Universit´ePierreetMarieCurie(Paris6)InstitutdeMath´ematiquesdeJussieu MichelWaldschmidt TheRiemannzetafunction CIMPA-ICTPResearchSchool,NesinMathematicsVillage2017 Artin L –functions,ArtinprimitiverootsConjectureandapplications

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ArtinL–functions, Artin primitive roots Conjecture and applications

CIMPA-ICTP Research School, Nesin Mathematics Village 2017

http://www.rnta.eu/Nesin2017/material.html

The Riemann zeta function Michel Waldschmidt

Universit´e Pierre et Marie Curie (Paris 6) Institut de Math´ematiques de Jussieu http://www.imj-prg.fr/~michel.waldschmidt/

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Content

The ring of arithmetic functions.

Dirichlet series.

Distribution of prime numbers : elementary results.

Abel’s Partial Summation Formula PSF.

Abscissa of convergence.

Euler Product.

Analytic continuation of the Riemann zeta function (I).

The Prime Number Theorem.

Analytic continuation of the Riemann zeta function (II).

Special values of the Riemann zeta function.

Functional Equation.

Riemann memoir.

Explicit formulae. Hadamard product Expansion.

The Riemann Hypothesis.

Diophantine problems

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The fundamental Theorem of arithmetic

n =Y

p

p^v^p⁽ⁿ⁾

ω(n) = X

p|n

1, Ω(n) =X

p|n

v_p(n).

Divisor function

τ(n) =σ₀(n) =d(n) = X

d|n

1.

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The fundamental Theorem of arithmetic

n =Y

p

p^v^p⁽ⁿ⁾

ω(n) = X

p|n

1, Ω(n) =X

p|n

v_p(n).

Divisor function

τ(n) =σ₀(n) =d(n) = X

d|n

1.

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Arithmetic functions f : N → C

Multiplicative function: f(ab) = f(a)f(b)if gcd(a,b) = 1.

Example : fork ∈C,σ_k(n) =P

d|nd^k.

Completely multiplicative function: f(ab) =f(a)f(b) for all a, b.

Example : fork ∈C,j^k =p_k :n7→n^k.

A multiplicative function is characterised by its values at prime powers :

f(n) = Y

p

f(p^v^p⁽ⁿ⁾).

A completely multiplicative function is characterised by its values at prime numbers :

f(n) = Y

p

f(p)^v^p⁽ⁿ⁾.

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Arithmetic functions f : N → C

d|nd^k.

f(n) = Y

p

f(p^v^p⁽ⁿ⁾).

f(n) = Y

p

f(p)^v^p⁽ⁿ⁾.

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Arithmetic functions f : N → C

d|nd^k.

f(n) = Y

p

f(p^v^p⁽ⁿ⁾).

f(n) = Y

p

f(p)^v^p⁽ⁿ⁾.

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Arithmetic functions f : N → C

d|nd^k.

f(n) = Y

p

f(p^v^p⁽ⁿ⁾).

f(n) = Y

p

f(p)^v^p⁽ⁿ⁾.

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Dirichlet series

D(f,s) =X

n≥1

f(n)n^−s.

Main example : Euler (1736), Riemann (1859) ζ(s) = X

n≥1

1 n^s·

Euler :ζ(2) = π²/6, ζ(4) =π⁴/90,ζ(2n)π⁻²ⁿ ∈Q(n ≥0), ζ(0) =−1/2, ζ(−1) = −1/12,

ζ(−2n) = 0 (n ≥1), ζ(−2n−1)∈Q (n≥0).

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Dirichlet series

D(f,s) =X

n≥1

f(n)n^−s.

n≥1

1 n^s·

ζ(−2n) = 0 (n ≥1), ζ(−2n−1)∈Q (n≥0).

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Dirichlet series

D(f,s) =X

n≥1

f(n)n^−s.

n≥1

1 n^s·

ζ(−2n) = 0 (n ≥1), ζ(−2n−1)∈Q (n≥0).

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Dirichlet series

D(f,s) =X

n≥1

f(n)n^−s.

n≥1

1 n^s·

ζ(−2n) = 0 (n ≥1), ζ(−2n−1)∈Q (n≥0).

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Dirichlet series

D(f,s) =X

n≥1

f(n)n^−s.

n≥1

1 n^s·

ζ(−2n) = 0 (n ≥1), ζ(−2n−1)∈Q (n≥0).

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Product of Dirichlet series

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

Convolution :

f ?g(n) = X

ab=n

f(a)g(b).

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The ring A of arithmetic functions

The ring of arithmetic functions is a commutative factorial ring A, the unity of which is the Kronecker delta function δ with

δ(n) =

(1 for n= 1, 0 for n>1.

The units are the arithmetic functionsf such that f(1)6= 0.

Exercise: check that the inverse g of a unit f is defined recursively by

g(1) =f(1)⁻¹, g(n) = −f(1)⁻¹ X

d|n,d<n

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

Iwaniec Chap. I

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The ring A of arithmetic functions

The ring of arithmetic functions is a commutative factorial ring A, the unity of which is the Kronecker delta function δ with

δ(n) =

(1 for n= 1, 0 for n>1.

The units are the arithmetic functionsf such that f(1)6= 0.

Exercise: check that the inverse g of a unit f is defined recursively by

g(1) =f(1)⁻¹, g(n) = −f(1)⁻¹ X

d|n,d<n

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

Iwaniec Chap. I

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Examples of arithmetic functions

The constant function 1=j⁰ :n 7→1.

The Kronecker delta functionδ.

The M¨obius function : µ.

Euler totient function :ϕ.

The functionj :j(n) =n for all n≥1.

The characteristic function of the squares :κ.

The lambda function :λ = (−1)^Ω. The von Mangoldt function :

Λ(n) =

(logp if n=p^m with p prime andm ≥1, 0 if n is not a prime power.

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Examples of arithmetic functions

Λ(n) =

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Examples of arithmetic functions

Λ(n) =

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Examples of arithmetic functions

Λ(n) =

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Examples of arithmetic functions

Λ(n) =

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Examples of arithmetic functions

Λ(n) =

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Examples of arithmetic functions

Λ(n) =

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Examples of arithmetic functions

Λ(n) =

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Exercises

µ ?1=δ, µ ?j =ϕ,

|µ|? κ=1, µ ? κ=λ,

|µ|?1= 2^ω, j^k ?1=σ_k, 1?1=τ, µ ?log = Λ.

M¨obius inversion formula :

g =1?f ⇐⇒ f =µ ?g.

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Exercises

µ ?1=δ, µ ?j =ϕ,

|µ|? κ=1, µ ? κ=λ,

|µ|?1= 2^ω, j^k ?1=σ_k, 1?1=τ, µ ?log = Λ.

M¨obius inversion formula :

g =1?f ⇐⇒ f =µ ?g.

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Examples of Dirichlet series

D(f,s) =X

n≥1

f(n)n^−s.

D(δ,s) = 1

D(1,s) =ζ(s) D(j^k,s) =ζ(s−k)

D(κ,s) =ζ(2s) D(µ,s) = 1 ζ(s)

D(τ,s) =ζ(s)² D(σ_k,s) =ζ(s −k)ζ(s) D(ϕ,s) = ζ(s −1)

ζ(s) D(|µ|,s) = ζ(s) ζ(2s) D(λ,s) = ζ(2s)

ζ(s) D(2^ω,s) = ζ(s)² ζ(2s)

D(log,s) =−ζ⁰(s) D(Λ,s) =−ζ⁰(s) ζ(s)

More exercises : Patterson p.10 exercise 1.3

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Examples of Dirichlet series

D(f,s) =X

n≥1

f(n)n^−s.

D(δ,s) = 1

D(1,s) =ζ(s) D(j^k,s) =ζ(s−k)

D(κ,s) =ζ(2s) D(µ,s) = 1 ζ(s)

ζ(s) D(|µ|,s) = ζ(s) ζ(2s) D(λ,s) = ζ(2s)

ζ(s) D(2^ω,s) = ζ(s)² ζ(2s)

D(log,s) =−ζ⁰(s) D(Λ,s) =−ζ⁰(s) ζ(s)

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Examples of Dirichlet series

D(f,s) =X

n≥1

f(n)n^−s.

D(δ,s) = 1

D(1,s) =ζ(s) D(j^k,s) =ζ(s−k)

D(κ,s) =ζ(2s) D(µ,s) = 1 ζ(s)

ζ(s) D(|µ|,s) = ζ(s) ζ(2s) D(λ,s) = ζ(2s)

ζ(s) D(2^ω,s) = ζ(s)² ζ(2s)

D(log,s) =−ζ⁰(s) D(Λ,s) =−ζ⁰(s) ζ(s)

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Examples of Dirichlet series

D(f,s) =X

n≥1

f(n)n^−s.

D(δ,s) = 1

D(1,s) =ζ(s) D(j^k,s) =ζ(s−k)

D(κ,s) =ζ(2s) D(µ,s) = 1 ζ(s)

ζ(s) D(|µ|,s) = ζ(s) ζ(2s) D(λ,s) = ζ(2s)

ζ(s) D(2^ω,s) = ζ(s)² ζ(2s)

D(log,s) =−ζ⁰(s) D(Λ,s) =−ζ⁰(s) ζ(s)

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Examples of Dirichlet series

D(f,s) =X

n≥1

f(n)n^−s.

D(δ,s) = 1

D(1,s) =ζ(s) D(j^k,s) =ζ(s−k)

D(κ,s) =ζ(2s) D(µ,s) = 1 ζ(s)

ζ(s) D(|µ|,s) = ζ(s) ζ(2s) D(λ,s) = ζ(2s)

ζ(s) D(2^ω,s) = ζ(s)² ζ(2s)

D(log,s) =−ζ⁰(s) D(Λ,s) =−ζ⁰(s) ζ(s)

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Examples of Dirichlet series

D(f,s) =X

n≥1

f(n)n^−s.

D(δ,s) = 1

D(1,s) =ζ(s) D(j^k,s) =ζ(s−k)

D(κ,s) =ζ(2s) D(µ,s) = 1 ζ(s)

ζ(s) D(|µ|,s) = ζ(s) ζ(2s) D(λ,s) = ζ(2s)

ζ(s) D(2^ω,s) = ζ(s)² ζ(2s)

D(log,s) =−ζ⁰(s) D(Λ,s) =−ζ⁰(s) ζ(s)

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The Prime Number Theorem.

Forx >0 let π(x) be the number of prime numbers ≤x :

π(x) = X

p≤x

1.

π(10) = 4,π(100) = 25, π(1000) = 168, π(10 000) = 1229.

Conjecture of Gauss (1792) and Adrien-Marie Legendre

(1798), proved by Hadamard and de la Vall´ee Poussin (1896) : PNT. For x → ∞,

π(x)∼ x logx· This is equivalent topn ∼nlogn.

Better approximation ofπ(x) : integral logarithm Li(x) =

Z x

2

dt logt·

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The Prime Number Theorem.

π(x) = X

p≤x

1.

π(10) = 4,π(100) = 25, π(1000) = 168, π(10 000) = 1229.

Z x

2

dt logt·

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The Prime Number Theorem.

π(x) = X

p≤x

1.

π(10) = 4,π(100) = 25, π(1000) = 168, π(10 000) = 1229.

Z x

2

dt logt·

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The Prime Number Theorem.

π(x) = X

p≤x

1.

π(10) = 4,π(100) = 25, π(1000) = 168, π(10 000) = 1229.

Z x

2

dt logt·

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The Prime Number Theorem.

π(x) = X

p≤x

1.

π(10) = 4,π(100) = 25, π(1000) = 168, π(10 000) = 1229.

Z x

2

dt logt·

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Distribution of prime numbers : elementary results.

θ(x) =X

p≤x

logp, ψ(x) = X

p^m≤x

logp =X

n≤x

Λ(n)

M(x) =X

n≤x

µ(n).

Theorem(elementary PNT) (P.L. Tchebyshev) : 1860’s

—cf. course of Francesco Pappalardi π(x) x

logx, θ(x)x, ψ(x)x for x → ∞.

The Prime Number Theoremπ(x)∼x/logx is equivalent θ(x)∼x and to ψ(x)∼x and to M(x) =o(x) as x → ∞.

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Distribution of prime numbers : elementary results.

θ(x) =X

p≤x

logp, ψ(x) = X

p^m≤x

logp =X

n≤x

Λ(n)

M(x) =X

n≤x

µ(n).

—cf. course of Francesco Pappalardi

π(x) x

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Distribution of prime numbers : elementary results.

θ(x) =X

p≤x

logp, ψ(x) = X

p^m≤x

logp =X

n≤x

Λ(n)

M(x) =X

n≤x

µ(n).

—cf. course of Francesco Pappalardi

π(x) x

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Elementary results on prime numbers.

Forn≥1,

2ⁿ≤ 2n

n

≤4ⁿ. Forn≥1 and p prime,

v_p(n!) =X

m≥1

n p^m

.

Forn≥1,

θ(2n)−θ(n)≤4 logn.

Consequence :

θ(x)≤C₁x, π(x)≥C₂ x logx·

Hindry p. 127–129

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Elementary results on prime numbers.

Forn≥1,

2ⁿ≤ 2n

n

v_p(n!) =X

m≥1

n p^m

.

Forn≥1,

Consequence :

Hindry p. 127–129

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Elementary results on prime numbers.

Forn≥1,

2ⁿ≤ 2n

n

v_p(n!) =X

m≥1

n p^m

.

Forn≥1,

Consequence :

Hindry p. 127–129

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Elementary results on prime numbers.

Forn≥1,

2ⁿ≤ 2n

n

v_p(n!) =X

m≥1

n p^m

.

Forn≥1,

Consequence :

Hindry p. 127–129

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Elementary results (continued)

For2≤y <x,

π(x)−π(y)≤ 1

logy(θ(x)−θ(y)).

θ(x)

logx ≤π(x)≤ θ(x)

logx + 2 log logx + x (logx)²·

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Elementary results (continued)

For2≤y <x,

π(x)−π(y)≤ 1

logy(θ(x)−θ(y)).

θ(x)

logx ≤π(x)≤ θ(x)

logx + 2 log logx + x (logx)²·

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Exercises

Deduce

X

p≤x

logp

p = logx +O(1) and

X

p≤x

1

p = log logx +C +O(1/logx)

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von Mangoldt function

ζ⁰(s)

ζ(s) =−X

n≥1

Λ(n) n^s

where Λ(n) =

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Exercise

Check

s→1lim

ζ(s)− 1 s−1

=γ where γ isEuler constant :

γ = lim

N→∞

1 1 +1

2 +1

3 +· · ·+ 1

N −logN.

Hint:

1 + 1 2 +1

3 +· · ·+ 1

N = 1 + Z N

1

([t]−t)dt t²

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Abel summation.

A₀ = 0, A_n=

n

X

m=1

a_m, a_n=A_n−An−1 (n≥1)

N

X

n=1

a_nb_n=

N−1

X

n=1

A_n(b_n−b_n+1) +A_Nb_N.

(telescopic sum)

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Abel summation.

A₀ = 0, A_n=

n

X

m=1

a_m, a_n=A_n−An−1 (n≥1)

N

X

n=1

a_nb_n=

N−1

X

n=1

A_n(b_n−b_n+1) +A_Nb_N.

(telescopic sum)

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Abel’s Partial Summation Formula PSF.

Leta_n be a sequence of complex numbers and A(x) =P

n≤xa_n. Let f be a function of classC¹. Then X

y<n≤x

a_nf(n) =A(x)f(x)− Z x

y

A(t)f⁰(t)dt.

Hindry p. 129–131

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Abel summation : an example.

Takean= 1 for all n, so that A(t) =btc. Then

N

X

n=M+1

f(n) = Z N

M

f(t)dt+ Z N

M

(t − btc)f⁰(t)dt

Takef(t) = 1/t. Then

N

X

n=1

1

n = logN +γ+O(1/N)

whereγ is Euler constant γ = 1−

Z ∞

1

{t}dt

t² where {t}=t − btc.

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Abel summation : an example.

Takean= 1 for all n, so that A(t) =btc. Then

N

X

n=M+1

f(n) = Z N

M

f(t)dt+ Z N

M

(t − btc)f⁰(t)dt

Takef(t) = 1/t. Then

N

X

n=1

1

n = logN +γ+O(1/N) whereγ is Euler constant

γ = 1− Z ∞

1

{t}dt

t² where {t}=t − btc.

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Abel summation for ζ .

Takea_n=n^−s for all n; assume Re(s)>1. An approximation of the series

X

n≥1

n^−s

is Z ∞

1

t^−sdt = 1 s−1· It is a fact that(s−1)ζ(s)→1 as s →1+.

We will see that if we remove this singularity of ζ ats = 1, the difference

ζ(s)− 1 s−1 becomes an entire function (analytic in C).

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Abel summation for ζ .

X

n≥1

n^−s

is Z ∞

1

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Abel summation for ζ .

X

n≥1

n^−s

is Z ∞

1

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Analytic continuation of ζ (s ) to Re(s ) > 0

(1−2^1−s)ζ(s)= 1^−s−2^−s+3^−s−4^−s+· · ·+(2n−1)^−s−(2n)^−s+· · ·

log(1−u) =−u 1 −u²

2 −u³

3 − · · · −uⁿ n − · · ·

1−1 2 +1

3−1

4 +· · ·= log 2.

s→1lim+

(1−2^1−s)ζ(s) = log 2

s→1lim₊(s −1)ζ(s) = 1

Hardy and Wright :lims→1P

ann^−s=P

ann⁻¹if the series in RHS converges.

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Analytic continuation of ζ (s ) to Re(s ) > 0

(1−2^1−s)ζ(s) = 1^−s−2^−s+3^−s−4^−s+· · ·+(2n−1)^−s−(2n)^−s+· · ·

log(1−u) =−u 1 −u²

2 −u³

3 − · · · −uⁿ n − · · ·

1−1 2 +1

3−1

4 +· · ·= log 2.

s→1lim+

(1−2^1−s)ζ(s) = log 2

s→1lim₊(s −1)ζ(s) = 1

ann^−s=P

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Analytic continuation of ζ (s ) to Re(s ) > 0

(1−2^1−s)ζ(s) = 1^−s−2^−s+3^−s−4^−s+· · ·+(2n−1)^−s−(2n)^−s+· · ·

log(1−u) =−u 1 −u²

2 −u³

3 − · · · −uⁿ n − · · ·

1−1 2 +1

3−1

4 +· · ·= log 2.

s→1lim+

(1−2^1−s)ζ(s) = log 2

s→1lim₊(s −1)ζ(s) = 1

ann^−s=P

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Analytic continuation of ζ (s ) to Re(s ) > 0

(1−2^1−s)ζ(s) = 1^−s−2^−s+3^−s−4^−s+· · ·+(2n−1)^−s−(2n)^−s+· · ·

log(1−u) =−u 1 −u²

2 −u³

3 − · · · −uⁿ n − · · ·

1−1 2 +1

3−1

4 +· · ·= log 2.

s→1lim+

(1−2^1−s)ζ(s) = log 2

s→1lim₊(s −1)ζ(s) = 1

ann^−s=P

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Analytic continuation of ζ (s ) to Re(s ) > 0

(1−2^1−s)ζ(s) = 1^−s−2^−s+3^−s−4^−s+· · ·+(2n−1)^−s−(2n)^−s+· · ·

log(1−u) =−u 1 −u²

2 −u³

3 − · · · −uⁿ n − · · ·

1−1 2 +1

3−1

4 +· · ·= log 2.

s→1lim+

(1−2^1−s)ζ(s) = log 2

s→1lim₊(s −1)ζ(s) = 1

ann^−s=P

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Analytic continuation of ζ (s ) to Re(s ) > 0

(1−2^1−s)ζ(s) = 1^−s−2^−s+3^−s−4^−s+· · ·+(2n−1)^−s−(2n)^−s+· · ·

log(1−u) =−u 1 −u²

2 −u³

3 − · · · −uⁿ n − · · ·

1−1 2 +1

3−1

4 +· · ·= log 2.

s→1lim+

(1−2^1−s)ζ(s) = log 2

s→1lim₊(s −1)ζ(s) = 1

ann^−s=P

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Analytic continuation of ζ (s ) (continued)

The functionζ(s)−1/(s−1)extends to an analytic function in Re(s)>0.

1 n^s =s

Z ∞

n

t^−s−1dt,

t

X

n=1

1 = [t].

ζ(s) =sX

n≥1

Z ∞

n

t^−s−1dt =s Z ∞

1

[t]t^−s−1dt

ζ(s) = s Z ∞

1

t^−sdt +s Z ∞

1

([t]−t)t^−s−1dt.

s Z ∞

1

t^−sdt = 1

s−1 + 1.

Hindry p. 139

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Analytic continuation of ζ (s ) (continued)

1 n^s =s

Z ∞

n

t^−s−1dt,

t

X

n=1

1 = [t].

ζ(s) =sX

n≥1

Z ∞

n

1

[t]t^−s−1dt

ζ(s) = s Z ∞

1

t^−sdt +s Z ∞

1

([t]−t)t^−s−1dt.

s Z ∞

1

t^−sdt = 1

s−1 + 1.

Hindry p. 139

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Analytic continuation of ζ (s ) (continued)

1 n^s =s

Z ∞

n

t^−s−1dt,

t

X

n=1

1 = [t].

ζ(s) =sX

n≥1

Z ∞

n

1

[t]t^−s−1dt

ζ(s) = s Z ∞

1

t^−sdt +s Z ∞

1

([t]−t)t^−s−1dt.

s Z ∞

1

t^−sdt = 1

s−1 + 1.

Hindry p. 139

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Analytic continuation of ζ (s ) (continued)

1 n^s =s

Z ∞

n

t^−s−1dt,

t

X

n=1

1 = [t].

ζ(s) =sX

n≥1

Z ∞

n

1

[t]t^−s−1dt

ζ(s) = s Z ∞

1

t^−sdt +s Z ∞

1

([t]−t)t^−s−1dt.

s Z ∞

1

t^−sdt = 1

s−1 + 1.

Hindry p. 139

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Analytic continuation of ζ (s ) (continued)

1 n^s =s

Z ∞

n

t^−s−1dt,

t

X

n=1

1 = [t].

ζ(s) =sX

n≥1

Z ∞

n

1

[t]t^−s−1dt

ζ(s) = s Z ∞

1

t^−sdt +s Z ∞

1

([t]−t)t^−s−1dt.

s Z ∞

1

t^−sdt = 1

s−1 + 1.

Hindry p. 139

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Analytic continuation of ζ (s ) (continued)

1 n^s =s

Z ∞

n

t^−s−1dt,

t

X

n=1

1 = [t].

ζ(s) =sX

n≥1

Z ∞

n

1

[t]t^−s−1dt

ζ(s) = s Z ∞

1

t^−sdt +s Z ∞

1

([t]−t)t^−s−1dt.

s Z ∞

1

t^−sdt = 1

s−1 + 1.

Hindry p. 139

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Abscissa of convergence

Letf(s) = P

n≥1a_nn^−s be a Dirichlet series converging at s₀. Then for anyC ≥1 it converges uniformly on the sector

n

s ∈C | Re(s−s₀)≥0, |s−s₀| ≤CRe(s−s₀)o

A Dirichlet series f(s) =P

n≥1a_nn^−s has an absissa of convergenceσ0 ∈R∪ {−∞,+∞}, such that the series converges for Re(s)> σ₀ and diverges forRe(s)< σ₀. The sum of the series is analytic in the half planeRe(s)> σ₀ and its derivatives are given by

f^(k)(s) =X

n≥1

a_n(−logn)^kn^−s.

Hindry p. 136

(71)

Abscissa of convergence

Letf(s) = P

n

s ∈C | Re(s−s₀)≥0, |s−s₀| ≤CRe(s−s₀)o

f^(k)(s) =X

n≥1

Hindry p. 136

(72)

Abscissa of convergence

Letf(s) = P

n

s ∈C | Re(s−s₀)≥0, |s−s₀| ≤CRe(s−s₀)o

f^(k)(s) =X

n≥1

Hindry p. 136

(73)

Absolute abscissa of convergence

The absolute abscissa of convergence σ_a is defined in the same way.

One can prove

σ₀ ≤σ_a ≤σ₀+ 1.

For the Riemann zeta function, σ₀ =σ_a = 1.

For the Dirichlet series

(1−2^1−s)ζ(s) = X

n≥1

(−1)ⁿ⁻¹n^−s,

we have σ₀ = 1 and σ_a = 0.

Riemann Hypothesis: the abscissa of convergence of the Dirichlet seriesD(µ,s) = 1

ζ(s) is 1/2.

(74)

Absolute abscissa of convergence

One can prove

σ₀ ≤σ_a ≤σ₀+ 1.

(1−2^1−s)ζ(s) = X

n≥1

(−1)ⁿ⁻¹n^−s,

ζ(s) is 1/2.

(75)

Absolute abscissa of convergence

One can prove

σ₀ ≤σ_a ≤σ₀+ 1.

(1−2^1−s)ζ(s) = X

n≥1

(−1)ⁿ⁻¹n^−s,

ζ(s) is 1/2.

(76)

Absolute abscissa of convergence

One can prove

σ₀ ≤σ_a ≤σ₀+ 1.

(1−2^1−s)ζ(s) = X

n≥1

(−1)ⁿ⁻¹n^−s, we have σ₀ = 1 and σ_a = 0.

ζ(s) is 1/2.

(77)

Absolute abscissa of convergence

One can prove

σ₀ ≤σ_a ≤σ₀+ 1.

(1−2^1−s)ζ(s) = X

n≥1

(−1)ⁿ⁻¹n^−s, we have σ₀ = 1 and σ_a = 0.

ζ(s) is 1/2.

(78)

Infinite products

Definition of the convergence of a product.

Remark.If z_n is any sequence of complex numbers and if one at least of thez_n vanishes, then the sequence

(z₀z₁z₂· · ·z_n)n≥0

converges to0.

Remark.

N→∞lim

N

Y

n=1

1− 1 n+ 1

= 0.

Hindry p. 135

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Infinite products

(z₀z₁z₂· · ·z_n)n≥0

converges to0.

Remark.

N→∞lim

N

Y

n=1

1− 1 n+ 1

= 0.

Hindry p. 135

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Infinite products

(z₀z₁z₂· · ·z_n)n≥0

converges to0.

Remark.

N→∞lim

N

Y

n=1

1− 1 n+ 1

= 0.

Hindry p. 135

(81)

Convergent infinite products

Definitions.An infinite product Q

n≥0(1 +un) is

convergent (resp. absolutely convergent) if there exists n₀ ≥0 such that |u_n|<1for n≥n₀ and the series P

n≥n0log(1 +u_n) is convergent (resp. absolutely convergent).

An infinite product of functionsQ

n≥0(1 +u_n(x)) is uniformly convergent on a compactK if there existsn₀ ≥0 such that

|un(z)|<1 for n≥n0 and all z ∈K and such that the series P

n≥n₀log(1 +u_n(z))is uniformly convergent on K.

(82)

Convergent infinite products

Definitions.An infinite product Q

n≥0(1 +un) is

convergent (resp. absolutely convergent) if there exists n₀ ≥0 such that |u_n|<1for n≥n₀ and the series P

n≥n0log(1 +u_n) is convergent (resp. absolutely convergent).

An infinite product of functionsQ

n≥0(1 +u_n(x)) is uniformly convergent on a compactK if there existsn₀ ≥0 such that

|un(z)|<1 for n≥n0 and all z ∈K and such that the series P

n≥n₀log(1 +u_n(z))is uniformly convergent on K.

(83)

Convergent infinite products

A product Q

n≥0(1 +u_n) is absolutely convergent if and only if the series P

n≥0|u_n| is convergent

A convergent infinite product is zero if and only if one of the factors is zero.

If the series P

n≥0log(1 +u_n(z))is uniformly convergent on every compact subset of an open setU, then the infinite productQ

n≥0(1 +u_n(z))is holomorphic on U.

(84)

Convergent infinite products

A product Q

If the series P

(85)

Convergent infinite products

A product Q

If the series P

(86)

Euler Product

The infinite product Q

p(1−p^−s) is uniformly convergent on any compact included in the half planeRe(s)>1. It defines an analytic function in this half plane and we have

ζ(s) = Y

p

1 1−p^−s·

ForX >0, letN(X)be the set of positive integers, the prime factors of which are ≤X. Then

Y

p≤X

1

1−p^−s = Y

p≤X

X

m≥0

1

p^ms = X

n∈N(X)

1 n^s.

Hindry p. 137

(87)

Euler Product

ζ(s) = Y

p

1 1−p^−s·

Y

p≤X

1

1−p^−s = Y

p≤X

X

m≥0

1

p^ms = X

n∈N(X)

1 n^s.

Hindry p. 137

(88)

Euler Product

ζ(s) = Y

p

1 1−p^−s·

Y

p≤X

1

1−p^−s = Y

p≤X

X

m≥0

1

p^ms = X

n∈N(X)

1 n^s.

Hindry p. 137

(89)

ζ (s ) 6= 0 for Re(s ) > 1

ForRe(s) =σ >1,

ζ(s)− Y

p≤X

1 1−p^−s

=

X

n6∈N(X)

1 n^s

≤X

n>X

1 n^s

=X

n>X

1 n^σ·

log(1−u) =−X

m≥1

u^m m·

log(1−p^−s) =−X

m≥1

p^−ms m

ζ(s) = exp X

p

X

m≥1

1 mp^ms

! .

(90)

ζ (s ) 6= 0 for Re(s ) > 1

ForRe(s) =σ >1,

ζ(s)− Y

p≤X

1 1−p^−s

=

X

n6∈N(X)

1 n^s

≤X

n>X

1 n^s

=X

n>X

1 n^σ·

log(1−u) =−X

m≥1

u^m m·

log(1−p^−s) =−X

m≥1

p^−ms m

ζ(s) = exp X

p

X

m≥1

1 mp^ms

! .

(91)

ζ (s ) 6= 0 for Re(s ) > 1

ForRe(s) =σ >1,

ζ(s)− Y

p≤X

1 1−p^−s

=

X

n6∈N(X)

1 n^s

≤X

n>X

1 n^s

=X

n>X

1 n^σ·

log(1−u) =−X

m≥1

u^m m· log(1−p^−s) =−X

m≥1

p^−ms m

ζ(s) = exp X

p

X

m≥1

1 mp^ms

! .

(92)

ζ (s ) 6= 0 for Re(s ) > 1

ForRe(s) =σ >1,

ζ(s)− Y

p≤X

1 1−p^−s

=

X

n6∈N(X)

1 n^s

≤X

n>X

1 n^s

=X

n>X

1 n^σ·

log(1−u) =−X

m≥1

u^m m· log(1−p^−s) =−X

m≥1

p^−ms m

ζ(s) = exp X

p

X

m≥1

1 mp^ms

! .

(93)

log ζ (s ) for Re(s ) > 1

logζ(s) =X

p

X

m≥1

1 mp^ms·

ζ⁰(s)

ζ(s) =−X

p

X

m≥1

logp p^ms ·

Recall the von Mangoldt function Λ(n) =

and the formula

ζ⁰(s)

ζ(s) =−X

n≥1

Λ(n) n^s ·

Hindry p. 138

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log ζ (s ) for Re(s ) > 1

logζ(s) =X

p

X

m≥1

1 mp^ms·

ζ⁰(s)

ζ(s) =−X

p

X

m≥1

logp p^ms ·

and the formula

ζ⁰(s)

ζ(s) =−X

n≥1

Λ(n) n^s ·

Hindry p. 138

(95)

log ζ (s ) for Re(s ) > 1

logζ(s) =X

p

X

m≥1

1 mp^ms·

ζ⁰(s)

ζ(s) =−X

p

X

m≥1

logp p^ms ·

and the formula

ζ⁰(s)

ζ(s) =−X

n≥1

Λ(n) n^s ·

Hindry p. 138

(96)

Euler products for multiplicative functions

Proposition: an arithmetic functionf is multiplicative (resp.

completely multiplicative) if and only if

D(f,s) =Y

p

∞

X

ν=0

f(p^ν)p^−νs

! ,

(resp.

D(f,s) =Y

p

∞

X

ν=0

f(p)p^−sν

!

=Y

p

1

1−f(p)p^−s·)

(97)

The Prime Number Theorem

Theorem. The integral Z ∞

1

(θ(t)−t)dt t² is convergent.

Corollary : θ(x)∼x as x → ∞.

Hindry p. 148

(98)

The Prime Number Theorem

Theorem. The integral Z ∞

1

(θ(t)−t)dt t² is convergent.

Corollary : θ(x)∼x as x → ∞.

Hindry p. 148

(99)

Laplace Transform

Leth(t) be a bounded piecewise continuous function. Then the integral

F(s) = Z ∞

0

h(u)e^−sudu

is convergent and defines a holomorphic function on the half plane Re(s)>0.

If this functionF can be analytically continued to a

holomorphic function on an open set containing the closed half plane Re(s)≥0, the the integral converges for s = 0 and

F(0) = Z ∞

0

h(u)du.

Hindry p. 148

(100)

Laplace Transform

Leth(t) be a bounded piecewise continuous function. Then the integral

F(s) = Z ∞

0

h(u)e^−sudu

is convergent and defines a holomorphic function on the half plane Re(s)>0.

If this functionF can be analytically continued to a

holomorphic function on an open set containing the closed half plane Re(s)≥0, the the integral converges for s = 0 and

F(0) = Z ∞

0

h(u)du.

Hindry p. 148

(101)

Consequence

Takeh(u) = θ(e^u)e^−u−1. Then the function F(s) =

Z ∞

0

h(u)e^−sudu = Z ∞

1

θ(t)−t t^s+2 dt

can be written as the sum of

− ζ⁰(s+ 1)

(s+ 1)ζ(s+ 1) − 1 s and a holomorphic function on Re(s)≥ −1/2.

The key point in the proof of the PNT is the following result of Hadamard and de la Vall´ee Poussin

Theorem. The functionζ(s) has no zero on the line Re(s) = 1.

Hindry p. 149

(102)

No zero of ζ (s ) on the line Re(s ) = 1

Trigonometric formula :

4 cosx +cos(2x) + 3 = 2(1 + cosx)² ≥0.

Consequence :

|ζ(σ+it)|⁴|ζ(σ+ 2it)ζ(σ)³ ≥1 forσ ≥1and t ∈R.

(103)

No zero of ζ (s ) on the line Re(s ) = 1

Trigonometric formula :

4 cosx +cos(2x) + 3 = 2(1 + cosx)² ≥0.

Consequence :

|ζ(σ+it)|⁴|ζ(σ+ 2it)ζ(σ)³ ≥1 forσ ≥1and t ∈R.

(104)

Euler Gamma function

The integral

Γ(s) = Z ∞

0

e^−tt^s−1dt

defines an analytic function on the half planeRe(s)>0which satisfies the functional equation

Γ(s+ 1) =sΓ(s).

The Gamma function can be analytically continued to a meromorphic function in the complex plane C with a simple pole at any integer≤0. The residue at s =−k (k ≥0) is (−1)^k/k!.