2 푝2 3 푝3 푝 prime 푝 prime 푝 prime 1 1 1 = 퐴 + 퐵 + 퐶 + 퐷 + ... 2 3 4 = 퐴 + 퐾

1 Í 푥푛 Í 1 for some 퐾 < 1. By letting 푥 → 1 in log 1−푥 = 푛=1 푛 we ﬁnd that 푛=1 푛 = log ∞. It follows that Í 1 퐴 = 푝 prime 푝 = log log ∞. Proof. From (10.10) and Proposition 10.20 we deduce that

∑︁ 1 ∑︁ 1 1 log 휁 (푠) − ≤ = 1. 푝푠 푘 푘 − 1 푝 prime 푘=2

rough draft: do not distribute 13.6 THE FUNCTIONAL EQUATION 81

From Taylor series expansion (13.13) we have for 푠 ∼ 1 and <(푠) > 1 that

∑︁ 1 1 푃(푠) = ∼ log 휁 (푠) ∼ log (13.14) 푝푠 푠 − 1 푝 prime

and thus the result.

Remark 13.28. Note the contrast: it holds that ∑︁ 1 1 ∼ log 푝푠 푠 − 1 푝 prime

휁 0 (푠) 1 by (13.14), but, as 휁 (푠) ∼ − 푠−1 , ∑︁ log 푝 1 ∼ 푝푠 푠 − 1 푝 prime by (10.8).

Theorem 13.29 (Mertens’ second theorem). It holds that

∑︁ 1 ∑︁ 휇(푘) 1 ∼ log log 푥 + 훾 + log 휁 (푘) + O . 푝 푘 log 푥 푝≤푥 푘=2

Í 휇(푘) Í 1 1 The Meissel–Mertens constant is 푀 B 훾 + 푘=2 푘 log 휁 (푘) = 훾 + 푝∈P log 1 − 푝 + 푝 ≈ 0.21649....

See Theorem 16.10 below for a proof based on the prime number theorem (PNT, Theo- rem 16.4 below). Remark 13.30. We have that log log 1010 000 ≈ 10.04 and the age of the universe is ≈ 4.3 × 1017 seconds.

13.6 THE FUNCTIONAL EQUATION

Riemann’s functional equation for the 휁-function is a reﬂection formula relating 휁 (푠) with 휁 (1 − 푠).

Theorem 13.31 (Functional equation). It holds that 휋푠 휁 (푠) = 2(2휋)푠−1 sin Γ(1 − 푠) 휁 (1 − 푠). 2 Further, with Riemann’s 휉-function

푠(푠 − 1) 푠 휉(푠) B Γ 휁 (푠) (13.15) 2 휋푠/2 2 we have that 휉(푠) = 휉(1 − 푠). (13.16)

1 Remark 13.32. Riemann actually considered the function Ξ(푧) B 휉 2 + 푖푧 .

Version: April 23, 2021 82 RIEMANN ZETA FUNCTION

Figure 13.4: Riemann, 1859

Proof. Recall from (8.8) the definition of the Jacobi theta function 휗 and define 3 ∑︁ 2 2 휓(푥) B 푥 휗0(푥) + 푥2 휗00(푥) = 2 푘2휋푥 − 3푘2휋푥 푒−푘 휋푥 (13.17) 2 푘=1 (see Figure 13.5a for the graph; cf. Riemann’s note, Figure 13.4). As Jacobi’s theta function, the function 휓 satisfies the functional equation (cf. (8.9)) 1 1 휓(푥) = √ 휓 ; (13.18) 푥 푥 indeed, we have from (13.17) that √ d √ d d 휓(푥) = 푥 푥3/2 휗0(푥) = 푥 푥3/2 휗(푥) d푥 d푥 d푥 and with (8.9) thus √ d d 1 휓(푥) = 푥 푥3/2 푥−1/2 휗 d푥 d푥 푥 √ d 1 1 1 = 푥 푥3/2 − 푥−3/2 휗 − 푥−5/2 휗0 d푥 2 푥 푥 √ d 1 1 1 1 = 푥 − 휗 − 휗0 d푥 2 푥 푥 푥 √ 1 1 1 1 1 1 = 푥 휗0 + 휗0 + 휗00 2푥2 푥 푥2 푥 푥3 푥 1 3 1 1 1 1 = √ 휗0 + 휗00 푥 2 푥 푥 푥2 푥 1 1 = √ 휓 , 푥 푥

rough draft: do not distribute 13.6 THE FUNCTIONAL EQUATION 83 i.e., (13.18). It holds that ∫ ∞ ∫ ∞ 2 푠 −1 ∑︁ 푠 −1 2 2 −푘2 휋푥 푥 2 휓(푥) d푥 = 푥 2 2 푘 휋 푥 − 3푘 휋 푥 푒 d푥 0 푘=1 0 ∞ 푠 −1 ∑︁ ∫ 푥 2 = 2푥2 − 3푥 푒−푥 d푥 푥 푠 푥← 2 2 푘2 휋 푘=1 0 푘 휋 휁 (푠) 푠 푠 = 2 Γ + 2 − 3 Γ + 1 휋푠/2 2 2 휁 (푠) 푠 푠 푠 푠 = Γ 2 + 1 − 3 휋푠/2 2 2 2 2 휁 (푠) 푠(푠 − 1) 푠 = Γ . 휋푠/2 2 2 Hence (cf. Table 12.1) ∫ ∞ 휁 (푠) 푠(푠 − 1) 푠 푠 −1 휉(푠) = Γ = 푥 2 휓(푥) d푥 (13.19) 푠/2 휋 2 2 0 ∫ ∞ 푠 − 3 1 = 푥 2 2 휓 d푥 0 푥 ∫ ∞ − 푠 + 3 −2 = 푥 2 2 휓 (푥) d푥 푥←1/푥 0 ∫ ∞ 1−푠 −1 = 푥 2 휓 (푥) d푥; 0 comparing with (13.19) reveals that 휉 is symmetric and thus the functional equation (13.16). However, convergence remains to be discussed. To this end observe that (13.17) implies that 휓 (푘) (푥) → 0 as 푥 → ∞ for every 푘 = 0, 1,... and further, 휓 (푘) (푥) → 0 as 푥 → 0 by (13.18). It follows that (13.18) holds for all 푠 ∈ C, the integral converges globally. Remark 13.33 (Fourier transform). With (13.19) we have that ∫ ∞ ∫ ∞ 푠 −1 푠−1 2 휉(푠) = 푥 2 휓(푥) d푥 = 푥 · 2 휓(푥 ) d푥 (13.20) 0 0 ∫ ∞ = 푒푥 (푠−1/2) · 2 푒푥/2휓푒2푥 d푥 −∞ ! ∫ ∞ 푠 − 1 = cos 2 푥 · 2푒푥/2 휓푒2푥 d푥, −∞ 푖 where 푥 ↦→ 2 푒푥/2 휓푒2푥 is symmetric. Remark 13.34 (Representations of cos 푡푥). The Fourier transform of the following representations derives from (6.8): sin 휋푡 sin 휋푡 ∑︁ 2푡 cos 푡푥 = − (−1)푘 cos 푘푥 (−휋 ≤ 푥 ≤ 휋) 휋푡 휋 푘2 + 푡2 푘=1 푡 푥 = lim 퐻2푛 √ (all 푥) 푛→∞ 2 푛 + − 휋푥 ∑︁ (2푥) 푗 휋(푡 − 푗) 푡 푗 2 = 푇 (cos 푥) = cos + 푡 cos 2 ; 푡 2 푗 2 푗 − 1 푗=1

Version: April 23, 2021 84 RIEMANN ZETA FUNCTION

4 0.8

0.6 2

0.4 20 40 60 80 100

0.2 -2

-4 0.5 1.0 1.5 2.0 2.5 2 (a) Graph of 2휓(푥 ), see (13.17) ( ) = ± 1 + (b) Riemann–Siegel function, 푍 푡 휁 2 푖푡

Figure 13.5: Function plots

퐻푛 is the Hermite polynomial and 푇푡 the generalized Chebyshev polynomial. Remark 13.35 (Riemann–Siegel theta function, 푍-function). To investigate the function 휁 on the 1 1 2 +푖푡 log 휋 critical line 푠 ∈ C: <(푠) = 2 it is convenient to deﬁne 휃(푡) B arg Γ 2 − 푡 2 (the principal branch of the function 휉) and Γ 1 + 푖푡 푖 휃 (푡) 1 − 푖푡 4 2 1 푍 (푡) B 푒 휁 + 푖푡 = 휋 2 휁 + 푖푡 . 2 Γ 1 + 푖푡 2 4 2 Note, that 푍 (푡) ∈ R and 휁 1 + 푖푡 = ±푍 (푡) for 푡 ∈ R. Figure 13.5b displays the graph of 푍. 2 푛 0( ) = − 1 0(− ) = (−1) 휁 (2푛+1)(2푛)! Proposition 13.36 (Explicit values). It holds that 휁 0 2 log 2휋 and 휁 2푛 22푛+1 휋2푛 . We give a further proof of the functional equation. ∞ ( ) = ∫ b푥c Second proof of Theorem 13.31. Recall from (13.5) that 휁 푠 푠 1 푥푠+1 d푥, thus 1 1 ∫ ∞ 푥 − b푥c − 1 휁 (푠) = + − 푠 2 d푥 (<(푠) > −1) 푠+1 푠 − 1 2 1 푥 ∫ ∞ 푥 − b푥c − 1 = −푠 2 d푥 (−1 < <(푠) < 0). 푠+1 0 푥 Now recall the sawtooth wave from (5.10) so that ∞ 푠 ∑︁ 1 ∫ ∞ sin 2휋푛푥 휁 (푠) = d푥 휋 푛 푥푠+1 푛=1 0 푠 ∑︁ (2푛휋)푠 ∫ ∞ sin 푥 = d푥 휋 푛 푥푠+1 푛=1 0 푠 휋푠 = (2휋)푠 휁 (1 − 푠) − Γ(−푠) sin , 휋 2 ∞ ∫ 푠−1 = Γ( ) 휋푠 and thus the result. We have used here 0 푥 sin 푥 d푥 푠 sin 2 , see Example 12.15 and Table 12.1.

rough draft: do not distribute 13.7 RIEMANNHYPOTHESIS (RH) 85

13.7 RIEMANNHYPOTHESIS (RH)

The ﬁrst non-trivial, or ’non-obvious’ zeros of the Riemann zeta are (cf. Figure 13.5b)

휌1 B 1/2 + 푖 · 14.134 725 141 734 693 790 ..., 휌2 B 1/2 + 푖 · 21.022 039 638 771 554 993 ..., 휌3 B 1/2 + 푖 · 25.010 857 580 145 688 763 ... and 휌4 B 1/2 + 푖 · 30.424 876 125 859 513 210 ....

The following formulation of the Riemann hypothesis (RH) quotes the second millennium problem, http://www.claymath.org/millennium-problems: The prime number theorem determines the average distribution of the primes. The Riemann hypothesis tells us about the deviation from the average. Formulated in Riemann’s 1859 paper, it asserts that all the ’non-obvious’ zeros of the zeta function are complex numbers with real part 1/2.

13.8 HADAMARDPRODUCT

Hadamards7 product follows from Weierstrass8 product theorem.

Theorem 13.37. It holds that

(log(2휋)−1− 훾 )푠 푒 2 Ö 푠 푠 휁 (푠) = 1 − 푒 휌 , 푠 휌 2(푠 − 1)Γ 1 + 2 휌

where the product is over the non-trivial zeros 휌 of 휁 and the 훾 again is the Euler–Mascheroni constant.

Corollary 13.38. It holds that

0 0 휁 (푠) 훾 1 1 Γ (푠/2 + 1) ∑︁ 1 1 = log 2휋 − − 1 − − + + , 휁 (푠) 푠 − Γ(푠/2 + ) 푠 − 휌 휌 2 1 2 1 휌

where the sum is among all nontrivial zeros 휌 of the Riemann zeta function. Theorem 13.39. It holds that 1 Ö 푠 휉(푠) = 1 − . 휌 2 휌 Theorem 13.40 (Li’s criterion). It holds that d 푧 ∑︁ 푛 log 휉 = 휆 + 푧 , d푧 푧 − 1 푛 1 푛=1

푛 푛 1 d 푛−1 ( ) = Í − − 1 where 휆푛 B (푛−1)! d푠푛 푠 log 휉 푠 푠=1 휌 1 1 휌 . The Riemann hypothesis is true iff 휆푛 > 0 for all 푛 = 1, 2,... 7Jacques Hadamard, 1865–1963, French 8Karl Weierstraß, 1815–1897, German

Version: April 23, 2021 86 RIEMANN ZETA FUNCTION

Î 푠 푧 Proof. Suppose that 푓 (푠) = 휌 1 − 휌 and deﬁne 휙(푧) B 푓 푧−1 . We have that

∑︁ 푧 ∑︁ (1 − 푧)휌 + 푧 log 휙(푧) = log 1 − = log (푧 − ) 휌 ( − 푧) 휌 휌 1 휌 1 − + 푧 ∑︁ 1 푧 휌 ∑︁ 1 = log = log 1 − 푧 1 − − log(1 − 푧) − 푧 휌 휌 1 휌

so that

0 − 1 푛+1 d 푧 d 휙 (푧) ∑︁ 1 1 휌 ∑︁ ∑︁ 1 log 푓 = log 휙(푧) = = − = 푧푛 1 − 1 − . d푧 푧 − 1 d푧 휙(푧) 1 − 푧 1 휌 휌 1 − 푧 1 − 휌 푛=0 휌

Now note that − 1 ≤ iff |휌 − | ≤ |휌| iff <휌 ≤ 1 . 1 휌 1 1 2 On the other hand, ! 1 d푛 1 d푛 ∑︁ 푠 푠푛−1 log 휉(푠) = 푠푛−1 log 1 − 푛 푛 (푛 − 1)! d푠 푠=1 (푛 − 1)! d푠 휌 휌 푠=1 푛 푛−푘 푘 1 ∑︁ ∑︁ 푛 d 푛−1 d 푠 = 푠 log 1 − (푛 − 1)! 푘 d푠푛−푘 d푠푘 휌 휌 푘=1 푠=1 푛 ! ∑︁ 1 ∑︁ 푛 푛 − 1 (푘 − 1)! = 1 − (푛 − 푘)! (푛 − 1)! 푘 푛 − 푘 (휌 − 1)푘 휌 푘=0 푛 ∑︁ ∑︁ 푛! 1 = 1 − 푘!(푛 − 푘)! (휌 − 1)푘 휌 푘=0 ∑︁ 1 푛 ∑︁ 휌 푛 = 1 − 1 + = 1 − . 휌 − 휌 − 휌 1 휌 1

Now replace 휌 ← 1 − 휌 is a zero as well and thus further

푛 푛 푛 1 d 푛−1 ∑︁ 1 ∑︁ 1 푠 log 휉(푠) = 1 − 1 + = 1 − 1 − , (푛 − ) 푠푛 휌 − 휌 1 ! d 푠=1 휌 1 휌

the assertion.

PROBLEMS

Exercise 13.1. Verify the identity (13.17).

rough draft: do not distribute Further results and auxiliary relations 14

14.1 OCCURRENCEOFCOPRIMES

6 ≈ Theorem 14.1. The probability of two numbers being coprime is 휋2 60.8%; more precisely: let copr(푥) B {(푚, 푛) ∈ N × N: 푚, 푛 ≤ 푥 and 푚 and 푛 are coprime}. Then |copr(푥)| 6 lim = . 푥→∞ 푥2 휋2 Proof. Deﬁne 2 퐴푘 (푥) B (푚, 푛) ∈ N : 푚, 푛 ≤ 푥 and gcd(푚, 푛) = 푘 Ð¤ and 퐴(푥) B 푘 ≤푥 퐴푘 (푥), the union of disjoint sets. Note that, for 푥 and 푘 ﬁxed, the map 푥 copr → 퐴 (푥) 푘 푘 (푚, 푛) ↦→ (푘 · 푚, 푘 · 푛)

2 Í Í 푥 is a bijection. Therefore, b푥c = 푘 ≤푥 |퐴푘 (푥)| = 푘 ≤푥 copr 푘 and by Corollary 9.25 thus Í 푥 2 |copr (푥)| = 푘 ≤푥 휇(푘) 푘 . 푥 2 푥 2 푥 Now note that 푘 − 푘 ≤ 2 푘 . Hence

∑︁ 푥 2 ∑︁ 1 |copr (푥)| − 휇(푘) ≤ 2푥 ≤ 2푥(1 + log 푥) 푘 푘 푘 ≤푥 푘 ≤푥 or

|copr (푥)| ∑︁ 휇(푘) 1 + log 푥 − ≤ 2 . (14.1) 푥2 푘2 푥 푘 ≤푥 Í∞ 휇(푘) = 1 = 6 → ∞ With (10.4) and (13.1) we ﬁnd that 푘=1 푘2 휁 (2) 휋2 . The result follows by letting 푥 in (14.1).

14.2 EXPONENTIALINTEGRAL

Deﬁnition 14.2. The exponential integral is

∫ 푧 푒푡 Ei(푧) B 푝.푣. d푡, −∞ 푡

where 푝.푣. denotes the Cauchy principal value.

87 88 FURTHER RESULTS AND AUXILIARY RELATIONS

Proposition 14.3. It holds that ∞ ∑︁ 푧푘 Ei(푧) = 훾 + log 푧 + . (14.2) 푘 · 푘! 푘=1 Proof. By integration by parts, (6.5) and (6.1), ∫ ∞ 푒−푡 ∫ ∞ ∫ ∞ log 휀 + d푡 = (1 − 푒−휀) log 휀 + 푒−푡 log 푡 d푡 −−−−→ 푒−푡 log 푡 d푡 = Γ0(1) = −훾. 휀 푡 휀 휀→0 0 It follows that ∞ ∫ −휀 푒푡 ∫ 푧 푒푡 − 1 ∑︁ ∫ 푧 푡푘−1 Ei(푧) ←−−−− d푡 − log 휀 + log 푧 + d푡 −−−−→ 훾 + log 푧 + d푡 휀→0 푡 푡 휀→0 푘! −∞ 휀 푘=1 0 and thus the result.

14.3 LOGARITHMICINTEGRALFUNCTION

Deﬁnition 14.4. The logarithmic integral is ∫ 푥 1 li(푥) B 푝.푣. d푡, 0 log 푡 for 푥 > 1 being the Cauchy principal value. Í∞ log푘 푧 Proposition 14.5. It holds that li 푧 = Ei(log 푧) = 훾 + log log 푧 + 푘=1 푘·푘! .

Asymptotic expansion. The logarithmic integral has the asymptotic series expansion 푥 1 2 푘! 푥 li(푥) = 1 + + + · · · + + O (14.3) log 푥 log 푥 log2 푥 log푘 푥 log푘+1 푥 as 푥 → ∞. To see this asymptotic expansion replace 푥 ← 푒푦, then

푦 ∫ 푒 1 ∫ ∞ 푒−푦푡 li (푒푦) = d푡 = 푒푦 d푡. 0 log 푡 푡←푒푦(1−푡) 0 1 − 푡 1 푛−1 푡 푛 By Taylor series expansion we have 1−푡 = 1 + 푡 + · · · + 푡 + 1−푡 , thus 푛 ∑︁ ∫ ∞ ∫ ∞ 푒−푦푡 푡푛 li (푒푦) = 푒푦 푡푘−1푒−푦푡 d푡 + 푒푦 d푡 1 − 푡 푘=1 0 0 푛 ∑︁ (푘 − 1)! = 푒푦 + 푅 (푦), 푦푘 푛 푘=1

∞ −푦푡 푛 ( ) = 푦 ∫ 푒 푡 = O / 푛+1 → ∞ where 푅푛 푦 푒 0 1−푡 d푡 1 푦 as 푦 . The expansion follows by replacing 푦 ← log 푥 again. More generally and by a similar reasoning it can be shown that ∫ 푥 d푡 푥 푘 (푘 + 1)푘 푘 + ℓ − 1 ℓ! 푥 = 1 + + + · · · + + O , (14.4) 푘 푘 2 ℓ 푘+ℓ+1 2 log 푡 log 푥 log 푥 log 푥 ℓ log 푥 log 푥 as 푥 → ∞.

rough draft: do not distribute 14.4 ANALYTIC EXTENSIONS 89

Remark 14.6. Differentiate (14.4) with respect to 푥 to see the identity. Remark 14.7. Compare with (14.3) and Gauß’ letter (Figure 1.1, second page) the extension 푥 푥 푥 1 1 = = 1 + + + ... . log 푥 − 1 1 log 푥 log 푥 log2 푥 log 푥 1 − log 푥

14.4 ANALYTIC EXTENSIONS

The following is from Zagier [22]. Theorem 14.8 (Analytic theorem, Newman [14]). Let 푓 (푡) (푡 ≥ 0) be bounded and locally inte- ∞ ( ) ∫ ( ) −푧 푡 <( ) grable function and suppose that the function 푔 푧 B 0 푓 푡 푒 d푡 ( 푧 > 0) extends holo- ∞ <( ) ≥ ∫ ( ) ( ) morphically to 푧 0. Then 0 푓 푡 d푡 exists and equals 푔 0 .

푇 −푧 푡 ( ) ∫ ( ) ∈ C Proof. For 푇 > 0 set 푔푇 푧 B 0 푓 푡 푒 d푡. The function 푔푇 is holomorphic for all 푧 . To prove the theorem we have to show that lim푇 →∞ 푔푇 (0) = 푔(0). Let 푅 > 0 be large and 퐶 the boundary of the region {푧 ∈ C: |푧| ≤ 푅, <(푧) ≥ −훿}, where 훿 > 0 is small enough (depending on 푅) so that 푔 is holomorphic in and on 퐶. By Cauchy’s theorem, 1 ∮ 푧2 1 푔(0) − 푔 (0) = 푔(푧) − 푔 (푧)푒푧푇 1 + d푧. (14.5) 푇 푇 2 2휋푖 퐶 푅 푧 (i) Note that for 푧 ∈ 퐶+ B {푧 ∈ 퐶 : |푧| = 푅 and <(푧) > 0}, ∫ ∞ ∫ ∞ −푇 <(푧) −푧푡 −푧푡 푒 |푔(푧) − 푔푇 (푧)| = 푓 (푡)푒 d푡 ≤ 퐵 푒 d푡 = 퐵 푇 푇 <(푧)

2 + 푧 1 = 1 + 푧 = 푧 + 푧 = 2<(푧) and further 1 푅2 푧 푧 푅2 푅2 푅2 푅2 , thus

2 푧 푇 푧 1 푇 <(푧) 2<(푧) 푒 1 + ≤ 푒 . 푅2 푧 푅2

+ 2퐵 Consequently, on 퐶 , the integrand in (14.5) is bounded by 푅2 . (ii) Now consider 푧 ∈ 퐶− B {푧 ∈ 퐶 : |푧| = 푅 and <(푧) < 0}. − (a) The function 푔푇 is an entire function (i.e., analytic everywhere) and the contour 퐶 in the integral (14.5) thus may be changed to 푧 ∈ C, <(푧) < 0 and |푧| = 푅. It holds that

∫ 푇 ∫ 푇 −<(푧) 푇 −푧 푡 −푧 푡 푒 |푔푇 (푧)| = 푓 (푡) 푒 d푡 ≤ 퐵 푒 d푡 = 퐵 0 −∞ |<(푧)|

2퐵 so that the integrand in (14.5) thus again is bounded by 푅2 . 2 ( ) 푧푇 + 푧 1 (b) The remaining integrand in (14.5) is the function 푔 푧 푒 1 푅2 푧 , which decreases uniformly on the remaining contour (푧 ∈ 퐶−, <(푧) = −훿), as 푇 increases. The total 4퐵+1 length of the contour is 2휋푅, thus it follows that |푔(0) − 푔푇 (0)| < 푅 for 푇 sufﬁciently large. The assertion follows, as 푅 > 0 is arbitrary.

Version: April 23, 2021 90 FURTHER RESULTS AND AUXILIARY RELATIONS

rough draft: do not distribute Prime counting functions 15

102 + 112 + 122 = 132 + 142.

Mental Arithmetic. In the Public School of S. Rachinsky

15.1 RIEMANNPRIMECOUNTINGFUNCTIONSANDTHEIRRELATION

Deﬁnition 15.1 (Prime-counting function). The prime-counting function is ∑︁ 휋(푥) B 1. 푝≤푥, 푝 prime

The Riemann prime-counting function is

∑︁ 휋푥1/푘 ∑︁ 1 퐽(푥) B = . (15.1) 푘 푘 푘=1 푝푘 ≤푥, 푝 prime

Different notation in frequent use is also 퐽 = Π.

1 Note that 휋(2) = 1, 휋(4) = 2, 퐽(4) = 2 2 and

휋(푝푛) = 푛 (15.2)

for the 푛th prime 푝푛 (푛 ∈ N). We also have that 푝 휋 ( 푝) = 푝, if 푝 is prime. Remark 15.2. As 휋(푥) = 0 for 푥 < 2 it is sufﬁcient in (15.1) to consider only 푥1/푛 ≥ 2 and to sum 1/ ( ) 휋 푥 푛 up to 푛 ≤ 1 = log (푥), i.e., 퐽(푥) = Ílog2 푥 . log푥 (2) 2 푛=1 푛

100 100

80 80

l i n l i n 60 ( ) 60 ( )

Out[37]= Out[47]=π(n) J(n) 40 n 40 n log(n) log(n)

20 20

100 200 300 400 500 100 200 300 400 500

Figure 15.1: Prime counting functions 휋 and 퐽

91 92 PRIMECOUNTINGFUNCTIONS

Theorem 15.3. It holds that ∑︁ 휇(푛) 휋(푥) = 퐽 푥1/푛 . (15.3) 푛 푛=1 Proof. Indeed, with (15.1),

∑︁ 휇(푑1) 1/ ∑︁ 휇(푑1) ∑︁ 1 1/ 퐽 푥 푑1 = 휋 푥 푑1푑2 푑1 푑1 푑2 푑1=1 푑1=1 푑2=1 ∑︁ 1 ∑︁ = 휋 푥1/푛 휇(푑 ) 푛 1 푛=1 푑1 푑2=푛 = 휋(푥), where we have employed (9.6) (the Dirichlet inverse of the Möbius function). Theorem 15.4 (Mellin transforms). It holds that ∫ ∞ 휋(푥) ( ) = log 휁 푠 푠 푠 d푥 0 푥(푥 − 1) ∫ ∞ 퐽(푥) = 푠 d푥 푠+1 0 푥 Í 1 and (recall the zeta prime function 푃(푠) = 푝 prime 푝푠 from (10.9)) ∫ ∞ 휋(푥) 푃(푠) = 푠 d푥. 푠+1 0 푥 Proof. By Riemann–Stieltjes integration by parts ∑︁ 1 log 휁 (푠) = − log 1 − 푝푠 푝 ∫ ∞ 1 = − − ( ) log 1 푠 d휋 푥 1 푥 ∞ 1 ∫ 푠 푠+1 = 휋(푥) 푥 d푥 1 1 1 − 푥푠 ∫ ∞ 1 = ( ) 푠 휋 푥 푠 d푥; 0 푥(푥 − 1) a slightly more explicit derivation is ∑︁ 1 log 휁 (푠) = − log 1 − 푝푠 푝 ∑︁ 1 = − 휋(푛) − 휋(푛 − 1) log 1 − 푛푠 푛=2 ∑︁ 1 1 = − 휋(푛) log 1 − − log 1 − 푛푠 (푛 + 1)푠 푛=2 ∑︁ ∫ 푛+1 푠 = 휋(푛) d푥 푥(푥푠 − 1) 푛=2 푛 ∫ ∞ 휋(푥) = 푠 푠 d푥. 2 푥(푥 − 1)

rough draft: do not distribute 15.2 CHEBYSHEVPRIMECOUNTINGFUNCTIONSANDTHEIRRELATION 93

Further ∫ ∞ 휋(푥) ( ) = log 휁 푠 푠 푠 d푥 0 푥(푥 − 1) ∫ ∞ 휋(푥) 1 = 푠 d푥 푠+1 −푠 0 푥 1 − 푥 ∫ ∞ 휋(푥) ∑︁ = 푠 푥푠 푥−푠푘 d푥 푥푠+1 0 푘=1 ∑︁ ∫ ∞ 휋(푥) = 푠 푥−푠푘 d푥 푥 푘=1 0 ∑︁ ∫ ∞ 휋 푥1/푘 1 = 푠 푥−푠 · 푥1/푘−1 d푥 1/ 1/푘 푥 푘 푘 푥←푥 푘=1 0 ∫ ∞ 퐽(푥) = 푠 d푥 푠+1 0 푥 and thus the result. The Mellin transformation of 푃 (cf. (10.9)) is immediate from

∫ ∞ 1 ∫ ∞ 1 ∫ ∞ 1 푃(푠) = d휋(푥) = − 휋(푥) d = 푠 휋(푥) d푥. 푠 푠 푠+1 1 푥 1 푥 1 푥

Proposition 15.5. It holds that (cf. (10.6))

∑︁ Λ(푛) 퐽(푥) = . (15.4) 푛 푛≤푥 log

1 log 푝 Λ(푛) ( ) = Í 푘 = Í 푘 = Í Proof. Indeed, 퐽 푥 푝 ≤푥 푘 푝 ≤푥 log 푝푘 푛≤푥 log 푛 .

15.2 CHEBYSHEVPRIMECOUNTINGFUNCTIONSANDTHEIRRELATION

Deﬁnition 15.6. The ﬁrst Chebyshev 1 function is

휋 (푥) ∑︁ ∑︁ 휗(푥) B log 푝 = log 푝푘 = log(푥#), (15.5) 푝≤푥, 푘=1 푝 prime

where Ö 푥# B 푝 푝≤푥, 푝 prime is the primorial. The second Chebyshev function is the summatory function of the von Mangoldt function (cf. (10.6)), ∑︁ ∑︁ 휓(푥) B Λ(푛) = log 푝; (15.6) 푛≤푥 푝푘 ≤푥, 푝 prime

Version: April 23, 2021 94 PRIMECOUNTINGFUNCTIONS

500

20 000 40 000 60 000 80 000 100 000 400 -100

300 -x200 Out[52]= x 200 ϑ( ) -300

100 -400

100 200 300 400 500 -500 (a) Plot of 휗 (b) Plot of 휗(푥) − 푥

Figure 15.2: Chebyshev function 휗

500

150 400 100

300 50 n Out[48]= Ψ(n) 200 20 000 40 000 60 000 80 000 100 000 -50 100 -100

-150 100 200 300 400 500 (a) Plot of 휓 (b) Plot of 휓(푥) − 푥

Figure 15.3: Chebyshev function 휓

rough draft: do not distribute 15.2 CHEBYSHEVPRIMECOUNTINGFUNCTIONSANDTHEIRRELATION 95

( ) ∼ ∫ 푥 ∫ 푝 1 = ∫ 푥 log 푝 = Remark 15.7. Note that we expect that 휗 푥 0+ log 푝 d 0 log 푡 d푡 0+ log 푝 d푡 푥.

Proposition 15.8. It holds that2 ∑︁ log 푥 휓(푥) = log 푝 (15.7) 푝≤푥, log 푝 푝 prime and ∑︁ 휓(푥) = 휗(푥) + 휗 푥1/2 + 휗 푥1/3 + · · · = 휗 푥1/푚 푚=1 and conversely, ∑︁ 휗(푥) = 휇(푚) 휓 푥1/푚 . 푚=1 Proof. Indeed,

∑︁ 1/푑 ∑︁ ∑︁ 1/푑 푑 휇(푑1) 휓 푥 1 = 휇(푑1) 휗 푥 1 2 푑1=1 푑1=1 푑2=1

∑︁ 1/푛 ∑︁ = 휗 푥 휇(푑1) 푛=1 푑1 푑2=푛 = 휗(푥), where we have employed (9.6) (the Dirichlet inverse of the Möbius function; cf. Theorem 15.3). Proposition 15.9 (Mellin transforms). It holds that (recall the deﬁnition of the zeta prime function (10.9)) d ∑︁ log 푝 ∫ ∞ 휗(푥) − 푃(푠) = = 푠 d푥 (15.8) d푠 푝푠 푥푠+1 푝 prime 1 and further (cf. Theorem 10.17) we have d 휁 0(푠) ∑︁ Λ(푛) ∫ ∞ 휓(푥) − log 휁 (푠) = − = = 푠 d푥 d푠 휁 (푠) 푛푠 푥푠+1 푛=2 1 and (cf. Proposition 10.16) ∑︁ Λ(푛) log 휁 (푠) = . 푛푠 log 푛 푛=2 Proof. We have that ∑︁ log 푝 ∫ ∞ d휗(푥) ∫ ∞ 휗(푥) −푃0(푠) = = = 푠 d푥. (10.9) 푝푠 푥푠 푥푠+1 푝 prime 1 1 From (10.2) (see also (10.17)) we deduce that

0 1 1 휁 (푠) d ∑︁ 1 ∑︁ 푝푠 log 푝 ∑︁ ∑︁ log 푝 ∑︁ Λ(푛) − = log 1 − = − = = 휁 (푠) d푠 푝푠 − 1 푝푘·푠 푛푠 푝 prime 푝 prime 1 푝푠 푝 prime 푘=1 푛=1

1Pafnuty Chebyshev, 1821–1894, Russian 2 log 푥 Note that log 푝 = log푝 푥.

Version: April 23, 2021 96 PRIMECOUNTINGFUNCTIONS

and thus the ﬁrst identity. The second identity follows from Abel’s summation formula (8.6),

휁 0(푠) ∑︁ Λ(푛) ∫ ∞ 1 ∑︁ ∫ ∞ 휓(푥) − = = d Λ(푘) = 푠 d푥 휁 (푠) 푛푠 푥푠 푥푠+1 푛=1 1 푘 ≤푥 1 | {z } 휓(푥)

as the Chebyshev function 휓 is the summatory function of Λ by (15.6).

15.3 RELATIONOFPRIMECOUNTINGFUNCTIONS

The extremely ingenious proof of the following proposition is due to Chebyshev.

Proposition 15.10 (Chebyshev). It holds that

휗(푥) < 푥 · log 4 (< 1.39 · 푥) (15.9)

for all 푥 > 0.

Î 푛 Proof. Employing the primorial (15.5) the statement is equivalent to 퐹푛 B 푛# = 푝≤푛 < 4 for all 푛 ∈ N which we shall verify by induction. The statement is true for 푛 = 0, 1, 2 and 3. 푛−1 푛 If 푛 is even, then 퐹푛 = 퐹푛−1 < 4 < 4 , as desired. Î 푘+1 If 푛 is odd, then 푛 = 2푘 + 1. By induction hypothesis 푝≤푘+1 < 4 . (푘+2)(푘+3)···(2푘+1) 2푘+1 Clearly, for every prime 푝 with 푘 + 2 ≤ 푝 ≤ 2푘 + 1, one has that 푝 | 1·2···푘 = 푘 and hence Ö 2푘 + 1 2푘 + 1 1 푝 ≤ = ≤ (1 + 1)2푘+1 = 4푘 . 푘 푘 + 1 2 푘+1<푝≤2푘+1 It follows that Ö Ö 푘+1 푘 2푘+1 푛 퐹푛 = 퐹2푘+1 = 푝 · 푝 < 4 · 4 = 4 = 4 , 푝≤푘+1 푘+1<푝≤2푘+1 the assertion.

Theorem 15.11 (Relations of the prime counting functions). The prime counting functions are related as follows: 푥 (i) 휋(푥) = 휗(푥) + ∫ 휗(푢) d푢 = 휗(푥) + O 푥 and log 푥 2 푢 log2 푢 log 푥 log2 푥 ( ) = ( ) − ∫ 푥 휋 (푢) = ( ) + O 푥 (ii) 휗 푥 휋 푥 log 푥 2 푢 d푢 휋 푥 log 푥 log 푥 .

푥 (iii) 퐽(푥) = 휓(푥) + ∫ 휓(푢) d푢 and log 푥 2 푢 log2 푢

( ) = ( ) − ∫ 푥 퐽 (푢) (iv) 휓 푥 퐽 푥 log 푥 2 푢 d푢. ( log 푛 if 푛 is prime, 1 Proof. Set 푎푛 B and 휙(푥) B . With Abel’s summation formula (8.6) 0 else log 푥 푥 (퐴(푥) = Í 푎 = 휗(푥)) it follows that 휋(푥) = Í 푎 휙(푛) = 휗(푥) + ∫ 휗(푢) d푢, thus (i), as 푛≤푥 푛 푛≤푥 푛 log 푥 1 푢 log2 푢

rough draft: do not distribute 15.3 RELATIONOFPRIMECOUNTINGFUNCTIONS 97 desired. As for the order of convergence recall from (15.9) that 휗(푢) ≤ log 4 , with (14.4) we 푢 log2 푢 log2 푢 푥 conclude that ∫ 휗(푢) d푢 = O 푥 . 2 푢 log2 푢 log2 푥 ( 1 if 푛 is prime, Applying Abel summation (8.6) again (with 푎푛 = , i.e., 퐴(푥) = 휋(푥) and 0 else Í ∫ 푥 휋 (푢) 휙(푥) = log 푥) gives 휗(푥) = 푛≤푥 푎푛 휙(푛) = 휋(푥) log 푥 − 푢 d푢, thus (ii). From (15.9) we have 1 ( ) = O ( ) ( ) = O 푥 ∫ 푥 휋 (푢) ≤ that 휗 푥 푥 , so we conclude from (i) that 휋 푥 log 푥 . Consequently, 2 푢 d푡 ∫ 푥 퐶 ∼ 퐶 푥 2 log 푢 d푢 log 푥 with (14.4), thus the order of convergence in (ii). 1 As for (iii) deﬁne 푎푛 B Λ(푛), then 퐴(푥) = 휓(푥) by (15.6). Set 휙(푥) = log 푥 . With (15.4) we 푥 obtain that 퐽(푥) = Í 푎 휙(푛) = 휓(푥) + ∫ 휓(푢) d푢. 푛≤푥 푛 log 푥 1 푢 log2 푢 Λ(푛) For (iv), applying Abel summation (8.6) (with 푎푛 = log 푛 , i.e., 퐴(푥) = 퐽(푥) and 휙(푥) = log 푥) ( ) = Í ( ) = ( ) − ∫ 푥 퐽 (푢) gives 휓 푥 푛≤푥 푎푛 휙 푛 퐽 푥 log 푥 1 푢 d푢, the result. Corollary 15.12. The following are equivalent:

푥 (i) 휋(푥) ∼ log 푥 for 푥 → ∞, (ii) 휗(푥) ∼ 푥 for 푥 → ∞ (cf. Remark 15.7),

푥 (iii) 퐽(푥) ∼ log 푥 for 푥 → ∞ and (iv) 휓(푥) ∼ 푥 for 푥 → ∞. Proof. We have from Theorem 15.11 that (푖) ⇐⇒( 푖푖) and (푖푖푖) ⇐⇒( 푖푣). It remains to verify that (푖) ⇐⇒( 푖푣). With (15.7) we have that

∑︁ log 푥 ∑︁ 휓(푥) = log 푝 ≤ log 푥 = 휋(푥) log 푥 푝 푝≤푥 log 푝≤푥 and further, for every 휀 > 0,

휓(푥) ≥ 휗(푥) ∑︁ ≥ log 푝 푥1−휀 <푝≤푥 ∑︁ ≥ log 푥1−휀 푥1−휀 <푝≤푥 = 휋(푥) − 휋 푥1−휀 ·(1 − 휀) log 푥. ≥ (1 − 휀) 휋(푥) − 푥1−휀 log 푥.

The equivalence (푖) ⇐⇒( 푖푣) thus is immediate.

Version: April 23, 2021 98 PRIMECOUNTINGFUNCTIONS

rough draft: do not distribute The prime number theorem 16

Fundamental for the prime number theorem (PNT) is that 휁 (푠) ≠ 0 for <(푠) = 1.

16.1 ZETA FUNCTION ON <(·) = 1

Theorem 16.1. For 휎 > 1 and 푡 ∈ R we have that

3 4 휁 (휎) 휁 (휎 + 푖푡) 휁 (휎 + 2푖푡) ≥ 1. (16.1)

Proof. Recall that |푒푧 | = 푒<(푧) . From Euler’s product formula (10.2) we deduce

∑︁ 1 |휁 (휎 + 푖푡)| = exp ©− log 1 − ª 푝 휎+푖 푡 ® 푝 prime « ¬ ∞ ∑︁ ∑︁ 1 = exp ©< © ªª ℓ 푝ℓ ( 휎+푖 푡) ®® 푝 prime ℓ=1 « « ¬¬ ∞ ∑︁ ∑︁ cos(ℓ 푡 log 푝) = exp © ª . ℓ 푝ℓ 휎 ® 푝 prime ℓ=1 « ¬ For any 휙 ∈ R the trigonometric identity

3 + 4 cos 휙 + cos 2휙 = 2 ·(1 + cos 휙)2 ≥ 0

holds true. It follows that

∞ 3 4 ∑︁ ∑︁ 3 + 4 cos(ℓ 푡 log 푝) + cos(2 ℓ 푡 log 푝) 휁 (휎) 휁 (휎 + 푖푡) 휁 (휎 + 2푖푡) = exp © ª ≥ 1, ℓ 푝ℓ 휎 ® 푝 prime ℓ=1 « ¬ the result. The following theorem is in essence due to Hadamard, its original proof has about 25 pages. Theorem 16.2. The Riemann 휁 function does not vanish for <(푠) ≥ 1, i.e., 휁 (푠) ≠ 0 whenever <(푠) ≥ 1. Proof. Euler’s product formula (10.2) converges for <(푠) > 1 and hence 휁 (푠) ≠ 0 for <(푠) > 1. It remains to show that 휁 (푠) ≠ 0 whenever <(푠) = 1. 1 푘 We have that 휁 (휎) ∼ 휎−1 from (13.13). Suppose that 휁 (1 + 푖푡) = 0, then 휁 (휎 + 푖푡) ∼ 푐(휎 − 1) 3 1 4 4푘 for 휎 ∼ 1, 푘 ≥ 1 and 푐 ≠ 0. From (16.1) it follows that 휎−1 푐 (휎 − 1) 휁 (휎 + 2푖푡) & 1, thus ( + ) ∼ 1 + 휁 휎 2푖푡 ( 휎−1)4푘−3 . This is a contradiction, as 휁 does not have a pole at 1 2푖푡, its only pole is at 푠 = 1.

99 100 THEPRIMENUMBERTHEOREM

VII. Ueber die Anzahl der Primzahlen unter e111er gegebenen GrÖsse. (Monatsberichte der Berliner Akademie, November 1859.)

Meinen Dank für die Auszeichnung, welche mir die Akademie durch die Aufnahme unter ihre Correspondenten hat zu Theil werden lassen, glaube ich am besten dadurch zu erkennen zu geben, dass ich von der hierdurch erhaltenen Erlaubniss baldigst Gebrauch mache durch Mittheilung einer Untersuchung über die Häufigkeit der Prim zahlen; ein Gegenstand, welcher durch das Interesse, welches Gauss und Di richl et demselben längere Zeit geschenkt haben, einer solchen Mittheilung vielleicht nicht ganz unwerth erscheint. Bei dieser Untersuchung diente mir als Ausgangspunkt die von Euler gemachte Bemerkung, dass das Product

__1 __ = E _ 1_ 1 n s , JI- 1 -- P" wenn für p alle Primzahlen, für n alle ganzen Zahlen gesetzt werden. Die Function der complexen Veriinderlichen s, welche durch diese beiden Ausdrücke, so lange sie convergiren, dargestellt wird, bezeichne ich durch ~(s ). Beide convergiren nur, so lange der reelle Theil von s grässer als 1 ist; es lässt sich indess leicht ein immer güHig blei bender Ausdruck der Function finden. Durch Anwendung der Gleichung

'" 'e- nx X S - 1 dx = 17 (s - 1) n' Jo erhält man zunächst Cf> n (S - 1) ~ (s) = J' . ~s- l (l x ex - 1 .o

rough draft: do not distribute

Figure 16.1: Riemann, 1859 16.2 THEPRIMENUMBERTHEOREM 101

The following proof is based on Newman [14], cf. also Zagier [22]. Theorem 16.3. The improper integral

∫ ∞ 휗(푥) − 푥 d푥 2 1 푥

∫ 푇 휗(푥)−푥 → ∞ is a convergent integral, i.e., the limit 1 푥2 d푥 exists, as 푇 . Proof. Recall from Euler’s product formula (or (15.8)) that

∑︁ log 푝 ∫ ∞ 휗(푥) ∫ ∞ 휗(푥) − 푥 푠 −푃0(푠) = = 푠 d푥 = 푠 d푥 + . 푝푠 푥푠+1 푥푠+1 푠 − 1 푝 prime 1 1

Recall from (10.8) (Theorem 10.17) that

휁 0(푠) d d ∑︁ 1 − = − log 휁 (푠) = log 1 − 휁 (푠) d푠 d푠 푝푠 푝 prime ∑︁ log 푝 ∑︁ log 푝 ∑︁ log 푝 = = + 푝푠 − 1 푝푠 푝푠 (푝푠 − 1) 푝 prime 푝 prime 푝 prime

and thus 1 휁 0(푠) 1 1 ∑︁ log 푝 푃0(푠) 1 − − − = − − 푠 휁 (푠) 푠 − 1 푠 푝푠 (푝푠 − 1) 푠 푠 − 1 푝 prime | {z } analytic for <(푠) ≥1 | {z } analytic for <(푠)>1/2 ∫ ∞ 휗(푥) − 푥 = d푥. 푠+1 1 푥

푃0 (푠) 1 By Theorem 16.2, 휁 (푠) ≠ 0 for <(푠) ≥ 1 and so − 푠 − 푠−1 extends to an analytic function in <(푠) ≥ 1, i.e.,

푃0(푧 + 1) 1 ∫ ∞ 휗(푥) − 푥 ∫ ∞ 푔(푧) B − − = d푥 = 휗푒푡 푒−푡 − 1 푒−푧푡 d푡 푧 + 1 푧 푥푧+2 1 0 | {z } 푓 (푡)

extends to an analytic function for <(푧) ≥ 0. By Chebyshev’s theorem (see (15.9)), the integrand 휗(푥) 푓 is bounded, as −1 < 푥 − 1 < 0.39. It follows from Theorem 14.8 that the improper integral ∞ ∞ ( ) = ∫ ( ) = ∫ 휗(푥)−푥 푔 0 0 푓 푡 d푡 1 푥2 d푥 exists.

16.2 THEPRIMENUMBERTHEOREM

Theorem 16.4 (Prime number theorem, PNT; Hadamard and de la Vallée Poussin,1 1896). It holds that 푥 휋(푥) ∼ log 푥 as 푥 → ∞. 1Charles Jean de la Vallée Poussin, 1866–1962, Belgian

Version: April 23, 2021 102 THEPRIMENUMBERTHEOREM

Proof. In view of Corollary 15.12 it is enough to verify that 휗(푥) ∼ 푥. To this end recall from ∫ ∞ 휗(푥)−푥 Theorem 16.3 that 1 푥2 d푥 converges. Assume, by contraposition, that 휗(푥) 푥, then there are 휆 > 1 and 휇 < 1 so that 휗(푥) > 휆푥 or 휗(푥) < 휇푥 for arbitrary large 푥 > 0. In the ﬁrst case, as 휗 is nondecreasing, 휗(푡) ≥ 휗(푥) > 휆 푥 for 푡 > 푥,

∫ 휆푥 휗(푡) − 푡 ∫ 휆푥 휆푥 − 푡 ∫ 휆 휆 − 푡 d푡 > d푡 = d푡 > 0, 2 2 2 푥 푡 푥 푡 푡←푥푡 1 푡 and in the second, 휗(푡) ≤ 휗(푥) < 휇푥 for 푡 < 푥, thus

∫ 푥 휗(푡) − 푡 ∫ 푥 휇푥 − 푡 ∫ 1 휇 − 푡 d푡 ≤ d푡 = d푡 < 0. 2 2 2 휇푥 푡 휇푥 푡 휇 푡

∫ ∞ 휗(푥)−푥 As 푥 is arbitrary large, it follows that the integral 1 푥2 d푥 does not converge. This contradicts Theorem 16.3.

1 +휀 Remark 16.5. The Riemann Hypothesis (see page 85) is equivalent to 휗(푥) = 푥 + O 푥 2 for every 휀 > 0.

Corollary 16.6. For every 휀 > 0 there is an 푛0 ∈ N so that there is a prime 푝 with

푛 < 푝 < (1 + 휀)푛 for all 푛 ≥ 푛0. Proof. By the prime number theorem,

(1+휀) 푥 휋 (1 + 휀)푥 log (1+휀) 푥 log 푥 lim = lim = (1 + 휀) lim = 1 + 휀. (16.2) 푥→∞ 푥→∞ 푥 푥→∞ 휋(푥) log 푥 log 푥 + log(1 + 휀) Therefore, there is 푛0 so that 휋 (1 + 휀)푥 > 휋(푥) for all 푥 > 푛0 and hence there is 푝 ∈ P between 푥 and (1 + 휀)푥.

16.3 CONSEQUENCESOFTHEPRIMENUMBERTHEOREM

Lemma 16.7. Let 푎푛 푛∈N and 푏푛 푛∈N are sequences with lim푛→∞ 푎푛 = lim푛→∞ 푏푛 = ∞ and 휋 푎푛 ∼ 휋 푏푛 , then also 푎푛 ∼ 푏푛 as 푛 → ∞. Proof. We show ﬁrst that lim sup 푎푛 ≤ 1. 푛→∞ 푏푛 Suppose this were false. Then there is 휀 > 0 and 푛푘 with 푛푘 → ∞ such that 푎푛푘 > (1 + 휀)푏푛푘 . Then 휋 푎푛푘 휋 (1 + 휀)푏푛푘 lim sup ≥ lim sup = 1 + 휀 푛→∞ 휋 푏푛푘 푛→∞ 휋 푏푛푘 by (16.2), which is a contradiction to the assertion. 푏푛 푎푛 By exchanging the roles of 푎 and 푏 it follows that lim sup ≤ 1 and thus lim →∞ = 푛 푛 푛→∞ 푎푛 푛 푏푛 1.

rough draft: do not distribute 16.3 CONSEQUENCESOFTHEPRIMENUMBERTHEOREM 103

Corollary 16.8. Let 푝푛 denote the 푛th prime (cf. (15.2)), then it holds that

푝푛 ∼ 푛 log 푛 (16.3) as 푛 → ∞. Proof. By the prime number theorem 푛 log 푛 푛 log 푛 푛 휋(푛 log 푛) ∼ = = ∼ 푛 = 휋(푝 ). log(푛 log 푛) log 푛 + log log 푛 log log 푛 푛 1 + log 푛 The assertion follows from the preceding lemma. Corollary 16.9 (Erdos˝ 2). It holds that 푝 푛+1 −−−−→ 1. 푝푛 푛→∞ ( + ) ( + ) Proof. Indeed, by (16.3), 푝푛+1 ∼ 푛 1 log 푛 1 −−−−→ 1. 푝푛 푛 log 푛 푛→∞ Theorem 16.10 (See Theorem 13.29 above). It holds that ∑︁ 1 ∼ log log 푥. 푝 푝≤푥 More generally, ∑︁ 1 1 ∼ log log 푥 + 푀 + O , 푝 푥 푝≤푥 log Í 휇(푘) 3 where 푀 B 훾 + 푘=2 푘 log 휁 (푘) = 0.261 497 212 ... is the Meissel –Mertens constant. Í∞ 1 Remark 16.11. Euler actually writes 푝 prime 푝 = log log ∞. Proof. We have from Theorem 15.11 that 휋(푛) = 푛 + O 푛 . Hence log 푛 log2 푛 푥 ∑︁ 1 ∑︁ 휋(푛) − 휋(푛 − 1) = 푝 푛 푝≤푥 푛=1 푥 푥−1 ∑︁ 휋(푛) ∑︁ 휋(푛) = − 푛 푛 + 1 푛=1 푛=0 푥−1 휋(푥) ∑︁ 휋(푛) = + 푥 푛(푛 + 1) 푛=1 푥−1 푥−1 휋(푥) ∑︁ 휋(푛) ∑︁ 휋(푛) = + − 푥 푛2 푛2 (푛 + 1) 푛=1 푛=1 푥−1 ∑︁ 휋(푛) = + O(1) 푛2 푛=1 푥−1 ∑︁ 1 1 = + O + O(1) 푛 log 푛 2 푛=1 푛 log 푛 = log log 푥 + O(1),

2Paul Erdos,˝ 1913–1996, Hungarian 3Ernst Meissel, 1826–1895, German astronomer

Version: April 23, 2021 104 THEPRIMENUMBERTHEOREM

0 where we have used that 1 = (log log 푥)0 and 1 = − 1 . 푥 log 푥 푥 log2 푥 log 푥 Remark 16.12. Mertens actually proved the statement without employing the prime number theorem.

rough draft: do not distribute Riemann’s approach by employing the zeta function 17

This section repeats Riemann’s path. Here we proceed formally and care about convergence later.

17.1 MERTENS’ FUNCTION

Assume that 푓 (휌) = 0, then the series expansion is

(푠 − 휌)2 푓 (푠) = 0 + 푓 0(휌)(푠 − 휌) + 푓 00(휌) + O(푠 − 휌)3, 2

or, assuming that 휌 is a simple zero (i.e., 푓 0(휌) ≠ 0),

1 1 푓 00(휌) = − + O(푠 − 휌)1, 푓 (푠) 푓 0(휌)(푠 − 휌) 2 푓 0(휌)2

1 the residue at 푠 = 휌 is 푓 0 (휌) . 1 Í 휇(푛) Í Recall from (10.4) that 휁 (푠) = 푛=1 푛푠 and from (13.7) that 푀(푥) = 푛≤푥 휇(푛). Assuming that all all zeros are simple it follows from Perron’s formula (12.4) that

∑︁∗ ∫ 푐+푖∞ 1 푥푠 1 ∑︁ 푥휌 ∑︁ 푥−2푛 푀∗ (푥) = 휇(푛) = d푠 = + + , 휁 (푠) 푠 휁 (0) 휌 · 휁 0(휌) −2푛 · 휁 0(−2푛) 푛≤푥 푐−푖∞ 휌: 휁 (휌)=0 푛=1

where the sum is along all zeros 휌 of the 휁-function on the critical strip.

17.2 CHEBYSHEVSUMMATORYFUNCTION 휓

Theorem 17.1 (Riemann-von Mangoldt explicit formula). For any 푥 > 1 it holds that

∑︁ 푥휌 휓∗ (푥) = 푥 − log 2휋 − (17.1) 휌 휌: 휁 (휌)=0 ∑︁ 푥휌 1 = 푥 − log 2휋 − − log 1 − 푥−2 . (17.2) 휌 2 휌: 휁 (휌)=0 nontrivial

∗ Í∗ Proof. The summatory function is 휓 (푥) = 푛≤푥 Λ(푛), cf. (15.6). Recall from (10.7) the Dirichlet 휁 0 (푠) Í Λ(푛) 휁 0 (푠) series 휁 (푠) = − 푛=1 푛푠 and the residue is Res 휁 (푠) , 푠 = 휌 = 1. With Perron’s formula (12.4)

105 106 RIEMANN’S APPROACH BY EMPLOYING THE ZETA FUNCTION

thus ∑︁∗ 휓∗ (푥) = Λ(푛) 푛≤푥 1 ∫ 푐+푖∞ 휁 0(푠) 푥푠 = − d푠 2휋푖 푐−푖∞ 휁 (푠) 푠 휁 0(0) ∑︁ 푥−2푛 ∑︁ 푥휌 = 푥 − − − 휁 (0) −2푛 휌 |{z} 푛=1 휌 pole |{z} 푠=0 | {z } |{z} trivial zeros nontrivial zeros 1 ∑︁ 푥휌 = 푥 − log 2휋 − log 1 − 푥−2 − 휌 2 휌

and hence the assertion.

17.3 RIEMANNPRIMECOUNTINGFUNCTION

Theorem 17.2. For 푥 > 1it holds that ∑︁ 퐽∗ (푥) = li(푥) − li(푥휌) (17.3) 휌: 휁 (휌)=0 ∑︁ ∫ ∞ d푡 = li(푥) − li(푥휌) − log 2 + . 푡(푡2 − 1) log 푡 휌: 휁 (휌)=0 푥 nontrivial

Proof. Recall that we have

∑︁∗ 1 ∑︁∗ Λ(푛) ∫ 푥 1 퐽∗ (푥) = = = d휓∗ (푡) 푘 log 푛 0 log 푡 푝푘 ≤푥 푛≤푥

and thus, with (17.1),

∫ 푥 휓∗0(푡) 퐽∗ (푥) = d푡 0 log 푡 푥 Í 휌−1 1 ∫ 1 − 휌 푡 − 3 = 푡 −푡 d푡 0 log 푡 ∑︁ ∫ ∞ d푡 = li(푥) − li(푥휌) − log 2 + . ( 2 − ) 휌 푥 푡 푡 1 log 푡

− d 휌 휌 푡휌 1 푡휌−1 Indeed, d푡 li(푡 ) = log(푡휌) = log 푡 .

17.4 PRIMECOUNTINGFUNCTION 휋

Remark 17.3. We note that li(푥휌) = Ei(휌 log 푥).

rough draft: do not distribute 17.4 PRIMECOUNTINGFUNCTION 휋 107

Theorem 17.4. For 푥 > 1 it holds that ∑︁ 휋(푥) = 푅(푥) − 푅(푥휌) 휌: 휁 (휌)=0 ∑︁ 1 1 휋 = 푅(푥) − 푅(푥휌) − + arctan log 푥 휋 log 푥 휌: 휁 (휌)=0 nontrivial where 푘 ∑︁ 휇(푛) 1 ∑︁ log 푥 푅(푥) B li 푥 /푛 = 1 + (17.4) 푛 푘! 푘 휁 (푘 + 1) 푛=1 푘=1 is Riemann’s 푅-function. The latter series is know as Gram1 series.

∗ Í 휇(푘) ∗ 1/푘 Proof. Recall from (15.3) that 휋 (푥) = 푘=1 푘 퐽 푥 and with (17.3) thus ∑︁∗ ∑︁ 1 1 휋 휋∗ (푥) = 1 = 푅(푥) − 푅(푥휌) − + arctan , log 푥 휋 log 푥 푘 ≤푥 휌 Proof of the Gram series (17.4), Riemann’s 푅-function. With (14.2) and (10.5) it holds that

∑︁ 휇(푛) ∑︁ 휇(푛) log 푥 푅(푥) = li 푥1/푛 = Ei 푛 푛 푛 푛=1 푛=1 ! ∑︁ 휇(푛) ∑︁ log푘 푥 = 훾 − log 푛 + log log 푥 + 푛 푘 푘! 푛푘 푛=1 푘=1 ∑︁ 휇(푛) ∑︁ 휇(푛) log 푛 ∑︁ log푘 푥 ∑︁ 휇(푛) = (훾 + log log 푥) − + . 푛 푛 푘! 푘 푛푘+1 푛=1 푛=1 푘=1 푛=1

1 0 − +... Now note that Í 휇(푛) = 1 , thus Í 휇(푛) = 0 and − Í 휇(푛) log 푛 = − 휁 (푠) = − (푠−1)2 = 푛=1 푛푠 휁 (푠) 푛=1 푛 푛=1 푛푠 휁 (푠)2 1 2 ( 푠−1 +... ) Í 휇(푛) log 푛 1 + O(푠 − 1) and thus 푛=1 푛 = −1 and the result (cf. also Exercise 9.2). For the rest see Riesel and Gohl [16].

1Jørgen Pedersen Gram, Danish actuary and mathematician

Version: April 23, 2021 108 RIEMANN’S APPROACH BY EMPLOYING THE ZETA FUNCTION

rough draft: do not distribute Further results 18

18.1 RESULTS

Theorem 18.1 (Dusart’s theorem, 1999). It holds that (see Gauß’ letter, Figure 1.1) 푥 푥 < 휋(푥) < log 푥 − 1 log 푥 − 1.1

for 푥 > 60184.

Theorem 18.2 (Rosser’s theorem1). It holds that

푝푛 > 푛 log 푛 (푛 ≥ 1).

The statement has been improved as follows:

Theorem 18.3. It holds that 푝 log 푛 + log log 푛 − 1 < 푛 < log 푛 + log log 푛 푛 and

푝 log log 푛 − 2 log2 log 푛 − 6 log log 푛 + 11 1 푛 = log 푛 + log log 푛 − 1 + − + 표 . 푛 log 푛 2 log2 푛 log2 푛

Theorem 18.4 (Voronin’s universality theorem2). Let 푈 be a compact subset of the strip {푧 ∈ C: 1/2 < <푧 < 1} such that the complement of 푈 is connected. Let 푓 : 푈 → C be continuous and holomorphic in the interior of 푈 which does not have any zeros in 푈. Then, for every 휀 > 0, there exists 훾 > 0 so that |휁 (푠 + 푖훾) − 푓 (푠)| < 휀

for all 푠 ∈ 푈.

Theorem 18.5 (Erdos–Kac˝ 3). The number of distinct primes of a random number is normally distributed. More precisely, for any 푎 < 푏,

( ) 푏 1 휔(푛) − log log 푛 1 ∫ 2 ≤ ≤ → −푡 /2 푛 푥 : 푎 < √︁ 푏 √ 푒 d푡 푥 log log 푛 2휋 푎 as 푥 → ∞.

1John Barkley Rosser, 1907–1989, US logician 2Sergei Michailowitsch Woronin, 1946–1997, Russian 3Mark Kac, 1914–1984, Polish

109 110 FURTHER RESULTS

1 Billingsley’s proof. Let 푋푝 be independent Bernoulli variables with 푃(푋푝 = 1) = 푝 . Set 푆푦 B Í E = Í 1 = + O( ) 2 var = Í 1 − 1 = 푝≤푦 푋푝, then 휇푦 B 푆푦 푝≤푦 푝 log log 푦 1 and 휎푦 B 푆푦 푝≤푦 푝 푝2 − log log 푦 + O(1) by Theorem 16.10. The central limit theorem implies 푆푦 휇푦 → N (0, 1) as 푦 → ∞. 휎푦 ( 1 if 푝 | 푛 Í Fix 푦 and let 1푝 (푛) B [푝 | 푛] = (the Iverson bracket) so that 휔푦 (푛) B | ≤ 1 = 0 else 푝 푛, 푝 푦 Í 1 푝≤푦 푝 (푛). We compare the moments of 휔푦 (푛) with those of 푆푦, i.e.,

푘 ! 1 ∑︁ ∑︁ 푘 − 1 ∑︁ 휔 (푛) − 휇 푘 − E 푆 − 휇 푘 = −휇 푘 푦 · 휔 (푛) 푗 − E 푆 푗 . 푥 푦 푦 푦 푦 푗 푦 푥 푦 푦 푛≤푥 푗=1 푛≤푥

The last term is ! ∑︁ 1 ∑︁ ∑︁ 1 j 푥 1 1 (푛)··· 1 (푛) − E 푋 ··· 푋 = − 푥 푝1 푝 푗 푝1 푝 푗 푥 퐿 퐿 푝1,..., 푝 푗 ≤푦 푛≤푥 푝1,..., 푝 푗 ≤푦 퐿Blcm( 푝1,..., 푝 푗 휋(푦) 푗 . 푥

( )− 푘 − 푘 This error bound implies that 1 Í 휔푦 푛 휇푦 ∼ E 푆푦 휇푦 ∼ E 푍 푘 as 푦 → ∞, where 푍 ∼ 푥 푛≤푥 휎푦 휎푦 1/log log log 푥 N(0, 1). If we choose 푦 B 푥 , then 휔(푛) − 휔푦 (푛) ≤ log log log 푥 and thus the result.

18.2 OPENPROBLEMS

Conjecture 18.6 (Goldbach’s4 conjecture). Every even integer greater than 2 can be expressed as the sum of two primes.

Conjecture 18.7 (Twin prime conjecture). There exist inﬁnity many twin primes (cf. Deﬁni- tion 2.19).

Problem 18.8 (Landau’s 4th problem). Are there inﬁnitely many primes of the form 푛2 + 1?

4Christian Goldbach, 1690–1764, German mathematician

rough draft: do not distribute Bibliography

[1] M. Abramowitz and I. A. Stegun. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Dover, New York, ninth Dover printing, tenth GPO printing edition, 1964. ISBN 0-486-61272-4. URL http://people.math.sfu.ca/~cbm/aands/ frameindex.htm. 40 [2] T. M. Apostol. Introduction to Analytic Number Theory. Springer New York, 1976. doi:10.1007/978-1-4757-5579-4.7 [3] H. Bateman. Tables of Integral Transforms. McGraw-Hill Book Company, Inc., New York- Toronto-London, 1954. 65 [4] P. Borwein, S. Choi, B. Rooney, and A. Weirathmueller, editors. The Riemann Hypothesis. Springer New York, 2008. doi:10.1007/978-0-387-72126-2.7 [5] H. Davenport. Multiplicative Number Theory. Springer New York, 1980. doi:10.1007/978- 1-4757-5927-3.7 [6] D. Duverney. Number Theory. WORLD SCIENTIFIC, 2010. doi:10.1142/7762. 25 [7] H. M. Edwards. Riemann’s Zeta Function. Academic Press, New York and London, 1974. URL https://books.google.at/books?id=ruVmGFPwN.7, 76 [8] O. Forster. Analytic number theory. Lecture notes, TU München, 2001. URL http://www. mathematik.uni-muenchen.de/~forster/v/ann/annth_all.pdf.7 [9] G. H. Hardy and M. Riesz. The General Theory Of Dirichlets Series. 1915. URL https: //archive.org/details/generaltheoryofd029816mbp/. 63 [10] G. H. Hardy and E. M. Wright. An introduction to the theory of numbers, 6th edition. 51(3): 283–283, 2010. doi:10.1080/00107510903184414.7 [11] H. Hasse. Ein Summierungsverfahren für die Riemannsche 휁-Reihe. Mathematische Zeitschrift, 32(1):458–464, 1930. doi:10.1007/BF01194645. 78 [12] H. Ivaniec and E. Kowalski. Analytic Number Theory, volume 53 of Colloquium Publications. American Mathematical Society, Providence, Rhode Island, 2004. ISBN 0-8218-3633-1. doi:10.1090/coll/053.7 [13] S. Khrushchev. Orthogonal Polynomials and Continued Fractions. Cambridge University Press, 2011. ISBN 0521854199. URL https://www.maths.ed.ac.uk/~v1ranick/papers/ khrushchev.pdf. 25 [14] D. J. Newman. Simple analytic proof of the prime number theorem. The American Mathe- matical Monthly, 87(9):693–696, 1980. doi:10.2307/2321853.7, 89, 101 [15] B. Riemann. Werke. Leipzig: Druck und Verlag von B. G. Teubner, 1876. 73 [16] H. Riesel and G. Gohl. Some calculations related to Riemann’s prime number formula. Mathematics of Computation, 24(112):969–983, 1970. doi:10.2307/2004630. 107

111 112 BIBLIOGRAPHY

Figure 18.1: Homer cubed

[17] J. Sander. Analytische Zahlentheorie. Lecture notes (in German), Universität Hannover, 2001. URL http://www.blu7.com/Skripte/Analytische_Zahlentheorie_SS01_Skript. pdf.7 [18] E. C. Titchmarsh. The Zeta-Function of Riemann. 1930.7

[19] E. C. Titchmarsh. The Theory of the Riemann Zeta function. Oxford University Press, Providence, Rhode Island, second edition, 1986. ISBN 0-19-853369-1.7, 59 [20] H. S. Wall. Analytic theory of continued fractions. Chelsea Publ., 1948. 25 [21] D. Zagier. Die ersten 50 Millionen Primzahlen. In Lebendige Zahlen, pages 39–73. Birkhäuser Basel, 1981. doi:10.1007/978-3-0348-5407-8.7 [22] D. Zagier. Newman’s short proof of the prime number theorem. The American Mathemati- cal Monthly, 104(8), 1997. doi:10.1080/00029890.1997.11990704.7, 89, 101

rough draft: do not distribute Index

B gcd, 10 Bernoulli number, 31 H polynomial, 31 Heaviside step function, 65 Bézout identity, 10 L C least common multiple, lcm,9 composite, 10 logarithmic integral, 88 coprime, 10 M D Möbius inversion, 54 Dirichlet multiple, 10 inverse, 52 product, 52 N series, 63 number Carmichael, 23 E Euclid’s lemma, 11 P Euler–Mascheroni constant, 36 prime, 10 exponential integral, 87 cousin, 13 sexy, 13 F twin, 13 Fermat prime gap, 13 liar, 23 prime number theorem, 101 pseudoprime, 23 prime-counting function, 휋, 91 function primes additive, 51 Fermat, 14 arithmetic, 51 Mersenne, 14 arithmetic function primorial, 93 divisor, 휎푘 , 52 Liouville, 휆, 51 R Möbius, 휇, 53 Riemann hypothesis, 85 number-of-divisors, 휎0, 52 Riemann-Siegel, 84 prime omega, big omega, Ω, 51 prime omega, 휔, 51 S sum-of-divisors, 휎, 52 Stieltes constants, 79 totient, 휑, 19 Stirling formula, 40 von Mangoldt, Λ, 59 summation Chebyshev, 휗, 휓, 93 Abel, 47 multiplicative, 51 Borel, 49 totally multiplicative, 51 by parts, 45 fundamental theorem of arithmetic, 12 Cesàro, 44 Euler, 44 G Euler–Maclaurin, 39 Gamma function, 35 Lambert, 46

113 114 INDEX

Poisson, 47 summatory function, 67

W Wallis’ product, 37

Z zeros non-obvious, 85 trivial, 79 zeta function, 57 prime 푃, 60

rough draft: do not distribute