lecture notes in analytic number...

Lecture Notes in Analytic Number Theory

Lectures by Dr. Sheng-Chi Liu

Throughout these notes, signifies end proof, and N signifies end ofexample.

Table of Contents

Table of Contents i

Lecture 1 Background and Introduction 11.1 Basic questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Landau’s problems . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Counting prime numbers . . . . . . . . . . . . . . . . . . . . . . . 3

Lecture 2 Different proofs of the infinitude of primes 32.1 Prime counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Lecture 3 Chebyshev’s idea continued 63.1 Primes in factorials . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Lecture 4 Lower bound of Chebyshev’s theorem 94.1 Chebyshev finalised . . . . . . . . . . . . . . . . . . . . . . . . . . 94.2 Averages of arithmetic functions . . . . . . . . . . . . . . . . . . 11

Lecture 5 Dirichlet convolution 125.1 Approximation of average functions by integrals . . . . . . . . . . 125.2 Dirichlet convolution . . . . . . . . . . . . . . . . . . . . . . . . . 14

Lecture 6 Mobius function 156.1 More convoluting . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Lecture 7 Mobius inversion formula 167.1 Mobius inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . 167.2 Multiplicative functions . . . . . . . . . . . . . . . . . . . . . . . 18

Lecture 8 Dirichlet series 198.1 Multiplicativity of Mobius function . . . . . . . . . . . . . . . . . 198.2 Dirichlet series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Notes by Jakob Streipel. Last updated July 29, 2019.

i

mailto:[email protected]

TABLE OF CONTENTS ii

Lecture 9 L-functions of convolutions 229.1 L-functions associated with Dirichlet convolutions . . . . . . . . 229.2 Dirichlet series and multiplicative functions . . . . . . . . . . . . 24

Lecture 10 Euler products 2510.1 L-series continued . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Lecture 11 Primes in arithmetic progressions 2811.1 Arithmetic progressions . . . . . . . . . . . . . . . . . . . . . . . 28

Lecture 12 Characters 2912.1 Analogue of Merten’s theorem . . . . . . . . . . . . . . . . . . . . 2912.2 Characters of finite abelian groups . . . . . . . . . . . . . . . . . 29

Lecture 13 Fourier analysis on finite abelian groups 3113.1 Orthogonality of characters . . . . . . . . . . . . . . . . . . . . . 3113.2 Fourier analysis on finite abelian groups . . . . . . . . . . . . . . 3213.3 Characters on cyclic groups . . . . . . . . . . . . . . . . . . . . . 3313.4 Dirichlet characters . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Lecture 14 Dirichlet characters 3314.1 L-functions attached to Dirichlet characters . . . . . . . . . . . . 33

Lecture 15 Mertens’ theorem for arithmetic progressions 3615.1 Proof finished . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3615.2 Dirichlet L-functions at 1 . . . . . . . . . . . . . . . . . . . . . . 37

Lecture 16 Landau’s lemma 3816.1 Landau’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Lecture 17 Riemann’s memoir 4017.1 Riemann’s memoir . . . . . . . . . . . . . . . . . . . . . . . . . . 4017.2 Review of Fourier analysis . . . . . . . . . . . . . . . . . . . . . . 41

Lecture 18 Poisson summation 4218.1 Fourier analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Lecture 19 The functional equation for ζ(s) 4519.1 Mellin transformation . . . . . . . . . . . . . . . . . . . . . . . . 4519.2 Gamma function . . . . . . . . . . . . . . . . . . . . . . . . . . . 4619.3 The functional equation for ζ(s) . . . . . . . . . . . . . . . . . . 47

Lecture 20 The functional equation for L-functions 4820.1 Theta series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4820.2 Trivial zeros of ζ(s) . . . . . . . . . . . . . . . . . . . . . . . . . . 4920.3 Functional equations of L-functions . . . . . . . . . . . . . . . . . 4920.4 Gauss sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Lecture 21 The functional equation for L-functions, continued 5121.1 Gauss sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

TABLE OF CONTENTS iii

Lecture 22 Functions of finite order 5422.1 Proof concluded . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5422.2 The Hadamard factorisation theorem . . . . . . . . . . . . . . . . 55

Lecture 23 Jensen’s formula 5623.1 Jensen’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 5623.2 Application of Jacobi’s formula . . . . . . . . . . . . . . . . . . . 58

Lecture 24 Hadamard factorisation theorem 5924.1 Hadamard factorisation . . . . . . . . . . . . . . . . . . . . . . . 59

Lecture 25 The infinite product for ζ(s) 6025.1 Hadamard factorisation finished . . . . . . . . . . . . . . . . . . . 6025.2 The infinite product for ζ(s) and the explicit formula . . . . . . . 6125.3 Infinite product for L(s, χ) . . . . . . . . . . . . . . . . . . . . . . 6225.4 Application to counting zeros of ζ(s) . . . . . . . . . . . . . . . . 62

Lecture 26 Counting zeros 6326.1 Application to counting zeros of ζ(s) . . . . . . . . . . . . . . . . 6326.2 Applications to countinjg zeros of L(s, χ) . . . . . . . . . . . . . 64

Lecture 27 Weil’s explicit formula 6627.1 Application to counting zeros of L(s, χ) . . . . . . . . . . . . . . 6627.2 Weil’s explicit formula . . . . . . . . . . . . . . . . . . . . . . . . 67

Lecture 28 Weil’s explicit formula 6828.1 Proof continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

Lecture 29 The theorem of Hadamard and de la Vallee-Poussin 7029.1 Zero free region . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

Lecture 30 Exceptional zeros 7330.1 Zero free region for L(s, χ) . . . . . . . . . . . . . . . . . . . . . . 73

Lecture 31 Landau’s theorem 7531.1 Exceptional zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . 7531.2 Siegel’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

Lecture 32 Siegel’s theorem 7832.1 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

Lecture 33 The Prime number theorem in Arithmetic progres-sions 80

33.1 Prime number theorem . . . . . . . . . . . . . . . . . . . . . . . . 80

Lecture 34 Bombieri-Vinogradov 8234.1 Prime number theorem in arithmetic progressions . . . . . . . . . 8234.2 The Bombieri-Vinogradov theorem . . . . . . . . . . . . . . . . . 83

Lecture 35 Bombieri-Vinogradov, continued 8535.1 The trivial case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8535.2 The nontrivial part . . . . . . . . . . . . . . . . . . . . . . . . . . 86

TABLE OF CONTENTS iv

Lecture 36 The Large Sieve 8736.1 The Large siece inequality . . . . . . . . . . . . . . . . . . . . . . 87

Lecture 37 Vaughan’s Identity 9037.1 Basic Mean Value Theorem, again . . . . . . . . . . . . . . . . . 90

BACKGROUND AND INTRODUCTION 1

Lecture 1 Background and Introduction

Number theory is the study of numbers, a natural starting point of which isthe study of the integers Z. The integers are equipped with addition andmultiplication—the opposite of addition, i.e. subtraction, doesn’t move us outof the integers, but the opposite of multiplication, being division, does. Henceanother type of numbers of great interest is the rational numbers Q.

On the other hand, to understand the integers one can, seemingly, settlefor understanding the prime numbers ℘, since with those we can build theintegers.

We want to study these prime numbers which, despite being the simplestbuilding blocks, turns out to be the hardest part of these sorts of questions.

1.1 Basic questions

It is true, and has been known for a very long time, that there are infinitelymany primes (more on this soon).

With that in mind, there are a few basic questions we will ask—and answer—in this course.

1. How many primes are there up to some number x? In other words, howdoes

π(x) = #{ p ∈ ℘ | p ≤ x }

behave as a function of x?

To answer this question precisely is very hard; there is essentially no optionapart from simply counting one by one. On the other hand, answeringthis question approximately, in a very precise sense, is entirely doable!In particular, the asymptotic behavious of this quantity is known as thePrime number theorem, which we will endeavour to prove in this course.

2. How many primes are there in a given arithmetic progression? That is tosay, given a, n > 0 with (a, n) = 1, how does

π(x; a, n) = #{℘ 3 p ≤ x | p ≡ a (mod n) }

behave? This is a much tougher question. That π(x; a, n) is infinite isDirichlet’s theorem. This, as well as understanding how this grows asymp-totically in x, is the second goal of this course.

1.2 Landau’s problems

At the 1912 International Congress of Mathematicians, Edmund Landau listedfour basic problems about prime numbers. All four of these remain unsolved tothis day. Listing them demonstrates how easy some of the fundamental problemsin number theory are to understand, and yet how inconceivably hard they areto solve.

1. Goldbach’s conjecture: Can every even integer greater than 2 be writtenas the sum of two primes?

Date: January 7th, 2019.

BACKGROUND AND INTRODUCTION 2

2. Twin prime conjecture: Are thjere infinitely many primes p such that p+2is prime?

3. Legendre’s conjecture: Does there always exist at least one prime betweenconsecutive perfect squares?

4. Are there infinitely many primes p such that p− 1 is a perfect square? Inother words, are there infinitely many primes of the form n2 + 1?

We’ll give some brief information about the current state and progress onthese. The biggest step toward the Goldbach conjecture is due to Chen, whoproves that for n sufficiently large, we can write 2n = p + q where q is eitherprime or the product of two primes.

There is also a weaker version of the Goldbach conjecture, naturally known asthe Weak Goldbach conjecture. This states that every odd number greater than5 can be written as the sum of three primes. This was proved by Vinogradovand Helfgott.

The Twin prime conjecture is on similar grounds as Goldbach’s conjecture:the best known is by Chen, proving that there are infinitely many primes p suchthat p+ 2 is either a prime or a product of two primes.

A related sort of problem, in a different direction, is asking about gapsbetween primes. In particular, are there infinitely many prime pairs with abounded gap M , i.e. is

${ (p, q) ∈ ℘× ℘ | 0 < |p− q| ≤M } =∞?

There has recently been tremendous progress in this problem. Some founda-tional results are due to Goldston–Pintz–Y1ld1r1m. Recently it was proven forM = 70 million by Yitang Zhang, and shortly thereafter a very similar resultwas proven by Maynard using a completely different method. The gap has sincebeen reduced to M = 246 by the Polymath Project, led by Tao. Under theassumption of the Elliot-Halberstam conjecture, M has been reduced to 6.

The closest known result to Legendre’s conjecture is due to Ingham, provingthat there is a prime between n3 and (n+ 1)3 for every sufficiently large n.

Finally, the best result on the road to the fourth of Landau’s problems isdue to Iwaniec, proving that there are infinitely many n ∈ Z such that n2 + 1 isa prime or a product of two primes.

1.3 Notation

We will use following notation when discussion asymptotic behaviour over andover. Let f(x) and g(x) be two functions on R, with g(x) 6= 0 for sufficientlylarge x.

We write f ∼ g to mean that

limx→∞

f(x)

g(x)= 1,

i.e. the values of the two functions are the same as x grows.We write f(x) = O(g(x)), or f(x)� g(x), to mean that |f(x)| ≤ Cg(x) for

some constant C and all x ∈ R, i.e. f is bounded by g (with some constant).

DIFFERENT PROOFS OF THE INFINITUDE OF PRIMES 3

Finally we write f(x) = o(g(x)) meaning that for any ε > 0 there exists aconstant N > 0 such that |f(x)| ≤ ε|g(x)| for all x ≥ N , or in other words g(x)is much bigger than f(x). We might also put it as

limx→∞

|f(x)||g(x)|

= 0.

1.4 Counting prime numbers

As claimed earlier, it is a fact that

Theorem 1.4.1. There are infinitely many prime numbers.

There are are many known proofs of this. The oldest among them is due toEuclid:

Proof. Suppose there aren’t infinitely many primes, meaning that there arefinitely many of them, say p1, p2, . . . , pn. Now consider the number m =p1p2 · · · pn + 1.

This number is not divisible by pi for any i, since its remainder is always1. Hence m = p1p2 · · · pn + 1 does not have a prime factor pi, which is acontradiction.

This is a very elegant proof. Indeed, it is too elegant, because it doesn’t tellus anything more!

Lecture 2 Different proofs of the infinitude ofprimes

2.1 Prime counting

Define the prime counting function

π(x) = #{ p ∈ ℘ | p ≤ x } =∑p≤x

1,

where when indexing a sum over p we mean that it sums only over prime num-bers.

The above theorem therefore says that π(x) → ∞ as x → ∞. A naturalquestion is then at what rate this goes to infinity.

Gauss (about 1792, around age 15) and Legendre (1798) gave a conjecturalasymptotic formula for π(x). It was proved by Hadamard and de la Vallee-Poissin independently in 1896, known as

Theorem 2.1.1 (Prime number theorem). As x→∞, we have

π(x) ∼ x

log(x).



Remark 2.1.2. In the 1830’s, Dirichlet reformulated the (then) Prime numberconjecture as

π(x) ∼ Li(x) :=

∫ x

2

dt

log(t).

Note that using integration by parts,

Li(x) =

∫ x

2

dt

log t=

t

log(t)

∣∣∣∣x2

−∫ x

2

− t

t log(t)2dt =

x

log(x)+

∫ x

2

1

log(t)2dt+O(1)

=x

log(x)+

x

log(x)2+

∫ x

2

1

log(t)2dt+O(1) =

x

log(x)+

x

log(x)2+O

(x

log(x)3

)and so forth, whence Li(x) ∼ x/ log(x).

Remark 2.1.3. The proof of the Prime number theorem actually gives

π(x) = Li(x) +O(x exp(−C√

log(x)))

for some constant C > 0.Additionally, assuming the Riemann hypothesis we get substantial power

savings, namelyπ(x) = Li(x) +O(x1/2 log(x)2).

Let us now prove the infinitude of the primes again, but this time with avery different method, courtesy of Euler.

Define the (Riemann) zeta function

ζ(s) =

∞∑n=1

1

ns.

As a historical anectode, this is named after Riemann despite Euler consideringit before Riemann was around because Euler did it only for s ∈ R, s > 1,whereas Riemann considered it for all of C, continued it analytically and founda functional equation for it.

Consider similarly for each prime p the sum

ζp(s) = 1 +1

ps+

1

p2s+ . . . =

∑α≥0

1

pαs= (1− p−s)−1

since it’s a geometric series, again assuming s > 1.Now by the unique factorisation of integers into primes,

ζ(s) =∏p

ζp(s) =∏p

(1− p−s)−1,

called the Euler product of ζ(s).Now by considering the sum for ζ(s) above as a Riemann sum for the integral

of 1/x2, we have

ζ(s) =

∞∑n=1

1

ns≥∫ ∞

1

1

xsdx =

1

s− 1,

whereby as s→ 1+, ζ(s)→∞.


Now if we suppose there are only finitely many primes, then

lims→1+

ζ(s) = lims→1+

∏p

(1− p−s)−1 =∏p

(1− p−1)−1,

is finite, which is a contradiction. Hence there are infinitely many primes.Now suppose we want to play this same game, but instead of counting the

primes all with weight 1, count them with weight 1/p.

Theorem 2.1.4 (Euler). The sum∑p

1

pdiverges.

Proof. For s > 1,

log ζ(s) =∑p

(1− p−s)−1 = −∑p

log(1− p−s).

Recalling the Taylor expansion of

log(1− x) = −(x+

x2

2+x3

3+ . . .

)(which is easy to see as the integral of 1/(1− x) = 1 + x+ x2 + . . .), we have

log ζ(s) =∑p

( 1

ps+O

( 1

p2s

)).

On the other hand

log ζ(s) ≥ log1

s− 1= − log(s− 1),

whence ∑p

(p−s +O(p−2s)) ≥ − log(s− 1)

which goes to ∞ as s→ 1+, so∑p

1

p+O(1) =∞.

A more interesting question, which we will answer later, is what the asymp-totic behaviour of ∑

p≤x

1

p.

Another approach is due to Chebyshev (1850),

Theorem 2.1.5. There exist constants 0 < c1 < c2 such that for x ≥ 2

c1x

log(x)≤ π(x) ≤ c2

x

log(x).

To prove this we need a bit of groundwork.

CHEBYSHEV’S IDEA CONTINUED 6

Definition 2.1.6 (p-adic valuation). Let p be a prime. For n ∈ Z \ { 0 } definethe p-adic valuation νp(n) to be the largest integer α ≥ 0 such that pα | n(i.e. pα+1 - n).

Moreover we define νp(0) = +∞.

Hence by factoring into prime factors, for any n we have

n =∏p

pνp(n).

Note also that νp(mn) = νp(m) + νp(n) for all m,n ∈ Z.Chebyshev’s idea is based on the property that n! is divisible by all the

primes less than or equal to n, and not by any other primes. We then have

n! =∏p≤n

pνp(n!),

which when taking logarithms becomes

log(n!) =∑p≤n

νp(n!) log p,

but on the other hand

log(n!) =

n∑k=1

log k = n log n− n+O(log n).

Lecture 3 Chebyshev’s idea continued

3.1 Primes in factorials

Continuing the calculations above, we need to get a sense of how νp(n!) behaves.To this end, note that by definition

νp(k) = max{α | pα | k } =∑α≥1pα|k

1,

which means that

νp(n!) =

n∑k=1

νp(k) =

n∑k=1

∑α≥1pα|k

1.

By switching the order of summation we get

νp(n!) =∑α≥1

∑1≤k≤npα|k

1 =∑α≥1

⌊n

pα

⌋.

By bxc we mean the largest integer less than or equal to x, or alternativelybxc = x− {x} where {x} is the fractional part of x. Crucial for our purposes isthat bxc ≤ x, so

νp(n!) ≤∑α≥1

n

pα=

n

p(1− p−1)=n

p+O

(n

p2

)Date: January 11th, 2019.


since 1/(1− p−1) = 1 + 1/p+O(1/p2).Hence

∑p≤n

νp(n!) log p ≤∑n≤p

log p

(n

p+O

(n

p2

))= n

∑p≤n

log p

p+O

n∑p≤n

log p

p2

.

Note that the sum in the error term is bounded by the same sum over all k ≤ ninstead of p ≤ n, which is the Riemann sum of a convergent integral, so theerror term is O(n).

Combining our estimates we therefore have

n log n− n+O(log n) ≤ n∑p≤n

log p

p+O(n),

and dividing by n gives us∑p≤n

log p

p≥ log n− 1 +O(1),

the right-hand side of which goes to ∞ as n → ∞. Hence there are infinitelymany primes.

We’re now ready to start proving Chebyshev’s theorem:

Proof. We start with the upper bound. First let us split π(x) into two sums:

π(x) =∑p≤x

1 =∑

p≤x1/2

1 +∑

x1/2<p≤x

1.

The first of these sums is trivially bounded by x1/2. For the second one, notethat ∑

x1/2<p≤x

1 ≤ 1

log(x1/2)

∑x1/2<p≤x

log p

since this way each term in the summand is at least 1. We will denote by ϑ(x)

ϑ(x) =∑p≤x

log p,

so that1

log(x1/2)

∑x1/2<p≤x

log p =1

log(x1/2)

(ϑ(x)− ϑ(x1/2)

).

The very clever idea of Chebyshev’s is to consider the binomial coefficient(2n

n

)=

(2n)!

n!n!.

Taking logarithms we have

log

(2n

n

)= log((2n)!)− 2 log(n!) =

2n∑k=1

log k − 2

n∑k=1

log k

= (2n) log(2n)− 2n+O(log(2n))− 2(n log n− n+O(log n))

= (log 2)2n+O(log(2n)).


On the other hand, by the Binomial theorem,

(1 + 1)2n =

2n∑k=0

>

(2n

n

)

so that(

2nn

)< 22n, meaning that log

(2nn

)< (log 2)2n.

It is also true that, by the same sort of argument as we used above,(

2nn

)is

divisible by primes in (n, 2n]. Hence

log

(2n

n

)=∑p≤2n

νp

((2n

n

))log p ≥

∑n<p≤2n

log p = ϑ(2n)− ϑ(n) < (log 2)2n,

whereby ϑ(2n)− ϑ(n) < (log 2)2n.Next, ϑ(x)− ϑ(x/2) ≤ (log 2)x for x ≥ 2 being real. To see this, for a given

x ≥ 2 take n to be the integer such that 0 ≤ x − 2n < 2, i.e. the biggest eveninteger less than x. Then 0 ≤ ϑ(x)− ϑ(2n) ≤ log x, and

0 ≤ ϑ(x)− ϑ(x/2)− (ϑ(2n)− ϑ(n)) ≤ log x,

wherein we bound the parenthesis by (log 2)2n < (log 2)x, so

ϑ(x)− ϑ(x/2) ≤ (log 2)x+O(log x).

With this in mind, let us write ϑ(x) as a telescoping sum:

ϑ(x) =∑

0≤k≤b log xlog 2 c

ϑ( x

2k

)− ϑ

( x

2k+1

)≤

∑0≤k≤b log x

log 2 c

((log 2)

x

2k+O(log(x/2k))

)= 2x log 2 +O((log x)2),

since the last sum is geometric.Returning to π(x), with this in hand we get

π(x) =∑

p≤x1/2

1 +∑

x1/2<p≤x

1 ≤ x1/2 +1

log(x1/2)

∑x1/2<p≤x

log p

≤ x1/2 +2

log x(ϑ(x)− ϑ(x1/2))

≤ x1/2 +2

log 2

(2(log 2)x+O((log x)2) +O(x1/2)

)= 4 log 2

x

log x+O(x1/2).

Hence we can take c2 large enough that π(x) ≤ c2x/ log x. It also establishesthat ϑ(x) ≤ cx for some c.

It is important and interesting to note that this does not tighten to c2 = 1,meaning that Chebyshev’s argument is not sufficient to prove the Prime numbertheorem.

LOWER BOUND OF CHEBYSHEV’S THEOREM 9

Lecture 4 Lower bound of Chebyshev’s theorem

4.1 Chebyshev finalised

To get a lower bound for Chebyshev’s theorem we need a more precise estimatefor νp(

(2nn

)). Recall how

νp

((2n

n

))= νp((2n)!)− 2νp(n!) =

2n∑k=1

νp(k)− 2

n∑k=1

νp(k)

=∑α≥1

⌊2n

pα

⌋− 2

∑α≥1

⌊n

pα

⌋=∑α≥1

(⌊2n

pα

⌋− 2

⌊n

pα

⌋)

=∑

1≤α≤ log 2nlog p

β

(n

pα

).

Here we mean

β(x) = b2xc − 2bxc =

{1, if {x} ∈ [1/2, 1)

0, if {x} ∈ [0, 1/2)

and the range of summation can be cut off at log(2n)/ log(p) since overwise theinteger parts above are 0 anyway.

The trivial upper bound here is log(2n)/ log p, naturally, since β(x) ≤ 1.Now recall

log

(2n

n

)=∑p≤2n

νp

((2n

n

))log p ≤

∑p≤2n

log plog 2n

log p= (log 2n)π(2n).

On the other hand,

log

(2n

n

)= log((2n)!)− 2 log(n!) =

2n∑k=1

log k − 2

n∑k=1

log k

= 2n log(2n)− 2n+O(log 2n)− 2(n log n− n+O(log n)),

hence

π(2n) ≥ (log 2)2n

log n+O(1),

and so for x ≥ 2,

π(x) ≥ (log 2)x

log x

(1 +O

(log x

x

)).

Ergo there exists some sufficiently large c1 so that π(x) ≥ c1x/ log x.



We can also extract from this a lower bound for ϑ(x):

ϑ(x) =∑p≤x

log p =∑

p≤x1/2

log p+∑

x1/2<p≤x

log p

≥ log 2 + π(x1/2) + log(x1/2)(π(x)− π(x1/2)

)≥ log 2 + c

x1/2

log x1/2+ c log(x1/2)

(x

log x− x1/2

log x1/2

)≥ cx+O(x1/2).

Hence there exists some sufficiently large c′ such that ϑ(x) ≥ c′x.

Theorem 4.1.1 (Mertens).∑p≤x

log p

p= log x+ o(1).

This result is, maybe surprisingly much easier than the Prime number the-orem, even though it would seem that counting the primes with the strangeweight log p/p would be harder than weight 1.

Proof. We have

log n! =

n∑k=1

log k = n log n− n+O(log n),

but on the other hand

log n! =∑p≤n

log p∑α≥1pα≤n

⌊n

pα

⌋.

The main term here comes from α = 1, since for α ≥ 2 we have∑p≤n


⌊n

pα

⌋≤∑p≤n


n

pα≤ n

∑p

log p

p2= O(n)

where for the last inequality we notice that the inner sum is n/(p2(1 − 1/p))since it’s geometric.

Moreover x=x+ {x}, so

∑p≤n

log p

⌊n

p

⌋+O(n) =

∑p≤n

log p

(n

p+O(1)

)+O(n)

= n∑p≤n

log p

p+O

(∑p≤n

log p)

+O(n).

But the error term is ϑ(n), which is order n, so this is

n∑p≤n

log p

p+O(N),

and dividing through by n gives us what we want.


4.2 Averages of arithmetic functions

Definition 4.2.1. An arithmetic function is a real or complex valued func-tion on N, i.e. f : N→ R or C.

Example 4.2.2. The constant function 1(n) = 1 for all n ∈ N is an arithmeticfunction. The delta function (at 1),

δ(n) =

{1, if n = 1

0, if n 6= 1

is another one we’ll use.Of particular interest is the characteristic function of the primes,

1℘(n) =

{1, if n ∈ ℘0, if n 6∈ ℘.

A final one we’ll have occasion to play with is the von Mangoldt function

Λ(n) =

{log p, if n = pα, α ≥ 1

0, if n 6= pα.N

Definition 4.2.3. Let f be an arithmetic function. The average functionMf (X) of f is the function defined on R≥0 by

Mf (X) =∑

1≤n≤X

f(n).

Note that this is not quite the usual arithmetic average; for that we’d divideby X. However if we know the behaviour of one, we automatically get thebehaviour of the other, so there is no need to complicate things further by thatextra division.

The main question we ask ourselves about such things is what, given anarithmetic function f , the behaviour of Mf (X) is as X →∞.

Example 4.2.4. Some of the functions we’re used to are averages of arithmeticfunctions. For instance, π(x) = M1℘(x). Similarly ϑ(x) = Mf (x) where f(n) =log(n)1℘(n).

Another function we will have reason to care about is

ψ(x) = MΛ(x) =∑pα≤x

log p. N

All of these three are closely related in the problem of counting primes.Indeed we would, and should, suspect that ϑ and ψ are close to one another,since one is counting primes with weight log p and the other is counting primepowers with the same weight. Indeed

Proposition 4.2.5. We have ψ(x) = ϑ(x) +O(x1/2 log x).

Hence by Chebyshev’s theorem, ψ(x) � ϑ(x) � x, by which we mean thatthey are the same order.

DIRICHLET CONVOLUTION 12

Proof. We write

ψ(x) = ϑ(x) +∑pα≤xα≥2

log p

So ∑pα≤xα≥2

log p =∑

p≤x1/2

log p

∑pα≤xα≥2

1

where the inside is bounded by log x/ log p, so

≤∑

p≤x1/2

log plog x

log p= O(log(x)x1/2),

where we’ve used the weaker bound; Chebyshev could give us something tighter.

Lecture 5 Dirichlet convolution

5.1 Approximation of average functions by integrals

In the special case where f : R→ R, the average Mf (X) is just a restriction toN, so we want to compare ∑

n≤x

f(n) ∼∫ x

1

f(t) dt.

Proposition 5.1.1 (Monotone comparison). Let f : R≥0 → R be continuous.Suppose f is monotone. Then we have

Mf (X) =

∫ X

1

f(t) dt+O(|f(1)|+ |f(X)|).

Proof. Suppose f is monotonically increasing. For n ≥ 2, we have∫ n

n−1

f(t) dt ≤ f(n) ≤∫ n+1

n

f(t) dt.

Summing this over 2 ≤ n ≤ bXc, we are done.If f is monotonically decreasing we do precisely the same, except the in-

equalities above are reversed.

Example 5.1.2. Take f(x) = log x, and this immediately gives us

∑n≤x

log n =

∫ X

1

log t dt+O(|log 1|+ |logX|) = X logX −X +O(logX)

with a little bit of integration by parts thrown in. N



Theorem 5.1.3 (Integration by parts). Let g be an arithmetic function and letf be a continuously differentiable function on [1, X]. Then

Mfg(X) =∑n≤x

f(n)g(n) = f(X)Mg(X)−∫ X

1

Mg(t)f′(t) dt.

Proof. By using integration by parts with Riemann-Stieltjes integrals,∑n≤x

f(n)g(n) =

∫ X

1−f(t) dMg(t) = f(t)Mg(t)

∣∣∣∣X1−−∫ X

1−f ′(t)Mg(t) dt

= f(X)Mg(X)−∫ X

1−Mg(t)f

′(t) dt.

Proposition 5.1.4. (i) ϑ(x) = (log x)π(x) +O

(x

log x

).

(ii) π(x) ∼ x

log xif and only if ϑ(x) ∼ x if and only if ψ(x) ∼ x.

Proof. (i) Writing

ϑ(x) =∑p≤x

log p =∑n≤x

log n1℘(n)

and then using the above, we get

ϑ(x) = (log x)π(x)−∫ x

1

π(t)(t) dt,

and by Chebyshev’s theorem∫ x

1

π(t)

tdt ≤ C

∫ x

1

1

t

t

log tdt = O

(x

log x

).

(ii) The first equivalence follows directly from the above, and the second equiv-alence we have proved before, namely Proposition 4.2.5.

Remark 5.1.5. Both Hadamard’s and de la Vallee-Poussin’s theorems of thePrime number theorem actually proved ψ(x) ∼ x.

Corollary 5.1.6 (Euler-Maclaurin formula). Let f be a C1 function on (0,∞).Let B1(x) = x− bxc − 1/2 = {x} − 1/2. For x > 1, we have

Mf (X) =∑n≤X

f(n) =

∫ X

1

f(t) dt+

∫ X

1

B1(t)f ′(t) dt−B1(1)f(1)−B1(X)f(X).

In particular∣∣∣∣∣Mf (X)−∫ X

1

f(t) dt

∣∣∣∣∣ ≤∫ X

1

|f ′(t)| dt+ |f(1)|+ |f(X)|.

Proof. Use summation by parts with g(n) = 1 and Mg(t) = btc.

Remark 5.1.7. This formula does not require f to be monotone.

Remark 5.1.8. The function B1(x) is called the first Bernoulli function andB1 := B1(0) is the first Bernoulli number .


5.2 Dirichlet convolution

Definition 5.2.1 (Dirichlet convolution). Let f and g be arithmetic functions.Define the Dirichlet convolution f ? g by

f ? g(n) =∑d|n

f(d)g(nd

)=∑ab=n

f(a)g(b).

Proposition 5.2.2. Dirichlet convlution

(i) is commutative, f ? g = g ? f ;

(ii) is associative, (f ? g) ? h = f ? (g ? h);

(iii) has an identity, namely the function δ(n) which is 1 if n = 1 and 0 other-wise;

(iv) sometimes has inverses. Specifically f is invertible under ?, i.e. thereexists some g such that f ? g = δ, if and only if f(1) 6= 0, in which caseits ?-inverse g is determined by the recurrence relation g(1) = 1/f(1),

g(n) = − 1

f(1)

∑d|nd>1

f(d)g(nd

).

Proof. (i) This follows directly by definition. If we switch f and g, the orderof summation is reversed, but the summands are the same.

(ii) We compute

(f ?g)?h(n) =∑ab=n

(f ?g)(a)h(b) =∑ab=n

∑cd=a

f(c)g(d)h(b) =∑bcd=n

f(c)g(d)h(b).

Doing precisely the same thing to f ? (g ? h)(n) we get the same sum, whencethe two are equal for all n and so equal as functions.

(iii) Note that f ?δ(n) is the sum of f(d)δ(n/d), but δ(n/d) = 1 onlt if n/d = 1,i.e. d = n, so f ? δ(n) = f(n) for all n.

(iv) For the forward direction, suppose f is ?-invertible. Then f ? g = δ forsome g, and evaluating this at n = 1 we get f ? g(1) = δ(1) = 1. Moreover

f ? g(1) =∑d|1

f(d)g

(1

d

)= f(1)g(1),

so we have f(1)g(1) = 1, which has a solution g(1) if and only if f(1) 6= 0.

For the converse direction, suppose f(1) 6= 0. Let us solve the equationf ? g = δ for g. First f(1)g(1) = 1, by the above, so g(1) = 1/f(1). For n > 1,

0 = δ(n) = f ? g(n) =∑d|n

f(d)g(nd

)= f(1)g(n) +

∑d|nd>1

f(d)g(nd

).

Solving this for g(n) gives us the recurrence we want.

MOBIUS FUNCTION 15

Remark 5.2.3. Since n/d < n for d > 1, g(n) is completely determined byf(1), f(2), . . . , f(n) and g(1), . . . , g(bn/2c).

Example 5.2.4. We define

d(n) := 1 ? 1(n) =∑d|n

1,

which counts the number of (positive) divisors of n. Similarly

d3(n) := 1 ? 1 ? 1(n) =∑abc=n

1

is the number of representations of n as a product of 3 positive integers. Ingeneral,

dk(n) := 1 ? 1 ? · · · ? 1︸︷︷︸k times

(n) =∑

a1a2···ak=n

1

is the number of representations of n as a product of k positive integers. N

Lecture 6 Mobius function

6.1 More convoluting

Proposition 6.1.1. We have log n = Λ ? 1(n), i.e.

log n =∑d|n

Λ(d).

Since Λ(n) is multiplicative, which we will explore more later, this veryneatly translates log into a multiplicative function.

Proof. Write n as its prime factorisation

n =∏p

pαp ,

and take logarithms, giving

log n =∑p

αp log p =∑p

log p(∑

αpα|n

1)

=∑pα|n

log p =∑d|n

Λ(d).

Definition 6.1.2 (Mobius function). The Mobius function µ is defined asthe ?-inverse of 1, i.e. µ(1) = 1 and

µ(n) = −∑d|nd>1

µ(nd

).

We will show later that this does in fact coincide with the definition of µcommonly found in elementary number theory texts, namely

µ(n) =

0 if n is not square free

(−1)r if n = p1p2 · · · pr, pi all distinct primes

1 if n = 1.


MOBIUS INVERSION FORMULA 16

Lecture 7 Mobius inversion formula

7.1 Mobius inversion

Recall that µ is the ? inverse of 1, i.e. µ ? 1 = δ.

Theorem 7.1.1 (Mobius inversion formula). Let f and g be arithmetic func-tions. Then the following identities are equivalent

(i) f(n) =∑d|n

g(d)

(ii) g(n) =∑d|n

µ(d)f(nd

).

Proof. Note that the first property says that f = g ? 1, and the second saysg = f ? µ. With this in mind the proof is obvious: f = g ? 1 if and only iff ? µ = g ? 1 ? µ if and only if f ? µ = g ? δ = g.

Example 7.1.2. In Proposition 6.1.1 we showed that

log n =∑d|n

Λ(d),

i.e. log n = Λ ? 1(n). This is equivalent to

Λ(n) = log ?µ(n) =∑d|n

µ(d) log(nd

). N

We can use this to count primes:

Theorem 7.1.3 (Mertens). (i)∑n≤x

Λ(n)

n= log x+O(1).

(ii)∑p≤x

log p

p= log x+O(1).

(iii)∑p≤x

1

p= log log x+O(1).

Proof. We write ∑n≤x

Λ(n)

n=∑p≤x

log p

p

∑pα≤xα≥2

log p

pα.

Now ∑pα≤xα≥2

log p

pα=

∑p≤x1/2

log p∑

2≤α≤ log xlog p

1

pα,

Date: January 23rd, 2019.


where the innermost sum is bounded by 1/(p2(1 − p−1)), where the geometricterm is bounded, whence this is∑

pα≤xα≥2

log p

pα�

∑p≤x1/2

log p

p2= O(1).

Hence the first two assertions are equivalent, and we have proved the second inTheorem 4.1.1.

We’ll give a direct proof of the first statement as well. Let us compute∑n≤x

log n

in two different ways. First, we know from before that it is x log x+O(x). Onthe other hand,∑

n≤x

log n =∑n≤x

∑d|n

Λ(d) =∑d≤x

Λ(d)∑

n≤bx/dc

1

=∑d≤x

Λ(d)x

d+O

∑d≤x

Λ(d) · 1

= x∑d≤x

Λ(d)

d+O(x),

since 0 ≤ x/d− bx/dc < 1 and the remainder is by Chebyshev.Combining these we see that

x∑d≤x

Λ(x)

d+O(x) = x log x+O(x),

which when dividing by x gives us∑d≤x

Λ(d)

d= log x+O(1).

Finally, for the last statement, let f(t) = 1/ log t and

g(n) =

{log pp if n = p prime

0 otherwise.

Then∑p≤x

1

p=∑n≤x

f(n)g(n) =

∫ x

2−f(t) dMg(t) = f(t)Mg(t)

∣∣∣∣x2−−∫ x

2−f ′(t)Mg(t) dt

=1

log x(log x+O(1)) +O(1) +

∫ x

2−

1

t(log t)2(log t+O(1)) dt

= O(1) +

∫ x

2−

1

t log tdt+O

(∫ x

2−

1

t(log t)2dt

)= log log x+O(1),

since the second integral is bounded.


7.2 Multiplicative functions

Definition 7.2.1. A nonzero arithmetic function f is called multiplicative if,for (m,n) = 1, we have

f(mn) = f(m)f(n).

Moreover f is called completely multiplicative if f(mn) = f(m)f(n) for allm and n.

Hence if we write n =∏pαp , for a multiplicative function f ,

f(n) =∏p

f(pαp),

so f is completely determined by its value on prime powers. Similarly, for acompletely multiplicative function f ,

f(n) =∏p

f(p)αp ,

meaning that f is completely determined by its value on primes.Note also that

f(1) = f(1 · 1) = f(1)f(1),

whence f(1) = 1 for all multiplicative functions. Hence they always have ?-inverses.

In the sequel we will use the notation pα ‖ n to mean that pα | n butpα+1 - n, i.e. pα divides n precisely.

Proposition 7.2.2. Suppose f and g are multiplicative functions. Then f ? gand the ?-inverse of f are both multiplicative.

Proof. Let (m,n) = 1. There is a bijection

{ d | d | mn } ←→ { (d1, d2) | d1 | m, d2 | n }

by d 7→ (gcd(d,m), gcd(d, n)) and d1d2 ← [ (d1, d2). Under this bijection, wecompute

(f ? g)(mn) =∑d|mn

f(d)g(mnd

)=∑d1|md2|n

f(d1d2)g

(m

d1

n

d2

)

=∑d1|md2|n

f(d1)f(d2)g

(m

d1

)g

(n

d2

)

=

∑d1|m

f(d1)g

(m

d1

)∑d2|n

f(d2)g

(n

dd

)= (f ? g)(m)(f ? g)(n).

Now let h be the ?-inverse of f . First h(1) = 1/f(1) = 1, and further

h(n) = −∑d|nd>1

f(d)h(nd

).

DIRICHLET SERIES 19

Let (m,n) = 1. We want to show that h(mn) = h(m)h(n), which we willprove by induction. Suppose for all (m′, n′) = 1 with m′n′ < mn we haveh(m′n′) = h(m′)h(n′). Then

h(mn) = −∑d|mnd>1

f(d)h(mnd

)= −

∑d1|md2|n

d1d2>1

f(d1d2)h

(m

d1

n

d2

)

=∑d1|md2|n

d1d2>1

f(d1)f(d2)h

(m

d1

)h

(n

d2

)

= −

∑d1|n

f(d1)h

(n

d1

)∑d2|m

f(d2)h

(m

d2

)+ f(1)f(1)h(m)h(n)

= −(f ? h)(m)(f ? h)(n) + h(m)h(n) = h(m)h(n)

since the first two factors at the end are δ(m) and δ(n) respectively.

Lecture 8 Dirichlet series

8.1 Multiplicativity of Mobius function

Remark 8.1.1. The previous proposition is not true if we replace multiplicativitywith complete multiplicativity. In other words, for f and g completely multi-plicative, neither f ? g nor the ?-inverse of f need be completely multiplicative(though of course by the proposition they must be multiplicative).

Example 8.1.2. The constant function 1 is completely multiplicative. Thusd(n) = 1 ? 1(n) is multiplicative (but in fact not completely multiplicative).Similarly dk(n) is multiplicative.

In particular d(pα) = α+ 1 since the divisors of pα are p0, p1, . . . , pα.

Hence if n =∏pαp , then

d(n) =∏p

d(pαp) =∏p

(αp + 1). N

Example 8.1.3. Since µ is the ?-inverse of 1 it is multiplicative. In particular,to compute µ(pα), let us consider µ ? 1(pα) = δ(pα), which is 0 if α ≥ 1 and 1if α = 0. On the other hand

µ ? 1(pα) =∑d|pα

µ(d) =∑

0≤β≤α

µ(pβ).

For α = 0, µ(1) = 1. For α = 1, µ(1)+µ(p) = 0, so µ(p) = −1. Furthermorefor α = 2, µ(1)+µ(p)+µ(p2) = 0, but the first two terms add to 0, so µ(p2) = 0,and in general µ(pα) = 0 for all α ≥ 2.


DIRICHLET SERIES 20

Hence if n =∏pαp ,

µ(n) =∏p

µ(pαp) =

0 if αp ≥ 2 for some p, i.e. n is not square free

(−1)k if n = p1p2 · · · pk, pi distinct

1 if n = 1.

N

8.2 Dirichlet series

Let f be an arithmetic function. The Dirichlet series associated with f is

L(s, f) =

∞∑n=1

f(n)

ns, s ∈ C.

Remark 8.2.1. This series might not converse. We need some control of the sizeof f .

Definition 8.2.2. An arithmetic function f is of polynomial growth if itsatisfies one of the following equivalent conditions:

(i) There exists a constant A ∈ R, depending on f , such that |f(n)| = O(nA).

(ii) There exists σ ∈ R such that the series L(σ, f) is absolutely convergent.

In this case we let

σf := inf{σ ∈ R | L(σ, f) converges absolutely } ∈ R ∪ {−∞}.

We call σf the abscissa of convergence of L(σ, f).

That (i) and (ii) are equivalent is pretty clear; in the forward direction our

series looks like∑

nA

ns , so it converges absolutely for Re s > 1 + A, and we canmake a similar argument backwards. Hence |f(n)| = O(nσf−1).

Proposition 8.2.3. Let f be an arithmetic function with polynomial growth.Let σf be its abscissa of convergence. Then for all σ > σf , the series L(s, f)converges absolutely and uniformly in the halfplane { s ∈ C | Re(s) > σ }.

In this domain,

L′(s, f) =∑n≥1

− log(n)f(n)

ns= L(s,− log ·f),

which also has abscissa of convergence σf .

Proof. Let σ > σf , and consider Re(s) ≥ σ. Then |f(n)/ns| ≤ |f(n)|/nσ.Summing this over n ≥ 1,∑

n≥1

∣∣∣∣f(n)

ns

∣∣∣∣ ≤∑n≥1

|f(n)|nσ

<∞

since L(σ, f) converges absolutely by definition since σ > σf . This bound isuniform for any Re(s) ≥ σ.

Finally by the absolute convergence we can switch the limit process in theseries and the derivative, and hence differentiate termwise.

DIRICHLET SERIES 21

An arithmetic function f determines its Dirichlet series uniquely, and a con-verse result is true as well.

Lemma 8.2.4. Let f and g be two arithmetic functions of polynomial growth.Suppose L(s, f) = L(s, g) for all s in

{ s ∈ C | Re(s) > max{σf , σg } }.

Then f = g.

Hence the function is uniquely determined by its Dirichlet series.

Proof. Without loss of generality we can assume g = 0 (by replacing f by f −g,i.e. move L(s, g) to the other side). So we have L(s, f) = L(s, g) = 0 for alls ∈ { s ∈ C | Re(s) > σf } (since σg = −∞). Let h(n) = f(n)/nσf−1. Then

L(s, f) =∑n≥1

f(n)

ns=∑n≥1

f(n)/nσf−1

ns−σf+1= L(s− σf + 1, h).

Hence L(s, h) = L(s+ σf − 1, f).For Re(s) > 1, Re(s + σf − 1) > σf , whence σh ≤ 1, so h(n) = O(n1−1) =

O(1).Now suppose h(n) 6= 0 for some n, and consequently let n0 be the smallest

such n. For Re(s) > 2,

0 =∑n≥n0

h(n)

ns=h(n0)

ns0

1 +ns0

h(n0)

∑n≥n0+1

h(n)

ns

.

Note that both of the h(n0) terms are nonzero by choice of n0, so

1 +ns0

h(n0)

∑n≥n0+1

h(n)

ns= 0.

But∑n≥n0+1

h(n)

ns�

∑n≥n0+1

1

nRe(s)�∫ ∞n0+1

1

tRe(s)dt =

1

(Re(s)− 1)(n0 + 1)Re(s)−1.

Hence

1 +ns0

h(n0)

∑n≥n0+1

h(n)

ns= 1 +O

(n

Re(s)0

h(n0)

1

(Re(s)− 1)(n0 + 1)Re(s)−1

)= 0

so

1 +O

(1

Re(s)− 1

)= 0

which is a contradiction as Re(s)→∞ since the remainder term vanishes.

L-FUNCTIONS OF CONVOLUTIONS 22

Lecture 9 L-functions of convolutions

9.1 L-functions associated with Dirichlet convolutions

Theorem 9.1.1. Let f and g be arithmetic functions of polynomial growth withabscissae of convergence σf and σg respectively. Then σf?g ≤ max{σf , σg }.Moreover for Re(s) = σ > max{σf , σg } we have

L(s, f)L(s, g) = L(s, f ? g).

Proof. For Re(s) > max{σf , σg },

∑n≥1

∣∣∣∣f ? g(n)

ns

∣∣∣∣ =

∞∑n=1

|∑ab=n

f(a)g(b)|

nRe(s)≤∞∑n=1

∑ab=n

|f(a)||g(b)||ab|Re(s)

=

∞∑1=1

∑ab=n

|f(a)||g(b)||a|Re(s)|b|Re(s)

=

( ∞∑a=1

|f(a)||a|Re(s)

)( ∞∑b=1

|g(b)||b|Re(s)

)<∞

so σf?g ≤ max{σf , σg }.For Re(s) > max{σf , σg } we have absolute convergence, so we can rearrange

terms in the series as desired, so

L(s, f)L(s, g) =

∞∑m=1

f(m)

ms

∞∑k=1

g(k)

ks=

∞∑n=1

1

ns

∑mk=n

f(m)g(k)

=

∞∑n=1

f ? g(n)

ns= L(s, f ? g).

Corollary 9.1.2. Suppose f is ?-invertible. Let g be the ?-inverse of f . Assumeσf , σg <∞. Then for Re(s) > max{σf , σg }, we have L(s, f)L(s, g) = 1.

Proof. By the last theorem, since f?g = δ, which is supported only on n = 1.

In particular L(s, f) does not vanis on

{ s ∈ C | Re(s) > max{σf , σg } }.

Example 9.1.3. For f = 1, we have

L(s, 1) =

∞∑n=1

1

ns= ζ(s),

converging on Re(s) > 1, so σ1 = 1. The Mobius function µ is the ?-inverse of1, so

L(s, 1)L(s, µ) = ζ(s)

∞∑n=1

µ(n)

ns= 1.

We have |µ(n)| ≤ 1 = O(n0), so σµ ≤ 1.



In fact it is equal to 1; let s = 1, so

∞∑n=1

|µ(n)|ns

≥∑p

1

p=∞

so σµ = 1.Hence L(s, 1)L(s, µ) = 1 for Re(s) > 1, and ζ(s) 6= 0 for Re(s) > 1. Also

1

ζ(s)= L(s, µ) =

∞∑n=1

µ(n)

ns

for Re(s) > 1. N

Example 9.1.4. Let d(n) = 1 ? 1(n). Then

L(s, d) = L(s, 1 ? 1) = L(s, 1)L(s, 1) = ζ(s)2

for Re(s) > 1. In general

ζ(s)k =

∞∑n=1

dk(n)

ns

for Re(s) > 1. N

Example 9.1.5. Euler’s ϕ-function is defined by

ϕ(n) = #{ 1 ≤ m ≤ n | (m,n) = 1 }.

This is multiplicative (which follows immediately from the Chinese remaindertheorem), and ϕ(n) = µ ? Id(n), where Id(n) = n. It’s easy to check this onprime powers.

Now

L(s, Id) =

∞∑n=1

n

ns= ζ(s− 1),

so σId = 2, and

L(s, ϕ) = L(s, µ)L(s, Id) =ζ(s− 1)

ζ(s)

for Re(s) > max{σµ, σId } = max{ 1, 2 } = 2.Hence

ζ(s− 1)

ζ(s)=

∞∑n=1

ϕ(n)

ns

for Re(s) > 2. N

Example 9.1.6. The von Mangoldt function Λ(n) = log ?µ(n), so

L(s,Λ) = L(s, log)L(s, µ).

Now

L(s, 1) = ζ(s) =

∞∑n=1

1

ns,


and so long as Re(s) > 1, so that we have absolute convergence, we can differ-entiate termwise, whence

ζ ′(s) =

∞∑n=1

− log n

ns

meaning that L(s, log) = −ζ ′(s). Then

L(s,Λ) =−ζ ′(s)ζ(s)

=

∞∑n=1

Λ(n)

ns

for Re(s) > 1. Note also that the fraction is a fraction of holomorphic functions,the bottom nonvanishing, so it is holomorphic and has an antiderivative, so

−(log ζ(s))′ =

∞∑n=1

Λ(n)

ns

for Re(s) > 1. N

9.2 Dirichlet series and multiplicative functions

Theorem 9.2.1. Let f be a multiplicative function of polynomial growth. Thenfor Re(s) = σ > σf , we have

(i) For all primes p,

Lp(s, f) :=∑α≥0

f(pα)

pαs

converges absolutely and uniformly in Re(s) ≥ σ.

This Lp(s, f) is called the local factor of f at p.

(ii) We have

L(s, f) =∏p

Lp(s, f),

which is an infinite product, so technically we mean

limP→∞

∏p≤P

Lp(s, f),

which converges uniformly in Re(s) ≥ σ.

(iii) Conversely, let f be an arithmetic funciton f such that σf < ∞ andf(1) = 1. Suppose

L(s, f) =∏p

Lp(s, f)

for Re(s) sufficiently large. Then f is multiplicative.

Proof. (i) This is trivial: the local factors are subseries of an absolutely anduniformly convergent series.

EULER PRODUCTS 25

(ii) The intuition is that we can factor n into primes, and collecting them asappropriate, the two sides are equal. More precisely, for the convergence, letP ≥ 2 and write p1 < p2 < . . . < pk ≤ P , the set of primes less than or equal toP . Then ∏

p≤P

Lp(s, f) =∑

α1,...,αk≥0

f(pα11 ) · · · f(pαkk )

(pα11 · · · p

αkk )s

=∑

α1,...,αk≥0

f(pα11 · · · p

αkk )

(pα11 · · · p

αkk )s

=∑n≥1

p|n⇒p≤P

f(n)

ns

Hence ∣∣∣L(s, f)−∏p≤P

Lp(s, f)∣∣∣ ≤∑

n>P

|f(n)|ns

→ 0

as P →∞ since the left-hand side becomes the sum over n with all prime factorsexceeding P , which is a smaller set than n > P , and the right-hand side is thetail of a convergent sum.

Lecture 10 Euler products

10.1 L-series continued

Proof continued. (iii) Let f be the multiplicative function defined by

f(n) =∏pα‖n

f(pα).

For σ > σf , |f(n)|/nσ = o(1), so for n large enough, say n > N , we have|f(n)|/nσ < 1.

Hence

|f(n)|nσ

=∏pα‖n

|f(pα)|pασ

=∏pα‖npα≤N

|f(pα)|pασ

∏pα‖npα>N

|f(pα)|pασ

= O(1)

with the implied constant depending on f , since the first product is a finiteproduct, hence a constant, and the second product is o(1) by choice of N .Therefore L(s, f) converges absolutely for Re(s) > σ + 1.

For such s,

L(s, f) =∏p

Lp(s, f) =∏p

∑α≥0

pα

pαs=∏p

Lp(s, f) = L(s, f)

so for Re(s) sufficiently large, the L-functions are equal, whence by Lemma 8.2.4f = f , and so f is multiplicative since f is multiplicative by construction.


EULER PRODUCTS 26

Corollary 10.1.1. If f is completely multiplicative with σf < ∞, then forRe(s) > σf ,

L(s, f) =∏p

(1− f(p)

ps

)−1

.

Proof. This follows immediately from f(pα) = f(p)α when f is completely mul-tiplicative, so

Lp(s, f) =∑α≥0

f(pα)

pαs=∑α≥0

f(p)α

psα=

(1− f(p)

ps

)−1

,

the last sum being geometric.

We call such a product representation an Euler product .

Proposition 10.1.2. Let f be multiplicative with σf < ∞. Let σ > σf . Thenthere exists P > 0 such that∏

p>P

Lp(s, f) =L(s, f)∏

p≤PLp(s, f)

does not vanish in Re(s) ≥ σ.Thus the zeros of L(s, f) in Re(s) ≥ σ are exactly the zeros of the local

factors Lp(s, f), p ≤ P .

Proof. Let σ > σf . By the previous Theorem 9.2.1,

L(s, f) =∏p

Lp(s, f)

for Re(s) ≥ σ. The right-hand side is interpreted as a convergent limit, so thereexists some P > 0 such that for p > P ,

|Lp(s, f)− 1| < 1

2

for Re(s) ≥ σ. Hence Lp(s, f) 6= 0 for p > P , Re(s) ≥ σ, since it is boundedaway from zero.

Example 10.1.3. For Re(s) > 1,

ζ(s) =

∞∑n=1

1

ns=∏p

(1− 1

ps

)−1

,

which is never 0, so ζ(s) 6= 0 for Re(s) > 1. Correspondingly

1

ζ(s)=∏p

(1− 1

ps

)is also nonvanishing in Re(s) > 1. N

EULER PRODUCTS 27

Example 10.1.4. For Re(s) > 1,

L(s, d) =

∞∑n=1

d(n)

ns= ζ(s)2 =

∏p

(1− 1

ps

)−2

.

Let f be the ?-inverse of d. Then

L(s, f) =1

ζ(s)2=∏p

(1− 1

ps

)2

=∏p

(1− 2

ps+

1

p2s

). N

Example 10.1.5. Let ϕ be the Euler ϕ-function. We have

L(s, ϕ) =ζ(s− 1)

ζ(s)=∏p

1− p−s

1− p1−s

for Re(s) > 2. N

Let us now explore a slightly more esoteric example, but that serves a pointin finding zeros between σf and σg, where f and g are each other’s ?-inverses.

Example 10.1.6. Let µs be the multiplicative function f defined by

µ2(pα) =

1, if α = 0

0, if α ≥ 2

−1, if α = 1, p 6= 2

−4, if α = 1, p = 2.

In other words, it is the usual Mobius function, except we’ve modified it specif-ically on n = 2. Note that |µ2(n)| ≤ 4, so L(s, µ2) converges absolutely forRe(s) > 1; σµ2

≤ 1. Moreover at s = 1,

|L(1, µ2)| =∞∑n=1

|µ2(n)|ns

≥∑p 6=2

1

p=∞,

so in fact σµ2 = 1.Now a natural question to ask is what the zeros of L(s, µ2) are for Re(s) > 1.

We compute the local factors

L2(s, µ2) =∑α≥0

µ2(2α)

2αs= 1− 4

2s,

and for p 6= 2,

Lp(s, µ2) =∑α≥0

µ2(pα)

pαs= 1− 1

ps,

meaning that

L(s, µ2) =

(1− 4

2s

)∏p 6=2

(1− 1

ps

),

where the first factor is 0 if s = 2, and the remaining factors are never zero forRe(s) > 1. Hence the only zero of L(s, µ2) on Re(s) > 1 is s = 2.

PRIMES IN ARITHMETIC PROGRESSIONS 28

Now let µ(−1)2 be the ?-inverse of µ2. It is multiplicative, since µ2 is, and

µ(−1)2 (pα) =

{1, if p 6= 2

4α, if p = 2.

Let σ > 2. For Re(s) > σ,

|L(s, µ(−1)2 )| ≤

∑n≥1

|µ(−1)2 (n)|nσ

=

(1− 4

2s

)−1 ∏p 6=2

(1− 1

ps

)−1

=1− 1/2σ

1− 4/2σζ(σ).

We chose σ > 2 because

L2(s, µ(−1)2 ) =

∑α≥0

µ(−1)2 (2α)

2αs=∑α≥0

4α

2sα=∑α≥0

(4

2s

)α,

which is geometric and converges for Re(s) < 2, so σµ(−1)2≤ 2, and moreover

∑n≥1

|µ(−1)2 (n)|n2

≥∑α≥0

(4

22

)α=∞,

so σµ(−1)2

= 2.

Hence L(s, µ2)L(s, µ(−1)2 ) = 1 for Re(s) > 2, and between σµ2 = 1 and

σµ(−1)2

= 2 we find the one zero at s = 2. N

Lecture 11 Primes in arithmetic progressions

11.1 Arithmetic progressions

Definition 11.1.1. An arithmetic progression is an infinite subset of Zsatisfying the property that there exists an integer q > 0 such that the distancebetween any two consecutive integers of this subset is q.

This integer q is called the modulus of the arithmetic progression. Hencean arithmetic progression of modulus q is of the form

Lq,a = a+ qZ,

with a ∈ Z.

Note that if a ≡ b (mod q), then Lq,a = Lq,b. Therefore the arithmeticprogressions of modulus q are indexed by the residue classes modulo q, i.e.Z/qZ.

Theorem 11.1.2 (Dirichlet). Let a, q ∈ Z, q > 0, with (a, q) = 1. Then the set℘q,a = ℘ ∩ Lq,a is infinite, i.e. there exist infinitely many primes p such thatp ≡ a (mod q).

Remark 11.1.3. The condition (a, q) = 1 is necessary. Indeed if (a, q) > 1, thenthere exists at most one prime p ≡ a (mod q), and such a prime p = (a, q). Thisis easy to see: everywhing in the arithmetic progression is of the form a + kq,but out of all of them we can factor (a, q) by definition.

Hence the residue classes containing infinitely many primes are (Z/qZ)×.

Date: February 1st, 2019.

CHARACTERS 29

As with the infinitude of the primes leading to us wondering about the Primenumber theorem, a natural question to ask here is this: what is the density ofthe set ℘q,a? In the same vein, let

π(x; q, a) = #(℘q,a ∩ [1, x]) = #{ p ≤ x | p ≡ a (mod q) }.

Theorem 11.1.4 (Landau). Let a, q > 0 with (a, q) = 1. Then

π(x; q, a) =1

ϕ(q)π(x)(1 + o(1)) ∼ π(x)

ϕ(q)

∑ x

ϕ(q) log x

as x→∞, the last being the Prime number theorem.

Now in particular, by definition, |(Z/qZ)×| = ϕ(q), so the primes are equidis-tributed amongst the residue classes modulo q.

Lecture 12 Characters

12.1 Analogue of Merten’s theorem

Theorem 12.1.1. Let a, q > 0, (a, q) = 1. Then we have

(i)∑n≤x

n≡a (mod q)

Λ(n)

n=

1

ϕ(q)log x+O(1).

(ii)∑p≤x

p≡a (mod q)

log p

p=

1

ϕ(q)log x+O(1).

(iii)∑p≤x

p≡a (mod q)

1

p=

1

ϕ(q)log log x+O(1).

We will prove (i) later. Moreover (i) implies (ii) implies (iii) in exactly thesame way as in Merten’s theorem.

Remark 12.1.2. Note that this theorem implies Dirichlet’s theorem, since forweighted sums over the primes in arithmetic progressions to diverge, there mustbe infinitely many primes in them.

12.2 Characters of finite abelian groups

Let G be a finite abelian group with identity denoted by 1.

Definition 12.2.1. A character χ of G is a group homomorphism χ : G→ C∗.

Remark 12.2.2. Such a character χ has the following properties:

(i) χ(ab) = χ(a)χ(b) for every a, b ∈ G since it is a homomorphism.

(ii) χ(1) = 1.

Date: February 4th, 2019.

CHARACTERS 30

(iii) χ(a)m = χ(am), which in turn is χ(1) = 1 if m = |G|. Hence every χ(a)is an mth root of unity, and |χ(a)| = 1. Thus χ : G → S1 ⊂ C, the unitcircle.

This also means that χ(a) = χ(a−1).

Let G denote the set of all characters of G. Then G is a group with multi-plcation defined by χ1 · χ2(a) = χ1(a)χ2(a). We call G the dual group of G,and its identity element is the trivial character χ0(a) = 1 for all a ∈ G.

Proposition 12.2.3. Let H be a subgroup of the finite abelian group G. Thenevery character of H extends to a character of G.

Proof. We prove it by induction on the index [G : H] = |G/H|. If [G : H] = 1,then G = H and we are done.

Assume that it is true for every subgroup H ′ of G with [G : H ′] < [G : H].Take g ∈ G \H, and let n be the smallest positive integer such that gn ∈ H

(which exists since gn = 1 in G/H for some n). Then we have our characterχ : H → C, but what should we define χ(g) to be? That’s hard to say, butcertainly χ(gn) = t ∈ C is well-defined, since gn ∈ H. Let H ′ = 〈H, g〉.Now wn = t for some w (actually n options), so pick one of them and defineχ′ : H ′ → C∗ by χ′(h) = χ(h) for h ∈ H, and χ′(g) = w.

For each h′ ∈ H ′, we have h′ = hgk for some h ∈ H and k ∈ Z, so χ′(h′) =χ(h)wk. This is well-defined, and clearly it is a character on H ′ and restricts toχ on H.

Now H < H ′ < G, with the first subgroup strict, so [G : H ′] < [G : H], soby our inductive hypothesis, χ′ extends to G.

Remark 12.2.4. The restriction ρ : G → H defined by χ 7→ χ|H is a homomor-phism. The proposition implies that ρ is surjective.

It’s kernel is, by definition,

ker ρ = {χ ∈ G | χ|H = χ0 },

which is the same as G/H, so we have the short exact sequence

{ 1 } G/H G H { 1 }.ρ

Proposition 12.2.5. Let G be a finite abelian group. Then |G| = |G|.

Proof. If G = 〈g〉 is cyclic, then χ ∈ G is determined by χ(g). Since χ(g) isa |G|th root of unity, and each |G|th root of unity determines a character χ,|G| = |G|.

Suppose G is not cyclic. Then there exists a nontrivial cyclic subgroup H ofG, and as above we have the short exact sequence

{ 1 } G/H G H { 1 }.ρ

Hence |G| = |G/H| · |H|.Now suppose the proposition is true for all finite abelian groups of order less

than n = |G|. Then

|G| = |G/H| · |H| = |G/H| · |H| = |G|,

FOURIER ANALYSIS ON FINITE ABELIAN GROUPS 31

where in the penultimate step we use the inductive hypothesis for G/H and ourargument about H being cyclic for the second term.

Remark 12.2.6. One can show that there is an isomorphism between G and G,however it is not canonical—it depends on the choice of generators.

That said, if G is isomorphic to its dual G, then similarly G must be isomor-

phic to its dualˆG, and it turns out there is a canonical isomorphism between

that and G.

Lecture 13 Fourier analysis on finite abelian groups

13.1 Orthogonality of characters

Let G be a finite abelian group. Let g ∈ G. Then g defines a character of G by

ˆg : G→ C∗

by ˆg(χ) = χ(g). So we have a homomorphism ε : G→ ˆG defined by g 7→ ˆg.

Proposition 13.1.1. The homomorphism ε : G→ ˆG is an isomorphism.

Proof. We have |G| = |G| = | ˆG|. Hence it suffices to show that ε is injective,

i.e. if g 6= 1, then ˆg 6= 1, so ˆg is not the trivial character on G. In other words,there exists some χ ∈ G such that ˆg(χ) 6= 1, i.e. χ(g) 6= 1.

Now let H = 〈g〉 ⊂ G. Define χ : H → C∗ by χ(g) = ξ 6= 1, with ξ an |H|throot of unity. Now extending χ to G finishes the proof.

Theorem 13.1.2. Let G be a finite abelian group of order n. Then

(i)∑g∈G

χ(g) =

{n if χ = χ0

0 if χ 6= χ0

,

(ii)∑χ∈G

χ(g) =

{n if g = 1

0 if g 6= 1.

Proof. For the first one, if χ = χ0, then∑g∈G

χ0(g) =∑g∈G

1 = n.

If χ 6= χ0, then there exists some a ∈ G with χ(a) 6= 1. Multiplying the sum byχ(a) we have

χ(a)∑g∈G

χ(g) =∑g∈G

χ(a)χ(g) =∑g∈G

χ(ag) =∑g∈G

χ(g)

since χ is a homomorphism, and moreover ag, as g runs over G, is just permutingthe arguments. Hence

(χ(a)− 1)∑g∈G

χ(g) = 0,


FOURIER ANALYSIS ON FINITE ABELIAN GROUPS 32

but by choice of a the parenthesis is nonzero, so the sum is 0.For the second orthogonality relation, note that∑

χ∈G

χ(g) =∑χ∈G

ˆg(χ),

so it reduces to the first case.

13.2 Fourier analysis on finite abelian groups

Let G be a finite abelian group. Let C(G) = { f : G → C }, which is a vectorspace over C.

A basis for C(G) is given by B = { δg | g ∈ G } where

δg(a) =

{1 if a = g,

0 if a 6= g.

Hence dimC C(G) = |G|, and we can write any f ∈ C(G) as

f(a) =∑g∈G

f(g)δg(a).

We can also define the Hermitian inner product 〈·, ·〉 on C(G) by

〈f, h〉 =1

|G|∑g∈G

f(g)h(g).

Then B is an orthogonal basis since

〈δg, δa〉 =1

|G|∑h∈G

δg(h)δa(h) =

{1|G| if g = a

0 if g 6= a.

Remark 13.2.1. All of these properties hold for any finite set G. In other words,we have not used the group structure of G.

In order to instead take advantage of the group structure of G, let χ1, χ2 ∈ G.Then

〈χ1, χ2〉 =1

|G|∑g∈G

χ1(g)χ2(g) =

{1 if χ1 = χ2

0 if χ1 6= χ2

by the orthogonality relation, since χ1(g)χ2(g) = (χ1χ2)(g) is another character,which is trivial if and only if they’re the same.

Now |G| = |G| = dimC C(G) so G is an orthonormal set spanning C(G), andhence an orthonormal basis.

This means that we also have a Fourier transform : for f ∈ C(G), f : G→C∗ by

f(χ) = 〈f, χ〉 =1

|G|∑g∈G

f(g)χ(g).

By orthogonality we then have the Fourier expansion

f(g) =∑χ∈G

f(χ)χ(g).

DIRICHLET CHARACTERS 33

Moreover we have the Parseval-Plancherel formula

‖f‖2 =1

|G|‖f‖2.

13.3 Characters on cyclic groups

Let G = 〈g〉 be a finite cyclic group of order n. Note that a character χ ∈ G iscompletely determined by χ(g); i.e., χ(g) = ξ is an nth root of unity and eachnth root of unity determined a character.

Proposition 13.3.1. Let G = 〈g〉 be a finite cyclic group of order n. Let µnbe the set of all nth roots of unity. Then ϕ : µn → G defined by ξ 7→ χξ, whereχξ(g), is an isomorphism.

Example 13.3.2. A finite cyclic group G of order n is isomorphic to Z/nZ =〈1〉, seen as an additive group. The characters of G are ψm where

ψm(1) = e2πim/n,

for m = 0, 1, 2, . . . , n− 1, noting that µn is a cyclic group of order n generatedby ξ = e2πi/n. N

13.4 Dirichlet characters

Definition 13.4.1. Let q ≥ 1 be an integer. The characters of the abeliangroup (Z/qZ)× are called Dirichlet characters of modulus q. The trivialcharacter is denoted by ξ0. This defines χ : (Z/qZ)× → C∗; we extend it toZ/qZ and then to Z by taking χ(a) = 0 if (a, q) 6= 1. In other words

χ : Z→ Z/qZ→ C∗

by

χ(n) =

{χ(n mod q) if (n, q) = 1

0 if (n, q) 6= 1.

Note two things: there are precisely ϕ(q) Dirichlet characters to the modulusq, and for the record (Z/qZ)× is not usually cyclic: it is cyclic only when q is2, 4, pk for p 6= 2, or 2pk (this is when there exist primitive roots modulo q).

Lecture 14 Dirichlet characters

14.1 L-functions attached to Dirichlet characters

Remark 14.1.1. Dirichlet characters χ are completely multiplicative, i.e., χ(mn) =χ(m)χ(n) for all m,n ∈ Z. Hence the Dirichlet series associated with χ has anEuler product

L(s, χ) =

∞∑n=1

χ(n)

ns=∏p

(1− χ(p)

ps

)−1

for Re(s) > 1 since |χ(n)| ≤ 1.



Proposition 14.1.2. Let χ (mod q) be a Dirichlet character.

(i) If χ = χ0, we have for Re(s) > 1,

L(s, χ0) =∏p|q

(1− 1

ps

)ζ(s)

which has an analytic continuation to Re(s) > 0 with a simple pole ats = 1.

(ii) If χ 6= χ0, then L(s, χ) converges uniformly on compact subsets in Re(s) >0 and thus it is a holomorphic function in Re(s) > 0.

Moreover we have for Re(s) = σ > 0 and X ≥ 2,

L(s, χ) =∑n≤X

χ(n)

ns+O

(q|s|σx−σ

),

with the convergence here being conditional.

Proof. (i) If χ = χ0, then

L(s, χ0) =∏p

(1− χ0(p)

ps

)−1

,

where χ0(p) is 1 on primes (p, q) = 1 and otherwise 0, so

L(s, χ0) =∏

(p,q)=1

(1− 1

ps

)−1

=∏p|q

(1− 1

ps

)∏p

(1− 1

ps

)−1

,

where the last factor is ζ(s). The first product is finite and defined for all s ∈ C.Hence the analytic continuation and the simple pole follows from same of ζ(s).

(ii) If χ 6= χ0, by orthogonality ∑a≤n<a+q

χ(n) = 0,

so if we letMχ(t) =

∑1≤n≤t

χ(n),

then |Mχ(t)| ≤ q uniformly, since each interval of length q is 0, and only the tailend can contribute at most q nonzero terms. (Indeed we can do sligthly better,in terms of ϕ(q), but it’s not necessary for our purposes.)

Hence∑1≤n≤X

χ(n)

ns=

∫ X

1−

1

tsdMχ(t) =

Mχ(t)

ts

∣∣∣∣X1−

+ s

∫ X

1−Mχ(t)t−s−1 dt.

For Re(s) = σ > 0, the first term is∣∣∣∣Mχ(X)

Xs

∣∣∣∣ ≤ q

Xσ→ 0


as X →∞. For the integral, consider the tail∣∣∣∫ ∞X

Mχ(t)t−s−1 dt∣∣∣ ≤ |s|q

σXσ→ 0

as X → ∞. Hence L(s, χ) converges uniformly on compact subsets in Re(s) >0.

Remark 14.1.3. Note that this is essentially the proof of Dirichlet’s test forconvergence; if ∣∣∣∑

n≤N

an

∣∣∣ < M

uniformly, and bn decreases monotonically to 0, then∑n≥0

anbn

converges.

With this we are finally equipped to prove Merten’s theorem for arithmeticprogressions.

Proof. We want to prove∑n≤X

n≡a (mod q)

Λ(n)

n=

1

ϕ(q)logX +O(1).

Now to filter the terms n ≡ a (mod q) out of the sum over all n, we average thesum over all characters χ (mod q), since

∑χ (mod q)

χ(b) =

{0 if b 6≡ 1 (mod q)

ϕ(q) if b ≡ 1 (mod q).

Hence ∑n≤X

n≡a (mod q)

Λ(n)

n=∑n≤X

Λ(n)

n

1

ϕ(q)

∑χ (mod q)

χ(a)χ(n),

since the latter sum is ϕ(q) if a ≡ n (mod q), and 0 otherwise. We rewrite thissum as

1

ϕ(q)

∑χ (mod q)

χ(a)∑n≤X

Λ(n)χ(n)

n=

1

ϕ(q)

∑χ (mod q)

χ(a)Sχ(X).

We split this into two sums, one for χ0 and one for χ 6= χ0. If χ = χ0,

Sχ0(X) =∑n≤X

Λ(n)χ0(n)

n=

∑n≤X

(n,q)=1

Λ(n)

n=∑n≤X

Λ(n)

n−∑p|q

log p∑α≥1pα≤X

1

pα.

Looking carefully at the last sum, the innermost sum is geometric and boundedby 1/(p(1−1/p)) ≤ 1, and with that bound in mind the sum over p | q becomesbounded by log q by bringing the sum inside the logarithm as a product. Hence

Sχ0(X) = logX +O(1) +O(log q) = logX +Oq(1)

MERTENS’ THEOREM FOR ARITHMETIC PROGRESSIONS 36

by Merten’s regular theorem.Now it remains to show that for χ 6= χ0,

Sχ(X) =∑n≤X

Λ(n)χ(n)

n= Oq(1).

Lecture 15 Mertens’ theorem for arithmetic pro-gressions

15.1 Proof finished

Proof continued. Carrying on from last time, we wish to show that for χ 6= χ0,Sχ(X) = Oq(1). To this end, consider

Tχ(X) =∑n≤X

(log n)χ(n)

n.

By Dirichlet’s test for convergence, this is Oq(1).Recall that

log n =∑d|n

Λ(d) =∑ab=n

Λ(a).

This means

Tχ(X) =∑n≤X

( ∑ab=n

Λ(n))χ(n)

n=∑a≤X

Λ(a)χ(a)

a

∑b≤X/a

χ(b)

b.

Looking at these carefully, we recognise the first sum as Sχ(X) and the innersum is L(1, χ) +O(qa/X) by Proposition 14.1.2. Hence the above is

Tχ(X) = Sχ(X)L(1, χ) +O

∑a≤x

Λ(a)χ(a)

a

qa

X

,

and the error term is� q

X

∑a≤X

Λ(a)� q

XX = q

where the second step is Chebyshev’s theorem.Hence

Tχ(X) = Sχ(X)L(1, χ) +Oq(1),

where the left-hand side is Oq(1) too, so if L(1, χ) 6= 0, then Sχ(X) = Oq(1).


MERTENS’ THEOREM FOR ARITHMETIC PROGRESSIONS 37

15.2 Dirichlet L-functions at 1

This is one of Dirichlet’s results:

Theorem 15.2.1 (Dirichlet). Let χ (mod q) be a nontrivial Dirichlet character.Then L(1, χ) 6= 0.

There are essentially two proofs of this, one using real analysis and one usingcomplex analysis. We will discuss the second here.

Proof. Consider

Lq(s) =∏

χ (mod q)

L(s, χ) = L(s, χ0)∏χ 6=χ0

L(s, χ).

Note that this is a finite product since there are finitely many Dirichlet charac-ters. If we enumerate those Dirichlet characters as χ0, χ1, . . . , χϕ(q)−1, we havein other words

Lq(s) = L(s, χ0 ? χ1 ? · · · ? χϕ(q)−1),

so

Lq(s) =

∞∑n=1

aq(n)

ns

with aq(n) = χ0 ?χ1 ? · · ·?χϕ(q)−1(n), with abscissa of convergence σq ≤ 1 sinceit is bounded by the maximum of σχ for all χ, all of which are 1.

We will show two properties later, which we will use here:

1. aq(n) ≥ 0 for all n ≥ 1, and

2. aq(nϕ(q)) ≥ 1 for all (n, q) = 1.

Evaluating our Lq(s) at σ = 1/ϕ(q), we get

∞∑n=1

aq(n)

nσ≥

∑(n,q)=1

aq(nϕ(q))

n≥

∑(n,q)=1

1

n=∞,

so σq ≥ 1/ϕ(q). Hence1

ϕ(q)≤ σq ≤ 1.

This contradicts the following lemma of Landau.

Lemma 15.2.2 (Landau). Let

L(s) =∑n≥1

f(n)

n

be a Dirichlet series with σf <∞ and f(n) ≥ 0 for all n ∈ N. Then L(s) doesnot admit a holomorphic continuation in a neighbourhood of s = σf .

The contradiction is as follows: By Landau’s theorem, Lq(s) does not admita holomorphic continuation in some neighbourhood of s = σf . But by Propo-sition 14.1.2, Lq(s) admits a meromorphic continuation to Re(s) > 0 with theonly possible simple pole at s = 1 (from χ0).

However if we suppose L(1, χ) = 0 for some χ 6= χ0, then we no longer havea pole at s = 1, so Lq(s) is holomorphic on Re(s) > 0, which contradicts withLandau’s lemma.

LANDAU’S LEMMA 38

Lecture 16 Landau’s lemma

We’ll finish up the proof of Dirichlet’s theorem by proving the three outstandingdetails: Landau’s lemma, and our two conditions on the coefficients aq(n).

16.1 Landau’s lemma

Lemma 16.1.1 (Landau). Let

L(s) =∑n≥1

f(n)

ns

be a Dirichlet series with σf <∞, and f(n) ≥ 0 for all n ∈ N. Then L(s) doesnot admit a holomorphic continuation in a neighbourhood of s = σf .

Proof. Without loss of generality, assume σf = 0 (else replace L(s) by L(s−σf )).Suppose that L(s) has a holomorphic continuation in an open disk D centred atσf = 0. Thus L(s) and all of its derivatives are holomorphic in D∪{ s |Re(s) >0 }. We will show that L(σ) converges for some σ ∈ D with σ < 0, whichcontradicts σf = 0, since f(n) ≥ 0 means that this convergence is absolute atσ.

First, we claim that ∑n≥1

f(n) = L(0).

In other words, what we naively would like the L(0) to be is, in fact, correct.Since f(n) ≥ 0,

L(σ) =∑n≥1

f(n)

nσ

is increasing as σ → 0+, and it is also bounded above by L(0). So by themonotone convergence theorem, L(σ)→ L(0) as σ → 0+. Now consider

∑n≤n

f(n) = limσ→0+

∑n≤N

f(n)

nσ≤ limσ→0+

∞∑n=1

f(n)

nσ= L(0).

On the other hand, for σ > 0,

L(σ) =∑n≥1

f(n)

nσ≤∑n≥1

f(n),

and as σ → 0+, the left-hand side goes to L(0), so putting this together

L(0) ≤∑n≥1

f(n) ≤ L(0),

so they’re equal.Secondly, we claim that L(σ) converges for some σ < 0, σ ∈ D.


LANDAU’S LEMMA 39

First, by assumption, L is holomorphic here. For Re(s) > 0,

L(k)(s) = (−1)k∑n≥1

f(n)(log n)k

ns.

Now importantly f(n)(log n)k ≥ 0, so the above argument again means that

L(k)(0) = (−1)k∑n≥1

f(n)(log n)k.

For σ ∈ D with σ < 0, we have the Taylor expansion

L(σ) =

∞∑k=0

L(k)(0)

k!σk =

∞∑k=0

(−1)k∑f(n)(log n)k

k!σk.

Switching the order of summation, which we can do since −σ ≥ 0, so we haveabsolute convergence and can rearrange the sum, we get

L(σ) =∑n≥1

f(n)∑k≥0

(−σ log n)k

k!=∑n≥1

f(n) exp(−σ log n) =∑n≥1

f(n)

nσ<∞.

With this done, we need to establish two things about the coefficients aq(n)of

Lq(s) =∏

χ (mod q)

L(s, χ) =

∞∑n=1

aq(n)

ns.

The two items it remains to show is that first, aq(n) ≥ 0 for all n, and second,aq(n

ϕ(q)) ≥ 1 for (n, q) = 1.Since aq(n) is multiplicative (being the convolution of many (completely)

multiplicative characters χ), it suffices to show aq(pk) ≥ 0 for all primes p,

k ≥ 0.We have

(Lq)p(s) =∏

χ (mod q)

Lp(s, χ) =∏χ

(1− χ(p)

ps

)−1

=∑k≥0

aq(pk)

pks.

Now if p | q¡ then (Lq)p(s) = 1, so

aq(pk) =

{1 if k = 0

0 if k 6= 0,

which is definitely nonnegative.If (p, q) = 1, then for notational convenience let z = p−s. We have |z| < 1/p

for Re(s) > 1. Let

E(z) =∏χ

(1− χ(p)z)−1 =∑k≥0

aq(pk)zk.

Taking logarithms,

logE(z) =∑χ

− log(1− χ(p)z) =∑χ

∑k≥1

χ(p)kzk

k

RIEMANN’S MEMOIR 40

where we’ve Taylor expanded the logarithm, since |z| < 1/p < 1. Switching theorder of summation, this is

logE(z) =∑k≥1

zk

k

∑χ

χ(pk) = ϕ(q)∑k≥1

pk≡1 (mod q)

zk

k.

Note that this is a power series with nonnegative coefficients, so taking expo-nentials to retrieve E(z), we get

E(z) = exp(logE(z)) = 1 + logE(z) +(logE(z))2

2!+ . . . ,

whence the coefficients of E(z) are also nonnegative, so aq(pk) ≥ 0.

For the second property, consider n = pm. We want aq(pmϕ(q)) ≥ 1, with

(p, q) = 1. Let ep be the order of p in (Z/qZ)×, i.e. pϕ(q) ≡ 1 (mod q). Thenpk ≡ 1 (mod q) if and only if ep | k, by Lagrange’s theorem. Hence ep | ϕ(q).

In the above computation, then,

logE(z) =ϕ(q)

ep

∑k

zepk

k=ϕ(q)

ep(− log(1− zep)) = log

((1− zep)−ϕ(q)/ep

),

so

E(z) = (1− zep)−ϕ(q)/ep =

(1

1− zep

)ϕ(q)/ep

=(1 + zep + z2ep + . . .

)ϕ(q)/ep.

The exponent is an integer, and ep | mϕ(q), so if h = mϕ(q), aq(pk) is the kth

coefficient, and hence aq(pk) ≥ 1.

Lecture 17 Riemann’s memoir

17.1 Riemann’s memoir

In 1860 Riemann published his one and only paper on number theory. In it heproved two results:

1. The Riemann zeta function ζ(s) has a meromorphic continuation to all ofC. It is holomorphic except for s = 1, which is a simple pole with residue1.

2. ζ(s) satisfies the functional equation

π−12 sΓ

(s2

)ζ(s) = π

12 (1−s)Γ

(1

2(1− s)

)ζ(1− s),

which is invariant under the transformation s 7→ 1− s.

Riemann further made several remarkable conjectures:

Date: February 22nd, 2019.

RIEMANN’S MEMOIR 41

1. ζ(s) has infinitely many zeros in the critical strip 0 < Re(s) < 1.

Note that the distribution of zeros is symmetric with respect to the realaxis and Re(s) = 1/2.

2. The number N(T ) of zeros of ζ(s) in the critical strip with 0 < Im(s) < Tsatisfies

N(T ) =T

2πlog

T

2π− T

2π+O(log T ),

proved by von Mangoldt in 1895. We will show N(T ) ∼ T log T .

3. The integral function

ζ0(s) = s(s− 1)π−12 sΓ

(s2

)ζ(s)

has the product representation

ζ0(s) = eA+Bs∏ρ

(1− s

ρ

)es/ρ,

where A and B are constants and ρ runs through the zeros of ζ(s) in thecritical strip. This was proved by Hadamard in 1893, and is a special caseof the Hadamard factorisation theorem. (Note that s− 1 comes from thepole, and s comes from the wish to make it invariant under s 7→ 1− s.)

4. There exists an explicit formula for π(x) − Li(x) with x > 1, involving asum over the zeroes of ζ(s). This was proved by von Mangoldt in 1895.

5. Finally, and most importantly, the Riemann hypothesis: the zeros ofζ(s) in the critical strip all lie on the critical line Re(s) = 1/2.

This, of course, remains unproven. There has been some progress, themost important of which is: Hardy, in 1914, showed that there are in-finitely many zeros on Re(s) = 1/2. (His argument also generalises toL(s, χ), and moreover to GL2 L-functions, but for higher dimensions itremains unknown.)

Then, in 1942, Selberg showed that there is a positive proportion of zerosof ζ(s) in the critical strip lying on the critical line. This is where the(very powerful) mollifier method comes from.

17.2 Review of Fourier analysis

Definition 17.2.1. Let (G,+) be an abelian topological group, i.e. (G,+) isan abelian group, G is a topological space, and moreover

G×G→ G

(a, b) 7→ a+ b

and

G→ G

a 7→ −a

POISSON SUMMATION 42

are continuous.A character ψ of G is a continuous group homomorphism ψ : G→ C×. Let

Hom(G,C×) denote the set of all such characters. It is an abelian group underthe multiplication of functions.

Definition 17.2.2. A character ψ is called unitary if ψ : G → S1 = { z ∈C× | |z| = 1 }. Let G denote the subgroup of Hom(G,C×) consisting of unitarycharacters.

Remark 17.2.3. If G is compact, then every character is unitary.

Proof. Let ψ : G → C× be a character. Then its image ψ(G) is compact sinceG is compact and ψ, by definition, is continuous. Hence ψ(G) is closed andbounded, since this characterises compact sets in C.

Now suppose there exists some ψ(g) = z with |z| 6= 1. Then ψ(gn) =ψ(g)n = zn ∈ ψ(G). Since |z| 6= 1, we have two options: if it exceeds 1, then asn increases, zn diverges to infinity, which contradicts ψ(G) being bounded. If|z| < 1, then zn converges to 0, and this contradicts ψ(G) being closed.

In the sequel we will write e(x) := exp(2πix).

Theorem 17.2.4. The map ϕ : (C,+)→ Hom(R,C×) defined by y 7→ ϕy, withϕy(x) := e(yx) is an isomorphism of groups. The restriction of ϕ to R gives an

isomoprhism of groups (R,+) ∼= R.

Lecture 18 Poisson summation

18.1 Fourier analysis

Proof. That this is a homomorphism is clear by definition.To see that it is injective, let y ∈ kerϕ. Then ϕy(x) = e(xy) = 1 for every

x ∈ R, which naturally requires y = 0.For surjectivity, given ψ ∈ Hom(R,C×) we need to show that ψ(x) = e(xy)

for some y ∈ C. To this end, let

Ψ(x) =

∫ x

0

ψ(t) dt.

Then

Ψ(x+ y) =

∫ x+y

0

ψ(t) dt =

∫ x

0

ψ(t) dt+

∫ x+y

x

ψ(t) dt.

In the latter integral, shift t 7→ t+ x, so∫ x+y

x

ψ(t) dt =

∫ y

0

ψ(t+ x) dt =

∫ y

0

ψ(t)ψ(x) dt

since ψ is a homomorphism. Hence the above becomes

Ψ(x+ y) =

∫ x

0

ψ(t) dt+ ψ(x)

∫ y

0

ψ(t) dt.



In other words,Ψ(x+ y) = Ψ(x) + ψ(x)Ψ(y).

We now fix y such that Ψ(y) 6= 0. This is possible since Ψ′(t) = ψ(t) isn’tidentically zero.

Taking derivatives with respect to x, this results in

ψ(x+ y) = ψ(x) + ψ′(x)Ψ(y),

where again we use the fact that ψ is a homomorphism to rewrite this as

ψ(x)ψ(y) = ψ(x) + ψ′(x)Ψ(y).

Rearranging this becomes

Ψ(y)ψ′(x) + (1− ψ(y))ψ(x) = 0,

which is a first order ordinary differential equation, and

ψ(x) = C exp(ux)

for some constants C, u ∈ C. Now ψ(0) = 1 so C = 1, i.e. ψ(x) = exp(ux), sowe have surjectivity!

Note that ϕy(x) is unitary if and only if y ∈ R, since e(xy) = exp(2πixy),and x ∈ R.

Corollary 18.1.1. The map

(Z,+)→ Hom(R/Z,C×) = R/Z

defined by n 7→ ϕn, ϕn(x) = e(nx), is an isomorphism of groups.

Proof. First, note that Hom(R/Z,C×) = R/Z since R/Z is compact.The injectivity part is identical to what we did for the previous theorem.For the onto part, consider ψ : R/Z→ C×. Extend it to ψ : R→ R/Z→ C×

periodically, i.e. ψ∣∣Z ≡ 1. Using the previous theorem., ψ = ϕy with y ∈ R by

unitarity. Moreover ψ(1) = ϕy(1) = e(y) = 1 since 1 ∈ Z, so y ∈ Z.

Definition 18.1.2. A function f : R→ C is said to be in the Schwartz classS(R) if f and all of its derivatives are rapidly decaying, meaning that for allA ≥ 0 and j ≥ 0,

f (j)(x)� (1 + |x|)−A.

In other words, f and its derivatives go to 0 very quickly.

Definition 18.1.3. Let f ∈ S(R). The Fourier transform of f is the function

f(y) =

∫Rf(x) e(−yx) dx,

for y ∈ R.


Remark 18.1.4. Because of the rapid decay, f(y) converves absolutely and uni-

formly for all y ∈ R. Hence f is infinitely differentiable with

f (j)(y) =

∫R

(−2πix)jf(x)e(−yx) dx.

Integrating this by parts, for y 6= 0, we have

f (j)(y) =1

2πiy

∫Rf ′(x)e(−yx) dx =

1

2πiyf ′(y) =

(1

2πiy

)jf (j)(y).

So for |y| ≥ 1,

f(y)� |y|−j∫R|f (j)(x)| dx� |y|−j

for every j ≥ 0. Hence

Proposition 18.1.5. The Fourier transform is a linear map from S(R) toS(R).

Let us note a few basic properties of the Fourier transformation. Let f ∈S(R).

1. Let h ∈ R, and g(x) = f(x+ h). Then g(y) = e(−hy)f(y).

2. f (j)(y) = (2πiy)j f(y).

3. Let λ ∈ R, λ 6= 0, and g(x) = f(λx). Then g(y) = 1|λ| f(y/λ).

4. The Fourier inversion formula saysˆf(x) = f(−x).

5. The Fourier transform of the Gaussian: if f(x) = e−πx2

, then f(y) = f(y).

One particular and very important property is

Theorem 18.1.6 (Poisson summation formula). Let f ∈ S(R). Then for x ∈R, ∑

n∈Zf(x+ n) =

∑n∈Z

f(n) e(nx).

Proof. Let

g(x) =∑n∈Z

f(x+ n).

Since f ∈ S(R), g(x) converges absolutely and uniformly for all x ∈ R. Hencewe can rearrange terms, differentiate or integrate under the summation, and soon, and g ∈ C∞(R), and it is periodic with period 1 by construction. Henceg(x) has a Fourier expansion,

g(x) =∑n∈Z

a(n) e(nx),

THE FUNCTIONAL EQUATION FOR ζ(S) 45

with

a(n) =

∫ 1

0

g(x) e(−nx) dx =

∫ 1

0

∑m∈Z

f(x+m) e(−nx) dx

=∑m∈Z

∫ 1

0

f(x+m) e(−n(x+m)) dx =∑m∈Z

∫ m+1

m

f(t) e(−nt) dt

=

∫Rf(t) e(−nt) dt = f(n).

Henceg(x) =

∑n∈Z

f(n) e(nx).

Corollary 18.1.7. For f ∈ S(R),∑n∈Z

f(n) =∑n∈Z

f(n).

Proof. Take x = 0 in the previous theorem.

This result is true in general for averaging over any lattice.

Lecture 19 The functional equation for ζ(s)

19.1 Mellin transformation

An important corollary to the Poisson summation formula is the following, doingthe same kind of averge but over an arithmetic sequence,

Corollary 19.1.1. Let q, a ∈ Z with q ≥ 1. For f ∈ S(R),∑n≡a (mod q)

f(n) =1

q

∑n∈Z

f

(n

q

)e

(qn

q

).

Proof. First note that ∑n≡q (mod q)

f(n) =∑n∈Z

f(qk + q).

Let g(x) = f(qx+ a), an affine shift. By the properties of the Fourier transfor-mation,

g(y) =1

qf

(y

q

)e

(ay

q

),

so ∑n∈Z

f(qk + q) =∑n∈Z

g(n) =∑n∈Z

g(n) =1

q

∑n∈Z

f

(n

q

)e

(an

q

).



Definition 19.1.2. Let f(x) ∈ C∞(R≥0) with rapid decay as x → ∞. TheMellin transform of f is

f(s) =

∫ ∞0

f(x)xsdx

x,

for s ∈ C, Re(s) > 0.

The integral converges uniformly on compact subsets of { s ∈ C |Re(s) > 0 }since f decays rapidly, and hence f(s) is a holomorphic function in Re(s) > 0.

Moreover, note that if f(x) = O(xN ) as x→ 0, then f(s) defines a holomor-phic function in the domain Re(s) > −N , by just studying the power of x inthe integrand.

Like the Fourier transform, the Mellin transform has an inversion formula:Suppose f ∈ S(R). For any y > 0,

f(y) =1

2πi

∫(σ)

f(s)y−s ds

for any σ > 0, where (σ) means the line σ + it for −∞ < t <∞.

Remark 19.1.3. The Mellin inversion formula and the Fourier inversion formulaare equivalent.

19.2 Gamma function

The Gamma function is defined as the Mellin transform of e−x, i.e.

Γ(s) :=

∫ ∞0

e−xxsdx

x,

for Re(s) > 0.This function, which turns out to be remarkably important, has many fas-

cinating properties. For instance,

1. Γ(s) admits a meromorphic continuation to C, which has simple poles ats = −n, n = 0, 1, 2, . . . with residues (−1)n/n!, and no other poles;

2. Γ(s+ 1) = sΓ(s);

3. Γ(s)Γ(1− s) = πsin(πs) ;

4. Γ(s)Γ(s+ 12 ) = 2

√π

22s Γ(2s);

5. Γ( 12 ) =

√π, and Γ(k + 1) = k! for k = 0, 1, 2, . . . (which is a consequence

of property 2);

6. Γ(s) has no zeros, i.e. Γ(s) 6= 0 for all s ∈ C; and finally (and incrediblyusefully),

7. Sterling’s formula: For any s > 0, −π + δ < arg s < π − δ,

log Γ(s) =

(s− 1

2

)log s− s+

1

2log(2π) +O

(1

|s|

)as |s| → ∞.


19.3 The functional equation for ζ(s)

Theorem 19.3.1. Letξ(s) = π−

s2 Γ(s

2

)ζ(s)

for Re(s) > 1. Then ξ(s) has a meromorphic continuation to C, holomorphicon C \ { 0, 1 }, and has simple poles as s = 0 ans s = 1 with residues −1 and 1respectively.

Moreover it satisfies the functional equation

ξ(s) = ξ(1− s).

Proof. By the definition of Γ,

Γ(s

2

)=

∫ ∞0

e−tt12 tdt

t.

Let t = n2πx. Then

π−12 sΓ

(s2

)n−s =

∫ ∞0

e−n2πxx

12 sdx

x.

Summing this over n, so as to develop a ζ(s) on the left-hand side, this becomes

π−s2 Γ(s

2

)ζ(s) =

∞∑n=1

∫ ∞0

e−n2πxx

12 sdx

x=

∫ ∞0

( ∞∑n=1

e−n2πx

)x

12xdx

x,

where switching the sum and integration is permitted since the rapid decay ofthe exponential gives us uniform convergence.

Let

w(x) =

∞∑n=1

e−n2πx,

so that

ξ(s) =

∫ ∞0

w(x)x12 sdx

x=

∫ 1

0

w(x)x12 sdx

x+

∫ ∞1

w(x)x12 sdx

x.

Riemann’s trick was splitting this integral into two, and now performing thechange of variables x 7→ 1/x in the first one

ξ(s) =

∫1

w

(1

x

)x−

12 sdx

x+

∫ ∞1

w(x)x12 sdx

x.

In order to make sense of w(x), we wish to extend it to a sum over all n ∈ Z,rather than n ≥ 1, so that we may use Poisson summation on it. To this end,

Definition 19.3.2. The theta series θ(x) is defined by

θ(x) =∑n∈Z

e−n2πx = 1 + 2

∑n≥1

e−n2πx = 1 + 2w(x),

so

w(x) =1

2(θ(x)− 1).

THE FUNCTIONAL EQUATION FOR L-FUNCTIONS 48

We’ll prove the following lemma for the theta series in a moment:

Lemma 19.3.3. θ(

1x

)= x1/2θ(x) for x > 0.

Taking the lemma for granted just now, we then get

w

(1

x

)=

1

2

(θ

(1

x

)− 1

)=

1

2(x1/2θ(x)− 1) =

1

2(x1/2(1 + 2w(x))− 1)

= −1

2+

1

2x1/2 + x1/2w(x).

Hence ∫ ∞1

w

(1

x

)x−

12 sdx

x=

∫ ∞1

(−1

2+

1

2x1/2 + x1/2w(x)

)x−

12 sdx

x

= −1

s+

1

s− 1+

∫ ∞1

w(x)x12 (1−s) dx

x.

Therefore

ξ(s) = −1

s+

1

s− 1+

∫ ∞1

w(x)(x

12 (1−s) + x

12 s) dxx.

Staring at this carefully we see that the first term above is defined for s 6= 0,the second is defined for s 6= 1, and the integral is defined for all s ∈ C sincew(x) decays rapidly. Hence ξ(s) has meromorphic continuation to C from theright-hand side, and ξ(s) = ξ(1−s) since the right-hand side is symmetric unders 7→ 1− s.

Lecture 20 The functional equation for L-functions

20.1 Theta series

It remains, form the last result, to prove

Lemma 20.1.1. θ( 1x ) = x1/2θ(x) for x > 0.

Proof. Let f(y) = e−y2πx. Then

f(u) =1√xe−πu

2/x,

so

θ(x) =∑n∈Z

f(n) =∑n∈Z

f(n) =1√x

∑n∈Z

e−πn2/x =

1√xθ

(1

x

),

by Poisson summation.

Corollary 20.1.2. The Riemann zeta function ζ(s) has analytic continuationto C\{ 1 }, has a simple pole at s = 1 with residue 1, and satisfies the functionalequation

ζ(1− s) = 2(2π)−sΓ(s) cos(πs

2

)ζ(s).

Date: March 1st, 2019.


Proof. Recall how

ξ(s) = π−s2 Γ(s

2

)ζ(s)

is holomorphic except at s = 0 and s = 1.At s = 0, it has a simple pole. Moreover, Γ( s2 ) has a simple pole at s = 0,

so ζ(s) must not have a pole there, since otherwise we have a double pole. Forthe same reason, ζ(0) 6= 0.

At s = 1, ξ(s) has a simple pole with residue 1, but Γ( 12 ) =

√π does not, so

ζ(s) must have a simple pole at s = 1. Moreover

1 = Ress=1

ξ(s) = π−12π

12 Ress=1

ζ(s),

so the residue of ζ(s) at s = 1 is 1.Finally ξ(s) = ξ(1− s) implies that

ζ(1− s) = π12−s

Γ( s2 )

Γ( 1−s2 )

ζ(s) = 2(2π)−sΓ(s) cos(πs

2

)ζ(s)

by manipulating the Gamma factors.

20.2 Trivial zeros of ζ(s)

Sinceξ(s) = π−

s2 Γ(s

2

)ζ(s)

is holomorphic in Re(s) < 0, and since Γ( s2 ) has simple poles as s = −2,−4,−6, . . .,it means that ζ(s) = 0 for s = −2,−4,−6, . . .. We call those the trivial zerosof ζ(s).

20.3 Functional equations of L-functions

Definition 20.3.1. Let χ be a Dirichlet character modulo q. Then there existsa minimal q∗ | q such that χ = χ0 · χ∗, with χ∗ a Dirichlet character modulo q∗

and χ0 the trivial character modulo q.This character χ∗ is uniquely determined by χ, and q∗ is called the conduc-

tor of χ. If q∗ = q, then χ is called a primitive character .We think of it in terms of the following commutative diagram:

ZqZ

Zq∗Z

C×

χχ∗

Proposition 20.3.2. The number of primitive characters modulo q is

ϕ∗(q) = ϕ ? µ(q) =∑d|q

µ(d)ϕ( qd

)= q

∏p‖q

(1− 2

p

) ∏p2‖q

(1− 1

p

)2

.


Proof. The number of Dirichlet characters modulo q is ϕ(q). Now for d | q,consider the diagram

ZqZ

ZdZ C×

χ

χ∗

Hence for every d | q, if we find the primitive characters modulo d and extendthem (as we know we can) to characters modulo q. In other words,

ϕ(q) =∑d|q

ϕ∗(d),

i.e. ϕ = ϕ∗ ? 1. By Mobius inversion, convolving with µ, we get ϕ∗ = ϕ ? µ.

Remark 20.3.3. Note that because of the first factor above, ϕ∗(q) = 0 if andonly if 2 | q and 22 - q, i.e. q ≡ 2 (mod 4). Hence primitive characters χ moduloq exist only when q 6≡ 2 (mod 4).

Proposition 20.3.4. For (n, q) = 1, we have∑∗

χ (mod q)

χ(n) =∑

d|(n−1,q)

ϕ(d)µ( qd

),

where by∑∗

we mean that we sum only over primitive characters.

Example 20.3.5. For q = 4, ϕ∗(4) = 1, so there is only one primitive Dirichlet

character modulo 4, namely χ4(n) = (−1)n−12 if 2 - n, else 0. Specifically

χ4(1) = −1, χ4(3) = 1, and otherwise it is 0. N

Example 20.3.6. For q = 8, ϕ∗(8) = 2, so there are two primitive characters.

The first one is χ8(n) = (−1)(n−1)(n+1)

8 if 2 - n, and the second one is χ4 ·χ8(n) =

(−1)(n−1)(n+5)

8 if 2 - n. N

20.4 Gauss sums

Definition 20.4.1. Let χ (mod q) be a Dirichlet character. Then

τ(χ) =∑

b (mod q)

χ(b) e

(b

q

)is called the Gauss sum of χ.

Note that this sum only sums over b ∈ (Z/qZ)× since otherwise χ(b) = 0.By orthogonality, for (a, q) = 1,

e

(a

q

)=

1

ϕ(q)

∑χ (mod q)

χ(a)τ(χ).

This formula is very useful since it transforms from an additive character toa multiplicative character. Note also that it is the Fourier expansion of theadditive character on (Z/qZ)×.

Proposition 20.4.2. Let χ be a primitive character modulo q. Then∑b (mod q)

χ(b) e

(ab

q

)= χ(a)τ(χ).

THE FUNCTIONAL EQUATION FOR L-FUNCTIONS, CONTINUED 51

Lecture 21 The functional equation for L-functions,continued

21.1 Gauss sums

Proposition 21.1.1. Let χ be a primitice character modulo q. Then∑b (mod q)

χ(b) e

(ab

q

)= χ(a)τ(χ).

Proof. We have two cases to consider: when (a, q) = 1, and when they’re notcoprime. Let us start with when they are.

Then∑b (mod q)

χ(b) e

(ab

q

)=

∑b∈(Z/qZ)×

χ(b) e

(ab

q

)=

∑b∈(Z/qZ)×

χ(ba−1) e

(aba−1

q

)

=∑

b∈(Z/qZ)×

χ(b)χ(a−1) e

(b

q

)= χ(a)τ(χ),

where we sum only over units since otherwise the charcater is 0 anyway, andwe perform the change of variable b 7→ ba−1 since that just permutes the termswe’re summing over.

If (a, q) 6= 1, then χ(a) = 0, so we need to show that∑b (mod q)

χ(b) e

(ab

q

)= 0.

Letting d = (a, q) and writing a = da1, q = dq1, with (a1, q1), so that∑b (mod q)

χ(b) e

(ab

q

)=

∑b (mod q)

χ(b) e

(a1b

q1

)

=∑

r (mod q1)

∑b (mod q)

b≡r (mod q1)

χ(b)

e

(a1r

q1

).

We claim, and will prove momentarily, that there exists some c ≡ q (mod q1)with (c, q) such that χ(c) 6= 1.

In that case,∑b (mod q)

b≡r (mod q1)

χ(b) =∑

b (mod q)b≡r (mod q1)

χ(bc) = χ(c)∑


χ(b)

by changing b 7→ bc, since c is a unit modulo q. Then

(1− χ(c))∑


χ(b) = 0,

Date: March 4th, 2019.


so the sub over the characters must be zero, since χ(c) 6= 1.Now let us prove the claim. Suppose there is no such c, i.e., χ(c) = 1 for

all c ≡ 1 (mod q1) with (c, q) = 1. Then for any m,n ∈ (Z/qZ)× with n ≡ m(mod q1), we have nm ≡ 1 (mod q1), where by m we mean the multiplicativeinverse modulo q1. Thus χ(nm) = 1, implying that χ(n) = χ(m), meaning thatχ factors modulo q1, so it is not primitive.

Proposition 21.1.2. Let χ be a primitive character modulo q. Then |τ(χ)| =√q.

Proof. We write

∣∣∣χ(a)τ(χ)∣∣∣2 =

∣∣∣∣∣∣∑

b (mod q)

χ(b) e

(ab

q

)∣∣∣∣∣∣2

=∑

b1 (mod q)b2 (mod q)

χ(b1)χ(b2) e

(a(b1 − b2)

q

).

Now summing this over a (mod q), we note that the right-hand side is nonzeroonly for b1 − b2 = 0 since by orthogonality∑

a (mod q)

e

(am

q

)=

{0 if q - m,q if q | m.

Hence|τ(χ)|2

∑a (mod q)

|χ(a)|2 = q∑

b (mod q)

|χ(b)|2,

where the sums are nonzero, so dividing through we get |τ(χ)|2 = q.

The proof of the functional equation for L(s, χ) is now very similar to thatof the Riemann zeta function.

Theorem 21.1.3. Let χ be a primitive character modulo q. Let

Λ(s, χ) = L∞(s, χ)L(s, χ)

be the completed L-function, where

L∞(s, χ) =( qπ

)s/2Γ

(s+ κ

2

)is called the archimedean factor of the (local) factor at infinity, and

κ =1

2(1− χ(−1)) =

{0 if χ is even,

1 if χ is odd.

A character χ being even means χ(−1) = 1, so that χ(−n) = χ(n), and oddcorrespondingly means χ(−1) = −1, so χ(−n) = −χ(n).

Then Λ(s, χ) has a holomorphic continuation to all s ∈ C and satisfies thefunctional equation

Λ(s, χ) = ε(χ)Λ(1− s, χ),

where

ε(χ) = iκτ(χ)√q

is called the root number of χ.


Note that by the previous proposition, |ε(χ)| = 1.

Proof. We separate the case where χ is even and where it’s odd. If it is even,we use the definition of Γ(s), from which we have

π−12 sq

12 sΓ

(1

2s

)n−s =

∫ ∞0

e−n2πtq t

12 sdt

t

for Re(s) > 0. (Cf. ζ(s) with q = 1.) Twisting by χ(n) and summing over n,we get

π−12 sq

12 sΓ

(1

2s

)L(s, χ) =

∫ ∞0

∑n≥1

χ(n)e−n2πtq

t12 sdt

t.

We switch the order of integration and summation since the exponential decayguarantees us absolute and uniform convergence for Re(s) > 1.

Now let

θχ(t) =

∞∑n=−∞

χ(n)e−n2πtq ,

so that

Λ(s, χ) =1

2

∫ ∞0

θχ(t)t12 tdt

t=

1

2

∫ 1

0

θχ(t)t12 tdt

t+

1

2

∫ ∞1

θχ(t)t12 tdt

t

=1

2

∫ ∞1

θχ

(1

t

)t−

12 tdt

t+

1

2

∫ ∞1

θχ(t)t12 tdt

t

We claim, and will prove momentarily, that

θχ

(1

t

)=

(t

q

)1/2

τ(χ)θχ(t).

With that as writ,

Λ(s, χ) =1

2

τ(χ)√q

∫ ∞1

θχ(t)t−12 (1−s) dt

t+

1

2

∫ ∞1

θχ(t)t12 sdt

t,

and

Λ(1− s, χ) =1

2

τ(χ)√q

∫ ∞1

θχ(t)t−12 sdt

t+

1

2

∫ ∞1

θχ(t)t12 (1−s) dt

t,

and comparing these we see that

Λ(s, χ) = ε(χ)Λ(1− s, χ),

where in this case, for even χ, ε(χ) = τ(χ)/√q. Note that as part of the

computation,τ(χ)τ(χ) = τ(χ)τ(χ) = |τ(χ)|2 = q.

FUNCTIONS OF FINITE ORDER 54

Lecture 22 Functions of finite order

22.1 Proof concluded

We left the proof with two outstanding issues: we had only considered χ beigneven, and we had a claim we hadn’t yet proved.

Claim. With θχ(t) =∑n∈Z

χ(n)θχe−n2πt

q , we have

θχ

(1

t

)=

(t

q

)1/2

τ(χ)θχ(t).

Proof. By definition,

τ(χ)θχ(t) =∑

b (mod q)

χ(b) e

(b

q

)∑n∈Z

χ(n)e−n2πtq .

Combining the exponentials and using Poisson summation, this becomes∑b (mod q)

χ(b)∑n∈Z

e−n2πtq + 2πibn

1 =∑

b (mod q)

χ(b)∑n∈Z

(qt

)1/2

e−(n+ b

q)2πq

t .

Multiplying and dividing the exponent by q, we get −(qn+b)2π/(qt), where theparenthesis then ranges over all integers n, so this becomes(q

t

)1/2∑n∈Z

χ(n)e−n2πqt =

(qt

)1/2

θχ

(1

t

).

Hence, for χ even, we are done.Now let us tackle the case where χ is odd, i.e. χ(−1) = −1.

Proof. Note that the method used above no longer works, since by χ being odd,θχ(t) is identically 0 for all t. Instead what we’ll do is replace s by s+ 1 in theexpression we play with for Γ, so

π−12 (s+1)q

12 (s+1)Γ

(1

2(s+ 1)

)n−s =

∫ ∞0

ne−n2πtq t

12 (s+1) dt

t,

where we’ve moved one of the n from n−(s+1) over to the other side, so as tomake the exponential, together with n, an odd function in n. We then proceedas usual: we twist by χ(n) and sum over n ≥ 1, getting

π−12 (s+1)q

12 (s+1)Γ

(1

2(s+ 1)

)L(s, χ) =

∫ ∞0

∑n≥1

χ(n)ne−n2πtt

t12 (s+1) dt

t,

where the summand is now an even function in n. Defining

θ′χ(t) =∑n∈Z

χ(n)ne−n2πtq ,


FUNCTIONS OF FINITE ORDER 55

the sum is 12θ′χ(t), and as before we write this as the integral from 0 to 1 plus

the integral from 1 to∞, then make a change of variables t 7→ 1/t in the former,

Λ(s, χ) =1

2

∫ ∞0

θ′χ(t)t12 (s+1) dt

t=

1

2

∫ 1

0

θ′χ(t)t12 (s+1) dt

t+

1

2

∫ ∞1

θ′χ(t)t12 (s+1) dt

t

=1

2

∫ 1

0

θ′χ

(1

t

)t−

12 (s+1) dt

t+

1

2

∫ ∞1

θ′χ(t)t12 (s+1) dt

t.

By the same kind of computation as before, with Poisson summation and all,

θ′χ

(1

t

)= −iq−1/2t3/2τ(χ)θ′χ(t),

by which

Λ(s, χ) =π1/2

2

(−iτ(χ)

q

∫ ∞1

θ′χ(t)t−12 (s+1) dt

t+

1

q1/2

∫ ∞1

θ′χ(t)t12 (s+1) dt

t

)which is the same as

ε(χ)Λ(1− s, χ).

22.2 The Hadamard factorisation theorem

The goal of the coming discussion is to express a holomorphic function on C as aninfinite product indexed by its zeros. To start this, we need some preliminaries.

Definition 22.2.1. A holomorphic function f : C → C is of finite order ifthere exists a constant α > 0 such that for any ε > 0, we have

|f(s)| � exp(|s|α+ε

)for all s ∈ C, with the implied constant depending on ε and f .

In this case we say that f is of order ≤ α. We say that f is of order α if itis of order ≤ α but not of order ≤ β for any β < α.

Example 22.2.2. Let Pn(s) = ansn + an−1s

n−1 + . . .+ a0, ai ∈ C and an 6= 0.This is of order 0. N

Example 22.2.3. The function f(s) = exp(s) is of order 1. N

Example 22.2.4. The function fn(s) = exp(Pn(s)) is of order n = degPn. N

Note also that fn(s), as defined above, is a function of order n which doesnot vanish in C. This is not a coincidence:

Proposition 22.2.5. A function f of order ≤ α which does not vanish on Cis of the form f(s) = exp(p(s)) for some polynomial p(s) of degree ≤ α. Hencef is of integral order deg p.

Proof. Since f(x) does not vanish on C, we can take its logarithm, i.e., g(s) :=log f(s) can be defined and is holomorphic on C. Hence it has a Taylor expansionat s = 0, i.e.

g(s) =

∞∑n=0

cnsn,

JENSEN’S FORMULA 56

with cn ∈ C. We claim that cn = 0 for n > α, from which the result follows.To prove the claim, note that since f is of order ≤ α,

Re g(s) = log|f(s)| ≤ Cε,f |s|α+ε = Cε,fRα+ε,

calling |s| = R. Writing cn = an + ibn, with an, bn ∈ R, and letting s = Re2πix,we have

Re g(Re2πix

)=∑n≥0

anRn cos(2πnx)− i

∑n≥1

bnR6n sin(2πnx).

Hence

anRn =

{∫ 1

0Re(g(Re2πix)) dx if n = 0,

2∫ 1

0Re(g(Re2πix)) cos(2πnx) dx if n ≥ 1

and

bnRn = 2

∫ 1

0

Re(g(Re2πix)) sin(2πinx) dx,

whence

|an|Rn ≤ 2

∫ 1

0

|Re(g(Re2πix))| · 1 dx ≤ 2Cε,fRα+ε

by bounding cos by 1. Hence for n > α,

|an| ≤ 2Cε,fRα−n+ε → 0

as R→∞, so an = 0.By the same argument, bn is 0 for n > α too.

Lecture 23 Jensen’s formula

23.1 Jensen’s formula

Remark 23.1.1. Note that in the proof of Proposition 22.2.5, it suffices to studya sequence of positive real numbers {Rn }∞n=1 with Rn → ∞ such that for alln ≥ 0, ε > 0, and |s| = Rn,

f(s)�f,ε exp(|s|α+ε).

This is because of the maximal principle, having control on the boundary of acircle implies having control inside the circle.

In order to state Jensen’s theorem we need a bit of set up. Let f : C → C.For R > 0, we define

Z(f,R) := { ρ ∈ C | f(ρ) = 0, |ρ| ≤ R }

andZ(f) := { ρ ∈ C | f(ρ) = 0 }.

We will also take sums and products over these sets—in this case we add ormultiply with multiplicity.



Theorem 23.1.2 (Jensen’s formula). Let R > 0. Let f be a holomorphicfunction in a neighbourhood of DR = { s ∈ C | |s| < R }. Suppose f does notvanish at s = 0 nor on CR = { s ∈ C | |s| = R }. Then∫ 1

0

log

∣∣∣∣f(Re2πit)

f(0)

∣∣∣∣ dt = log∏

ρ∈Z(f,R)

R

|ρ|=

∑ρ∈Z(f,r)

logR

|ρ|.

Note that like usual residue calculus with Cauchy’s redisue theorem on f ′/f ,this also counts zeros, but not with weight 1.

Proof. Write

f(s) = F (s)∏

ρ∈Z(f,R)

(s− ρ)

where F (s) is a holomorphic function that doesn’t vanish in DR. Hence

log

∣∣∣∣f(s)

f(0)

∣∣∣∣ = log

∣∣∣∣F (s)

F (0)

∣∣∣∣+∑

ρ∈Z(f,R)

log

∣∣∣∣s− ρρ∣∣∣∣.

Looking at the first term,

log

∣∣∣∣F (s)

F (0)

∣∣∣∣ = Re logF (s)

F (0).

The inside of this real part is a holomorphic function in a neighbourhood of DR,and vanishes at s = 0, so taking Re2πit = z,∫ 1

0

logF (Re2πit)

F (0)dt =

1

2πi

∫CR

logF (z)

F (0)

dz

z= 0

since the integrand is a holomorphic function, so we can contract the integrationto a point.

For the second term, note that∣∣∣∣Re2πit − ρρ

∣∣∣∣ =R

|ρ|

∣∣∣e2πit − ρ

R

∣∣∣ =R

|ρ|

∣∣∣∣1− ρ

Re2πit

∣∣∣∣meaning that

log

∣∣∣∣Re2πit − ρρ

∣∣∣∣ = logR

|ρ|+ log

∣∣∣∣1− ρ

Re2πit

∣∣∣∣ = logR

|ρ|+ Re log

(1− ρ

Re2πit

).

As above, the latter is a holomorphic function vanishing at z = 0, so∫ 1

0

log

(1− ρ

Re2πit

)dt = 0,

so when all the dust settles only the sum of log(R/|ρ|) remains.


23.2 Application of Jacobi’s formula

Theorem 23.2.1. Let f be a holomorphic function of order ≤ α. Let

N(f,R) =∑

ρ∈Z(f,R)

1,

i.e., the number of zeros of f of modulus ≤ R, with multiplicity.

(i) For all R > 0, ε > 0, we have

N(f,R)�f,ε Rα+ε +

{1, if f(0) = 0,

0, if f(0) 6= 0.

(ii) The series ∑ρ∈Z(f)

1

1 + |ρ|α+ε

converges.

Proof. Suppose f(0) 6= 0 and f(s) 6= 0 on C2R. For ρ ∈ Z(f,R), log(2R/ρ) ≥log 2, so

(log 2)N(f,R) ≤∑

ρ∈Z(f,R)

log2R

|ρ|≤

∑ρ∈Z(f,2R)

log2R

|ρ|=

∫ 1

0

log

∣∣∣∣f(2Re2πit)

f(0)

∣∣∣∣ dt�∫ 1

0

(2R)α+ε dt = (2R)α+ε � Rα+ε,

where for the integral we use Jensen’s formula, and the bound is by the order.Note that if f(s) = 0 for some s ∈ C2R, then we consider R < R′ < R + δ

for some δ > 0, such that f(s) 6= 0 for all s ∈ C2R′ . This is possible since f isa holomorphic function, meaning that if it has an accumulation point of zeros,then it is identically zero. Thus

N(f,R) ≤ N(f,R′)� (R′)α+ε � Rα+ε.

Now suppose f(0) = 0, with the zero of order m. Then

N(f,R) = N

(f(s)

sm, R

)+m�f R

α+ε + 1.

For the series,

∑ρ∈Z(f,R)

1

1 + |ρ|α+ε=

∫ R

0

1

1 + rα+εdN(f, r)

=N(f, r)

1 + rα+ε

∣∣∣∣R0

−∫ R

0

N(f, r)rα−1+ε

(1 + rα+ε)2dr

=N(f,R)

1 +Rα+ε−∫ R

0

N(f, r)rα−1+ε

(1 + rα+ε)2dr.

By the first part, this is O(1), and letting R→∞ we are done.

HADAMARD FACTORISATION THEOREM 59

Lecture 24 Hadamard factorisation theorem

24.1 Hadamard factorisation

Theorem 24.1.1. Let f be a holomorphic function of order at most 1 such thatf(0) 6= 0. Then we have

f(s) = A exp(bs)∏

ρ∈Z(f)

(1− s

ρ

)es/ρ

where A = f(0), b = f ′(0)/f(0), and the product converges uniformly on com-pact subsets of C.

Proof. Let K ⊂ C be a compact subset. Then K ∩ Z(f) is a finite set (sinceotherwise f = 0, since a holomorphic function with an accumulation point ofzeros must be zero everywhere). For s ∈ K and ρ ∈ Z(f) \K, we have(

1− s

ρ

)es/ρ = 1 +OK

(1

|ρ|2

)by expanding the exponential as a Taylor series.

Note that an infintie product∏

(1 + an), with an ≥ 0, converges if and onlyif∑an converges, and since ∑

ρ∈Z(f)

1

|ρ|2

converges by Jensen’s formula,∏ρ∈Z(f)

(1− s

ρ

)es/ρ

converges uniformly on K (since our estimate above is independent of s).Set

h(s) =∏

ρ∈Z(f)

(1− s

ρ

)es/ρ.

This is holomorphic and has exactly the same zeros as f(s). Hence f(s)/h(s) isholomorphic and does not vanish on C.

We claim that f(s)/h(s) has order ≤ 1.Then, by Proposition 22.2.5,

f(s)

h(s)= exp(a+ bs)

for some a, b ∈ C.Now for all ρ ∈ Z(f), (

1− s

ρ

)es/ρ

∣∣∣∣s=0

= 1,

so h(0) = 1, whence f(0) = exp(a) = A.


THE INFINITE PRODUCT FOR ζ(S) 60

Similarly,d

ds

(1− s

ρ

)es/ρ

∣∣∣∣s=0

= 0

so h′(0) = 0, so b = f ′(0)/f(0).

It remains to prove the claim, meaning we need to show

log

∣∣∣∣f(s)

h(s)

∣∣∣∣ = O(|s|1+ε).

Proof. Writing the logarithm as

log|f(s)| − log|h(s)|,

we note first that f(s) is order at most 1 by assumption, so

log|f(s)| � R1+ε

on |s| = R for all R > 0.It remains to show that − log|h(s)| � R1+ε on a sequence of R → ∞,

|s| = R. The intelligent idea is this: Since∑ρ∈Z(f)

1

|ρ|2<∞,

the total length of all intervals

Lecture 25 The infinite product for ζ(s)

25.1 Hadamard factorisation finished

The missing piece from our proof of Hadamard factorisation is to show that for

h3(s) =∏|ρ|>2R

(1− s

ρ

)es/ρ,

we have− log|h3(s)| � R1+ε

for R = |s|.Write this as

− log|h3(s)| =∑|ρ|>2R

− log

∣∣∣∣(1− s

ρ

)es/ρ

∣∣∣∣ =∑|ρ|>2R

− log(1 +O(|s/ρ|2)

)since (

1− s

ρ

)es/ρ =

(1− s

ρ

)(1 +

s

ρ+O

(∣∣∣∣ sρ∣∣∣∣2))

= 1 +O(|s/ρ|2),



by expanding the exponential as a Taylor series.Now expanding this logarithm as a Taylor series in turn, we have

− log|h3(s)| �∑|ρ|>2R

∣∣∣∣ sρ∣∣∣∣2 =

∑|ρ|>2R

1

|ρ|1+ε

R1−ε

|ρ|1−εR1+ε � R1+ε

since the first term in the latter sum converges since the order is ≤ 1, and thesecond term is bounded by (1/2)1+ε since |ρ| > 2R.

Corollary 25.1.1. Let f be a holomorphic function of order at most 1 suchthat f(0) 6= 0. For s ∈ C \ Z(f), we have

d

ds(log f(s)) =

f ′(s)

f(s)= b+

∑ρ∈Z(f)

(1

ρ− 1

ρ− s

),

with b = f ′(0)/f(0). Moreover∑ρ∈Z(f)

(1

ρ− 1

ρ− s

)converges uniformly on compact subsets K ⊂ C \ Z(f).

Proof. The first part is Hadamard factorisation on f ′(s)/f(s). For the secondpart, note that if s ∈ K,

1

ρ− 1

ρ− s=

−sρ(ρ− s)

= OK

(1

|ρ|2

)since s is in a bounded set.

25.2 The infinite product for ζ(s) and the explicit formula

Letξ0(s) = s(1− s)ζ(s) = s(1− s)π− s2 Γ

(s2

)ζ(s).

Proposition 25.2.1. ξ0(s) has order at most 1.

Proof. Since ξ(s) = ξ(1− s), we also have ξ0(s) = ξ0(1− s), so by symmetry wemay assume Re(s) ≥ 1/2.

Bounding each factor above one at a time, we have

|π− s2 | = e−s2 log π � exp

(|s|1+ε

).

Next, ∣∣∣Γ(s2

)∣∣∣� exp(|s|1+ε

)by Stirling’s formula.

Finally,

ζ(s) =s

s− 1− s

∫ ∞1

{t}t1+s

dt

for Re(s) > 0. The first term is bounded, with the pole at s = 1 cancelled bythe factor (1− s) from ξ0(s). The integral is convergent since Re(s) > 0.

Hence s(1− s)ζ(s)� exp(|s|ε), hence ξ0(s) has order at most 1.


Corollary 25.2.2. (i) ξ0(s) = A exp(bs)∏

ρ∈Z(ξ0)

(1− s

ρ

)esρ ,

(ii)∑

ρ∈Z(ξ0)

1

|ρ|1+ε<∞.

Remark 25.2.3. The zeros of ξ0(s) are precisely the nontrivial zeros of ζ(s), sincesΓ( s2 ) has no zeros and the zero of (1 − s) is cancelled by the pole of ζ(s) ats = 1.

25.3 Infinite product for L(s, χ)

Let χ (mod q) be a primitive character, and

Λ(s, χ) =( qπ

) 12 (s+κ)

Γ

(s+ κ

2

)L(s, χ),

satisfyingΛ(s, χ) = ε(χ)Λ(1− s, χ)

with

κ =

{0, if χ is even,

1, if χ is odd,

and |ε(χ)| = 1.

Proposition 25.3.1. Λ(s, χ) has order at most 1.

Proof. By essentially the same proof as for ζ(s), except here we have

L(s, χ) = s

∫ ∞1

S(t)

t2+1dt

for Re(s) > 0, with

S(t) =∑n≤t

χ(n),

and |S(t)| ≤ q. This integral is convergent, so � exp(|s|ε).

25.4 Application to counting zeros of ζ(s)

By Corollary 25.1.1,

ξ′0ξ0

(s) =1

s+

1

s− 1 ζ′∞ζ∞

(s) +ζ ′

ζ(s) = b+

∑ρ∈Z(ξ0)

(1

ρ− 1

ρ− s

)

converges uniformly on compact subsets of C \ Z(f).Note that ξ0(s) takes real values on the real line, and it is holomorphic, so

by Schwartz reflection principle ξ0(s) = ξ0(s).Hence Z(ξ0) is invariant under complex conjugation, so∑

ρ∈Z(ξ0)

(1

ρ− 1

ρ− s

)=

1

2

∑ρ∈Z(ξ0)

(1

ρ+

1

ρ− 1

ρ− s− 1

ρ− s

).

COUNTING ZEROS 63

Writing ρ = β + iγ, 0 ≤ β ≤ 1, γ ∈ R, and

1

ρ+

1

ρ=

2β

|ρ|2,

we have ∑ρ∈Z(ξ0)

(1

ρ+

1

ρ

)<∞,

whereby1

2

∑ρ∈Z(ξ0)

(1

s− ρ+

1

s− ρ

).

converges uniformly on compact subsets of C \ Z(f).

Proposition 25.4.1. For every s ∈ Z(ξ0),

ξ′0ξ0

(s) =1

2

∑ρ∈Z(ξ0)

(1

s− ρ+

1

s− ρ

)+O(1)

and

−ζ′

ζ(s) =

1

s+

1

s− 1− 1

2

∑ρ∈Z(ξ0)

(1

s− ρ+

1

s− ρ

)+ζ ′∞ζ∞

(s) +O(1).

Lecture 26 Counting zeros

26.1 Application to counting zeros of ζ(s)

Corollary 26.1.1. Let N(T ) := #{ ρ = β + iγ ∈ Z(ξ0) | |γ| ≤ T }. For T ≥ 1,we have

(a) N(T + 1)−N(T )� log(2 + T ),

(b) N(T )� T log(2 + T ).

Proof. We have

−ζ′

ζ(s) =

1

s+

1

s− 1+ζ ′∞ζ∞

(s)− 1

2

∑ρ∈Z(ξ0)

(1

s− ρ+

1

s− ρ

)+O(1).

Letting s = 2 + iT , we have the series representation

−ζ′

ζ(s) =

∞∑n=1

Λ(n)

ns<∞

since we are in the region of absolute convergence. Also,

ζ ′∞ζ∞

(s)� log(2 + T )


COUNTING ZEROS 64

by Stirling’s formula.Hence

1

2

∑ρ∈Z(ξ0)

(1

s− ρ+

1

s− ρ

)� log(2 + T ).

Now writing ρ = β + iγ, with 0 ≤ β ≤ 1, we have

Re

(1

s− ρ

)=

2− β|s− ρ|2

≥ 1

(2− β)2 + |γ − T |2≥ 1

4 + |γ − T |2≥ 1

5δ|γ−T |≤1.

Therefore

1

5

∑|γ±T |≤1

1 ≤∑

ρ∈Z(ξ0)

Re

(1

s− ρ+

1

s− ρ

)� log(2 + T ),

where the left-hand side sum, of course, is N(T + 1)−N(T − 1), which in turnis an upper bound for N(T + 1)−N(T ), so

N(T + 1)−N(T )� log(2 + T ).

For the second part, simply write N(T ) as a telescoping sum,

N(T ) =

T∑n=1

(N(n)−N(n− 1))� T log(2 + T ).

Remark 26.1.2. From the proof we note, in particular, that∑ρ∈Z(ξ0)

1

4 + |γ ± T |2� log(2 + T ),

where ρ = β + iγ.Secondly, moving things over,

ξ′0ξ0

(s) =1

2

∑|γ±T |≤1

(1

s− ρ+

1

s− ρ

)+O(log(2 + |T |)),

where s = σ + iT 6∈ Z(ξ0), −1 ≤ σ ≤ 2.

26.2 Applications to countinjg zeros of L(s, χ)

We wish to perform the same sort of estimates for L(s, χ).Let χ be a primitive character modulo q ≥ 1. Let κ = 1

2 (1− χ(−1)), and

L∞(s, χ) :=

(π

q

)− s2Γ

(s+ κ

2

).

Then Λ(s, χ) = L∞(s, χ)L(s, χ) = ε(χ)Λ(1− s, χ). Note that

Z(Λ(s, χ)) ⊂ { s ∈ C | 0 ≤ Re(s) ≤ 1 }

is exactly the set of nontrivial zeros of L(s, χ).

COUNTING ZEROS 65

By Hadamard factorisation,

Λ′

Λ(s, χ) =

L′∞L∞

(s, χ) +L′

L(s, χ) = b(χ) +

∑ρ∈Z(Λ(s,χ))

(1

ρ− 1

ρ− s

),

where

b(χ) =Λ′

Λ(0, χ).

This causes trouble, since unlike for ζ it depends on a sum over zeros.Note that

Λ(s, χ) = ε(χ)Λ(1− s, χ) = εχΛ(1− s, χ),

so if ρ ∈ Z(Λ(s, χ)), then 1− ρ ∈ Z(Λ(s, χ)). Thus

b(χ) =Λ′

Λ(0, χ) = −Λ′

Λ(1, χ) = −b(χ)−

∑ρ′∈Z(Λ(s,χ))

(1

ρ′− 1

ρ′ − 1

).

Writing ρ = 1− ρ′ ∈ Z(Λ(s, χ)), this becomes

−b(χ)−∑

ρ∈Z(Λ(s,χ))

(1

1− ρ+

1

ρ

).

Combining this we have

Re b(χ) = −1

2

∑ρ∈Z(Λ(s,χ))

(1

1− ρ+

1

ρ

)

since b(χ) + b(χ) = 2 Re b(χ), so

Re b(χ) = −1

2

∑ρ∈Z(Λ(s,χ))

Re

(1

ρ+

1

ρ

)= −

∑ρ∈Z(Λ(s,χ))

Re1

ρ.

Thus

ReΛ′

Λ(s, χ) =

∑ρ∈Z(Λ(s,χ))

Re1

s− ρ.

Summarising,

Proposition 26.2.1. For s 6∈ Z(Λ(s, χ)),

(a) ReΛ′

Λ(s, χ) =

∑ρ∈Z(Λ(s,χ))

1

s− ρ, and

(b) ReL′

L(s, χ) =

∑ρ∈Z(Λ(s,χ))

Re1

s− ρ− Re

L′∞L∞

(s, χ).

WEIL’S EXPLICIT FORMULA 66

Lecture 27 Weil’s explicit formula

27.1 Application to counting zeros of L(s, χ)

Corollary 27.1.1. Let N(T, χ) := #{ ρ = β + iγ ∈ Z(Λ(s, χ)) | |γ| ≤ T }. ForT ≥ 1, we have

(a) N(T + 1, χ)−N(T, χ)� log(q(2 + T )), and

(b) N(T, χ)� T log(q(2 + T )).

Proof. For s = 2+ iT , we have the (absolutely) convergent series representation

L′

L(s, χ) =

∞∑n=1

Λ(n)χ(n)

ns<∞,

andL′∞L∞

(s, χ)� log(q(2 + T ))

by Stirling’s formula. By Proposition 26.2.1, we hence have∑ρ∈Z(Λ(s,χ))

Re1

s− ρ� log(q(2 + T )).

Note that

Re1

s− ρ=

σ − β|s− ρ|2

≥ 1

4 + |T − γ|2≥ 1

5δ|T−γ|≤1,

writing ρ = β+ iγ. For s = 2− iT we get the same result, but |T −γ| is replacedby |T + γ|. Hence

1

5

∑|T±γ|≤1

1 ≤∑ρ

Re1

s− ρ� log(q(2 + T )),

so

N(T + 1, χ)−N(T, χ) ≤ N(T + 1, χ)−N(T − 1, χ)� log(q(2 + T )).

The second part, as before, is just a telescoping sum.

Remark 27.1.2. We have∑ρ∈Z(Λ(s,χ))

1

4 + |γ ± T |2� log(q(2 + T )).

Moreover,Λ′

Λ(s, χ) =

∑|γ±T |≤1

1

s− ρ+O(log(q(2 + T )))

for s = σ + iT 6∈ Z(Λ(s, χ)), −1 ≤ σ ≤ 2.



Proof. The second part is a little bit more delicate in this case than for ζ(s).For s = 2 + iT ,

Λ′

Λ(s, χ) = b(χ) +

∑ρ

(1

ρ+

1

2 + iT − ρ

),

where the left-hand side is � log(q(2 + T )). Thus

Λ′

Λ(s, χ) =

∑ρ

(1

s− ρ− 1

(2 + iT )− ρ

)+O(log(q(2 + T )))

=∑

|γ±T |≤1

1

s− ρ−

∑|γ±T |≤1

1

2 + iT − ρ+

∑|γ±T |>1

(1

s− ρ− 1

2 + iT − ρ

)+O(log(q(2 + T ))).

Now estimating the sums,∑|γ±T |≤1

1

2 + iT − ρ�

∑|γ±T |≤1

1� log(q(2 + T )),

and since1

s− ρ− 1

2 + iT − ρ=

s− σ(s− ρ)(2 + iT − ρ)

� 1

|T ± γ|2.

Summing both sides, we get � log(q(2 + T )), meaning that all of the sums wedon’t care about get baked into the same error term.

27.2 Weil’s explicit formula

Theorem 27.2.1. Let f ∈ C∞c ((0,∞)), (i.e., infinitely differentiable with com-pact support) and let

f(s) =

∫ ∞0

f(x)xsdx

x

be its Mellin transform. Let

f(x) =1

xf

(1

x

).

We have∑n≥1

(f(n)+f(n))Λ(n) = f(1)+f(0)+1

2πi

∫( 12 )

(ζ ′∞ζ∞

(s) +ζ ′∞ζ∞

(1− s))f(s) ds−

∑ρ∈Z(ξ0)

f(ρ).

Proof. First we claim that

1

2πi

∫( 32 )

f(s)ξ′0ξ0

(s) ds =1

2πi

∫( 32 )

(1

s+

1

s− 1+ζ ′∞ζ∞

(s)

)f(s)−

∑n≥1

Λ(n)f(n).

Since ξ0(s) = s(1− s)ζ∞(s)ζ(s), and

−ζ′

ζ(s) =

∑n≥1

Λ(n)

ns

in the region of absolute convergence,

1

2πi

∫( 32 )

f(s)ξ′0ξ0

(s) ds =1

2πi

∫( 32 )

f(s)

(1

s+

1

s− 1+ζ ′∞ζ∞

(s)

)ds−

∑n≥1

Λ1

2πi

∫( 32 )

f(s)n−s ds.

The final integral is the Mellin inversion of f(n), whence we have our claim.


Lecture 28 Weil’s explicit formula

28.1 Proof continued

We have ∑ρ∈Z(ξ0)

f(ρ) =1

2πi

∫( 32 )

f(s)ξ′0ξ0

(s) ds− 1

2πi

∫(− 1

2 )

f(s)ξ′0ξ0

(s) s.

By making the change of variables s 7→ 1− s in the second integral, we switch

from f(s) 7→ f(1 − s),ξ′0ξ0

(s) 7→ ξ′0ξ0

(1 − s) = − ξ′0

ξ0(s), and finally the line of

integration moves to 32 . Hence by the claim this becomes

1

2πi

∫( 32 )

(1

s+

1

s− 1+ζ ′∞ζ∞

(s)

)f(s) ds−

∑n≥1

Λ(n)f(n)+

1

2πi

∫( 32 )

(1

s+

1

s− 1+ζ ′∞ζ∞

(s)

)f(1− s) ds−

∑n≥1

Λ(n)f(n),

where we’ve used

˜f(s) =1

2πi

∫ ∞0

f(x)xsdx

x=

1

2πi

∫ ∞0

1

xf

(1

x

)xsdx

x,

which if we switch 1x 7→ t becomes∫ ∞

0

tf(t)t−sdt

t= f(1− s).

Moving the line of integration from ( 32 ) to ( 1

2 ) we pick up a pole at s = 1,

with residues f(1) in the first integral and f(0) in the second, so our expressionbecomes

1

2πi

∫( 32 )

(1

s+

1

s− 1

)(f(s) + f(1− s)) ds =

f(1) + f(0) +1

2πi

∫( 12 )

(1

s+

1

s− 1

)(f(s) + f(1− s)) ds.

This last integral is 0, since if we switch s 7→ 1 − s, we get negative the sameintegral, so it is equal to negative itself.

Theorem 28.1.1. Let χ be a Dirichlet character modulo q. Then, with f asabove, ∑

n≥1

(χ(n)f(n) + χ(n)f(n))Λ(n) =

1

2πi

∫( 12 )

(L′∞L∞

(s, χ) +L′∞L∞

(1− s, χ)

)f(s) ds−

∑ρ∈Z(Λ(s,χ))

f(ρ).

Proof. The proof is the same as above, except with L∞ in place of ζ∞.

Date: April 1st, 2019.


Theorem 28.1.2 (Hadamard, de la Vallee-Poussin). There exist an absoluteconstant c > 0 such that ζ(s) 6= 0, s = σ + it, in the region

σ ≥ 1− c

log(2 + |t|).

We will prove this later on. For now, remark that the value of c is actuallycomputable. For instance, Habiba and Kadin computed

c =1

5.69693

in 2005. People are often improving this ever so slightly.

Corollary 28.1.3. There exists a constant c > 0 such that for f ∈ C∞c ((0,∞)),X ≥ 2 large,∑

n≥1

Λ(n)f( nX

)= X

∫ ∞0

f(t) dt+Of

(X exp(−c

√logX)

).

We call this a smooth sum since f is smooth. One can ‘unsmooth’ the sum,obtaining ∑

n≤X

Λ(n) = X +O(X exp(−c′√

logX))

for some c′ > 0, whence∑

Λ(n) ∼ X, which is equivalent to the Prime numbertheorem.

Remark 28.1.4. If supp f = [1, 2], we get 1 ≤ nX ≤ 2, i.e., X ≤ n ≤ 2X.

Proof of Corollary 28.1.3. Let g(t) = f( tX ). Then g(s) = f(s)Xs, so

g(t) =1

tg

(1

t

)=

1

tf

(1

tX

),

which is zero for t ≥ 1 and X large since f has compact support, and 1tX ≈ 0.

Thus by Weil’s explicit formula∑n≥1

Λ(n)g(n) = g(1)+g(0)+1

2πi

∫( 12 )

(ζ ′∞ζ∞

(s)− ζ ′∞ζ∞

(1− s))g(s) ds−

∑ρ∈Z(ξ0)

g(ρ),

whence ∑n≥1

Λ(n)f( nX

)=

f(1)X + f(0) +1

2πi

∫( 12 )

(ζ ′∞ζ∞

(s)− ζ ′∞ζ∞

(1− s))f(s)Xs ds−

∑ρ∈Z(ξ0)

f(ρ)Xρ.

The first term is the integral we want, with the second term being Of (1). Forthe sum over zeros,∑

ρ

f(ρ)Xρ ≤∑

ρ=β+iγ

|f(ρ)|Xβ ≤ X∑ρ

|f(ρ)|X−c

log(2+|γ|)

THE THEOREM OF HADAMARD AND DE LA VALLEE-POUSSIN 70

by the zero free region, where the exponential in the end is

exp

(− x logX

log(2 + |γ|)

).

We attack this final sum by splitting the sum over zeros with log(2 + |γ|) ≤√logX and those with log(2 + |γ|) >

√logX. The former becomes

exp(−c√

logX)∑ρ

|f(ρ)|

wherein the sum converges by exponential decay of the Mellin transform. Thelatter is bounded by

�∑ρ

|f(ρ)|

and the summand is� 1/|γ|s by rapid decay—we can take any exponent. Sucha sum converges since the L-function is of order ≤ 1.

Lecture 29 The theorem of Hadamard and de laVallee-Poussin

29.1 Zero free region

Surprisingly and remarkably, most of the following discussion is a consequenceof the following elementary trigonometric inequality: for all θ ∈ R,

3 + 4 cos(θ) + cos(2θ) ≥ 0.

To see that this is true, note simply that the left-hand side is equal to 2(1 +cos(θ))2, which is of course nonnegative.

To recover the proper zero free region, we need to first establish that thereare no zeros on the line (1).

Theorem 29.1.1. ζ(1 + it) 6= 0 for all t ∈ R.

Proof. For s = σ + it, t 6= 0, σ > 1, we have

log ζ(s) =∑p

∞∑n=1

1

npnσe−itn log p.

Taking real parts,

Re log ζ(s) =∑p

∞∑n=1

1

npnσcos(tn log p).

Calling tn log p = θ, we apply the inequality, getting

3 Re log ζ(σ) + 4 Re log ζ(σ + it) + Re log ζ(σ + 2it) ≥ 0.

Date: April 3rd, 2019.


More to the point,

3 log|ζ(σ)|+ 4 log|ζ(σ + it)|+ log|ζ(σ + 2it)| ≥ 0.

Taking exponentials,

|ζ(σ)|3|ζ(σ + it)|4|ζ(σ + 2it)| ≥ 1.

Now ζ(σ) = 1σ−1 + O(1) as σ → 1+. Suppose ζ(σ + it) = 0 for some

t 6= 0. This implies |ζ(σ + it)|4 = Ot((σ − 1)4) as σ → 1+, and also that|ζ(σ + 2it)| = Ot(1) as σ → 1+. Hence

|ζ(σ)|3|ζ(σ + it)|4|ζ(σ + 2it)| → 0

as σ → 1+, which is a contradiction to it being bounded below by 1. Hencethere can’t be any zeros on Re(s) = 1.

Theorem 29.1.2. There exists a constant c > 0 such that

Z(ξ0) ⊂{s ∈ C

∣∣∣∣ Re(s) ≤ 1− c

log(2 + |Im(s)|)

},

i.e., ζ(s) 6= 0 for

Re(s) > 1− c

log(2 + |Im(s)|).

Proof. For σ > 1, we have the series representation

−ζ′

ζ(σ + it) =

∞∑n=1

Λ(n)

nσ+it,

where as above we write the imaginary exponent n−it = e−it logn.Hence taking real parts, we have

Re

(−ζ′

ζ(σ + it)

)=∞∑n=1

Λ(n)

nσcos(t log n).

Hence, in our trigonometric inequality,

3

(−ζ′

ζ(σ)

)+ 4 Re

(−ζ′

ζ(σ + it)

)+ Re

(−ζ′

ζ(σ + 2it)

)≥ 0.

Recall that by Hadamard factorisation, for s = σ + it ∈ Z(ξ0),

−ζ′

ζ(s) =

1

s+

1

s− 1+ b−

∑ρ∈Z(ξ0)

(1

s− ρ+

1

ρ

)+ζ ′∞ζ∞

(s).

For σ > 1,

−ζ′

ζ(σ) ≤ 1

σ − 1+ C1

because of the pole.


For s = σ + it, σ > 1, t 6= 0, we have

−Re

(ζ ′

ζ(σ + it)

)≤ C2 log(2 + |t|)−

∑ρ

Re

(1

s− ρ+

1

ρ

),

where the logarithm is from the arhimedian factor, bounded with Stirling’sformula. In particular,

Re1

s− ρ=

σ − β|s− ρ|2

≤ 1

|s− ρ|2,

so the sum is bounded.Applying this for σ + it, we get

−Re

(ζ ′

ζ(σ + it)

)≤ C2 log(2 + |t|),

and for σ + 2it,

−Re

(ζ ′

ζ(σ + 2it)

)≤ C3 log(2 + |t|)

as well. However we don’t want the first of those two estimates above quiteas-is, as it gets rid of the information from the zeros. Instead let us detect zerosone at a time, by fixing ρ0 = β0 + iγ0 ∈ Z(ξ0), and taking s = σ+ it with t = γ0.Then

Re1

s− ρ= Re

1

σ + iγ0 − (β0 + iγ0)=

1

σ − β0,

so if we get rid of the sum over zeros, except keeping only this one term fromρ0, we have

−Re

(ζ ′

ζ(σ + it)

)≤ C2 log(2 + |t|)− 1

σ − β0.

Thus our basic inequality becomes

3

(1

σ − 1+ C1

)+ 4

(C2 log(2 + |t|)− 1

σ − β0

)+ C3 log(2 + |t|) ≥ 0

for t = γ0 and all σ > 0. Thus

4

σ − β0− 3

σ − 1≤ C4 log(2 + |t|)

for all σ > 1. Taking in particular σ = 1 + δlog(2+|t|) , for some δ > 0, we get

β0 ≤ 1 +δ

log(2 + |t|)− 4δ

(δC4 + 3) log(2 + |t|)≤ 1− c

log(2 + |t|)

for some c > 0 by taking δ small enough.

EXCEPTIONAL ZEROS 73

Lecture 30 Exceptional zeros

30.1 Zero free region for L(s, χ)

Theorem 30.1.1. There exista a constant c > 0 such that if χ is a Dirichletcharacter modulo q, then L(s, χ) has no zeros in a region defined by

σ > 1− c

log(q(2 + |t|)),

writing s = σ + it, unless χ is a real quadratic character in which case L(s, χ)has at most one (necessarily real) zero β < 1 in this region. Such a zero iscalled an exceptional zero or a Siegel zero.

Proof. The strategy is essentially the same, we just need to take a little bit ofextra care sometimes. Letting s = σ + it with σ > 1 we again have the seriesrepresentation

−L′

L(s, χ) =

∞∑n=1

Λ(n)

nσχ(n)e−it logn.

Note that |χ(n)e−it logn| = 1 for (n, q) = 1 (else 0), so let us write χ(n)e−it logn =eiθ, so that cos(θ) = Re(χ(n)e−it logn), and correspondingly e2iθ = χ2(n)e−i2t logn,and so cos(2θ) = Re(χ2(n)e−2it logn).

We also need to be a little bit careful with the constant 3 in our basictriginometric inequality: we’ll write 1 = χ0(n) for (n, q) = 1 (and we don’tcare about other n, since they contribute nothing). Hence, multiplying by theappropriate factors and summing, the inequality becomes

−3L′

L(σ, χ0) + 4 Re

(−L′

L(σ + it, χ)

)+ Re

(−L′

L(σ + 2it, χ2)

)≥ 0.

As before we want an upper bount for this. Note first of all that if χ = χ0, then

L(s, χ) = ζ(s)∏p|q

(1− p−s),

so the result follows from Theorem 29.1.2 for ζ(s).Now consider the case where χ is complex, in which case χ2 is nontrivial, so

for σ > 1,

−L′

L(σ, χ0) =

∞∑n=1

χ0(n)Λ(n)

nσ≤ −ζ

′

ζ(σ) ≤ 1

σ − 1+ C1.

Secondly, by Hadamard factorisation,

−L′

L(s, χ) =

1

2log

q

π+

1

2

Γ′

Γ

(s+ κ

2

)− b(χ)−

∑0≤Re(ρ)≤1

(1

s− ρ− 1

ρ

).

We showed before that

Re b(χ) = −∑ρ

1

ρ,

Date: April 5th, 2019.

EXCEPTIONAL ZEROS 74

so taking real parts we have

Re

(−L′

L(s, χ)

)=

1

2log

q

π+

1

2Re

Γ′

Γ

(s+ κ

2

)−∑ρ

Re1

s− ρ.

By Stirling’s formula for the Gamma factors and by nonnegativity of the lastsum, we have

Re

(−L′

L(σ + 2it, χ2)

)≤ C2 log(q(2 + |t|))

for χ2 6= χ0.For the middle term, as before, we isolate a zero: given ρ0 = β0 + iγ0, take

s = σ + it with σ > 0 and t = γ0. Then keeping precisely that term from theabove sum,

Re

(−L′

L(σ + it, χ)

)≤ C2 log(q(2 + |t|))− 1

σ − β0.

Inserting these and going through the motions,

β0 ≤ 1− c

log(q(2 + |t|)),

for some c > 0 and t = γ0.Secondly, and more intricately, we need to consider the case where χ is real.

The above bound works only for χ2 6= χ0, but now we in fact have preciselyχ2 = χ0, so we need to do something else.

There are three cases to consider now. First, we are dealing with zeros atgreat height, say |γ0| ≥ 6(1 − β0). Take s = σ + it with t = γ0 again, andestimate

Re

(−L′

L(σ + 2it, χ2)

)≤ C3 log(q(2+|t|))+Re

1

(σ + 2it)− 1≤ C3 log(q(2+|t|))+ σ − 1

(σ − 1)2 + 144(1− β0)2

where the last term comes from the term 1s−1 , and the estimate stems from

Re1

(σ + 2it)− 1=

σ − 1

(σ − 1)2 + 4γ20

≤ σ − 1

(σ − 1)2 + 144(1− β0)2.

Therefore

3

(1

σ − 1+ C1

)+4

(C2 log(q(2 + |t|))− 1

σ − β0

)+

(σ − 1

(σ − 1)2 + 144(1− β0)2+ C3 log(q(2 + |t|))

)≥ 0.

Taking σ = 1 + 6(1− β0), it eventually follows that

β0 ≤ 1− c

log(q(2 + |t|)).

Secondly, suppose we are concerned with a low-lying zero, say 0 < |γ0| <6(1 − β0). Since χ is rael, ρ0 = β0 + iγ0 being a zero implies ρ0 = β0 − iγ0 isone too. Taking s = σ since the height t is quite small now, we have

−L′

L(σ, χ) ≤ −Re

(1

σ − ρ0

)−Re

(1

σ − ρ0

)+C5 log q =

−2(σ − β0)

(σ − β0)2 + γ20

+C5 log q.

LANDAU’S THEOREM 75

Since t = 0 and χ real, χ2 = χ0, our inequality becomes simply

−L′

L(σ, χ0)− L′

L(σ, χ) ≥ 0.

For the first term the old bound works just fine, and for the second term we usethe above. Taking σ = 1 + δ(1 − β0), and then later δ = 13 (for example), weget the result we want.

Finally, suppose γ0 = 0, so ρ0 = β0 ∈ R. This is where we might get anexceptional zero.

Suppose there are two real zeros, say, 0 ≤ β0 ≤ β1 < 1. Then

−L′

L(σ, χ) ≤ − 1

σ − β0− 1

σ − β1+ C5 log q ≤ − 2

σ − β0+ C5 log q,

so1

σ − 1+ C1 −

2

σ − β0+ C5 log q ≥ 0.

Taking σ = 1 + δ(1− β0),

1

1− β0

(1

δ+

2

δ + 1

)+ C6 log q ≥ 0,

which if we take δ = 2 gives us

β0 ≤ 1− c

log q,

for some c > 0.Now this gives the same sort of zero free region, except for the notable hitch

that it bounds β0—we might still have the hypothetical β1 to the right of it.Note that there can only ever be one exceptional zero, since otherwise we

can repeat this argument with β1 in place of β0.

A question one now asks is this: For which q ∈ N is there a real quadraticcharacter χ (mod q) such that L(s, χ) has an exceptional zero?

There are no known examples. Moreover, if we assume the Riemann hy-pothesis for L(s, χ), then such q cannot exist.

Indeed, Landau shows that such a q is very, very rare if it exists.

Lecture 31 Landau’s theorem

31.1 Exceptional zeros

Theorem 31.1.1 (Landau). There exists a constant c > 0 such that if χ1

(mod q1) and χ2 (mod q2) are real quadratic characters such that χ1χ2 is non-trivial, then L(s, χ1)L(s, χ2) has at most one real zero β such that

1− c

log q1q2< β < 1.



Proof. We have

1 + χ1(n) + χ2(n) + χ1χ2(n) = (1 + χ1(n))(1 + χ2(n)) ≥ 0,

so for σ > 1, multiply by Λ(n)/nσ and sum over n, getting

−ζ′

ζ(σ)− L′

L(σ, χ1)− L′

L(σ, χ2)− L′

L(σ, χ1χ2) ≥ 0.

Now assume L(s, χi) has an exceptional zero βi, and assume β1 ≤ β2. Wewant an upper bound for the left-hand side above, and we have some from ourprevious discussion:

−ζ′

ζ(σ) ≤ 1

σ − 1+ c1,

as well as

−L′

L(σ, χi) ≤ −

1

σ − βi+ c2 log qi,

and finally

−L′

L(σ, χ1χ2) ≤ c3 log q1q2

since by assumption χ1χ2 6= χ0. Hence

1

σ − 1− 1

σ − β1− 1

σ − β2+ c4 log q1q2 ≥ 0.

Since β1 ≤ β2, we have moreover

1

σ − 1− 2

σ − β1+ c4 log q1q2 ≥ 0.

Taking σ = 1 + 2(1− β1), we get

β1 ≤ 1− c

log q1q2,

for some c > 0.

Note that this means that if β1 = β2, then neither zero is in this specialregion.

Corollary 31.1.2. There exists a constant c > 0 such that∏χ (mod q)

L(s, χ)

has at most one zero in the region

σ > 1− c

log q(2 + |t|),

for s = σ + it. If such a zero exists, then it is real and the associated characterχ is quadratic.


Corollary 31.1.3 (Page). There exista a constant c > 0 such that for everyQ ≥ 1, ∏

q≤Q

∏∗

χ (mod q)

L(s, χ),

where by the asterisk we mean a product over primitive characters, has at mostone zero in the region

σ < 1− c

logQ(2 + |t|),

for s = σ + it. If such a zero exists, then it’s real and the associated characteris quadratic.

31.2 Siegel’s theorem

Siegel’s theorem is about bounding the zero-free region away from 1, and essen-tially the idea is to leverage upper bounds of the L-function at 1.

Theorem 31.2.1. Let χ 6= χ0 be a quadratic character modulo q. ThenL(1, χ)� q−1/2.

Proof. Define τ(n) := 1?χ(n). We have showed before (in past homework) thatτ(n) ≥ 0 for all n and τ(n) ≥ 1 if n is a perfect square.

Consider ∑n≥1

τ(n)e−nx .

We can easily get a lower bound for this:∑n≥1

τ(n)e−nx ≥

∑m≥1

τ(m2)e−m2

x ≥∑m≥1

e−m2

x � x1/2

by comparing with the integral.To get asymptotics for it, note that by Mellin inversion

e−x =1

2πi

∫(2)

Γ(s)X−s ds,

so ∑n≥1

τ(n)e−nx =

1

2πi

∫(2)

∑n≥1

τ(n)

ns

Γ(s)Xs ds,

where the inner sum is ζ(s)L(s, χ) by the definition of τ(n). Shifting the lineof integration from (2) to (− 1

2 ) picks up residues from ζ(s) at s = 1, namelyXL(1, χ), and from Γ(s) at s = 0, namely ζ(0)L(0, χ).

Hence the sum is

L(1, χ)X + ζ(0)L(0, χ) +1

2πi

∫(− 1

2 )

ζ(s)L(s, χ)Γ(s)Xs ds.

We have ζ(0) = − 12 , and L(0, χ) ≥ 0 since it’s positive at 1, and we translate

by the functional equation. By studying the completed L-function for L(s, χ)at 3/2, using Stirling, and then using the functional equation we translate to

SIEGEL’S THEOREM 78

−1/2 and get that L(s, χ)� q|t| on this line. Hence by exponential decay, theentire integral is � qX−1/2.

ThusX1/2 � L(1, χ)X +O(qX−1/2).

Taking X = q, we deduce L(1, χ)� q−1/2.

Corollary 31.2.2. There exists a cosntant c > 0 such that if χ (mod q) is aquadratic character with L(s, χ) having an exceptional zero β, then

β ≤ 1− c

q1/2(log q)2.

Proof. We can show L′(σ, χ)� (log q)2 for

1− 1

log q≤ σ ≤ 1.

By the mean value theorem,

q−1/2 � L(1, χ) = L(1, χ)− L(β, χ) = L′(σ, χ)(1− β)� (1− β)(log q)2,

soβ ≤ 1− c

q1/2(log q)2,

for c > 0.

Theorem 31.2.3 (Siegel’s theorem). For any ε > 0, there exists a constantC(ε) > 0 such that

L(1, χ) ≥ C(ε)

qε

for all primitive quadratic characters χ (mod q).

Remark 31.2.4. This constant is ineffective—we don’t know how to compute anumerical value for ε < 1/2.

Lecture 32 Siegel’s theorem

32.1 Proof

Proof. Let χ1 (mod q1) and χ2 (mod q2) be two primitive characters. Define

τ(n) = 1 ? χ1 ? χ2 ? χ1χ2(n),

so thatL(s, τ) = ζ(s)L(s, χ1)L(s, χ2)L(s, χ1χ2).

We have (essentially) showed before that τ(n) ≥ 0 for all n.We claim that for any ε > 0, there exists a primitive quadratic character

χ1 (mod q1) and β with 1 − ε < β < 1 such that L(β, τ) ≤ 0 for all primitivequadratic characters χ2 (mod q2), with q2 6= q1.


SIEGEL’S THEOREM 79

Remark 32.1.1. We do not know how to compute this q1, which is what makesthis result ineffective.

We prove this in two cases. First, suppose there are no zeros on [1 − ε, 1]of L(s, χ) for any quadratic character χ. Then take any χ1 (mod q1). Wehave L(β, χ) > 0 since L(1, χ) > 0 for quadratic characters χ 6= χ0. LikewiseL(s, χ2) > 0, and L(s, χ1χ2) > 0 since χ1χ2 6= χ0 because q1 6= q2.

Moreover ζ(β) < 0 since it is approximately 1/(s−1) near 1, so L(β, τ) < 0.Secondly, suppose there exists χ (mod ()q) with a real zero β ∈ [1 − ε, 1].

Take χ1 = χ, so that L(β, χ1) = 0. Then of course L(β, τ) = 0 too.With this in hand we are ready to prove Siegel’s theorem. Take χ1, q1, and

β as in the claim. Let

λ = Ress=1

L(s, τ) = L(1, χ1)L(1, χ2)L(1, χ1χ2).

Then1

e≤∑n≥1

τ(n)

nβe−

nx =

1

2πi

∫(2)

L(s+ β, τ)Γ(s)Xs ds

by Mellin inversion. Moving the line of integration from (2) to (−β), we pickup poles at s = 1 − β and s = 0 with residues λΓ(1 − β)X1−β and L(β, τ)respectively. Thus

1

e≤ λΓ(1− β)X1−β + L(β, τ) +

1

2πi

∫(−β)

L(s+ β, τ)Γ(s)Xs ds,

where by choice of β, L(β, τ) ≤ 0 by the claim.We have

L(it, χ)� ((2 + |t|)q)1/2+ε

for χ (mod q), andζ(it)� (2 + |t|)1/2+ε,

soL(it, τ)� (2 + |t|)2+ε(q1q2)1+ε.

Note that −β is close to −1, and Γ(s) has a simple pole at s = −1, soΓ(−β) = O(1/(1− β)), meaning that the integral above is

O

((q1q2)1+εX−β

1− β

),

and therefore

1

e≤ λΓ(1− β)X1−β +O

((q1q2)1+εX−β

1− β

),

having dropped L(β, τ) since it’s nonpositive. Now if λX1−β ≥ (q1q2)1+εX−β ,so λX ≥ (q1q2)1+ε, so since

λ = L(1, χ1)L(1, χ2)L(1, χ1χ2)� (q1q2)−1,

we haveX

q1q2> (q1q2)1+ε,

THE PRIME NUMBER THEOREM IN ARITHMETIC PROGRESSIONS 80

meaning that

1� λX1−β

1− β

provided (q1q2)2+ε � X, guaranteeing that

λΓ(1− β)X1−β ≥ O(

(q1q2)1+εX−β

1− β

)so that we have the asymptotics we want.

Now for χ 6= χ0 (mod q), L(1, χ)� log q, so

λ� L(1, χ2)(log q1)(log q1q2).

Take X = (q1q2)2+ε, which may be a new epsilon, and insert into the bound onλ above. Then

L(1, χ2)� 1

(log q1)(log q1q2)· (q1q2)−(2+ε)(1−β)(1− β)

Now q1 and β depend on ε, so we get

L(1, χ2) ≥ C(ε)q−(2+ε)(1−β)2 (log q2)−1.

Moreover, (1− β) < ε, so (2 + ε)(1− β) < 3ε, and log q � qε, so

L(1, χ2) ≥ C(ε)

qε.

Corollary 32.1.2. For any ε > 0, there exists a constant C(ε) > 0 such thatif χ (mod q) is a quadratic character such that L(s, χ) has an exceptional zeroβ, then

β ≤ 1− C(ε)

qε.

Proof. The same as Corollary 31.2.2.

Lecture 33 The Prime number theorem in Arith-metic progressions

33.1 Prime number theorem

Recall how we proved the ordinary Prime number theorem by

π(x) ∼ x

log x⇔ ϑ(x) =

∑p≤x

log p ∼ x ⇔ ψ(x) =∑n≤x

Λ(n) ∼ x.

We prove this last step by smoothing the sum over all n with a smoothingfunction, using the explicit formula to write in terms of the Zeta function, andthen use the zero-free region of ζ(s).


THE PRIME NUMBER THEOREM IN ARITHMETIC PROGRESSIONS 81

The strategy for arithmetic progressions is largely the same. For a and qwith (a, q) = 1,

π(x; q, a) =∑p≤x

p≡a (mod q)

1 ∼ π(x)

ϕ(q)

is equivalent to

ϕ(x; q, a) =∑p≤x

p≡a (mod q)

log p ∼ x

ϕ(q),

is equivalent to

ψ(x; q, a) =∑n≤x

n≡a (mod q)

Λ(n) ∼ x

ϕ(q),

which we will prove.Let f ∈ C∞c ((0,∞)), (a, q) = 1. Define

ψf (x; q, a) :=∑n≤x

n≡a (mod q)

Λ(n)f(nx

).

By the orthogonality relations,

ψf (x; q, a) =1

ϕ(q)

∑χ (mod q)

χ(a)ψf (x, χ),

whereψf (x, χ) :=

∑n≥1

Λ(n)χ(n)f(nx

).

For χ = χ0 we get

ψf (x, χ0) =∑

(n,q)=1

Λ(n)f(nx

)=∑n≥1

Λ(n)f(nx

)−

∑(n,q)>1

Λ(n)f(nx

).

Studying each sum separately,∑(n,q)>1

Λ(n)f(nx

)�

∑n�x

(n,q)>1

Λ(n) =∑p|q

∑pα<<x

log p� (log x)∑p|q

log p = (log x)(log q).

On the other hand,∑n≥1

Λ(n)f(nx

)= x

∫ ∞0

f(t) dt+Of (x exp(−c√

log x))

for some c > 0 by Corollary 28.1.3.Hence

ψf (x; q, a) =x

ϕ(q)

∫ ∞0

f(t) dt+1

ϕ(q)

∑χ 6=χ0 (mod q)

χ(a)ψf (x, χ)+Of

(1

ϕ(q)(x exp(−c

√log x) + log x log q)

).

Proposition 33.1.1. Suppose q � exp( 12

√log x).

BOMBIERI-VINOGRADOV 82

(a) If χ 6= χ0 such that L(s, χ) has no exceptional zeros, then there exists aconstant c > 0 such that ψf (x, χ)�f x exp(−c

√log x) uniformly in χ.

(b) If χ 6= χ0 such that L(s, χ) has an exceptional zero β ∈ R, then thereexists a constant c > 0 such that

ψf (x, χ) = −f(β)Xβ +Of (x exp(−c√

log x))

uniformly in χ.

Proof. Choose x large enough such that f( 1nx ) = 0 for all n ≥ 1. By the explicit

formula,

ψf (x, χ) =1

2πi

∫( 12 )

(L′∞L∞

(s, χ)− L′∞L∞

(1− s, χ))f(s)xs ds−∑

ρ∈Z(Λ(s,χ))

f(ρ)xρ.

The integral is � x1/2. By Theorem 30.1.1,∑′

ρ∈Z(Λ(s,χ))

|f(ρ)| � exp

(− c log x

log q(2 + γ)

),

with Im(ρ) = γ.Then ∑′

ρ∈Z(Λ(s,χ))log(q(2+|γ|))≤

√log x

|f(ρ)| � exp(−c√

log x),

and ∑′

ρ∈Z(Λ(s,χ))log(q(2+|γ|))>

√log x

|f(ρ)| �∑|f(ρ)| �

∑ 1

|γ|2�∑ 1

|γ|3/21

|γ|1/2.

Corollary 33.1.2. Suppose q � exp( 12

√log x). Then

ψ(x; q, a) =x

ϕ(q)

∫ ∞0

f(t) dt− 1

ϕ(q)

∑χ

χ(a)f(β)xβ +Of (x exp(−c√

log x))

for some c > 0, where the sum is over χ with exceptional zero β.

Lecture 34 Bombieri-Vinogradov

34.1 Prime number theorem in arithmetic progressions

We can now finish the prime number theorem:

Theorem 34.1.1 (Siegel-Walfisz theorem). Let C < 0 (effective) as in Corol-lary 33.1.2. Then for all A > 0, q � (log x)A, (a, q) = 1, and X ≥ 2, wehave

ψf (x; q, a) =x

ϕ(q)

∫ ∞0

f(t) dt+Of,A(x exp(−c√

log x)),

where the implied constant depending on A and f is ineffective.



The ineffectiveness comes from us using Siegel’s theorem in the proof.

Proof. We have

xβ = x exp(−(1− β) log x) ≤ x exp(−C(ε)q−ε log x)

by Siegel’s theorem. Since q � (log x)A,

xβ � x exp(−C(ε)(log x)1−Aε).

Taking ε = 1/(3A), we get

xβ � x exp(−C(ε)(log x)2/3).

Thus also:

Theorem 34.1.2 (Siegel-Walfisz). Suppose A > 0. For q � (log x)A, (a, q) =1, x ≥ 2, we have

ψ(x; q, a) =x

ϕ(q)+OA(exp(−c

√log x)),

for some c > 0, which is equivalent with the Prime number theorem in arithmeticprogressions.

Another way of saying this:

Theorem 34.1.3 (Siegel-Walfisz, variant). Suppose A > 0. For q � (log x)A

and χ (mod q), we have

ψ(x, χ)− δ(χ)x�A x exp(−c√

log x)

for some c > 0, where

δ(χ) =

{1, if χ = χ0,

0, if χ 6= χ0.

34.2 The Bombieri-Vinogradov theorem

There are two questions we want to ask about all of this.

• In the proof we use that there is no real zeros of L(s, χ) near 1 (i.e.,Siegel’s theorem). What is the error term if we assume the GeneralisedRiemann Hypothesis (GRH), i.e., all zeros of L(s, χ) in 0 ≤ Re(s) ≤ 1 lieon Re(s) = 1

2 .

• Can we enlarge the range of q?

Under GRH we can deduce that

ψ(x; q, a) =x

ϕ(q)+O(x1/2(log x)2)

uniformly for all q.To have this make asymptotic sense, note that ϕ(q) ∼ q, so

x

q> x1/2(log x)2


implies

q <x1/2

(log x)2,

meaning q < x1/2.A remarkable result is that we can achieve the GRH error term on average:

Theorem 34.2.1 (Bombieri-Vinogradov). For any A > 0, Q ≤ X1/2,∑q≤Q

max(a,q)=1

maxy≤X

∣∣∣∣ψ(y; q, a)− y

ϕ(q)

∣∣∣∣�A X(logX)−A +X1/2Q(logX)4.

Remark 34.2.2. (i) The implied constant is ineffective since we use Siegel’stheorem in the proof. However, in 2013, Akbary and Hambrook managedto prove Bombieri-Vinogradov without relying on Siegel’s theorem.

(ii) Taking X1/2(logX)−A−4 ≤ Q ≤ X1/2, then the second term in the theo-rem is larger, so the sum is bounded by X1/2(logX)4. The theorem saysthat if we average over q, this bound is as good as GRH for all χ (mod q),q ≤ X1/2(logX)−A−4.

(iii) For Q > X1/2, we use the trivial bound

ψ(X; q, a) =∑n≤X

n≡a (mod q)

Λ(n) ≤ (logX)

(X

q+ 1

),

so the sum is

�∑q≤Q

(X

q+ 1

)logX � X(logX)(logQ) +Q(logX),

which is a better bound than the theorem.

Proof. By orthogonality,

ψ(X; q, a) =1

ϕ(q)

∑χ (mod q)

χ(a)ψ(X,χ).

Hence ∣∣∣∣ψ(y; q, a)− y

ϕ(q)

∣∣∣∣ =1

ϕ(q)

∣∣∣∣∣∣∑

χ (mod q)

χ(a)ψ(y, χ)− δ(χ)y

∣∣∣∣∣∣.Getting rid of the character, since it is of modulus 1, we may potentially losesome cancellation. We get

≤ 1

ϕ(q)

∑χ (mod q)

|ψ(y, χ)− δ(χ)y|.

So it suffices to bound∑q≤Q

1

ϕ(q)

∑χ (mod q)

maxy≤X|ψ(y, χ)− δ(χ)y|.

BOMBIERI-VINOGRADOV, CONTINUED 85

We can moreover reduce this to primitive characters. If χ (mod q) is inducedby a primitive character χ∗ (mod q∗), q∗ | q, then

ψ(y, χ) = ψ(y, χ∗) +O

∑p|qp-q∗

(log p)∑

k≤ log ylog p

1

.

To see this, note that

Z Zq∗Z C

ZqZ

χ∗

χ

Evaluating at n = pk, p | q, p - q∗, we have χ(n) = 0 but χ∗(n) 6= 0, and thelog p accounts for Λ(pk), and the inner sum counts how many of them thereare.

Lecture 35 Bombieri-Vinogradov, continued

35.1 The trivial case

The remainder at the end of the last discussion is of order (log y)(log q), so thesum is∑

q≤Q

1

ϕ(q)

∑q∗|q

∑∗

χ∗ (mod q∗)

maxy≤X|ψ(y, χ∗)− δ(χ∗)y|+O(Q(log q)(logX)).

Writing q = q∗r,∑q∗≤Q

∑r≤Q/q∗

1

ϕ(q∗r)

∑∗

χ∗ (mod q∗)

maxy≤X|ψ(y, χ∗)− δ(χ∗)y|. (35.1.1)

Note that ϕ(q∗r) ≥ ϕ(q∗)ϕ(r). Moreover∑q≤Q

1

ϕ(q)� logQ

since ∑q≤Q

1

ϕ(q)≤∏p≤Q

∞∑n=0

1

ϕ(pn)=∏p≤Q

(1 +

∞∑n=1

1

pn − pn−1

)

=∏p≤Q

(1− 1

p

)−1(1 +

1

p(p− 1)

)� logQ.

Thus for (35.1.1) we have the bound

�∑q≤Q

logQ

ϕ(q)

∑∗

χ (mod q)



BOMBIERI-VINOGRADOV, CONTINUED 86

Let R = (logX)A+4. We will splot the range of q according to whetherq ≤ R, for which we can use Siegel’s theorem, and R < q ≤ Q, for which weneed new machienery, namely the Large Sieve and Polya’s inequality.

Claim. With definitions as above,∑q≤R

logQ

ϕ(q)

∑∗

χ (mod q)

maxy≤X|ψ(y, χ)− δ(χ)y| �A R(logX)X exp(−c

√logX)

for some c > 0. Note that this moreover is

� X(logX)−K

for any K > 0.

Proof. We consider two cases: y ≤√X, and

√X < y ≤ X. First, when

y ≤√X,

|ψ(y, χ)− δ(χ)y| � y ≤√X

trivially since we can bound the left-hand side by ψ(y) ∼ y. Hence∑q≤R

logQ

ϕ(q)

∑∗

χ (mod q)

√X � (logQ)

√XR� (logX)

√XR.

Second, when√X < y ≤ X, q ≤ R = (logX)A+4 � (log y)A+4, so by

Siegel-Walfisz,

|ψ(y, χ)− δ(χ)y| �A y exp(−c√

log y)� X exp(−c√

logX).

Hence∑q≤R

logQ

ϕ(q)

∑∗

χ (mod q)

X exp(−c√

logX)� (logQ)X exp(−c√

logX)R,

and logQ� logX.

35.2 The nontrivial part

We now need to deal with∑R<q≤Q

logQ

ϕ(q)

∑∗

χ (mod q)


Note that in this range of q, or in particular q > 1, there are no trivial primitivecharacters modulo q, so δ(χ) = 0, and we really only need to deal with

logQ∑

R<q≤Q

1

ϕ(q)

∑∗

χ (mod q)

maxy≤X|ψ(y, χ)|. (35.2.1)

Theorem 35.2.1 (Basic mean value theorem). Let

T (X,Q) :=∑q≤Q

q

ϕ(q)

∑∗

χ (mod q)

maxy≤X|ψ(y, χ)|.

Then for Q ≥ 1, X ≥ 2, we have

T (X,Q)� (X +X5/6Q+X1/2Q2)(logXQ)3.

THE LARGE SIEVE 87

We will prove this at some point, but for now let us assume it true.Let

f(q) =1

ϕ(q)

∑∗

χ (mod q)

maxy≤X|ψ(y, χ)|,

the inner summand above, so that (35.2.1) is

(logQ)∑

R<q≤Q

f(q) = (logQ)∑

R<q≤Q

1

qf(q)q.

Writing this as a Stieltje’s integral,

(logQ)

∫ Q

R

1

td(∑q≤t

f(q)q),

where the weight is T (X, t). Integrating by parts gives us

(logQ)

(1

tT (X, t)

∣∣∣∣QR

+

∫ Q

R

1

t2T (X, t) dt

)Using the Mean value theorem above, this is

� (logQ)

(Q−1(X +X5/6Q+X1/2Q2)(logXQ)3 +

∫ Q

R

t−2(X +X5/6t+X1/2t2)(logXt)3 dt

)and by bounding logQ� logX and logXt� logXQ,

� Q−1(X+X5/6Q+X1/2Q2)(logX)4 +(XR−1 +X5/6 logQ+X1/2Q)(logX)4,

which we can finally bound as

� X(logX)−A +X1/2Q(logX)4

since R = (logX)A+4.

Theorem 35.2.2 (Polya-Vinogradov inequality). Let χ 6= χ0 (mod q), withq > 1. Then ∣∣∣ M+N∑

n=M+1

∣∣∣ ≤ 2q1/2(log q).

Note that trivially the sum is ≤ q, but we know that if we sum over a fullset of residues it is 0, so we should expect a lot of cancellation. This theoremreveals just how much!

Lecture 36 The Large Sieve

36.1 The Large siece inequality

For N,Q > 1, we have

∑q≤Q

q

ϕ(q)

∑∗

χ (mod q)

∣∣∣∣∣M+N∑n=M+1

anχ(n)

∣∣∣∣∣2

� (N +Q2)1/2M+N∑n=M+1

|an|2


THE LARGE SIEVE 88

for any { an } ⊂ C.A consequence of this is

Theorem 36.1.1. ∑q≤Q

q

ϕ(q)

∑∗

χ (mod q)

maxy

∣∣∣ M∑m=1

N∑n=1

mn≤y

ambnχ(mn)∣∣∣

� (log 2MN)(M +Q2)1/2(N +Q2)1/2

(M∑m=1

|am|2)1/2( N∑

n=1

|bn|2)1/2

Note that without the condition mn ≤ y, the sum splits (since χ(mn) =χ(m)χ(n) by complete multiplicativity), and by Cauchy’s inequality

∑q≤Q

q

ϕ(q)

∑∗

χ (mod q)

∣∣∣ M∑m=1

N∑n=1

ambnχ(mn)∣∣∣

≤

∑q≤Q

q

ϕ(q)

∑∗

χ (mod q)

∣∣∣∣∣M∑m=1

amχ(m)

∣∣∣∣∣21/2∑

q≤Q

q

ϕ(q)

∑∗

χ (mod q)

∣∣∣∣∣N∑n=1

bnχ(n)

∣∣∣∣∣21/2

We can then use the Large sieve at each factor separately, getting

� (M +Q2)1/2

(M∑m=1

|am|2)2

(N +Q2)1/2

(N∑n=1

|bn|2)2

So dropping the condition mn ≤ y gives us something quite close. We need agood way to detect this condition mn ≤ y.

Lemma 36.1.2. For any T > 0, β > 0, α ∈ R, and β 6= |α|, we have∫ T

−Teitα

sin(tβ)

πtdt = δβ(α) +O

(1

T |β − |α||

),

where

δβ(α) =

{1, if |α| < β,

0, if |α| > β.

Proof sketch. First, ∫ ∞−∞

eitαsin(tβ)

πtdt = δβ(α).

This is essentially a consequence of studying∫(c)

ysds

s

for c > 0, and then letting c→ 0.Secondly, by integration by parts∫ ∞

T

eitαsin(tβ)

πtdt� 1

T |β − |α||,

and similarly for the lower tail.

THE LARGE SIEVE 89

Now to use this lemma to detect mn ≤ y, we take β = log y and α = logmn,except the lemma only applies if β 6= |α|. We can get around this: since mn ∈ Z,without loss of generality we can take y = k + 1

2 , k ∈ Z, so 0 ≤ k ≤MN . Thus

δβ(logmn) =

∫ T

−Teit log(mn) sin(t log y)

πtdt+O

(1

T |β − |α||

)=

∫ T

−T(mn)it

sin(t log y)

πtdt+O

(1

T |β − |α||

).

Note that the term (mn)it is of modulus 1, which will come in handy soon.Note also that

|β − |α|| = |log y − logmn| =∣∣∣∣log

mn

y

∣∣∣∣.Hence

M∑m=1

N∑n=1

mn≤y

ambnχ(mn) =

M∑m=1

N∑n=1

ambnχ(mn)δβ(logmn)

=

∫ T

−T

(M∑m=1

ammit

)(N∑n=1

bnnit

)sin(t log y)

πtdt+O

(T−1

M∑m=1

N∑n=1

|ambn|∣∣∣∣log

mn

y

∣∣∣∣−1).

We need a few estimates. First, |log x| � min{ 1, |x− 1| } for all x > 0, andtherefore∣∣∣∣log

mn

y

∣∣∣∣� min

{1,

∣∣∣∣mny − 1

∣∣∣∣ }� min

{1,

∣∣∣∣y + 12

y− 1

∣∣∣∣ }� 1

y� 1

MN.

Also,|sin(t log y)| ≤ min{ 1, |t log y| } ≤ min{ 1, |t| log 2MN },

where we’ve taken 2MN to avoid the possible issue when M = N = 1.Therefore

maxy

∣∣∣∣∣∣∣∣M∑m=1

N∑n=1

mn≤y

ambnχ(mn)

∣∣∣∣∣∣∣∣�∫ T

−T

∣∣∣∣∣M∑m=1

amm−itχ(m)

∣∣∣∣∣∣∣∣∣∣N∑n=1

bnn−itχ(n)

∣∣∣∣∣min

{1

|t|, log 2MN

}dt

+O

(T−1MN

M∑m=1

N∑n=1

|ambn|

),

and by Cauchy the sums in the error term is

�

(M∑m=1

|am|2)1/2

M1/2

(N∑n=1

|bn|2)1/2

N1/2.

Therefore the final error term is

O

T−1M3/2N3/2

(M∑m=1

|am|2)1/2( N∑

n=1

|bn|2)1/2

VAUGHAN’S IDENTITY 90

and we’ll pick an appropriate T toward the end.Now average over ∑

q≤Q

q

ϕ(q)

∑∗

χ (mod q)

,

bring it into the integral, since only the sums with χ in them depend on q, andso by the large sieve

∑q≤Q

q

ϕ(q)

∑∗

χ (mod q)

maxy

∣∣∣ M∑m=1

N∑n=1

mn≤y

ambnχ(mn)∣∣∣

� (M +Q2)1/2(N +Q2)1/2

(M∑m=1

|am|2)1/2( N∑

n=1

|bn|2)1/2 ∫ T

−Tmin

{1

|t|, log 2MN

}dt+

+∑q≤Q

q

ϕ(q)

∑∗

χ (mod q)

T−1(MN)3/2

(M∑m=1

|am|2)1/2( N∑

n=1

|bn|2)1/2

.

This last term is

� T−1(MN)3/2

(M∑m=1

|am|2)1/2( N∑

n=1

|bn|2)1/2 ∑

q≤Q

q,

with the final sum being � Q2, so taking T = (MN)3/2 we get the error wewant.

Finally the remaining integral is∫ T

−Tmin

{1

|t|, log 2MN

}dt� log(2MN)

since T = (MN)3/2. Note that therefore it is important we didn’t take T →∞at first to det just δβ(α), with no error, since then our upper bound is justinfinity.

Lecture 37 Vaughan’s Identity

37.1 Basic Mean Value Theorem, again

Lemma 37.1.1 (Vaughan’s Identity). Suppose u > 0, v > 0, y ≥ 2, andf : N→ C. Then ∑

n≤y

Λ(n)f(n) = S1 − S2 − s3 + s4

Date: April 22nd, 2019.


where

S1(f) =∑m≤u

µ(m)∑n≤ y

m

f(mn) log(n),

S2(f) =∑m≤uv

Cm∑n≤ y

m

f(mn), where Cm =∑k≤u`≤vkl=m

Λ(k)µ(`),

S3(f) =∑m>u

∑n>vmn≤y

∑k|mk>u

Λ(k)

µ(n)f(mn),

S4(f) =∑n≤v

Λ(n)f(n).

Remark 37.1.2. Note how S1(f) and S2(f) are bilinear forms of the type∑m,n amcmn.

If the length of the sum over m isn’t too big, then we can handle it. So we wantu and v small compared to y.

Next, S3(f) is a bilinear form of the type∑m,n ambncmn. Here we can use

the Large Sieve.Finally, S4(f) is of the same form as the original sum, but it’s shorter.

Proof. Consider the purely algebraic identity

−ζ′

ζ(s) = G(s)(−ζ ′(s))− F (s)G(s)ζ(s)− (−ζ ′(s)− F (s)ζ(s))(G(s)− 1

ζ(s)) + F (s),

where

F (s) =∑n≤u

Λ(n)

ns

and

G(s) =∑n≤v

µ(n)

ns.

Call the first terms D1(s), D2(s), D3(s), and D4(s), respectively, and writethem as Dirichlet series

Dj(s) =

∞∑n=1

aj(n)

ns

for j = 1, 2, 3, 4. Since the left-hand side above has coefficients Λ(n), we get theidentity

Λ(n) = a1(n)− a2(n)− a3(n) + a4(n),


where consequently

a1(n) =∑md=nd≤v

µ(d) log n,

a2(n) =∑

mdr=nm≤ud≤v

Λ(m)µ(d),

a3(n) =∑mk=um>uk>1

Λ(m)∑d|kd≤v

µ(d),

a4(n) =

{Λ(n), n ≤ u0, n > u.

Adding these up and weighing by f(n), we get∑Λ(n)f(n) =

∑a1(n)f(n)−

∑a2(n)f(n)−

∑a3(n)f(n) +

∑a4(n)f(n),

which is the identity sought.

We have an outstanding result ot prove.We want to show that

T (X,Q) :=∑q≤Q

q

ϕ(q)

∑∗

χ (mod q)

maxy≤X|ψ(y, χ)| � (X +X5/6Q+X1/2Q2)(logXQ)3,

whereψ(y, χ) =

∑n≤y

Λ(n)χ(n).

Proof of Basic mean value theorem. We have two cases. First, if Q2 > X, thenapply the Large Sieve with M = 1, a1 = 1, N = bXc, and bn = Λ(n), getting

T (X,Q)� (logX)Q(X +Q2)1/2

bXc∑n=1

Λ(n)s

1/2

The penultimate term is � (Q2)1/2 = Q, and the final term we can bound by

�

∑pα≤X

(log q)2

1/2

�((logX)2X

)1/2.

So in all, this expression is bounded by

� (logX)2Q2X1/2.

For the second case, Q2 ≤ X, we split into y ≤ u2 and u2 < y ≤ X. If wecheat a bit and let u = v = min{Q2, X1/3, XQ−2 } (which are properly acquired


by doing the calculations with u and v and then optimising), we get, for y ≤ u2,via the Large Sieve,

� Q(u2 +Q2)1/2(log u2)

∑n≤u2

Λ(n)2

1/2

� Q(u2 +Q2)1/2(log u2)(log(u2)u)1/2 � Q(uQ)(log u)2u

� (QX2/3 +Q2X1/3)(logX)2.

Finally for u2 < y ≤ X, apply Vaughan’s theorem with

f(n) =

{χ(n), n ≤ y0, n > y.

Then it suffices to bound

Tj :=∑q≤Q

q

ϕ(q)

∑∗

χ (mod q)

maxu2<y≤X

|Sj(χ)|.

lecture notes in analytic number...

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