Content
Artin L–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
Université Pierre et Marie Curie (Paris 6)
Institut de Mathématiques de Jussieu
http://www.imj-prg.fr/~michel.waldschmidt/
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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Arithmetic functions f : N ! C
The fundamental Theorem of arithmetic
Multiplicative function : f (ab) = f (a)f (b) if gcd(a, b) = 1.
n=
Y
p
Example : for k 2 C,
!(n) =
1,
p|n
Divisor function
⌧ (n) =
0 (n)
⌦(n) =
=
P
d|n
dk.
Completely multiplicative function : f (ab) = f (a)f (b) for all
a, b.
vp (n)
p
X
k (n)
Example : for k 2 C, j k = pk : n 7! nk .
X
A multiplicative function is characterised by its values at prime
powers :
Y
f (n) =
f (p vp (n) ).
vp (n).
p|n
= d(n) =
X
p
A completely multiplicative function is characterised by its
values at prime numbers :
Y
f (n) =
f (p)vp (n) .
1.
d|n
p
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Dirichlet series
Product of Dirichlet series
D(f , s) =
X
f (n)n s .
n 1
D(f , s)D(g , s) = D(f ? g , s).
Main example : Euler (1736), Riemann (1859)
Convolution :
X 1
⇣(s) =
·
ns
n 1
Euler : ⇣(2) = ⇡ 2 /6, ⇣(4) = ⇡ 4 /90, ⇣(2n)⇡ 2n 2 Q (n
⇣(0) = 1/2, ⇣( 1) = 1/12,
⇣( 2n) = 0 (n 1), ⇣( 2n 1) 2 Q (n 0).
f ? g (n) =
X
f (a)g (b).
ab=n
0),
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The ring A of arithmetic functions
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Examples of arithmetic functions
The constant function 1 = j 0 : n 7! 1.
The ring of arithmetic functions is a commutative factorial ring
A, the unity of which is the Kronecker delta function with
(
1 for n = 1,
(n) =
0 for n > 1.
The Kronecker delta function .
The Möbius function : µ.
Euler totient function : '.
The function j : j(n) = n for all n
1.
The characteristic function of the squares : .
The units are the arithmetic functions f such that f (1) 6= 0.
Exercise: check that the inverse g of a unit f is defined
recursively by
X
g (1) = f (1) 1 , g (n) = f (1) 1
f (n/d)g (d) (n > 1).
The lambda function :
= ( 1)⌦ .
The von Mangoldt function :
(
log p if n = p m with p prime and m
⇤(n) =
0
if n is not a prime power.
d|n,d<n
1,
Iwaniec Chap. I
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Exercises
Examples of Dirichlet series
D(f , s) =
X
f (n)n
s
.
n 1
µ?1= ,
D( , s) = 1
µ ? j = ',
|µ| ?  = 1,
µ?= ,
jk ? 1 =
|µ| ? 1 = 2! ,
1 ? 1 = ⌧,
k,
µ ? log = ⇤.
Möbius inversion formula :
g =1?f
()
f = µ ? g.
D(1, s) = ⇣(s)
D(j k , s) = ⇣(s
D(, s) = ⇣(2s)
D(µ, s) =
D(⌧, s) = ⇣(s)2
⇣(s 1)
D(', s) =
⇣(s)
⇣(2s)
D( , s) =
⇣(s)
D(
1
⇣(s)
= ⇣(s k)⇣(s)
⇣(s)
D(|µ|, s) =
⇣(2s)
⇣(s)2
D(2! , s) =
⇣(2s)
⇣ 0 (s)
D(log, s) =
k)
k , s)
D(⇤, s) =
⇣ 0 (s)
⇣(s)
More exercises : Patterson p.10 exercise 1.3
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The Prime Number Theorem.
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Distribution of prime numbers : elementary results.
For x > 0 let ⇡(x) be the number of prime numbers  x :
X
⇡(x) =
1.
✓(x) =
X
log p,
(x) =
px
px
⇡(10) = 4, ⇡(100) = 25, ⇡(1000) = 168, ⇡(10 000) = 1229.
Conjecture of Gauss (1792) and Adrien-Marie Legendre
(1798), proved by Hadamard and de la Vallée Poussin (1896) :
PNT. For x ! 1,
x
⇡(x) ⇠
·
log x
M(x) =
X
X
log p =
p m x
X
⇤(n)
nx
µ(n).
nx
Theorem (elementary PNT) (P.L. Tchebyshev) : 1860’s
— cf. course of Francesco Pappalardi
⇡(x) ⇣
This is equivalent to pn ⇠ n log n.
Better approximation of ⇡(x) : integral logarithm
Z x
dt
Li(x) =
·
2 log t
x
,
log x
✓(x) ⇣ x,
(x) ⇣ x
for x ! 1.
The Prime Number Theorem ⇡(x) ⇠ x/ log x is equivalent
✓(x) ⇠ x and to (x) ⇠ x and to M(x) = o(x) as x ! 1.
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Elementary results on prime numbers.
For n
1,
n
2 
For n
✓
◆
 4n .
X
⌫
2n
n
1 and p prime,
vp (n!) =
m 1
For n
Elementary results (continued)
For 2  y < x,
⇡(x)
n
.
pm
⇡(y ) 
1
(✓(x)
log y
✓(y )).
1,
✓(2n)
✓(n)  4 log n.
✓(x)
✓(x)
x
 ⇡(x) 
+
·
log x
log x + 2 log log x (log x)2
Consequence :
✓(x)  C1 x,
⇡(x)
C2
x
·
log x
Hindry p. 127–129
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Exercises
von Mangoldt function
Deduce
X log p
px
and
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X1
px
p
p
⇣ 0 (s)
=
⇣(s)
= log x + O(1)
where
= log log x + C + O(1/ log x)
⇤(n) =
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(
log p
0
X ⇤(n)
n 1
ns
if n = p m with p prime and m
if n is not a prime power.
1,
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Exercise
Abel summation.
Check
✓
lim ⇣(s)
s!1
where
1
s
is Euler constant :
1
◆
=
A0 = 0,
An =
n
X
am ,
an = An
An
1
(n
1)
m=1
1 1 1
1
+ + + ··· +
N!1 1
2 3
N
= lim
N
X
log N.
N
X1
a n bn =
An (bn
bn+1 ) + AN bN .
n=1
n=1
Hint:
1 1
1
1 + + + ··· + = 1 +
2 3
N
Z
N
([t]
1
t)
dt
t2
(telescopic sum)
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Abel’s Partial Summation Formula PSF.
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Abel summation : an example.
Take an = 1 for all n, so that A(t) = btc. Then
N
X
Let an be
Pa sequence of complex numbers and
A(x) = nx an . Let f be a function of class C 1 . Then
X
y <nx
an f (n) = A(x)f (x)
Z
x
f (n) =
n=M+1
A(t)f 0 (t)dt.
Z
N
f (t)dt +
M
Z
N
(t
M
btc)f 0 (t)dt
Take f (t) = 1/t. Then
y
N
X
1
n=1
where
Hindry p. 129–131
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n
= log N +
is Euler constant
Z 1
dt
=1
{t} 2
t
1
+ O(1/N)
where
{t} = t
btc.
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Abel summation for ⇣.
Take an = n
of the series
s
Analytic continuation of ⇣(s) to Re(s) > 0
for all n ; assume Re(s) > 1. An approximation
X
n s
(1 21
s
)⇣(s) = 1
s
s
2
n 1
is
Z
1
1
s
t dt =
1
s
1
log(1
u) =
·
1
1)⇣(s) ! 1 as s ! 1+ .
It is a fact that (s
We will see that if we remove this singularity of ⇣ at s = 1,
the di↵erence
1
⇣(s)
s 1
becomes an entire function (analytic in C).
1
=s
ns
⇣(s) = s
Z
1
n 1
⇣(s) = s
1
s
t
X
dt,
1
s
s 1
t dt + s
1
1
t
dt = s
n
Z
t s dt =
1
Z
([t]
[t]t
s 1
s 1
s
1
+· · ·+(2n 1)
u3
3
···
un
n
s
(2n)
s
+· · ·
···
1
+ · · · = log 2.
4
21
s
)⇣(s) = log 2
1)⇣(s) = 1
Hardy and Wright : lims!1
P
P
an n s =
an n 1 if the series in RHS converges. 22 / 74
P
A Dirichlet series f (s) = n 1 an n s has an absissa of
convergence 0 2 R [ { 1, +1}, such that the series
converges for Re(s) > 0 and diverges for Re(s) < 0 .
The sum of the series is analytic in the half plane Re(s) >
and its derivatives are given by
X
f (k) (s) =
an ( log n)k n s .
dt
dt.
1
1
s
P
Let f (s) = n 1 an n s be a Dirichlet series converging at s0 .
Then for any C 1 it converges uniformly on the sector
n
o
s 2 C | Re(s s0 ) 0, |s s0 |  C Re(s s0 )
1
t)t
4
Abscissa of convergence
1 = [t].
1
1 1
+
2 3
lim (s
n=1
1
Z
s 1
u2
2
s!1+
1) extends to an analytic function
n
XZ
Z
t
u
1
lim (1
Analytic continuation of ⇣(s) (continued)
1/(s
s
s!1+
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The function ⇣(s)
in Re(s) > 0.
+3
+ 1.
0
n 1
Hindry p. 139
Hindry p. 136
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Absolute abscissa of convergence
Infinite products
The absolute abscissa of convergence a is defined in the same
way.
One can prove
0  a  0 + 1.
For the Riemann zeta function,
For the Dirichlet series
(1
21 s )⇣(s) =
0
=
X
a
Definition of the convergence of a product.
Remark. If zn is any sequence of complex numbers and if one
at least of the zn vanishes, then the sequence
(z0 z1 z2 · · · zn )n
= 1.
0
converges to 0.
( 1)n 1 n s ,
Remark.
n 1
lim
we have 0 = 1 and a = 0.
Riemann Hypothesis : the abscissa of convergence of the
1
Dirichlet series D(µ, s) =
is 1/2.
⇣(s)
N!1
N ✓
Y
n=1
1
1
n+1
◆
= 0.
Hindry p. 135
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Convergent infinite products
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Convergent infinite products
Q
A product P n 0 (1 + un ) is absolutely convergent if and only if
the series n 0 |un | is convergent
Q
Definitions. An infinite product n 0 (1 + un ) is
convergent (resp. absolutely convergent) if there
P exists n0 0
such that |un | < 1 for n n0 and the series n n0 log(1 + un )
is convergent (resp. absolutely convergent).
A convergent infinite product is zero if and only if one of the
factors is zero.
Q
An infinite product of functions n 0 (1 + un (x)) is uniformly
convergent on a compact K if there exists n0 0 such that
|u
Pn (z)| < 1 for n n0 and all z 2 K and such that the series
n n0 log(1 + un (z)) is uniformly convergent on K .
P
If the series n 0 log(1 + un (z)) is uniformly convergent on
every compact
subset of an open set U, then the infinite
Q
product n 0 (1 + un (z)) is holomorphic on U.
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28 / 74
⇣(s) 6= 0 for Re(s) > 1
Euler Product
Q
The infinite product p (1 p s ) is uniformly convergent on
any compact included in the half plane Re(s) > 1. It defines
an analytic function in this half plane and we have
⇣(s) =
Y
p
For Re(s) =
⇣(s)
1
1 p
Y
pX
·
s
> 1,
1
1 p
pX
1
1 p
s
n62N (X )
log(1
X 1
X 1
1

=
·
s
ns
n
n
n>X
n>X
X um
u) =
m 1
For X > 0, let N (X ) be the set of positive integers, the prime
factors of which are  X . Then
Y
s
X
=
log(1
YX 1
X 1
=
=
.
ms
s
p
n
pX m 0
Xp
p s) =
m 1
n2N (X )
⇣(s) = exp
·
ms
m
XX
p
Hindry p. 137
m
m
1
ms
mp
1
!
.
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log ⇣(s) for Re(s) > 1
log ⇣(s) =
Euler products for multiplicative functions
XX
p
⇣ 0 (s)
=
⇣(s)
m
1
·
mp ms
1
X X log p
p
m 1
p ms
Proposition : an arithmetic function f is multiplicative (resp.
completely multiplicative) if and only if
!
1
Y X
⌫
⌫s
D(f , s) =
f (p )p
,
·
Recall the von Mangoldt function
(
log p if n = p m with p prime and m
⇤(n) =
0
if n is not a prime power.
and the formula
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p
(resp.
1,
D(f , s) =
1
Y X
p
⇣ 0 (s)
=
⇣(s)
X ⇤(n)
n 1
ns
⌫=0
⌫=0
f (p)p
s ⌫
!
=
Y
p
1
1
f (p)p
s
·)
·
Hindry p. 138
31 / 74
32 / 74
The Prime Number Theorem
Theorem. The integral
Z
1
(✓(t)
1
t)
Laplace Transform
Let h(t) be a bounded piecewise continuous function. Then
the integral
Z 1
F (s) =
h(u)e su du
dt
t2
0
is convergent and defines a holomorphic function on the half
plane Re(s) > 0.
is convergent.
If this function F 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
Z 1
F (0) =
h(u)du.
Corollary : ✓(x) ⇠ x as x ! 1.
Hindry p. 148
0
Hindry p. 148
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Consequence
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No zero of ⇣(s) on the line Re(s) = 1
Take h(u) = ✓(e u )e u 1. Then the function
Z 1
Z 1
✓(t) t
su
F (s) =
h(u)e du =
dt
t s+2
0
1
Trigonometric formula :
4 cos x + cos(2x) + 3 = 2(1 + cos x)2
0.
can be written as the sum of
⇣ 0 (s + 1)
(s + 1)⇣(s + 1)
1
s
Consequence :
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 Vallée Poussin
Theorem. The function ⇣(s) has no zero on the line
Re(s) = 1.
|⇣( + it)|4 |⇣( + 2it)⇣( )3
for
1
1 and t 2 R.
Hindry p. 149
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Euler Gamma function
The integral
(s) =
Euler Gamma function
Z
1
t s 1
e t

1
(s) = e s t s
s
dt
0
defines an analytic function on the half plane Re(s) > 0 which
satisfies the functional equation
(s + 1) = s (s).
1
0
(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!.
1
+
s
Z
1
e t t s dt =
0
1
(s + 1).
s
(s + n + 1)
·
s(s + 1) · · · (s + n)
Remark. From (1) = 1 we deduce (n + 1) = n! for n
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(1/2) =
p
Hence
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Euler Gamma function
⇡
(1/2) =
0.
Z
1
e tt
1/2
dt = 2
0
Z
Exercises:
1
e
x2
dx.
(s) (1
0
Z 1Z 1
1
2
2
2
(1/2) =
e x y dxdy
4
Z0 1 Z0 ⇡/2
2
=
e r rdrd✓
0
=
=

s) =
⇣s ⌘ ✓s + 1◆
= 21
2
2
⇡
·
sin ⇡s
s
✓ ◆
1
(s).
2
0
1
e
2
r2
1
0
ns n!
·
n!1 s(s + 1) · · · (s + n)
(s) = lim
⇡
2
⇡
·
4
1
= se
(s)
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s
1 ⇣
Y
n=1
s⌘
1+
e
n
s/n
.
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Analytic continuation of ⇣(s) (continued)
Analytic continuation
For Re(s) > 1,
1
⇣(s) =
(s)
1
et
Z
1
0
1
e
Z
1
dt
,
1 t
ts
et
0
=
1
X
Lemma. Let f 2 C 1 (R>0 ) be a fast decreasing function at
infinity. Then the function defined for Re(s) > 0 by the
integral
Z 1
1
dt
L(f , s) =
f (t)t s
(s) 0
t
has an analytic continuation to C.
nt
e
,
Special values :
n 1
Under the assumptions of the lemma, for n
L(f , n) = ( 1)n f (n) (0).
nt s dt
(s)
t
= s ·
t
n
Colmez p. 41
Colmez p. 40
41 / 74
Bernoulli numbers
f0 (t) =
t
et
1
·
satisfies the hypotheses of the Lemma. Define (Bn )n
f0 (t) =
X
n 0
B1 =
42 / 74
http://www.bernoulli.org/
The function
B0 = 1,
0 we have
1
,
2
1
B2 = ,
6
Bn
0
by
tn
·
n!
B4 =
1
,
30
B6 =
1
,
42
B3 = B5 = B7 = · · · = 0.
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44 / 74
⇣(s) as an integral
⇣(s) =
Values at negative integers
1
(s)
Z
1
0
The Riemann zeta function ⇣(s) has a meromorphic
continuation to C which is analytic in C \ {1}, with a simple
pole at s = 1 with residue 1.
ts 1
dt
et 1
For n a positive integer,
⇣( n) = ( 1)n
Poles :
For (s) : s = 0, 1, 2, . . . residue ( 1)n /n! at s = n.
Z 1 s 1
t
For
dt : s = 1, 0, 1, 2, . . . residue Bn+1 /(n + 1)!
t
e
1
0
at s = n.
In particular ⇣( n) 2 Q for n
⇣(0) =
1
,
2
⇣( 1) =
⇣( 5) =
Hence the poles cancel except for s = 1.
1
,
252
Bn+1
·
n+1
0 and ⇣( 2n) = 0 for n
1
,
12
⇣( 3) =
⇣( 7) =
1
,
120
1
·
240
45 / 74
The sum of the inverses of the roots of a polynomial f with
f (0) = 1 is f 0 (0) : for
1 be a positive integer. Then
⇣(2n) =
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“Proof” of ⇣(2) = ⇡ 2/6, following Euler
Values at positive even integers
Theorem. Let n
1.
1 + a1 z + a2 z 2 + · · · + an z n = (1
1
(2i⇡)2n
B2n
·
2
(2n)!
we have ↵1 + · · · + ↵n =
Write
sin x
=1
x
In particular ⇣(2n)/⇡ 2n is a rational number.
Examples :
⇣(2) = ⇡ 2 /6 (The Basel problem).
⇣(4) = ⇡ 4 /90, ⇣(6) = ⇡ 6 /945, ⇣(8) = ⇡ 8 /9450.
The Basel problem, first posed by Pietro Mengoli in 1644, was
solved by Leonhard Euler in 1735, when he was 28 only.
47 / 74
↵1 z) · · · (1
↵n z)
a1 .
x2 x4
+
3!
5!
x6
+ ···
7!
Set z = x 2 . The zeroes of the function
p
sin z
z
z2 z3
p
=1
+
+ ···
3! 5!
7!
z
are ⇡ 2 , 4⇡ 2 , 9⇡ 2 , . . . hence the sum of the inverses of these
numbers is
X 1
1
=
·
2⇡2
n
6
n 1
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Remark
Let
Completing Euler’s proof
Y✓
sin x
=
1
x
n 1
2 C. The functions
f (z) = 1 + a1 z + a2 z 2 + · · ·
and
e f (z) = 1 + (a1 + )z + · · ·
have the same zeroes, say 1/↵i .
The sum
P
i
↵i cannot be at the same time
a1 and
a1
◆
.
X 1
x2
1
+ · · · =)
= ·
2
2
6
n ⇡
6
n 1
sin x
=1
x
z
x2
n2 ⇡ 2
http://en.wikipedia.org/wiki/Basel problem
.
Evaluating ⇣(2). Fourteen proofs compiled by Robin Chapman.
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Another proof (Calabi)
50 / 74
Another proof (Calabi)
X
1
=
x 2n y 2n .
x 2y 2
n 0
1
P. Cartier. – Fonctions polylogarithmes, nombres polyzêtas et
groupes pro-unipotents. Sém. Bourbaki no. 885 Astérisque
282 (2002), 137-173.
Z
0
1Z
Z
1
0
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1
0
Z
1
0
x 2n dx =
0
1
·
2n + 1
X
dxdy
1
=
·
2
2
2
1 x y
(2n
+
1)
n 0
x=
Z
1
dxdy
=
1 x 2y 2
sin u
,
cos v
Z
y=
sin v
,
cos u
0u⇡/2, 0v ⇡/2, u+v ⇡/2
dudv =
⇡2
·
8
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Completing Calabi’s proof of ⇣(2) = ⇡ 2/6
From
X
n 0
one deduces
The function ⇡cotg(⇡z)
Proposition : for z 2 C \ Z,
⇡2
1
=
·
(2n + 1)2
8
1
1 X
⇡cotg⇡z = +
z n=1
X 1
X 1
X
1
=
+
·
2
2
n
(2n)
(2n + 1)2
n 1
n 1
n 0
X
n 1
⇣(2) =
1
X
n=1
1
1X 1
=
·
2
(2n)
4 n 1 n2
2z
z2
n2
=
✓
2
1
1
+
z +n z n
1
X
◆
.
⇣(2n)z 2n 1 .
n=1
1
e 2i⇡z + 1
2i⇡
1 X (2i⇡z)n
⇡cotg⇡z = i⇡ 2i⇡z
= i⇡+ 2i⇡z
= i⇡+
Bn
·
e
1
e
1
z n=0
n!
X 1
4X
1
⇡2
=
=
·
2
2
n
3
(2n
+
1)
6
n 1
n 0
Colmez p. 42
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Functional equation
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Non trivial zeroes
An entire function (analytic in C) is defined by
⇠(s) = s(s
1)⇡
s/2
(s/2)⇣(s).
Denote by Z the multiset of zeroes (counting multiplicities) of
⇣(s) in the critical strip 0 < Re(s) < 1 and by Z+ the multiset
of zeroes (counting multiplicities) of ⇣(s) in the critical strip
0 < Re(s) < 1 with positive imaginary part.
Then
Z+ = Z [ {1 ⇢ | ⇢ 2 Z }.
⇠(0) = ⇠(1) = 1.
Theorem (Riemann) :
⇠(s) = ⇠(1
s).
Iwaniec Chap. 7
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Hadamard product expansion
Explicit formula for the logarithmic derivative
Explicit formula :
s(1
s)⇡
s/2
(s/2)⇣(s) =
e
Bs
Y✓
1
⇢2Z
s
⇢
◆
e s/⇢
with
B=
1X
1
=
2 ⇢2Z ⇢(1 ⇢)
We can write
Y✓
Bs
e
1
⇢2Z
s
⇢
◆
e
s/⇢
1
1 + log(4⇡) =
2
2
=
Y✓
1
⇢2Z+
s
⇢
◆✓
1
⇣ 0 (s)
1
= log ⇡
⇣(s)
2
0.023095 . . .
s
1
⇢
◆
1 0 (s/2)
2 (s/2)
1
s
1
s
1
+
X
⇢2Z
1
s
⇢
·
.
Patterson p.33-34, Iwaniec Chap. 8
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Poisson formula
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Poisson formula
Corollary. The theta series
For f 2 L1 (R) let fˆ be its Fourier transform :
Z +1
fb(y ) =
f (x)e 2i⇡xy dx.
1
✓(u) =
e
⇡un2
n2Z
satisfies the functional equation , for u > 0 :
p
✓(1/u) = u✓(u).
P
Assume that the function x 7! n2Z f (x + n) is continuous
with bounded variation on [0, 1] ; then
X
X
f (n) =
fb(m).
n2Z
X
m2Z
For Re(s) > 1,
⇠(s) = s(s
Recall van der Corput Formula (first course by Francesco
Pappalardi).
1)
Z
1
0
(✓(u)
1)u s/2
du.
2u
Colmez p. 51
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The Riemann Memoir (1859).
The Riemann Memoir (1859) (continued).
On the number of primes less than a given magnitude (9p.)
I
I
I
I
The
⇣(s) defined by the Dirichlet series
P function
s
n
has
an analytic continuation to the whole
n 1
complex plane where it is holomorphic except a simple
pole at s = 1 with residue 1.
The following functional equation holds
✓
◆
1 s
s/2
(s 1)/2
⇡
(s/2)⇣(s) = ⇡
⇣(1 s).
2
The following product formula holds
s(s
1)⇡
s/2
(s/2)⇣(s) =
Y✓
1
⇢
I
The following prime number formula holds
X[
X x⇢
[
(x) =
⇤(n) = x
log(2⇡)
⇢
⇢
nx
I
The Riemann zeta function ⇣(s) has simple zeroes at
s = 2, 4, , 6, . . . which are called the trivial zeroes,
and infinitely many non-trivial zeroes in the critical strip
of the form ⇢ = + i with 0   1 and 2 R.
◆
✓
1
log 1
2
1
x2
◆
.
The Riemann Hypothesis. Every non-trivial zero of ⇣(s) is on
the critical line Re(s) = 1/2 :
⇢=
1
+i .
2
http://www.claymath.org/sites/default/files/ezeta.pdf
Iwaniec Chap. 5
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The Riemann Hypothesis.
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Notes by Riemann
The complex zeroes of the Riemann zeta function ⇣(s) in the
critical strip 0 < Re(s) < 1 lie on the critical line
Re(s) = 1/2 :
s 2 C, 0 < Re(s) < 1 and ⇣(s) = 0
s
⇢
=)
Non–trivial zeroes :
1
2
Re(s) = 1/2.
3
4
= 14.134725 . . .
= 21.022039 . . .
= 25.01085 . . .
= 30.42487 . . .
http://oeis.org/wiki/Riemann_zeta_function
Equivalent statement:
⇡(x) = Li(x) + O(x 1/2 log x)
as x ! 1.
Asymptotic expansion :
x X n!
x
x
Li(x) '
'
+
+ ···
n
log x n 0 (log x)
log x (log x)2
Riemann hypothesis for the Möbius function :
M(x) = O(x 1/2+✏ )
as x ! 1.
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Non trivial zeroes of ⇣(s)
The asymptotic formula for N(T )
Hardy (1914) : infinitely many non-trivial zeroes of ⇣(s) are on
the critical line.
Let N(T ) be the number of zeroes ⇢ = + i of ⇣(s) in the
rectangle
0 < < 1,
0 <  T.
Norman Levinson (1974) : 1/3 of the non-trivial zeroes of
⇣(s) are on the critical line.
Then
N(T ) =
Brian Conrey (1989) improved this further to two-fifths
(precisely 40.77 %) are real. The present record seems to
belong to S. Feng, who proved that at least 41.28% of the
zeroes of the Riemann zeta function ⇣(s) are on the critical
line.
for T
T
T
log
+ O(log T )
2⇡
2⇡e
2.
Iwaniec Chap. 9
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Zero free region for ⇣(s)
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Diophantine problem
De la Vallée Poussin (1896) :
>1
Conjecture. The numbers
c
log(2 + |t|)
⇡, ⇣(3), ⇣(5), . . . , ⇣(2n + 1), . . .
for an absolute constant c > 0.
(x) = x + O x exp( c
p
are algebraically independent.
log x) .
Apéry (1978) : ⇣(3) 62 Q
Korobov and Vinogradov (1957)
>1
c(log t)
2/3
,
t
(x) = x + O exp( c(log x)3/5 (log log x)
Rivoal (2000) : infinitely many ⇣(2n + 1) are irrational ;
the numbers ⇣(2n + 1) span a Q–vector space of infinite
dimension.
2
1/5
) ,
x
Zudilin (2004) : At least one of the 4 numbers
⇣(5), ⇣(7), ⇣(9), ⇣(11) is irrational.
3
Iwaniec Chap. 11
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Multizeta values (MZV)
⇣(s) is a period
For s integer
2,
Z
⇣(s) =
Euler
⇣(s1 , . . . , sk ) =
X
n1 >n2 >···>nk
Induction
Z
1
s1
n
·
· · nksk
1 1
for s1 , . . . , sk positive integers with s1
2.
1>t1 >t2 ···>ts >0
t1 >t2 ···>ts >0
dt2
dts
···
t2
ts
dt1
dts
···
t1
ts
1
1
·
1
1
·
dts
·
1 ts
X tn
dts
1
=
s
1 ts
n
n 1
1
1
·
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MZV are periods
⇣(2, 1) =
Proof.
We have
Z t2
0
dt3
=
1 t3
and
hence
Z
1>t1 >t2 >t3 >0
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Linear relations among MZV
Z
1>t1 >t2 >t3 >0
X tn
1
2
n 1
n
Z
, next
Z
1
0
As a consequence, multiple zeta values satisfy a lot of
independent linear relations with integer coefficients.
Example
Product of series :
dt1
dt2
dt3
·
·
·
t1 1 t2 1 t3
t1m 1 dt1 =
t1
0
t2n 1 dt2
t2
1
=
X tm
1
m>n
m
⇣(2)2 = 2⇣(2, 2) + ⇣(4)
,
Product of integrals :
1
,
m
⇣(2)2 = 2⇣(2, 2) + 4⇣(3, 1)
X 1
dt1
dt2
dt3
·
·
=
= ⇣(2, 1)
t1 1 t2 1 t3 m>n 1 m2 n
Hence
⇣(4) = 4⇣(3, 1).
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Conjecture Rohrlich–Lang
Any algebraic dependence relation among the numbers
(2⇡) 1/2 (a) with a 2 Q lies in the ideal generated by the
standard relations :
(1) Translation :
(a + 1) = a (a),
Artin L–functions, Artin primitive roots Conjecture and applications
CIMPA-ICTP Research School,
Nesin Mathematics Village 2017
http://www.rnta.eu/Nesin2017/material.html
(2) Reflexion :
(a) (1
a) =
⇡
.
sin(⇡a)
The Riemann zeta function
(3) Multiplication : for any positive integer n, we have
◆
n
Y1 ✓
k
a+
= (2⇡)(n 1)/2 n na+(1/2) (na).
n
k=0
Michel Waldschmidt
Université Pierre et Marie Curie (Paris 6)
Institut de Mathématiques de Jussieu
(Universal odd distribution).
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http://www.imj-prg.fr/~michel.waldschmidt/
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