Math 539 :: Analytic Number Theory
Fall 2005
Lecture Notes
Course Taught by Dr. Greg Martin
Notes Prepared by Desmond Leung
December 9, 2005
First version :: December 2nd, 2005
(Lecture 1)
φ(n) = #{1 ≤ a ≤ n : (a, n) = 1}.
φ(n)
= “probability” that a “random chosen” integer is relatively prime to n.
n
φ(n)
n
has a (limiting) distribution function for every α ∈ (0, 1).
φ(n)
1
< α exists.
lim # n ≤ x :
x→∞ x
n
Prime Number Theorem
π(x) := #{p ≤ x : p is prime} ∼
x
.
log x
Riemann Asserted :: If ζ(s) = 0 and s 6= −2, −4, −6, . . . then Re(s) = 12 . (Also known as
the Riemann Hypothesis)
We’re trying to understand arithmetic functions f : N → C.
1
Notations/Examples ::
0(n) = 0 ∀n
1(n) = 1 ∀n

 1 if n = 1,
e(n) =
 0 if n > 1.
φ(n) = #{1 ≤ k ≤ n : (k, n) = 1}
σ(n) = sum of positive divisors of n
X
=
d.
d|n
τ (n) = number of positive divisors of n
X
=
1.
d|n

0
µ(n) = (Möbius) :: µ(n) =
if n is not squarefree,
 (−1)k if n is the product of k distinct divisors.

 1 if n is squarefree,
|µ|(n) =
 0 if not.
ω(n) = # of distinct prime divisors of n
X
=
1.
p|n
Ω(n) = # of prime factors of n, counted with multiplicity
X
=
k.
pk kn
Reality Check :: µ(n) = (−1)ω(n) e(Ω(n) + 1 − ω(n)).
We’re interested in these sets of questions about arithmetic functions.
• Range of values?
• (Best case scenario) Distribution of values?
• Average value - usually
1
x
P
n≤x
f (n).
Also correlations between them.
2
Example :: ω(n)
• Range :: {0} ∪ N.
But ω(n) ≤ n.
√
ω(n) ≤ τ (n) ≤ 2 n.
ω(n) ≤ Ω(n) ≤
log n
.
log 2
• Average Value ::

X
ω(n) =
n≤x
X


X

n≤x
p|n

X X  X x .
1 =
1
1

=
p
p≤x
n≤x
p|n
p≤x
Use O-notation for errors. O(f (x)) denotes an unspecified functions g(x) satisfying |g(x)| ≤ C|f (x)|
for some C > 0. Thus,
X
ω(n) =
n≤x
X x
p
p≤x
=x
X1
p≤x
p
+ O(1)
+
X
O(1)
p≤x
(Triangle Inequality)
!
X1
X
=x
+O
1 .
p
p≤x
p≤x
Turns out that
X1
p≤x
p
= log log x + O(1).
Therefore,
X
ω(n) = x(log log x + O(1)) + O(π(x))
n≤x
= x log log x + O(x) + O(π(x))
= x log log x + O(x)
We conclude “ω(n) is log log n on average.”
!
Since
X
ω(n) ∼
n≤x
X
n≤x
3
log log n .
Notation :: f (x) ∼ g(x)
if
limx→∞
f (x)
g(x)
= 1.
Because
X
ω(n) = x log log x + O(x),
n≤x
we have
P
ω(n)
=1+O
x log log x
n≤x
1
log log x
.
Hence,
P
lim
x→∞
ω(n)
= 1.
x log log x
n≤x
So,
X
ω(n) ∼ x log log x.
n≤x
(Lecture 2)
Question :: How frequent are squarefree numbers?

 1 if n is squarefree,
|µ|(n) = µ2 (n) =
 0 otherwise.
Let l(n) denote the largest d such that d2 |n. We’ll use

 1 if k = 1
X
µ(d) =
= e(k).
 0 if k > 1
d|k
Therefore,


 1 if l(n) = 1 
X
|µ|(n) =
=
µ(d)
 0 if l(n) > 1 
d|l(n)
X
=
µ(d).
d: d2 |n
4
Now we look at the summatory function


X
X X
X
X

|µ|(n) =
µ(d) =
µ(d)
1
n≤x
n≤x
=
X
√
d≤ x
d: d2 |n
µ(d)
jxk
√
d≤ x
d2
=
X
µ(d)
√
d≤ x
n≤x
d2 |n
x
d2

+ O(1)

X µ(d)
X

=x
+
O
|µ(d)| .
d2
√
√
d≤ x
d≤ x
The error term (Trivially?) is


O
X
√
d≤ x
√
1 = O( x).
The sum in the main term converges (by comparison with the sum


∞
X
X
µ(d)
µ(d) 
x
−
.
2
d
d2
√
d=1
d≥ x
Moreover,
X
X |µ(d)|
X 1
µ(d)
≤
≤
.
√ d2 2
2
√ d
d>√x d
d> x
d> x
Now,
5
P
1
),
d2
and so becomes
By Integral comparison test, we get
Z ∞
X 1
dt
1
< √ 2 =−
2
t
√ d
b xc t
d> x
Notice that btc ≥
t
2
∞
√
b xc
1
= √ .
b xc
for t ≥ 1. So
1
1
√ ≤
√ =O
b xc
(1/2) x
We now have
X
|µ|(n) = x
n≤x
(We’ll evaluate
More generally, if F (n) =
P
X
n≤x
d|n
∞
X
µ(d)
d=1
P∞
µ(d)
d=1 d2
d2
1
√
x
.
√
+ O( x).
the next time we see it.)
f (d), then
F (n) = x
X f (d)
d≤x
d
6
!
+O
X
d≤x
|f (d)| .
For some sequence {an }, using a power series as a generating function is helpful.
∞
X
{an }
an x n ,
n=0
since
∞
X
!
an x n
∞
X
!
bn x n
=
X
n=0
i+j=n
n=0
n=0
!
∞
X
This motivates the “convolution” :: {an } ∗ {bn } =
ai b j
.
nP
o
i+j=n ai bj .
In Number Theory, its more common to use Dirichlet series,
∞
X
an
{an }
n=1
We see that
∞
X
an
!
ns
n=1
∞
X
bn
ns
n=1
!
ns
.
!
∞
X
1
=
ns
n=1
X
ac b d .
cd=n
This suggests defining the Dirichlet Convolution of two arithmetic functions.
(f ∗ g)(n) =
X
f (c)g(d) =
cd=n
For example, (f ∗ 1)(n) =
P
c|n
f (c)1
n
c
X n
f
g(d).
d
d|n
=
P
c|n
f (c).
(Lecture 3)
X
φ(n) = n
but
φ∗1=?
d|n
Notation :: Given α ∈ R, arithmetic function f , let T α f denote the function
T α f (n) = nα f (n),
and let T = T 1 . Also, for k ∈ Z≥0 , let Lk f (n) = (log n)k f (n), L = L1 .
Example ::
L(f ∗ g)(n) = (log n)(f ∗ g)(n))
X
= log n
f (c)g(d)
cd=n
=
X
(log c + log d)f (c)g(d)
cd=n
= (Lf ) ∗ g + (Lg) ∗ f.
7
Derivation of Arithmetic Functions
One can also show :: T α (f ∗ g) = T α f ∗ T α g.
X
φ(d) = n,
d|n
by Möbius Inversion,
φ(n) =
X
d|n
n
µ(d) .
d
And so
φ(n) X µ(d)
=
.
n
d
d|n
Using the convolution notation ∗,
φ ∗ 1 = T1
φ = T1 ∗ µ
T −1 φ = 1 ∗ T −1 µ
Facts about ∗ ::
• (f ∗ g) ∗ h = f ∗ (g ∗ h)
• If f ∗ g = 0, then f = 0 or g = 0 (Integral Domain)
• f has an inverse (∃ g with f ∗ g = e) if and only if f (1) 6= 0 (Unique Maximal Ideal)
Recall :: If F (n) =
X
P
d|n
f (d), then
F (n) =
n≤x
XX
f (d) =
n≤x d|n
=x
f (d)
d≤x
X f (d)
d≤x
X
d
X
1=
n≤x
d|n
X
d≤x
f (d)
jxk
d
!
X
+O
|f (d)| .
d≤x
Therefore,
X φ(n)
n≤x
n
!
X µ(d) =x
+O
d d
d≤x
d≤x
!!
!
∞
X µ(d)
X µ(d) X µ(d) =x
+O
+O
d2 d .
2
d
d=1
d>x
d≤x
X µ(d)/d
8
We had before,
X µ(d) =O 1 ,
d2 x
d>x
also,
x
Z
X µ(d) X 1
≤
<1+
d d
d≤x
1
d≤x
Therefore,
X φ(n)
n≤x
n
=x
∞
X
µ(d)
d=1
d2
dt
= 1 + log x.
t
+ O(log x).
General Fact :: If f (n) is multiplicative, and
∞
X
|f (d)| < ∞,
d=1
then
∞
X
f (d) =
Y
(1 + f (p) + f (p2 ) + · · · ).
p
d=1
Proof. Consider
Y
X
(1 + f (p) + f (p2 ) + · · · ) =
p≤y
f (n).
(Smooth Numbers)
n∈N
p|n⇒p≤y
P
To prove that the right hand side converges to ∞
n=1 f (n), we look at
X
X
X
X
∞
≤
=
f
(n)
|f (n)| → 0 as y → ∞.
f
(n)
−
f
(n)
n>y
n=1
n∈N
n∈N
∃ p|n with p>y
p|n⇒p≤y
(By hypothesis). Going back...
X φ(n)
n≤x
Taking f (d) =
n
=x
∞
X
µ(d)
d2
d=1
+ O(log x).
µ(d)
,
d2
∞
X
µ(d)
d=1
d2
µ(p) µ(p2 )
=
1 + 2 + 4 + ···
p
p
p
Y
1
=
1− 2 .
p
p
Y
9
But also,
∞
X
Y
1
π2
1
1
=
=
1 + 2 + 4 + ···
2
6
d
p
p
p
d=1
!
−1
Y
Y
1
1
=
=
1− 2
.
1
p
1
−
2
p
p
p
We conclude that
∞
X
µ(d)
d=1
d2
=
6
.
π2
Notation ::

 1 , if n = pj (p prime, j ∈ N)
j
κ(n) =
 0, if ω(n) ≥ 2 or n = 1.
(“von Mangoldt Λ function”)

 log p if n = pj ,
Λ(n) = Lκ(n) =
0
else.
We also define several summatory functions ::
π(x) =
X
p≤x
1,
θ(x) =
X
κ(n) =
∞
X
π(x1/j )
j=1
n≤x
j
,
ψ(x) =
X
Λ(n).
n≤x
(Lecture 4)
Notation :: f (x) g(x) means f (x) = O(g(x)). Also, f (x) g(x) means g(x) f (x).
Note :: If we write f g, then both f and g should be non-negative.
(Pros of )
f g(x) + h(x) g1 (x) + h1 (x) j(x).
(Pros of O)
f (x) = g(x) + O(x1/2 ).
10
(Less writing)
Conjecture on Distribution of Prime Pairs ::


Y p − 1  x

#{p ≤ x : p + 2k is also prime} ∼ 2C2 
,

p − 2  log2 x
p|k
p>2
where C2 = “Twin primes constant”
C2 =
Y
p
1
1−
(p − 1)2
.
So we can say
#{p ≤ x : p + 2k is prime} k
x
.
log2 x
k or Ok means the implied constant may depend on k.
The standard interpretation of
#{p ≤ x : p + 2k is prime} x
log2 x
is provably false.
Lemma ::
X
1 ∗ Λ = L1 ⇒
Λ(d) = log n.
d|n
Note that
Λ(n) =
X
µ(d) log
X
X
n
= log n
µ(d) −
µ(d) log d.
d
d|n
d|n
d|n
But
log n
X
µ(d) → 0.
d|n
So
Λ(n) = −
X
µ(d) log d.
d|n
Proof.
X
Λ(d) =
d|n
X
log p =
X
k log p
pk kn
(p,r)
pr |n
=
X
log pk = log n.
pk kn
11
Bounds for multiplicative functions
Example :: For φ(n), we can show that φ(n) δ n1−δ for any δ > 0. Proof. Lets bound
from below.
Y φ(pk )
Y pk−1 (p − 1)
φ(n)
=
=
n1−δ
pk(1−δ)
pk−kδ
pk kn
pk kn
Y
Y
1 kδ
1
−kδ
/p
=
1−
p .
=
1−
p
p
k
k
p kn
p kn
Thus,
φ(n) Y
1 δ
≥
1−
p ≥
n1−δ
p
Y
p|n
p|n
1
1−
p
pδ .
(1− p1 )pδ <1
We conclude that
φ(n)
≥
n1−δ
Y
p
1
1−
p
pδ = C(δ).
(1− p1 )pδ <1
This proves φ(n) ≥ C(δ)n1−δ .
This argument can yield explicit constants as well.
Example :: δ =
1
,
20
1 1/20
p = 7, 1 −
7
≈ 0.945,
7
1
p = 11, 1 −
111/20 ≈ 1.025.
11
Therefore,
Y
φ(n)
1 1/20
≥
1−
p
n19/20
p
p≤7
1
1
1
1
= 1−
1−
1−
1−
(2 · 3 · 5 · 7)1/20
2
3
5
7
= 0.298 . . .
12
We conclude that φ(n) > 0.298n19/20 .
(Lecture 5)
Warm-up Question :: Which is bigger,
X
φ(n)2005
X
or
n≤x
n2004 ?
n≤x
Answer 1 :: We showed that
φ(n) δ n1−δ ,
Take δ =
1
,
4010
for any δ > 0.
then
φ(n) n1−1/4010
1
1
φ(n)2005 n2005− 2 = n2004 2
Z x
X
X
1
2005
2004 21
φ(n)
n
≈
t2004 2 dt
n≤x
0
n≤x
2005 21
X
n2004
t
Z
≈
.
x
t2004 dt x2005 .
0
n≤x
Answer 2 ::
X
φ(n)2005 ≥
n≤x
X
(p − 1)2005
p≤x
X x 2005
p2005 ≥
2
p≤x
x/2<p≤x
x X
x2005
1 = x2005 π(x) − π
.
2
X
x/2<p≤x
Assuming π(x) ∼
x
,
log x
we get
X
2005
φ(n)
x
2005
n≤x
x
x/2
−
log x log (x/2)
.
Note that
x/2
x/2
x/2
.
=
=
1
log(x/2)
log x − log 2
(log x) 1 + O log x
13
Fact :: If f (x) → 0, then
1
= 1 + O(f (x)).
1 + O(f (x))
Since
1
= 1 − g(x) + g(x)2 − g(x)3 + · · ·
1 + g(x)
We know |g(x)| ≤ Cf (x), then |g(x)|k ≤ C k f (x)k . So
| − g(x) + g(x)2 − g(x)3 + · · · | ≤ |Cf (x)| + |C 2 f (x)2 | + · · ·
|Cf (x)|
=
≤ 2|Cf (x)|,
1 − |Cf (x)|
when x 1.
Therefore,
x/2
x/2
=
log(x/2)
log x
1+O
Example :: Discuss the convergence of
1
log x
P
n≤x
x
=
+O
2 log x
x
log2 x
µ(n)
.
n
Method 1 :: Bound n uniformly. ie., n1 ≤ 1, so
X µ(n) X
6
|µ(n)| ∼ 2 x.
≤1·
n π
n≤x
n≤x
Method 2 :: Split into dyadic blocks,
X
U <n≤2U
In this range,
1
n
≤ u1 . So
X µ(n) 1
≤
n U
U <n≤2U
X
µ(n)
.
n
|µ(n)|
U <n≤2U
1
Q(2U )
(Q(2U ) − Q(U )) ≤
∼
U
U
12
= 2 1.
π
=
More precisely,
Q(x) =
√
6
x
+
O(
x).
π2
14
6
2U
π2
U
.
So
√
1
1 6
6
(Q(2U ) − Q(U )) =
2U − 2 U + O( U )
U
U π2
π
6
1
.
= 2 +O √
π
U
Now,
k−1
X |µ(n)|
X
= 1+
n
k
k=0
n≤2
X
2k <n≤2k+1
|µ(n)|
n
k−1 X
6
1
= 1+
+O √
2
π
2k
k=0
(Convergent geometric series)
6
= 1 + 2 k + O(1).
π
In other words,
X |µ(n)|
n≤x
Conclusion ::
P
n
µ(n)
n
n
=
6 log x
+ O(1).
π 2 log 2
does not converge absolutely.
(Lecture 6)
Q(x) :=
X
|µ(n)| = # of square-free integers ≤ x.
n≤x
Warm Up :: If h = 1 ∗ j, then
X
h=x
h≤x
Example 1 :: Show that
P
d≤x
X j(d)
d≤x
µ(d)
d
d
!
+O
X
|j(d)| .
d≤x
is a bounded function of x.
15
Solution. Using that 1 ∗ µ = e, we have
x
X µ(d)
d≤x
d
!
=
X
e(n) + O
X
n≤x
|µ(d)|
d≤x
= 1 (if x ≥ 1) + O(x) = O(x)
⇒
X µ(d)
d≤x
d
Example 2 :: Let h = τ , τ (n) =
X
= O(1).
P
d|n
1 = (1 ∗ 1)(n). Then
τ (n) = x
X1
d≤x
d
!
+O
X
1·1
d≤x
= x(log x + O(1)) + O(x)
= x log x + O(x).
What do graphs of these functions look like?
Q(x) =
X
n≤x
16
|µ|(n)
Q(x) −
T (x) =
6
x
π2
X
n≤x
17
τ (n)
T (x) − x log x
1
T (x) − x log x − x (Top),
7
1
T (x) − x log x − x (Bottom)
6
18
T (x) = x log x − cx
x
can have small error with xd − 1, say x =
5000, d = 2. but if x = 5000, d = 2500, then xd = 2, so xd − 1 = 1, big error.
Recall
P
n≤x
τ (n) =
P
d≤x
d
, then
x
d
Lets consider
X
(f ∗ g)(n) =
XX
f (i)g(j)
n≤x ij=n
n≤x
=
X
c≤x
=
X
f (c)
X
g(d)
d≤ xc
f (c)G
c≤x
x
c
.
Notation ::
F (x) =
X
f (n),
G(x) =
n≤x
X
n≤x
19
g(n).
X
g(d)F
d≤x
20
x
d
X
f (c)G
c≤x
x
c
Dirichlet’s Hyperbola Method
We can group these lattice points in the following way...
21
This turns out to be
X
n≤x
f ∗ g(n) =
X
c≤y
f (c)G
x
c
+
X
d≤x/y
22
g(d)F
x
− F (y)G
.
d
y
x
Dirichlet’s Divisor Problem
X
τ (n) =
n≤x
X
1 ∗ 1(n).
n≤x
Using
X
1(n) = bxc,
n≤x
then
x
T (x) =
+
− byc
1(d)
c
d
y
c≤y
d≤x/y
X x
X x
x
=
− O(1) +
+ O(1) − (y + O(1))
+ O(1)
c
d
y
c≤y
d≤x/y
X 1
X1
x
x
x
+ O(y) + x
+O
=x
− y + O(y) + O
.
c
d
y
y
y
c≤y
X jxk
jxk
X
d≤x/y
(Lecture 7)
...We will do an aside here....
Lemma 1 :: Fix m ∈ N. Set
Z
x
F (x) =
m−1
x
Z x
1
{t} {x}
dt −
dt,
−
2
t
t m−1
m−1 t
where {t} = t − btc. Then

 0 if x ∈ (m − 1, m),
F (x) =
 1 if x = m.
m
Proof. Start by noting
Z x
Z
x
m−1
1
dt −
t
m−1
x
Z
m−1
Z x
{t}
t − (m − 1)
dt =
dt
when m − 1 < x ≤ m, so
2
t
t2
m−1
x
Z x
{t}
m−1
−(m − 1) dt =
dt −
.
t2
t2
t
m−1
m−1
Thus, if m − 1 < x < m, we have
x
x
t {t} m − 1 +
=− = 0.
F (x) = −
t
t
t m−1
m−1
23
If x = m, then
m
m
{t} + m − 1 (m − 1) (m − 1) (m − 1)
F (m) = −
+
=−
=−
t
t
m
m−1
m−1
m−1
1
(m − 1) m
+
= .
=−
m
m
m
Lemma 2 :: When x ≥ 1, we have
x Z x
X 1 Z x1
{t}
{t} =
dt −
−
dt.
2
u
t
t
t
1
1
1
1,n≤x
Proof. Apply Lemma 1 with m = x = 2, m = x = 3, . . . , m = x = bxc, and the extra
piece m = bxc + 1, x = x. Then
Z
2
Z
Z
bxc
+··· +
+
1
3
Z
x
+
bxc−1
2
Z
x
=
bxc
etc...
1
Proposition (Lemma 3.13 in B. & D.) :: For x ≥ 1,
X1
1
= log x + γ + O
,
n
x
n≤x
where
γ = lim
x→∞
X1
− log x
n
n≤x
!
Z
∞
=1−
1
{t}
dt ≈ 0.577215 . . .
t2
(Euler’s Constant)
Proof.
x Z x
Z x
X1
X 1
1
{t} {t}
=1+
= 1+
dt −
−
dt
2
n
n
t
t
t
1
1
1
n≤x
1<n≤x
Z ∞
Z ∞
{x}
{t}
{t}
= 1 + log x −
−
dt
+
dt
x
t2
t2
1
x
Z ∞
Z ∞
{t}
{x}
{t}
= log x + 1 −
dt + O
+
dt .
t2
x
t2
1
x
The error term is
1
+
x
Z
x
∞
1
1
dt .
2
t
x
24
P
We were asymtotically evaluating T (x) =
n≤x
τ (n). We had shown using Dirichlet’s
hyperbola method that for any 1 ≤ y ≤ x,
T (x) = x
X 1
x
x
x
+ O(y) + x
+O
− y + O(y) + O
.
c
d
y
y
y
X1
c≤y
c≤(x/y)
By proposition,
1
x
x
x
T (x) = x log y − γ + O
+ x log + γ + O
−x+O y+
y
y
y
y
x
.
= x log x + (2γ − 1)x + O y +
y
√
Since y + xy is minimized at y = x. we conclude that
√
T (x) = x log x + (2γ − 1)x + O( x).
This leads to the Dirichlet Divisor Problem.
Note on Minimizing Error Terms :: We often encounter error terms of the form
O(I(x)) + D(x)), when I is increasing and D is decreasing. We could use calculus to find
the exact minimum but we can also use
max{I(x), D(x)} ≤ I(x) + D(x) ≤ 2 max{D(x), I(x)},
and max{I(x), D(x)} is minimized when I(x) = D(x) (ie. y =
Let’s do one more example of the hyperbola method.
Example :: Let s(n) be the indicator function of squares.

 1 if n ∈ N2
s(n) =
 0 else.
We have
X
√
s(n) = b xc.
n≤x
25
x
y
⇒y=
√
x).
So lets try to evaluate
X
µ2 ∗ s(n).
n≤x
First try (Dead End) ::
X
µ2 ∗ s(n) =
X
S(x) =
X
n≤x
µ2 (c)S
c≤x
x
c
, where
s(n).
n≤x
So
X
2
µ (c)S
x
c
c≤x
r x
=
µ (c)
c
c≤x
r
X
x
2
=
µ (c)
+ O(1)
c
c≤x
X
2
√ X µ2 (c)
√ + O(x).
= x
c
c≤x
(The error term is x but the main term is
√
x)
Second try (Dead End) ::
X
(µ2 ∗ s)(n) =
n≤x
X
s(d)Q
d≤x
=
X
s(d)
d≤x
x
d
6 x
+O
π2 d
r x
d
√ X s(d)
6 X s(d)
√
+O
= 2x
x
π d≤x d
d
d≤x
The error term is (setting d = l2 )
√ X 1
√
√
√
x
x log x x log x.
√ l
l≤ x
26
!
.
(Lecture 8)
Let’s recall some facts ::
µ2 ∗ s, where µ2 is the indicator function of squarefrees, and s is the indicator function of
squares. Hence every number can be represented as such.
X
(µ2 ∗ s)(n),
n≤x
with (c, d) : cd = n, µ2 (c) = 1, and s(d) = 1. We can draw a chart....
r
0 1 2 3 4 5 ···
µ2 (pr )
1 1 0 0 0 0 ···
s(pr )
1 0 1 0 1 0 ···
µ2 ∗ s(pr ) 1 1 1 1 1 1 · · ·
µ2 ∗ s(pr ) =
X
µ2 (c)s(d).
cd=pr
Riemann-Stieltjes Integrals
Goal :: Use our understanding of
P
n≤x f (n) to gain understanding of
various smooth functions g.
Examples ::
X
n≤x
f (n)nα ,
X
f (n) log n,
n≤x
27
X f (n)
.
log n
n≤x
P
n≤x
f (n)g(n) for
Definition :: (F, G) is a compatible pair of functions if
• Both F and G are locally of bounded variation (Difference of two increasing functions).
• One of F, G is right-continuous and the other left-continuous.
Think!! Two standard situations ::
• F, G are both smooth.
• One is smooth and the other is a summatory function
P
n≤x
a(n).
Definition :: The Riemann-Stieltjes Integral
Z b
F (t)dG(t),
a
is defined to be the limit of
N
X
F (ξi )(G(xi ) − G(xi−1 ))
i=1
over all partitions a = x0 < x1 < x2 < · · · < xN −1 < xN = b and each ξi ∈ [xi−1 , xi ].
Fact :: If (F, G) is a compatible pair of functions, then
Z b
F (t)dG(t)
exists.
a
Two most important classes of examples ::
1. If G(x) =
P
g(n), then
n≤x
Z
b
X
F (t)dG(t) =
a
2. If G(x) =
Rx
c
F (n)g(n)
say for F smooth.
a<n≤b
g(t)dt, then
Z b
F (t)dG(t) (A Riemann-Stieltjes Integral)
a
Z
simply equals
b
F (t)g(t)dt. (A Riemann Integral)
a
28
Theorem :: (“Summation by Parts”)
If (F, G) are compatible pair of functions, then
b Z b
Z b
G(t)dF (t).
F (t)dG(t) = F (t)G(t) −
a
a
a
- Most Important Incarnation.
Let G(x) =
X
g(n), and let F be smooth, then
n≤x
(∗)
X
a<n≤b
b Z b
G(t)F 0 (t)dt.
F (n)g(n) = F (t)G(t) −
a
a
Proof of (∗). Note that
b
Z
F 0 (t)dt.
F (n) = F (b) −
n
Thus,
X
X
F (n)g(n) =
F (t)dt
0
g(n) F (b) −
n
a<n≤b
a<n≤b
b
Z
= F (b)(G(b) − G(a)) −
X
Z
g(n)
!
b
F 0 (t)
= F (b)(G(b) − G(a)) −
n
Z
F 0 (t)dt
n
a<n≤b
Z
b
X
g(n) dt
a<n≤t
b
F 0 (t)(G(t) − G(a))dt
a
Z b
Z b
0
F 0 (t)G(t)dt
= F (b)(G(b) − G(a)) + G(a)
F (t)dt −
a
a
Z b
= F (b)(G(b) − G(a)) + G(a)(F (b) − F (a)) −
F 0 (t)G(t)dt.
= F (b)(G(b) − G(a)) −
a
(Lecture 9)
Partial Summation ::
Z b
a
b Z b
F (t)dG(t) = F (t)G(t) −
G(t)dF (t).
If F is differentiable, and G(x) =
a
a
P
g(n), then
b Z b
X
F (n)g(n) = F (t)G(t) −
G(t)F 0 (t)dt.
n≤x
a
a<n≤b
29
a
Example 1 :: Find an asymptotic formula for
H(x) =
X φ(n)
.
n2 log n
n≤x
Solution. Let
X φ(n)
6
= 2 x + O(log x).
n
π
1<n≤x
G(x) =
Let
F (x) =
1
.
x log x
Then
Z
H(x) =
=
=
=
=
x
1
dG(t)
3/2 t log t
x
Z x
1
G(t) −
G(t)d
t log t 3/2
t log t
3/2
Z x
G(x)
G(3/2)
G(t)(−(t log t)−2 (log t + 1))dt
−
−
x log x 3/2 log 3/2
3/2
Z x 1
6
1
1
+
dt
O
+ O(1) +
t + O(log t)
log x
π2
t2 log t t2 log2 t
3/2
Z x
Z x 1
1
1
6
dt
+
O
+ +
O(1) + 2
dt.
π 3/2 t log t
t log2 t t2 t2 log t
3/2
The whole error term is
Z
∞
O
3/2
dt
t log2 t
= O(1).
So
x
6
6
H(x) = 2 log log t + O(1) = 2 log log x + O(1).
π
π
3/2
P
1
Example 2 :: Define ζ(α) = ∞
n=1 nα (valid for α > 1). Also, if Re(α) > 1, then
∞ ∞
X
1 X
1
=
.
nα Re(α)
n
n=1
n=1
So ζ(α) converges absolutely if Re(α) > 1.
Remark on Example 1 :: Note that
Z x
1
H(x) =
dG(t) for any 1 ≤ c < 2.
c t log t
30
So we often write
Z
x
1
H(x) = lim+
dG(t)
→0
2− t log t
Z x
1
dG(t).
=
2− t log t
Write F (x) =
1
,
xα
g(n) = 1, so G(x) = bxc. Then
Z ∞
1
ζ(α) =
dbtc
α
1− t
∞ Z ∞
1
btc btcd α
= α −
t 1−
t
−
Z1 ∞
dt
= 0−0+α
btc α+1
t
1−
Z ∞
dt
=α
(t − {t}) α+1 . (Note the change from 1− to 1)
t
1
Splitting the integral
Z
∞
Z
∞
dt
{t} α+1
t
1
1 1−α
Z ∞
1−α
1
dt
∞
−
−α
{t} α+1
=α
1−α 1−α
t
1
Z ∞
α
dt
=
−α
{t} α+1 .
α−1
t
1
ζ(α) = α
dt
−α
tα
Notes ::
1. This expression is an analytic function of α in the region
{Re(α) > 0} − {1}, hence it provides an analytic continuation of ζ(α).
2. Note that
1
α
1
ζ(α) −
=
−
−α
α−1
α−1 α−1
Z ∞
dt
= 1−α
{t} α+1 .
t
1
Z
1
∞
{t}
dt
tα+1
So ζ(α) has a simple pole at α = 1, with Residue 1. Moreover,
Z ∞
1
dt
lim ζ(α) −
=1−1
{t} 2 = γ,
α→1
α−1
t
1
31
where γ is the Euler’s constant. So
1
+ γ + γ1 (α − 1) + γ2 (α − 1)2 + · · ·
α−1
ζ(α) =
(Lecture 10)
Example 3 :: Using partial summation to get a better upper bound than dyadic blocks.
X µ(d) log d
d
d≤x
.
Recall
X µ(d)
d≤x
d
X |µ(d)|
1 and
d≤x
d
log x.
Method 1 ::
X
2k−1 ≤d<2k
µ(d) log d
k
d
X
2k−1 ≤d<2k
|µ(d)|
k log 2k k 2 .
d
Thus
k
X µ(d) log d
X
j 2 k3
d
k
j=1
d<2
X µ(d) log d
d
d≤x
log3 x.
Method 2 :: Let
M (x) =
X µ(d)
d≤x
d
= O(1).
Then
X µ(d) log d
d≤x
d
Z
x
=
log t dM (t)
1−
x
Z x
dt
M (t)
= M (t) log t −
t
−
−
1 Z x 1 dt
log x + O
log x.
1− t
Define
T (x) =
X
log n,
ψ(x) =
n≤x
X
n≤x
32
Λ(n).
Now,
Λ(n) =
X
d|n
n
µ(d) log ,
d
or
Λ = µ ∗ L1.
Note that
Z
x
x
Z
log tdt + logbxc.
log tdt < T (x) <
1
1
Therefore, T (x) = x log x − x + O(log x). Since Λ = µ ∗ L1, we have
ψ(x) =
X
µ(d)T
x
d
d≤x
=
X
µ(d)
d≤x
x
d
log
= (x log x − x)
x x
x − + O log
d d
d
X µ(d)
d≤x
d
−x
X µ(d) log d
d
d≤x
+O
x
|µ(d)| log
d
d≤x
X
!
.
Idea :: Replace µ(d) by some other sequence {ad } with properties we might like.
1. Some finite set D such that ad 6= 0 ⇒ d ∈ D.
2.
P
µ(d)
d
3.
P
ad log d
d
d∈D
d∈D
= 0.
≈ −1.
What can we say if we use ad ?
•
X
d∈D
ad T
x
d
= (by the same series of manipulation)
= (x log x − x)
X ad
d
d∈D
= 0−x
X ad log d
d∈D
d
−x
X ad log d
d∈D
d
x
+O
|ad | log
d
d∈D
!
X
+ O(log x · max{|ad |}),
d∈D
where we can replace the big-O term with O{ad } (log x) (True when x > max(D)).
33
•
X
ad T
x
d
d∈D
=
X
d∈D
=
X
ad
log n
n≤x/d
X
X X
ad
d∈D
Λ(m).
n≤x/d m|n
Using n = ml,
=
X
ad Λ(m) =
=
Λ(m)E
x
m
m≤x
E(y) =
X
Λ(m)
m≤x
dlm≤x
X
X
ad =
X
ad
d∈D
dl≤y
dl≤x/m
,
where
jy k
d
X
.
The closer E(y) is to 1, the closer this is to ψ(x).
Key Fact ::
E(y) =
X
ad
d∈D
=y
with
ad
d∈D d
d
d
X ny o
ad
−
,
d
d
d∈D
X
d∈D
This is periodic in y, with period LCM(D).
What we do look for is {ad } such that
1. ad 6= 0 ⇒ d ∈ D.
ad
d∈D d
P
3.
P
d∈D
n y o
→ 0. So
E(y) = −
2.
−
X ad
d∈D
P
y
= 0.
ad log d
d
4. E(y) = −
≈ −1.
P
d∈D
ad
y
d
≈ 1.
34
ad
ny o
d
.
ad
(Lecture 11)
Summary :: Take any sequence {ad } supported on a finite set D such that
X ad
= 0.
d
d∈D
Define
E(x) =
X
ad
jxk
d
d∈D
=−
X
ad
nxo
d∈D
d
.
We derived
•
X
d∈D
ad T
x
d
=x
X ad log d
d
d∈D
x
|ad | log
+O
d
d∈D
X
!
.
•
X
ad T
x
d
d∈D
=
X
Λ(k)E
x
k≤x
k
.
We also know
T (x) = x log x − x + O(log x).
This is Chebyshev’s Method. We can choose any {ad } to get numerical bounds.
Lets try a1 = 1, a2 = −2, ad = 0 for d ≥ 3.
E(y) = −{y} + 2
ny o
2
is periodic with period 2.

 0 if 0 ≤ y < 1,
E(y) =
 1 if 1 ≤ y < 2.
So
X
d∈D
ad T
x
d
X
=
Λ(k).
k≤x
bx/kc is odd
In particular, its ≤ ψ(x), and also ≥ ψ(x) − ψ(x/2).
x
X
= −x(− log 2) + O(log x)
ad T
d
d∈D
= x log 2 + O(log x).
35
In other words,
x log 2 + O(log x) ≤ ψ(x).
Also,
x log 2 + O(log x) ≥ ψ(x) − ψ(x/2),
(x/2) log 2 + O(log x) ≥ ψ(x/2) − ψ(x/4),
..
.
Adding O(log x) of these inequalities together,
1
1 1
x log 2 1 + + + · · · + s.t. + O(log2 x) ≥ ψ(x).
2 4
2
!
∞
X
1
1
+O
+ O(log2 x) ≥ ψ(x).
x log 2
k
2
x
k=0
x(2 log 2) + O(log2 x) ≥ ψ(x).
The functions
ψ(x) =
X
Λ(n),
θ(x) =
n≤x
X
log p,
and π(x) =
p≤x
p≤x
are closely related.
ψ(x) =
X
log p +
p≤x
X
log p +
p2 ≤x
X
log p + · · ·
p3 ≤x
= θ(x) + θ(x1/2 ) + θ(x1/3 ) + · · ·
K
X
log x
1/k
.
=
θ(x ), where K =
log
2
k=1
Note that θ(y) ≤ y log y trivially, and so
ψ(x) = θ(x) + O
K
X
!
x1/k log x1/k
k=2
√
√ = θ(x) + O K x log x
= θ(x) + O(x1/K log2 x).
36
X
1,
Note that
Z
x
π(x) =
2−
1
dθ(t).
log t
Summing by parts,
x
Z x
1
θ(t) θ(t)d
π(x) =
−
log t 2−
log t
2−
Z x
θ(x)
θ(t)
=
+
2 dt.
log x
2 t log t
If we have, for example, θ(x) ≤ Cx, then
Z x
Ct
Cx
+
π(x) ≤
2 dt
log x
2 t log t
Z √x Z x ! C
Cx
+ √
+
dt
≤
log x
log2 t
2
x
Z √x
Z x
C
Cx
C
≤
+
2 dt + √
2 √ dt
log x
x
x log
2 log 2
C √
Cx
4C
+
x
≤
x+
2
log x log 2
log2 x
Cx
x
≤
+O
.
log x
log2 x
In particular, Chebyshev’s bounds for ψ(x) imply that
x
x
x log 2
2x log 2
+O
+O
≤ π(x) ≤
.
log x
log x
log2 x
log2 x
(Lecture 12)
We already know
ψ(x) x. (“is of order of magnitude x”)
That is, ψ(x) x and ψ(x) x. This implies
π(x) x
.
log x
We also recall
T (x) =
X
log n = x log x + O(x).
n≤x
37
Merten’s Formulas
1.
X Λ(n)
= log x + O(1).
n
n≤x
2.
X log p
= log x + O(1).
p
p≤x
Proof. We write, using L1 = 1 ∗ Λ,
X
T (x) =
log n = x
n≤x
X Λ(d)
d
d≤x
!
+O
X
Λ(d) .
d≤x
By Chebyshev’s Bounds, this is
x
X Λ(d)
d
d≤x
+ O(x).
Comparing to T (x) = x log x + O(x), we have
x
X Λ(d)
d
d≤x
= x log x + O(x),
so divide through by x.
To derive 2, note that
X Λ(n)
n
n≤x
−
X log p
p≤x
p
=
X log p
.
pr
p,r
r≥2
pr ≤x
The sum over r is a geometric series, so this is bounded by
X
log p
p
∞
X
X log p
X log n
1
=
1.
r
2
p
p(p
−
1)
n
r=2
p
n
Note that from 1,
X Λ(n)
n
(∗)
n
Z
x
1
dψ(t)
1− t
x
Z x
ψ(t) 1
=
−
ψ(t)d
t 1−
t
−
Z x1
dt
= O(1) +
ψ(x) 2 .
t
1
=
38
The partial summation argument we saw Lecture 11 shows that
x
ψ(x)
.
+O
π(x) ∼
log x
log2 x
Therefore, the Prime Number Theorem
π(x) ∼
x
log x
lim sup
ψ(x)
≥ 1 and
x
is equivalent to ψ(x) ∼ x.
We can now prove ::
x→∞
lim inf
x→∞
ψ(x)
≤ 1.
x
Suppose, for the sake of contradiction, that
lim inf
x→∞
ψ(x)
> 1.
x
Choose c so that
1 < c < lim inf
x→∞
ψ(x)
.
x
By the definition of lim inf, there is some x0 such that
ψ(x)
>c
x
for all x > x0 .
Therefore,
Z
x
1
Z x0
Z x
dt
ψ(t)
dt
ψ(x) 2 >
dt +
c
2
t
t
t
1
x0
= O(1) + c(log x − log x0 )
= c log x + O(1).
But this is impossible by (∗),
Z
x
ψ(x)
1
X Λ(n)
dt
=
+ O(1)
t2
n
n≤x
= log x + O(1).
Therefore, lim inf x→∞
ψ(x)
x
≤ 1. A similar argument establishes the lim supx→∞ bound.
39
Remark :: This implies that if
π(x)
x/ log x
has a limit, then the limit is 1.
Merten’s Formulas (Con’t) ::
3.
X1
p
p≤x
∼ log log x + b + O
4.
Y
p≤x
−1
1
1−
p
1
log x
for some b ∈ R.
,
= ec log x + O(1),
for some constant c.
Proof. Write
L(x) =
X log p
p
p≤x
,
and define
R(x) = L(x) − log x.
So R(x) 1 from 2. Then
X1
p≤x
Z
x
1
dL(t)
p
2− log t
x
Z x
L(t) 1
=
dt
−
L(t) −
log t 2−
t log2 t
2−
Z x
L(x)
(log t + R(t))
=
+
dt
log x
t log2 t
2
Z x
Z x
dt
R(t)
R(x)
= 1+
+
+
2 dt.
log x
2 t log t
2 t log t
=
Since the last integral converges as x → ∞ (since R(t) 1), we have
Z ∞
Z ∞
X1
1
R(t)
R(t)
+ log log t − log log 2 +
=1+O
dt.
2 dt −
p
log t
t log t
t log2 t
2
x
p≤x
Define
Z
b = 1 − log log 2 +
2
∞
R(t)
dt,
t log2 t
and noting that
Z
x
∞
R(t)
dt log2 t
Z
∞
dt
t log2 t
x
∞
1
= 1 .
= −
log t x
log x
40
We conclude
X1
p≤x
p
= log log x + b + O
1
log x
.
(Lecture 13)
So now we know
•
X1
1
= log x + γ + O
.
n
x
n≤x
•
X1
p
p≤x
= log log x + b + O
1
log x
.
We wish to prove
Y
p≤x
1
1−
p
−1
= eγ log x + O(1).
(Merten’s Formula)
1. Prove ec log x + O(1) for some constant c (easy).
2. Prove c = γ (annoying).
Proof of 1.
log
Y
p≤x
1
1−
p
−1
Note that
−1
log(1 − t)
=
In particular,
−1
log(1 − t)
=t+
t2 t3
+ + ···
2
3
1
=
log 1 −
p
p≤x
∞ k
X
t
k=1
−1=
−1
X
k
∞ k
X
t
k=2
.
∞
X
tk
=t ·
.
k
k+2
k=1
2
This power series converges (to an analytic function) for |t| < 1, hence is bounded uniformly
on |t| ≤ 1/2. Therefore,
1
log 1 −
p
−1
1
− p
2
1
p
41
uniformly in p.
So
1
log 1 −
p
p≤x
X
−1
Thus
1
log 1 −
p
p≤x
X
−1
1
+
log 1 −
=
−
p p≤x
p
p≤x
−1
X1 X
1
=
+
log 1 −
−
p
p
p
p≤x
X1
−1
=
log log x + b + O
where
d=
X
p
X
1
log 1 −
p
−1
1
−
p
1
log x
!
=
1
p
!
1
p
!
X 1
+O
p2
p>x
1
+d+O
,
x
XX 1
.
kpk
p k≥2
Setting c = b + d,
= log log x + c + O
Exponentiating,
Y
p≤x
1
1−
p
−1
log log x c O( log1 x )
=e
ee
1
log x
c
.
= e log x 1 + O
1
log x
since et = 1 + O(t) for |t| ≤ 1/2.
Proof of 2. We have
X
1
log 1 −
p
p≤x
−1
1
= log log x + c + O
log x
X 1
1
=
+ (c − γ) + O
.
n
log x
n≤log x
Unmotivated Step :: For all δ ∈ (0, 1/2), define
−1 !
Z ∞
X
1
dx,
Aδ = δ
x−1−δ
log 1 −
p
1
p≤x
!
Z ∞
X 1
Bδ = δ
x−1−δ
dx,
n
1
n≤log x
Z ∞
Cδ = δ
x−1−δ (c − γ)dx,
Z1 ∞
1
−1−δ
Dδ = δ
x
O min 1,
dx.
log x
1
Then Aδ = Bδ + Cδ + Dδ for all δ ∈ (0, 1/2).
42
,
!
.
Lemma 1 :: Aδ = log ζ(1 + δ) + O(δ) as δ → 0+ .
Lemma 2 :: log ζ(1 + δ) = log δ −1 + O(δ).
Lemma 3 :: Bδ = log δ −1 + O(δ).
Lemma 4 :: Cδ = c − δ. (Proved!)
Lemma 5 :: Dδ δ log δ −1 .
Given all these lemmas,
log δ −1 +O(δ) = log δ −1 +O(δ)+c−γ+O(δ log δ −1 ) or c−γ = O(δ+δ log δ −1 ) = O(δ log δ −1 ).
Taking limδ→0+ yields c − γ = 0.
Helpful observation ::
−δ ∞
Z ∞
x
−1−δ
= w−δ
δ
x
dx = δ
−δ w
w
for all δ, w > 0.
Proof of Lemma 1.
−1 !
1
Aδ = δ
x−1−δ
log 1 −
dx
p
1
p≤x
−1 Z ∞
X
1
=
log 1 −
δ
x−1−δ dx,
p
p
p
Z
∞
X
by Tonelli’s Theorem. So
−1
1
Aδ =
log 1 −
p−δ
p
p
XX 1
=
p−δ .
k
kp
p k≥1
X
43
On the other hand,
∞
X
Y
1
1
1
ζ(1 + δ) =
=
1 + 1+δ + 2(1+δ) + · · ·
n1+δ
p
p
n=1
p
−1
Y
1
.
=
1 − 1+δ
p
p
So
log ζ(1 + δ) =
X
log 1 −
p
=
p
k≥1
p1+δ
1
XX
−1
1
k(p1+δ )k
.
Thus
X X 1 p−δ − p−kδ
1
(Aδ − log ζ(1 + δ)) =
.
δ
kpk
δ
p k≥1
We need to show this is O(1). We do this by showing limδ→0+ exists. This limit is
lim+
δ→0
X X 1 p−δ − p−kδ
XX 1
p−δ − p−kδ
=
lim
kpk
δ
kpk δ→0+
δ
p k≥2
p k≥2
XX 1
(− log p)p−δ + (k log p)p−kδ
lim
kpk δ→0+
1
p k≥2
XX 1
(k − 1) log p.
=
k
kp
p k≥2
(L’Hôspital) =
This is a finite number. It’s
XX 1
log p
pk
p k≥2
X log p
X log n
=
1.
2
p(p
−
1)
n
p
n
≤
(Lecture 14)
Proof of Lemma 2. We saw before that
lim ζ(α) −
α→1
1
α−1
44
= γ.
In particular, for δ near 0,
1
+ O(1)
δ
1
= (1 + O(δ)).
δ
ζ(1 + δ) =
Therefore,
1
+ log(1 + O(δ))
δ
= log δ −1 + O(δ).
log ζ(1 + δ) = log
...We were in the middle of proving
−1
Y
1
1−
= eγ log y + O(1).
p
p≤y
Proof of Lemma 3.
Z
∞
x−1−δ
Bδ = δ
1
X 1
n
n≤log x
!
dx
Z ∞
∞
X
1
=
δ
x−1−δ dx
n
n
e
n=1
∞
X
1 n −δ
=
(e )
n
n=1
= log(1 − e−δ )−1 .
Since et = 1 + t + O(t2 ) for t near 0, this gives
Bδ = log(1 − (1 − δ + O(δ 2 ))−1
= log(δ + O(δ 2 ))−1 = − log δ(1 + O(δ))
= − log δ − log(1 + O(δ))
= log δ −1 + O(δ).
Proof of Lemma 5.
Z
Dδ δ
∞
x
1
−1−δ
1
min 1,
log x
45
dx.
We need Dδ = o(1) as δ → 0. We estimate this over 3 ranges separately.
(a)
e
Z
x
δ
−1−δ
1
1
min 1,
log x
Z
e
x−1−δ dx
1 −δ e
x
=δ
−δ 1
dx = δ
= 1 − e−δ
= 1 − (1 + O(δ)) = O(δ).
(b)
Z
e1/δ
δ
x
−1−δ
min 1,
e
1
log x
dx
e1/δ
1
≤δ
dx = δ log log x
x log x
e
e
−1
= δ(log δ − 0) = δ log δ −1 .
Z
e1/δ
(c)
Z
∞
δ
x
min 1,
−1−δ
e1/δ
1
log x
Z
∞
x−1−δ dx
dx < δ
e1/δ
= δ(e1/δ )−δ = δ/e.
Thus Dδ = O(δ log δ −1 ) = o(1) as δ → 0.
Reminder :: The notation f (x) = o(g(x)) means
lim
f (x)
= 0 as x → whatever.
g(x)
Example ::
x
x
⇔ π(x) =
(1 + o(1))
log x
log x
x
x
=
+o
.
log x
log x
π(x) ∼
I prefer explicit error terms like
x
π(x) =
+O
log x
46
x
log2 x
.
It turns out that we get better error terms by writing
π(x) = li(x) + O(· · · ),
where
Z
x
li(x) =
2
dt
.
log t
Proposition :: For all n ≥ 3,
e−γ n
φ(n) ≥
log log n
1+O
1
log log n
,
and this is best possible.
Proof. Suppose for a (optimistic) moment that n is of the form
n=
Y
p.
p≤y
Then
Y
φ(n)
=
1−
n
p|n
Y
=
1−
p≤y
=
1
p
1
p
=
eγ
1
log y + O(1)
1
eγ log y 1 + O log1 y
1
1
= γ
1+O
.
e log y
log y
Note that
log n =
X
log p = θ(y) y
p≤y
by Chebyshev’s estimate. (ie. cy < θ(y) < Cy for constants c, C). So log log n = log θ(y) =
log y + O(1), or log y = log log n + O(1). This yields (check!)
1
1
= γ
1+O
.
e log log n
log log n
47
Now we consider general n. Let y be the ω(n)-th prime, and set
m=
Y
p.
p≤y
We see ::
•
n≥
Y
p≥
Y
p = m.
p≤y
p|n
•
Y
1
1
φ(m)
φ(n) Y
=
.
1−
≥
1−
=
n
p
p
m
p|n
p|m
Therefore,
φ(m)
1
1
φ(n)
≥
= γ
1+O
n
m
e log log m
log log m
1
1
≥ γ
1+O
.
e log log n
log log n
A similar proof shows ::
log n
ω(n) ≤
log log n
1+O
1
log log n
which is also best possible.
We saw that the minimal order of φ(n) is
n
.
eγ log log n
This means
φ(n) ≥ (1 + o(1))
eγ
or
φ(n) &
n
eγ log log n.
48
n
,
log log n
,
Much earlier I claimed that
log n
ω(n) ≤
+O
log log n
log n
log log2 n
,
(Here, log log2 n = (log log n)2 ) follows by a similar argument... We’ll use the fact that
ω(x)
ω(x)
π(x) =
+O
log x
log2 x
or
x
ω(x)
π(x) −
.
log x
log2 x
Proof. Recall the argument that
2ω(n) n .
Y 2
Y 2
2ω(n)
=
≤
.
n
pk
p
k
1/
p kn
p≤2
Equivalently,
ω(n) log 2 ≤ log n +
X
(log 2 − log p)
p≤21/
= log n + (log 2)π(21/ ) − · ω(21/ )
ω(21/ )
1/
= log n + (log 2) π(2 ) −
log 21/
21/
= log n + O
log2 21/
= log n + O(2 21/ ).
We now choose
=
log 2
.
log log n
log 2
1
log log n/ log 2
ω(n) log 2 ≤
log n + O
2
log log n
log log2 n
log 2 log n
1
=
+O
log n .
log log n
log log2 n
Dividing by log 2 finishes the proof.
49
A similar argument will show
log n
log τ (n) ≤ log 2
+O
log log n
log n
log log2 n
,
or
τ (n) ≤ 2(1+o(1)) log n/ log log n n .
Yet,
2(1+o(1)) log n/ log log n exp(logb n),
for 0 < b < 1.
Lemma ::
X
(ω(n) − log log x)2 x log log x.
(Variance)
n≤x
Proof. We expand the left-hand side to
X
ω(n)2 − 2 log log x
n≤x
X
ω(n) + bxc log log2 x.
n≤x
We’ve already seen that
X
ω(n) = x log log x + O(x).
n≤x
X
ω 2 (n) = x log log2 x + O(x log log x).
n≤x
Therefore,
X
(ω(n) − log log x)2 = (x log log2 x + O(x log log x))
n≤x
− 2 log log x(x log log x + O(x))
+ x log log2 x + O(log log2 x)
= 0 · x log log2 x + O(x log log x).
It turns out the same estimate holds for
X
(ω(n) − log log n)2 .
n≤x
50
Now let S = {n ∈ N : |ω(n) − log log n| > (log log n)3/4 }. Then
X
1 ≤
n≤x
n∈S
X (ω(n) − log log n)2
n≤x
(log log n)3/2
. (log log x)−3/2
X
(ω(n) − log log n)2
n≤x
−3/2
(log log x)
x log log x
x
=
.
(log log x)1/2
We’ve shown
ω(n) = log log n + O((log log n)3/4 ),
for almost all n.
(“almost all” means “on a set of density 1.”)
for density δ ⇒
X
f (n) ∼ δ.
n≤x
Comment :: Averages of Ω(n) are the same as for ω(n).
X
X x x x + 2 + 3 + ···
Ω(n) =
p
p
p
n≤x
p≤x
X 1
∼x
p−1
p≤x
X 1
1
+O
=x
.
2
p
p
p≤x
(Lecture 15)
We saw that
X
(ω(n) − log log n)2 x log log x.
n≤x
This implies that if E(n) goes to infinity faster than
√
log log n, then almost all integers n
will satisfy
|ω(n) − log log n| < E(n).
(In fact it’s true that
P
n≤x (ω(n)
51
− log log n)2 ∼ x log log x.)
Thinking in terms of
average(ω(n)) ∼ log log n.
variance(ω(n)) ∼ log log n.
We define
ω(n) − log log n
√
<b .
S(a, b) = n ∈ N : a <
log log n
We have the Erdös-Kac Theorem ::
1
#{n ≤ x : n ∈ S(a, b)}
x→∞ x
= density of S(a, b)
Z b
1
2
= √
e−t /2 dt = Normal Distribution!
2π a
lim
Analogy :: Xp a random variable, with
Pr (Xp = 1) =
1
,
p
1
Pr (Xp = 0) = 1 − .
p
Then
!
E
X
Xp
!
∼ log log x ∼ σ
p≤x
2
X
Xp .
p≤x
This analogy is imperfect, because
1
1 x
1 1 x
#{n ≤ x : p|n} =
= −
.
x
x p
p x p
In the random variable analogy, Xp and Xq are independent. However,
1
{n ≤ x : p|n and q|n}
x
1
1 x
=
−
pq x pq
1 1 x
1 1 x
6
=
1− −
−
.
p x p
q x q
52
Probabilistic Heuristics
1. Squarefree Numbers
n is a squarefree ⇔ for all primes p, p2 - n.
Heuristics ::
• “Probability” that p2 - n is 1 −
1
p2
.
• These events are “independent.”
Therefore, the “probability” that n is squarefree should be
Y
1
1
6
1− 2 =
= 2.
p
ζ(2)
π
p
(Got the right answer!)
2. Prime numbers
n is prime ⇔ for all primes p ≤
√
n, p - n. The “probability” that p - n is 1 − p1 ,
so we get the prediction that n is prime with probability
Y 1
1
√
∼ γ
1−
p
e log n
√
p≤ n
= 2e−γ
1
.
log n
This would imply the prediction
π(x) ∼ 2e−γ
x
,
log x
but this is provably false from Chebyshev (Note that 2e−γ ≈ 1.12).
3. Prime values of Polynomials
Example :: f (x) = x2 + 1.
Question :: Given a prime p, what’s the “probability” that p|f (n) = n2 + 1?.
The number of solutions to n2 + 1 ≡ 0 (mod

1



−1
1+
= 2

p


0
p) is
if p = 2,
if p ≡ 1 (mod 4),
if p ≡ 3 (mod 4).
53
Thus the probability that p|f (n) is
1
p
1+
−1
p
.
So we might predict something like
Y
√
p≤
1
1−
p
1+
−1
p
f (n)
as the probability that f (n) is prime.
From example 2, we’re suspicious...
So instead, lets predict
#{n ≤ x : n2 + 1 is prime}
−1
x Y
1
−1
1
∼
1−
1+
1−
2 log x p
p
p
p
Y
Y
x
p−2
p
.
=
2 log x
p−1
p−1
p≡1 (mod 4)
p≡3 (mod 4)
In general, if f is irreducible, and
σ(p) = #{a (mod p) : f (a) ≡ 0 (mod p)},
then
−1 Y
x
1
σ(p)
#{n ≤ x : f (n)is prime} ∼
1−
1−
.
deg(f ) log x p
p
p
(Lecture 16)
Definition :: A Dirichlet Series is an expression of the form
A(s) =
∞
X
an n−s
for some sequence {an }. (Think arithmetic function)
n=1
Convention :: In Analytic Number Theory, we tend to use s (rather than z) as the
complex variable, writing s = σ + it, where σ, t ∈ R.
Goal #1 :: To understand the region of convergence of a Dirichlet series A(s).
54
Lemma :: Suppose that A(s0 ) converges, where s0 = σ0 + it0 . Then for any H > 0, the
series A(s) converges uniformly in the sector
S = {s = σ + it : σ ≥ σ0 , |t − t0 | ≤ H(σ − σ0 )}.
Proof. Define
R(u) =
X
an n−s0 .
n>u
Then given > 0, there exists M such that |R(u)| < for all u ≥ M . (Sequences of 1 and
M + 1 and 1 + N are Cauchy sequences). Then
Z N
N
X
−s
an n =
us0 −s d(−R(u))
n=M +1
M
= −R(u)u
N Z
+
N
s0 −s M
55
M
R(u)(s0 − s)us0 −s−1 du.
Therefore,
Z N
N
X
−s σ0 −σ
σ0 −σ
|R(u)|uσ0 −σ−1 du
an n ≤ |R(M )|M
+ |R(N )|N
+ |s0 − s|
M
n=M +1
Z ∞
uσ0 −σ−1 du
< · 1 + · 1 + |s0 − s|
M
|s0 − s|
= 2 + M σ0 −σ
σ − σ0
|s0 − s|
1+
.
σ − σ0
This shows that {
PN
n=1
an n−s } is a Cauchy sequence. Hence A(s) converges. Moreover,
for s ∈ S,
|s0 − s| ≤ |t0 − t| + (σ − σ0 )
≤ (H + 1)(σ − σ0 ),
and so the upper bound is S . Hence A(s) converges uniformly on S.
Theorem :: Any Dirichlet series A(s) has an abscissa of convergence σc ∈ R ∪ {−∞, ∞}
with the properties ::
• A(s) converges if σ > σc ,
• A(s) doesn’t converge if σ < σc .
56
Moreover, A(s) is locally uniformly convergent in {s : σ > σc }. Consequently, A(s) is
analytic on {s : σ > σc }.
Proof. Define σc = inf{Re(s) : A(s) converges}. Obviously, A(s) diverges if σ < σc . If
σ > σc , then there exists s0 with σ > σ0 > σc such that A(s0 ) converges. Then take
H>
t − t0
,
σ − σ0
and apply Lemma.
57
Example :: Define
A(s) =
∞
X
(−1)n−1 n−s
n=1
= 1−
1
1
1
+ s − s + ···
s
2
3
4
First of all, A(s) converges absolutely precisely where
∞
X
n=1
n−1 −s
|(−1)
n |=
∞
X
n−σ = ζ(σ)
converges,
n=1
that is, on {s : σ > 1}.
However, A(σ) (σ ∈ R) is an alternating series of terms decreasing in absolute value when
σ > 0. Thus A(σ) converges for σ > 0; by the Theorem we concludes that A(s) converges
on {s : σ > 0}. (Note :: A(s) diverges for σ ≤ 0 simply because the summand doesn’t
tend to zero).
58
Definition :: The abscissa of absolute convergence, σa , of A(s) is the abscissa of the
convergence of
|A|(s) =
∞
X
|an |n−s .
n=1
Example Above ::
A(s) =
∞
X
(−1)n−1 n−s ,
σc = 0,
n=1
Theorem :: We have
σa − 1 ≤ σc ≤ σa .
Proof. σc ≤ σa is obvious. Suppose > 0, then
∞
X
an n−(σc +)
converges.
n=1
Hence an n−(σc +) → 0 as n → ∞. Then
∞
X
an n−(σc +1+2)
n=1
59
σa = 1.
converges absolutely by comparison with
∞
X
n−(1+) .
n=1
This shows σa − 1 ≤ σc .
(Lecture 17)
Remark ::
∞
X
(−1)n−1 n−s =
n=1
=
X
X
n−s −
n even
n odd
∞
X
X
n−s − 2
n=1
= ζ(s) − 2
n−s
n−s
n even
∞
X
(2m)−s
m=1
= ζ(s) − 21−s
∞
X
m−s = ζ(s)(1 − 21−s ).
m=1
(Manipulations valid for σ > 1.)
Therefore,
∞
X
1
ζ(s) =
(−1)n−1 n−s
1−s
1−2
m=1
for σ > 1.
Note ::
1 − 21−s = 0 ⇔ 21−s = 1
⇔ exp((1 − s)(log 2)) = 1
⇔ (1 − s) log 2 = 2πik, k ∈ Z
2πik
, k ∈ Z.
⇔ s=1−
log 2
So this provides another analytic continuation of ζ(s) to
2πik
{σ > 0}\ 1 +
, k∈Z .
log 2
60
Remark on General Dirichlet Series
P
−s
−s
Given A(s) = ∞
as
n=1 an n , let F be the first integer with aF 6= 0. Then A(s) ∼ aF F
P∞
σ → ∞. In particular, if n=1 an n−s = 0 for all σ > K, say. Then an = 0 for all n ∈ N.
It follows that
∞
X
an n
−s
=
n=1
∞
X
bn n−s
for σ > K ⇒ {an } = {bn }.
n=1
Theorem :: (Landau, 1905)
A(s) =
∞
X
an n−s .
n=1
Suppose σc is finite and an ≥ 0 for all n ∈ N. Then A(s) has a singularity at σc .
The Idea ::
Proof. By replacing an with an n−σc , we may assume σc = 0. Suppose A(s) is analytic at
0, that is, analytic on some {|s| < δ}. Expand A(s) in a power series around s = 1.
∞
X
1 (k)
A (1)(s − 1)k . (Power Series Expansion)
A(s) =
k!
k=0
61
P∞
an n−s , for σ > σc , then since A(s) is locally uniform converP
−s
gent, then we can differentiate term-by-term. ie., A0 (s) = ∞
n=1 (− log n)an n .
Sideshow :: If A(s) =
n=1
Thus,
(k)
∞
X
k
A (1) = (−1)
(log n)k an n−1 .
n=1
Therefore,
∞
∞
X
X
an
1
k
k
A(s) =
(s − 1) (−1)
logk n.
k!
n
n=1
k=0
Now the power series A(s) converges in a disk (about s = 1) of radius at least
√
it converges at s = −δ 0 , say, for 0 < δ 0 < 1 + δ 2 − 1.
√
1 + δ 2 so
∞
∞
X
(1 + δ 0 )k X an
A(−δ ) =
(log n)k .
k!
n
n=1
k=0
0
Now, everything in sight is non-negative, hence
A(−δ 0 ) =
=
=
∞
∞
X
an X 1
((1 + δ 0 ) log n)k
n k=0 k!
n=1
∞
X
an
n=1
∞
X
n=1
n
0
e(1+δ ) log n
∞
an 1+δ0 X
0
n
=
an n−(−δ ) .
n
n=1
But this asserts that the Dirichlet series
P∞
n=1
an n−s converges at s = −δ 0 , contradicting
σc = 0. Thus A(s) must have a singularity at s = 0.
(Lecture 18)
Recall :: If g is a multiplicative function, and
either
∞
X
|g(n)| < ∞ or
n=1
then
Y
(1 + |g(p)| + |g(p2 )| + · · · ) < ∞,
p
∞
X
n=1
g(n) =
Y
(1 + g(p) + g(p2 ) + · · · ).
p
62
Let F (s) =
P∞
n=1
f (n)n−s , where f is multiplicative, and σ > σa . Then
∞
X
|f (n)n−s | < ∞
n=1
by definition of σa . But f (n)n−s = (T −s f )(n) is multiplicative, hence by the “recall,”
∞
X
f (n)n
−s
=
Y
n=1
p
f (p) f (p2 )
1 + s + 2s + · · ·
p
p
,
for σ > σa .
This is called the Euler product for F (s).
Note :: If f is totally multiplicative, then the factors in the Euler product are geometric
series, and so
∞
X
f (n)n
−s
=
Y
n=1
p
f (p)
1− s
p
−1
.
Examples ::
(a)
ζ(s) =
∞
X
n−s =
Y
(1 − p−s )−1 ,
σ > 1.
p
n=1
(b)
∞
X
n=1
µ(n)n
−s
=
Y
p
1
1 − s + 0 + 0 + ···
p
Y
1
=
(1 − p−s ) =
,
ζ(s)
p
σ > 1.
Personal Notes ::
• In the function with µ, σa is exactly 1 because there are enough squarefrees so that
it cannot get anymore left (justified by a partial summation argument)
• Infinite products converges implies the product being non-zero - think in terms of
log-land (product goes to zero ⇒ log ⇒ −∞.)
63
(c)
∞
X
2
µ (n)n
−s
=
Y
n=1
p
=
Y 1 − p−2s
p
=
1
1+ s
p
1 − p−s
Y
Y
(1 − p−s )−1 (1 − p−2s )
p
p
=
ζ(s)
,
ζ(2s)
σ > 1.
(d) Define the Liouville lambda-function
λ(n) = (−1)Ω(n) .
(Recall Ω(n) counts the number of prime factors of n with multiplicity)
Then
∞
X
λ(n)n−s =
Y
(1 − p−s + p−2s − p−3s + · · · )
p
n=1
Y
=
(1 + p−s )−1
p
=
Example ::
η(n) =
∞
X
ζ(2s)
,
ζ(s)
σ > 1.
(−1)n−1 n−s . (Dirichlet Eta function)
n=1
Recall that σa = 1, σc = 0 for η(s) and that η(s) = (1 − 21−s )ζ(s).
Note that
(−1)n−1

 −1 if 2|n,
=
 1 if 2 - n.
That is, it’s the multiplicative function f defined by

 −1 if p = 2,
f (pk ) =
 1 if p > 2.
64
Therefore, for σ > 1,
η(s) = (1 − 2−s − 2−2s − · · · )
Y
(1 + p−s + p−2s + · · · )
p>2
(∗)
=
−s
Y
2
1−
(1 − p−s )−1 .
1 − 2−s p>2
Notes ::
• This actually equals to
ζ(s) 1 −
2−s
1 − 2−s
(1 − 2−s ) = ζ(s)(1 − 2−s − 2−s ),
but we know this already...
• The infinite product in (∗) converges exactly where
P
p
p−s converges, which is σ > 1.
In particular, it is false to say η(s) = (Euler product) for σ > σc .
We’ve use the fact that
Y
(1 + xn ) converges
⇔
n
X
log(1 + xn ) converges.
n
(This is from the definition of convergence of infinite product)
X
xn converges.
⇔
n
(Since log(1 + xn ) = xn + O(x2n ))
Personal Note :: This power series expansion is only good when xn is small, that said,
if xn isn’t small, then it won’t converge anyways...
More Examples ::
∞
X
n=1
(n,q)=1
n−s


 0 if ∃ p|q, p|n, 
=
n−s
 1 if not.

n=1
Y
Y
=
(1 − p−s )−1 (1 + 0 + 0 + · · · )
∞
X
p-q
p|q
Y
= ζ(s) (1 − p−s ),
p|q
65
σ > 1.
This is easy to generalize to
f (n)n−s .
P
(n,q)=1
Non-Euler product stuff
Suppose
F (s) =
∞
X
f (n)n−s
for σ > σc .
n=1
Then
•
∞
X
α
(T f )(n)n
−s
=
n=1
∞
X
(f (n)nα )n−s
n=1
= F (s − α),
for σ > σc +Re(α).
•
∞
X
k
(L f )(n)n
−s
=
n=1
∞
X
(f (n) logk n)n−s
n=1
= (−1)k F (k) (s),
for σ > σc .
Example ::
−ζ 0 (s) = −
∞
X
!0
1(n)n−s
n=1
=
=
∞
X
n=1
∞
X
(L1)(n)n−s
(log n)n−s ,
valid for σ > 1.
n=1
!
Y
log ζ(s) = log
(1 − p−s )−1
p
=
X
log(1 − p−s )−1
p
∞
X X
(p−s )k
=
k
p
k=1
=
∞
X
κ(n)n−s ,
n=1
66
!
for σ > 1.
ζ 0 (s)
−
= −(log ζ(s))0
ζ(s)
∞
X
=
(log n)κ(n)n−s
n=1
=
∞
X
Λ(n)n−s .
n=1
(Lecture 19)
Theorem :: Suppose
A(s) =
∞
X
a(n)n
−s
and B(s) =
n=1
∞
X
b(n)n−s ,
n=1
both converge absolutely at s. Then the Dirichlet series
∞
X
(a ∗ b)(n)n−s
n=1
converge absolutely at s to A(s)B(s).
Proof.
∞
X
(a ∗ b)(n)n−s =
n=1
=
∞
X
X
n=1
cd=n
∞
X
!
a(c)c
c=1
a(c)b(d)(cd)−s
−s
∞
X
b(d)d−s · 1,
d=1
and this is valid (in reverse?) by absolutely convergence.
Note :: If A, B are just convergent. ie., A(s) = B(s) = η(s). η converges for σ > 0, but
η ∗ η(s) = η 2 (s) = converges for σ > 1/4. (Information is lost when taken the convolution).
Example 1 :: The convolution identity 1 ∗ µ = e corresponds to
ζ(s) ·
1
= 1.
ζ(s)
Example 2 ::
A(s) =
∞
X
n=1
67
(σ > 1)
τ (n)n−s .
We know τ (n) n , and so σc = σa = 1. Also by the Theorem,
so A(s) = ζ 2 (s).
τ = 1 ∗ 1,
(σ > 1)
This gives an analytic continuation of A(s) to {σ > 0}\{1}. We have
ζ(s) =
1
+ γ + O(s − 1) near s = 1.
s−1
Then
2
1
+ γ + O(s − 1)
A(s) =
s−1
1
2γ
=
+
+ O(1).
(s − 1)2 s − 1
So A(s) has a double pole at s = 1. Also,
Ress=1 A(s) = 2γ.
(Not 1!)
Note :: Notice that for simple poles,
Ress=s0 f (s) = lim (s − s0 )f (s).
s→s0
But for higher order poles, the residue exists but the limit does not.
Example 3 ::
B(s) =
∞
X
σ(n)n−s .
n=1
We have
σ(n) =
X
or σ = 1 ∗ T 1.
d,
d|n
Therefore, B(s) = ζ(s)ζ(s − 1).
Since
∞
X
T 1(n)n
n=1
(σ > 2)
−s
=
∞
X
n·n
n=1
Remark :: Ress=2 B(s) = ζ(2).
68
−s
=
∞
X
n=1
n−(s−1) .
Example 4 :: The Dirichlet series Identity
0 ζ (s)
ζ(s) = −ζ 0 (s),
−
ζ(s)
corresponds to
∞
X
!
∞
X
Λ(n)n−s
!
n−s
∞
X
=
or Λ ∗ 1 = L1,
L1(n)n−s
,
n=1
n=1
n=1
!
X
or
Λ(d) = log n.
d|n
Example 5 ::
C(s) =
∞
X
φ(n)n−s .
n=1
Solution 1. Note
X
or φ ∗ 1 = T 1.
φ(n) = n,
d|n
So C(s)ζ(s) = ζ(s − 1), or
C(s) =
ζ(s − 1)
. (σ > 2)
ζ(s)
Solution 2. Note
φ(n) X µ(d)
=
,
n
d
d|n
(depends on primes dividing n, not the prime powers)
so T −1 φ = 1 ∗ T −1 µ. This gives
C(s + 1) = ζ(s)
1
.
ζ(s + 1)
(Lecture 20)
Remarks From The Past :: If
F (s) =
∞
X
−s
f (n)n ,
G(s) =
n=1
∞
X
g(n)n−s ,
n=1
H(s) = F (s)G(s) =
∞
X
n=1
69
(f ∗ g)(n)n−s ,
and
for σ > max{σa (F ), σa (G)}. In other words,
σa (H) ≤ max{σa (F ), σa (G)}.
Note :: This might not be an equality!
Example ::
F (s) = ζ(s),
G(s) =
1
. Then H(s) = 1.
ζ(s)
Then σa (F ) = 1, σa (G) = 1, but σa (H) = −∞.
(Verify)
Analogy :: Let ordp (n) = k if and only if pk kn. Then
ordp (a + b) ≥ min{ordp (a), ordp (b)}.
Note that this might not be an equality, but the smallest two values among {ordp (a), ordp (b), ordp (a + b)}
are the same (this is somewhat related to the p-adic metric...).
Example 1 :: Define
k
∞ X
n
Dk (s) =
n−s . (k ∈ N)
φ(n)
n=1
Show that σa = σb = 1. But that Dk can be meromorphically continued to {σ > 0}, with
its only pole a simple pole at s = 1. Calculate Ress=1 Dk (s).
Solution. Notice that
n
φ(n)
k
≤ ((eγ + o(1)) log log n)k
k, n .
Therefore, σa = σc = 1.
The function (n/φ(n))k is multiplicative.
70
Lets experiment ::
r
pr
φ(pr )
0
k
µ(pr )
∗
1
1
p
p−1
1
1
2
k
p
p−1
-1
p
p−1
k
−1
···
3
k
p
p−1
k
···
0
0
···
0
0
···
So we define fk (n) to be the multiplicative function satisfying
 k
 p
− 1 if r = 1,
p−1
r
f (p ) =

0
if r > 1.
Let
Fk (s) =
∞
X
fk (n)n−s .
n=1
Then
n
φ(n)
Dk (s)
k
∗ µ(n) = fk (n),
1
= Fk (s),
ζ(s)
so
σ > 1.
Since Fk converges absolutely for σ > 1, we have the Euler product
Y
fk (p) fk (p2 )
Fk (s) =
1 + s + 2s + · · ·
p
p
p
Y
=
(1 + f (p)p−s ), σ > 1.
p
71
Where does Fk converge?
We have
(1 − x)k = 1 + kx + Ok (x2 )
= 1 + Ok (x) uniformly for |x| ≤ 1/2.
Thus
fk (p) =
p
p−1
k
−1=
1
1−
p
−k
1
.
− 1 = Ok
p
⇔
X
fk (p)p−s
So
Y
(1 + fk (p)p−s ) converges
p
converges.
p
But
X
p
|fk (p)p−s | k
X1
p
p
p−σ
converges for σ > 0.
Therefore, Dk (s) = ζ(s)Fk (s) is an analytic continuation of Dk to {σ > 0}, except for a
simple pole at s = 1.
The residue is
Ress=1 ζ(s)Fk (s) = lim(s − 1)ζ(s)Fk (s)
s→1
= lim(s − 1)ζ(s) lim F (s)
s→1
s→1
= Ress=1 ζ(s)Fk (1)
∞
X
fk (n)
= 1 · Fk (1) =
n
n=1
!!
k
Y
1
p
1+
−1
.
=
p
p
−
1
p
72
For example, take k = 1.
=
=
=
=
!
∞
X
n −s
= F1 (1)
Ress=1
n
φ(n)
n=1
Y
1
p
1+
−1
p p−1
p
Y
1
1
−
1+
p
−
1
p
p
Y p2 − p + 1 p(p − 1)
p
=
Y
Y p3 + 1
p6 − 1
=
p(p2 − 1)
p(p2 − 1)(p3 − 1)
p
p
=
Y
1 − p−6
ζ(2)ζ(3)
=
.
−2
−3
(1 − p )(1 − p )
ζ(6)
p
Hold on!
ζ(2)ζ(3)
ζ(6)
was the average value of
n
!
φ(n)
Lets look at
ζ(2s)ζ(3s)
. (Not really a connection to the previous example...)
ζ(6s)
General Remark :: If
A(s) =
∞
X
an n−s ,
n=1
then for k ∈ N,
A(ks) =
∞
X
an n−ks
n=1
=
∞
X
bm m−s ,
n=1

 a if m = nk ,
n
where bm =
 0 otherwise.
Therefore,
ζ(2s)ζ(3s)
=
ζ(6s)
∞
X
n=1

n−s 
1 if n = square,
0 else.
= (From Homework)
 
∗
X
1 if n = cube,
0 else.
n−s .
n power-full
73
 
∗
µ(m) if n = m6 ,
0
else.


Where are the poles of
ζ(2s)ζ(3s)
?
ζ(6s)
• Whenever ζ(ρ) = 0, this function has a pole at ρ/6, however, there are none with
Re(ρ/6) > 1/6.
• Pole at s = 1/2, Ress=1/2 = ζ(3/2)/ζ(3).
• Pole at s = 1/3, Ress=1/3 = ζ(2/3)/ζ(2).
(Lecture 21)
Goal :: Write down some contour integral involving
A(s) :=
∞
X
an n−s
that relates to
n=1
X
an .
n≤x
Convention :: When we write
Z
c+iT1
Z
c+i∞
or
c−iT1
c−i∞
we mean a contour integral over a vertical line (segment).
Lemma :: (Sound Bite Version) Let c > 0. Then



1 if y > 1,

Z c+i∞ s

1
y
ds = 0 if 0 < y < 1,

2πi c−i∞ s


 1 if y = 1.
2
Note ::
Z
c+i∞
Z
means
c+iT2
lim
T1 ,T2 →∞
c−i∞
.
c−iT1
If the integral converges absolutely, then we can let T1 , T2 → ∞ however we like. Given
this remark, the y = 1 case of the lemma is technically a lie ::
Z c+iT
1
1
1
lim
ds = .
T →∞ 2πi c−iT s
2
74
Lemma :: (Explicit Version) Let c, y, T > 0. Define
Z c+iT s
1
y
Ic (y, T ) =
ds.
2πi c−iT s
Then
• If y > 1, then
|Ic (y, T ) − 1| < y min 1,
c
1
πT log y
.
• If 0 < y < 1, then
|Ic (y, T )| < y min 1,
c
1
πT log y −1
.
• If y = 1, then
Ic (y, T ) −
c
1 <
.
2
πT
Proof. (y > 1) Consider
1
2πi
I
ys
ds around the following contour ::
s
By the residue theorem,
1
2πi
I
ys
ys
ds = Ress=0 = 1.
s
s
75
Also, the right-hand segment gives exactly Ic (y, T ). On the bottom segment,
Z c−iT s Z c
1
y yδ
ds
≤
dδ
2πi
−u−iT s
−u |δ − iT |
(change of variable s = δ − iT )
c
Z c
1
1 y δ δ
<
y dδ =
2πT −∞
2πT log y −∞
yc
.
=
2π log y
We get the same estimate for the top segment. On the left-hand segment:
Z −u+iT s Z T
1
y 1
y −u
ds ≤
dt
2πi
s 2π −T | − u + it|
−u−iT
y −u
<
2T.
2πu
Therefore,
|Ic (y, T ) − 1| < 2
yc
y −u
+
2T.
2πT log y 2πu
Taking u → ∞ yields
|Ic (y, T ) − 1| ≤
yc
.
πT log y
Consider instead
1
2πi
I
ys
ds.
s
On this contour ::
76
The length of the arc is < 2πR. On this arc, |y s | ≤ y c , and 1/|s| = 1/R. Therefore,
Z
s
1
y
< 1 2π · R · y c · 1 = y c .
ds
2πi
2π
R
arc s
Remark :: The case 0 < y < 1 is very similar, only using the contours
(The left diagram :: No pole inside, we get 0 instead of 1)
The case y = 1 actually 1st year calculus:
Z c+iT s
Z T
Z T
1
1
1
i dt
1
1
1
ds =
=
+
dt = · · ·
2πi c−iT s
2πi −T c + it
2π 0
c + i c − it
In the sound Bite version, put y = x/n:



 1 if 0 < n < x,
Z c+i∞ s
1
x ds 
= 0 if n > x > 0,

2πi c−i∞ n
s


 1 if n = x > 0.
2
Therefore,
∞
X
n=1
an
1
2πi
Z
c+i∞
c−i∞
x s ds X 1
=
an + ax if x ∈ N .
n
s
2
n≤x
77
This leads to:
Perron’s Formula :: (Sound Bite version) Let
A(s) =
∞
X
an n−s
n=1
have abscissa of absolute convergence σa . Let x > 0 and c > max{0, σa }. Then
Z c+i∞
X0
1
xs
A(s) ds =
an ,
2πi c−i∞
s
n≤x
where
P0
n≤x
means, if n ∈ N, include 12 ax in the sum instead of ax .
Perron’s Formula :: (Explicit Version) Let
A(s) =
∞
X
an n−s .
n=1
Let x > 0 and C > max{0, σa }, and T > 0. Then
Z c+iT
X0
s
1
x
an −
A(s) ds
2πi
s
c−iT
n≤x
X
xc
c
ax
V
+
|an | min 1,
+
if n ∈ N. .
T
T |x − n|
T
x/2<n<3x/2
Here,
V =
∞
X
|an |n−c .
n=1
78
(Lecture 22)
Proof of Perron’s Formula (Explicit Version). Let
A(s) =
∞
X
an n−s ,
n=1
let c > max{0, σa }, and let x, T > 0. Then
1
an −
2πi
n≤x
X0
Z
c+iT
c−iT


x




1 if n > 1,


∞


s
X
x
x
an 0 if 0 < n < 1,
A(s) ds =


s


n=1


 1 if x = 1.

2
n
!
Z c+iT X
∞
xs
1
an n−s
ds
−
2πi c−iT
s
 n=1



1
Z c+iT s 
∞

 

X
1
x ds 

=
an  0 −
,


2πi
n
s


c−iT


n=1

1

2
valid by absolute converges of A(s).
By the lemma from Lecture 21, the left-hand side
(
)
∞
x c
X
1
c
|an |
min 1, + |ax | if n ∈ N .
n
T
T log nx n=1
log x 1, and hence their contribution
have
The terms in the sum with n ≤ x2 or n ≥ 3x
2
n
is
X
|x−n|≥x/2
∞
c X
x
T
n=1
x c 1
|an |
·
n
T
|an |
nc
xc
=
V.
T
For the terms x < n < 3x/2, use the “1” for the smallest n, and for the others (by a power
series expansion),
x−n
log
= log 1 −
n
n
x
79
−1
n−x
.
x
Therefore,
X
|an |
x c
n
x/2<n<3x/2
n6=x
(
1
min 1, T log
)
c
x
n
2
1,
X
x/2<n<3x/2
n6=x
x
T |n − x|
.
Final Answer :: Left-hand side
X
xc
V + 2c
T
x/2<n<3x/2
n6=x
Example 1 :: Express
P0
n≤x
x
|an | min 1,
T |n − x|
c
+ |ax | if n ∈ N .
T
τ (n)2 as a contour integral.
Solution. Let
B(s) =
∞
X
τ 2 (n)n−s .
n=1
Lets try to write B(s) in terms of ζ, for example. Since τ 2 is multiplicative, we look at
values on prime powers. (∗µ here would be “like” a “difference operator”...)
0 1
2
3
τ 2 (pr )
1 4
9
16 25 36 · · ·
τ 2 ∗ µ(pr )
1 3
5
7
9
11 · · ·
τ 2 ∗ µ ∗ µ(pr )
1 2
2
2
2
2
···
τ 2 ∗ µ ∗ µ ∗ µ(pr )
1 1
0
0
0
0
···
τ 2 ∗ µ ∗ µ ∗ µ ∗ µ(pr ) 1 0 -1
0
0
0
···
80
4
5
···
r
Hence for σ > 1,
B(s)
1
ζ(s)
4
=
∞
X
(τ 2 ∗ µ ∗ µ ∗ µ ∗ µ)(n)n−s
n=1
=
Y
p
0
1
1 + s − 2s + 0 + · · ·
p
p
=
1
.
ζ(2s)
Hence
B(s) =
ζ 4 (s)
,
ζ(2s)
and so
1
τ (n) =
2πi
n≤x
X0
2
Note ::
∞
X
Z
c+i∞
c−i∞
ζ 4 (s) xs
ds. (c > 1)
ζ(2s) s
τ 2 (n)n−s =
n=1
ζ 4 (s)
.
ζ(2s)
The other rows give ::
∞
X
ζ 3 (s)
=
τ (n2 )n−s .
ζ(2s)
n=1
∞
X
ζ 2 (s)
=
2ω(n) n−s .
ζ(2s)
n=1
∞
X
ζ(s)
µ2 (n)n−s .
=
ζ(2s)
n=1
Example 2 :: Express
ψ0 (x) =
X0
Λ(n)
n≤x
using explicit Perron Formula.
Solution. We know that
∞
X
ζ0
Λ(n)n−s = − (s).
ζ
n=1
Therefore, for c > 1,
Z c+iT 0 s
1
ζ
x
ψ0 (x) −
− (s)
ds
2πi c−iT
ζ
s
∞
X
xc X |Λ(n)|
x
Λ(x)c
c
+2
|Λ(n)| min 1,
+
if x ∈ N .
T n=1 nc
T |x − n|
T
x/2<n<3x/2
n6=x
81
Using Λ(n) ≤ log n,
0 X
ζ
xc
− (c) + 2c log x
T
ζ
x/2<n<3x/2
n6=x
x
min 1,
T |x − n|
+
c log x
.
T
The sum is
x/2
X
x
x log x
1+2
1+
,
T
k
T
k=1
and so
Z c+iT 0
1
ζ
xs
ψ0 (x) −
− (s)
ds
2πi c−iT
ζ
s
0 ζ
x log2 x
c log x
xc
c
− (c) + 2 log x +
+
.
T
ζ
T
T
(Lecture 23)
In Lecture 22, we showed
Z c+iT 0 s
x
1
ζ
ψ0 (x) =
Λ(n) =
ds
− (s)
2πi c−iT
ζ
s
n≤x
c 0 x
ζ
x log x
c
c
+O
− (c) + 2 log x 1 +
+ log x ,
T
ζ
T
T
X0
for c > 1, T > 0. So in particular, lets stipulate 1 < T ≤ x and choose
c=1+
1
,
log x
so xc = ex. Then
1
ψ0 (x) =
2πi
Z
c+1/ log x+iT
c+1/ log x−iT
s
ζ0
x
x log2 x
− (s)
ds + O
,
ζ
s
T
since
1
ζ0
− (1 + δ) ∼ .
ζ
δ
Reason :: −ζ 0 /ζ has a simple pole of residue 1 at s = 1, so
−
ζ0
1
∼
ζ
s−1
82
near s = 1.
At a minimum, we’d like to know that ζ(s) 6= 0 for σ = 1. (In fact, this statement is
equivalent to the prime number theorem π(x) ∼ x/ log x). To see this, we’ll use
log ζ(s) =
∞
X
κ(n)n−s ,
n=1
where κ(pr ) = r−1 and κ(n) = 0 otherwise (σ > 1).
Consider ::
Re(3 log ζ(σ) + 4 log ζ(σ + it) + log ζ(σ + 2it)).
Since log ζ converges absolutely for σ > 1,
= Re
∞
X
!
κ(n)n−σ 3 + 4nit + n
−2it
n=1
=
∞
X
κ(n)n−σ (3 + 4 cos(−t log n) + cos(−2t log n)) .
n=1
Note that
3 + 4 cos θ + cos 2θ = 3 + 4 cos θ + (2 cos2 θ − 1)
= 2(1 + cos θ)2 ≥ 0.
We conclude that
Re(3 log ζ(σ) + 4 log ζ(σ + it) + log ζ(σ + 2it)) ≥ 0.
And so
ζ(σ)3 |ζ(σ + it)4 ||ζ(σ + 2it)| ≥ 1. (σ > 1)
Suppose ζ(1 + it) = 0, then ζ(σ + it) σ − 1 for σ > 1, σ near 1. We know that
ζ(σ) ∼
1
σ−1
and ζ(σ + 2it) 1.
83
Then
ζ 3 (σ)|ζ 4 (σ + it)||ζ(σ + 2it)| 1
(σ − 1)4 · 1 = σ − 1.
3
(σ − 1)
This is a contradiction as σ → 1+ . Hence ζ(1 + it) 6= 0.
To extend this argument, we need results from classical complex analysis: Section 8.4 of
Bateman and Diamond.
Borel-Carathéodory Lemma :: Suppose that f (z) is analytic in {|z| < R}, f (0) = 0,
and satisfies Re(f (z)) ≤ U there. Then for {|z| < r}, where 0 < r < R, we have
|f (z)| ≤
2r
U.
R−r
In fact,
2R
|f (k) (z)|
≤
U.
k!
(R − r)k+1
Lemma 8.10 :: Let f be analytic on {|z − z0 | ≤ r}, with f (z0 ) 6= 0. Suppose that
f (z) M
f (z0 ) < e ,
for {|z − z0 | ≤ r}. Then for {|z − z0 | ≤ r/3}, we have
f0
X 1 36M
,
(z) −
<
f
z − ρ
r
ρ
where ρ runs over the zeros of f in {|z − z0 | < r/2}.
84
(Lecture 24)
Consequence of Lemma 8.10 (and the Borel-Carathéodory Lemma)
Proposition :: Let |t| ≥ 6/7, 6/7 ≤ δ ≤ 10/7. Then
X 1
ζ0
(s) =
+ O(log T ),
ζ
s
−
ρ
ρ
where the sum is over
3
8
.
ρ : ζ(ρ) = 0, + it − ρ ≤
7
7
Notation :: We let T = |t| + 4.
This can be proved by applying Lemma 8.10 to
8
f (z) = ζ z + + it ,
7
The error term O(log T ) uses ::
• ζ(s) T
(See Lemma 8.4 in B. & D.).
• ζ(8/7 + it) 1
(Euler Product).
85
6
r= .
7
Remark :: If δ > 1, then each
1
s−ρ
has positive real part.
Zero-free Region for ζ(s)
Theorem :: Suppose ρ1 = 1 − δ + it satisfies
0<δ≤
1
,
10
6
|t| ≥ ,
7
and ζ(ρ1 ) = 0.
Then there exists an absolute constant c such that
δ>
c
.
log T
Proof. Let s0 = 1 + 4δ, s1 = 1 + 4δ + it, s2 = 1 + 4δ + 2it. Then
ζ0
ζ0
ζ0
Re −3 (s0 ) − 4 (s1 ) − (s2 )
ζ
ζ
ζ
∞
X
=
Λ(n)n−1−4δ (3 + 4 cos(t log n) + cos(2t log n)) ≥ 0.
n=1
On the other hand,
0
• − ζζ (s0 ) is real, and
ζ0
1
1
− (s0 ) =
+ O(1) =
+ O(1).
ζ
s0 − 1
4δ
86
•
!
X 1
ζ0
Re − (s1 ) = Re −
+ O(log T )
ζ
s1 − ρ
ρ
1
+ O(log T )
≤ −Re
s − ρ1
1
= − + O(log T ).
5δ
•
!
X 1
ζ0
Re − (s2 ) = Re −
+ O(log T ) ≤ 0 + O(log T ).
ζ
s
−
ρ
2
ρ
Therefore,
ζ0
ζ0
ζ0
0 ≤ Re −3 (s0 ) − 4 (s1 ) − (s2 )
ζ
ζ
ζ
1
1
≤3
+ O(1) + 4 − + O(log T ) + O (log T ) .
4δ
5δ
This gives
1
≤ O(log T ),
2δ
or δ 1
.
log T
We’ll need the following estimate ::
Lemma 8.11 in B. & D. :: If c > 0 is such that ζ(s) 6= 0 in the region
1−δ <
c
.
log T
Then
ζ0
(s) log T
ζ
in the region
2c
c
s:1−δ <
, |s − 1| >
3 log T
3 log T
Recall ::
1
ψ0 (x) =
2πi
Z
1+ log1 x +iT
1+ log1 x −iT
ζ 0 xs
− (s) ds + O
ζ
s
87
.
x log2 x
T
.
Consider the rectangle with vertices
1+
1
± iT,
log x
1−
c
± iT.
2 log T
Inside the rectangle, ζ(s) has no zeros, and so
0
I
ζ
xs
x1
1
ζ 0 xs
= x.
− (s) ds = Ress=1 − (s) ds =
2πi
ζ
s
ζ
s
1
On the top and bottom segments ::
1
2πi
Z
1+ log1 x ±iT
c
1+ 2 log
T
±iT
ζ 0 xs
1
− (s) ds (log T )
ζ
s
T
Z
1+ log1 x
xσ dσ
−∞
1+ 1
log T xσ log T
=
T log x −∞
x log T
.
=
T log x
On the left side ::
Z 1− c +iT
Z 1− c +iT
2 log T
2 log T
c
1
ζ 0 xs
1− 2 log
T
− (s) ds (log T )x
c
2πi 1− 2 logc T −iT
ζ
s
−iT
1− 2 log
T
ds .
s
Furthermore,
Z T
Z ds <2
s
0
1
2
dt
log T.
+t
We conclude that
c
x log2 x
x log T
ψ0 (x) = O
+x+O
+ O (log T )2 x1− 2 log T
T
T log x
−c
1
2
2
log
T
+x
= x + O x log x
.
T
We choose T so that
−c
1
= x 2 log T .
T
c
log T =
log x
2 log T
r
c log x
⇒ log T =
.
2
88
Final Formula ::
r
2
ψ0 (x) = x + O x log x · exp −
c log x
2
!!
.
(Lecture 26)
Two main Achievements from Lecture 25
1. There exists c > 0 such that if ζ(σ + it) = 0, then
1−σ >
c
,
log τ
where τ = |t| + 4.
2.
r
ψ0 (x) = x + O x log2 x · exp −
c log x
2
!!
.
Road from ψ0 (x) to π(x)
1. Let c <
p
c/2 (note that it’s from the old c to a new c) and note that

0
if x ∈
/ N,
ψ(x) − ψ0 (x) =
 1 Λ(x) log x if x ∈ N.
2
So
ψ(x) = x + O(x · exp(−c
p
log x)).
2. From
√
ψ(x) ≥ θ(x) ≥ ψ(x) − 2ψ( x)
√
√
θ(x) = ψ(x) + O(ψ( x)) = ψ(x) + O( x).
√
Thus θ(x) = x + O(x · exp(−c log x)).
3. Going from
θ(x) =
X
log p to π(x) =
p≤x
X
1.
p≤x
is partial summation.
Z
x
π(x) =
2−
1
dθ(t) =
log t
Z
x
2−
89
dt
+
log t
Z
x
2−
1
d(θ(t) − t).
log t
We define
x
Z
dt
.
log t
Li(x) =
2
Note that
Z x
2−
x Z x
1
θ(t) − t (θ(t) − t)
d(θ(t) − t) =
+
dt
log t
log t 2
t log2 t
2
√
Z x
p
exp(−c log t)
O(x · exp(−c log x)) +
dt.
log2 t
2
If we split this integral into
Z
y
x
Z
+
2
,
y
√
where y = x exp(−c log x), we find that
π(x) = Li(x) + O(x · exp(−c
p
log x)).
4. Weaker but simpler statements
x Z x Z x
dt
t 1
Li(x) =
=
−
td
log t 2
log t
2 log t
2
Z x
dt
x
+ O(1) +
=
2 .
log x
2 log t
We can deduce that
x
Li(x) =
+O
log x
Integrating by parts again gives
Z x
Z x
dt
dt
2 +2
3
2 log t
2 log t
x
log2 x
···
.
···
and so on
In fact, for any K ∈ N,
Li(x) =
K
X
(k − 1)!x
k=1
logk x
+ OK
x
logK+1 x
.
Note that
x · exp(−c
x
p
log x) k,c
90
log
k+1
x
for any k ∈ N.
Riemann’s Functional Equation for ζ(s) (1859)
Define
s
1
−s/2
ξ(s) = s(s − 1)π
Γ
ζ(s).
2
2
Riemann Proved ::
• ξ(s) is entire.
• ξ(1 − s) = ξ(s).
(Lecture 27)
Lemma :: Define
θ(x) =
X
2 πx
e−n
,
(x > 0)
n∈Z
then
√
1
θ
= xθ(x).
x
(Modular Form of Weight 1/2, whatever this means...)
Proof. Using Poisson Summation Formula ::
If f (x) is smooth, f, f 0 integrable, then
X
f (n) =
n∈Z
X
fb(m),
m∈Z
where
Z
∞
f (x)e−2πixt dx.
fb(t) =
−∞
−u2 π/x
So take f (u) = e
.
X
X
1
θ
=
f (n) =
fb(m)
x
n∈Z
m∈Z
XZ ∞
2
=
e−u π/x e−2πium du.
m∈Z
−∞
91
Set u = vx,
XZ ∞
1
2
θ
=
e−v πx e−2πivxm xdv
x
m∈Z −∞
Z ∞
X
2
−π 2 mx
=x
e−πx(v+m) dv,
e
−∞
m∈Z
since −π 2 mx − πx(v + im)2 = −v 2 πx − 2πivxm.
Note ::
•
Z
∞
2
e−v dv =
√
π.
−∞
•
∞
Z
−λv 2
e
r
dv =
−∞
π
λ
for λ > 0.
It turns out that
∞
Z
−λ(v+im)2
e
r
dv =
−∞
π
λ
So
r
X
1
π
−π 2 mx
e
θ
=x
x
πx
m∈Z
√
= x · θ(x).
Proof of... We start from
Γ
s
2
Z
∞
e−t ts/2
=
0
dt
t
(Re(s) > 0).
Changing t = n2 πx, and note that
dt
n2 πdx
dx
= 2
=
.
t
n πx
x
So
Γ
s
2
Z
=
∞
2 πx
e−n
0
92
ns π s/2 xs/2
dx
,
x
or
π
−s/2 −s
n Γ
where
ω(x) −
s
2
∞
X
Z
∞
0
xs/ 2 ω(x)
=
0
2 πx
e−n
=
n=1
dx
,
x
θ(x) − 1
.
2
Splitting the integral,
π
−s/2
Γ
s
2
∞
Z
x
ζ(s) =
s/2
1
dx
ω(x) +
x
Z
1
xs/2 ω(x)
0
dx
.
x
In the second integral, set
u=
1
,
x
du = −
1
dx,
x2
so
du
dx
=− .
u
x
Hence the second integral is
Z
∞
u
−s/2
1
1 du
ω
.
u u
The Lemma gives
√
1
1√
1
ω
= u · ω(u) +
u− .
u
2
2
Hence
π
−s/2
Z ∞
√
dx
1√
1 du
ζ(s) =
x ω(x) +
u−s/2 ( u · ω(u) +
Γ
u+ )
2
x
2
2 u
Z1 ∞
Z1 ∞
Z ∞
1−s
s
1
dx
1
du
=
xs/2 ω(x) +
u 2 ω(u) +
u− 2 − 2 du
x
u
2 1
1
1
Z ∞
s
1
−
u− 2 −1 du
2 1
Z
s
∞
s/2
and − 1s . Hence
Z ∞
s
1−s
s
1
1
dx
−s/2
ζ(s) =
(x 2 + x s ) ω(x) +
π
Γ
− ,
2
x
s−1 s
1
The last two integrals are
1
s−1
or
1
ξ(s) = s(s − 1)
2
Z
∞
s
(x 2 + x
1
1. The right-hand side converges for all s ∈ C.
2. ξ(1 − s) = ξ(s).
93
1−s
2
) ω(x)
dx 1
+ .
x
2
N (T ) = #{ρ in critical strip : 0 < Im(ρ) < T, ζ(ρ) = 0}
T
T
=
log
+ O(log T ).
2π
2πe
(Lecture 28)
Conjectures from Riemann’s Memoir :: (Exerpt from Davenport)
1. Asymptotic Formula
N (T ) = #{s : 0 ≤ σ ≤ 1, |t| ≤ T ; ζ(s) = 0}
T
T
∼
log
+ O(log T ).
2π
2πe
2. Hadamard Factorization
A+Bs
ξ(s) = e
Y ρ: ζ(ρ)=0
94
s
1−
ρ
es/e .
Similar to ::
∞ Y
z
z
1+
sin(πz) = πz
1−
n
n
n=1
Y
z z/n
= πz
1−
e .
n
n∈Z\{0}
We have ::
1
ξ(s) = s(s − 1)
2
Z
1
blah + ,
2
so eA = ξ(0) = 1/2.
It turns out that
γ
1
B = − − 1 + log(4π)
2
2
X
1
=−
,
ρ
ρ: non-trivial
zeros of ζ
where we interpret
P
ρ
as
X
lim
T →∞
.
ρ: |Im ρ| ≤ T
3. Explicit Formula :: (analogous to)
ψ0 (x) = x −
X
ρ: ζ(ρ)=0
0≤Re(ρ)≤1
1
xρ
1
− log 2π − log 1 − 2 ,
| {z } 2
ρ
x
ζ 0 (0)/ζ(0)
for x > 0.
4. Riemann Hypothesis . . .
Example of using Perron’s Formula ::
P
Determine n≤x τ (n)2 asymptotically.
We saw that
∞
X
τ (n)2 n−s =
n=1
95
ζ 4 (s)
.
ζ(2s)
Hence, by Perron’s formula :: for 1 < c < 2,
1
τ (n) =
2πi
n≤x
X0
2
+
∞
xc X
ζ 4 (s) xs
ds + O
|τ (n)2 |n−c
ζ(2s)
s
T
c−iT
n=1
X
x
τ (x)2
2
|τ (n) | min 1,
+
if x ∈ N
T |x − n|
T
Z
c+iT
x/2<n<3x/2
n6=x
In this example :: Use τ (n) n , and we’ll allow all O− or − constants to depend
on . Taking c = 1 + 1/ log x as before, we get ::
The error term is
4
xζ 1+
T ζ 2+
The sum is x log x
T
1
log x
2
log x
+x
X
2
min 1,
x/2<n<3x/2
x
T |x − n|
+
x
.
T
as before; also,
ζ 4 (1 + δ)
δ −4
∼
+ O(δ −3 )
ζ(2(1 + δ))
ζ(2)
from Laurent expansion. Thus the error is
x1+
T
for T ≤ x.
(Lecture 29)
In the example started on Lecture 28, we have shown ::
1
τ (n) =
2πi
n≤x
X
2
Z
1+ log1 x +iT
1+ log1 x −iT
ζ 4 (s) xs
ds + O
ζ(2s) s
x1+
T
,
valid for 1 ≤ T ≤ x and any > 0.
(Recall that in this example, all O− and − constants can depend on )
Plan from here ::
96
- Look at
1
2πi
ζ 4 (s) xs
ds
ζ(2s) s
I
over the rectangle with vertices
1+
1
± iT
log x
and
1
+ ± iT.
2
- Calculate
Ress=1
ζ 4 (s) xs
ζ(2s) s
.
To do the first part, we need the following estimates ::
•
∞
∞
1 X
µ(n) X −2s
=
|n | = ζ(2σ),
≤
ζ(2s) n2s n=1
n=1
1
for σ > .
2
In particular, for σ ≥ 1/2 + , we have
1 1
1.
ζ(2s) ≤ ζ 2 2 + • We’ll use the estimate :: for 1/2 ≤ σ ≤ 1,

 |t|1−σ log |t| if |t| ≥ 2,
ζ(s)  1
if |t| ≤ 2.
1−σ
(Also ζ(s) log |t| for σ ≥ 1 (Lemma 8.4))
Horizontal sides
Z 1+1/ log x+iT 4
Z 1+1/ log x 4
ζ (s) xs
1
|ζ (s)| · 1 · xσ
ds dσ
2πi 1/2++iT
ζ(2s) s
T
1/2+
Z
Z
4 σ
1 1
1 1+1/ log x
1−σ
T
log T x dσ +
(log T )4 xσ dσ. (T ≥ 2)
T 1/2+
T 1
These integrals are
1
x σ
T 3 log4 T (x/T 4 )σ T log T
dσ =
4
log(x/T 4 )
1/2+ T
1/2+
3
4
Z
1
97
(T 4 ≤ x).
If T < x1/5 say, then
1
1
1
<
.
4
1/5
log(x/T )
log x
log x
So this is
x 1/2+ x
T 3 log4 T
−
log x
T4
T4
T 3 log4 T x
x1+
<
.
log x T 4
T
The second integral is
log4 T 1 1+1/ log x
x1+
x
.
T log x
T
For the left-hand side ::
1
2πi
Z
1/2++iT
1/2+−iT
Z T
ζ 4 (s) xs
|ζ(1/2 + + it)|4
1/2+
ds x
dt
ζ(2s) s
|1/2 + + it|
−T
4
Z T
Z 2
4 dt
1
dt
+
t1−(1/2+) log t
1 − (1/2 + ) 1/2 + t
2
0
Z T
1 + (log x)4
t1−4 dt
2
T
x t2−4 1+
2 − 4 2
x T 2−4 x T 2 .
Now we know
1
τ (n) =
2πi
n≤x
X
2
I
ζ 4 (s) xs
ds + O
ζ(2s) s
x1+
2
+x T .
T
From Complex Analysis we conclude
4
x
X
ζ (s) xs
2
2
+O x
+T
.
τ (n) = Ress=1
ζ(2s) s
T
n≤x
The Laurent Expansions of
•
4
1
ζ (s) =
+ γ + γ1 (s − 1) + · · ·
s−1
1
4γ
?
?
=
+
+
+
+ O(1).
4
3
2
(s − 1)
(s − 1)
(s − 1)
s−1
4
98
•
1
1
=
+?(s − 1)+?(s − 1)2 + · · ·
sζ(2s)
ζ(2)
•
xs = x(xs−1 ) = xe(s−1) log x
∞
X
(log x)k
=x
(s − 1)k
k!
k=0
x(log x)2
x(log x)3
2
(s − 1) +
(s − 1)3 + O((s − 1)4 ).
= x + x log x(s − 1) +
2
6
Multiplying these together yields
ζ 4 (s) xs
x
1
1
=
+ ··· +
4
ζ(2s) s
ζ(2) (s − 1)
s−1
x log3 x
2
+?x log x+?x log x+? + O(1).
6ζ(2)
So
Ress=1
ζ 4 (s) xs
= xP (log x),
ζ(2s) s
where P (y) is a cubic polynomial with leading coefficient
1
1
= 2.
6ζ(2)
π
(Lecture 30)
X
n∈Z
X
1
π
1
= lim
=√ .
n + 1/3 T →∞ n∈Z n + 1/3
3
|n|≤T
We saw
1.
1
τ (n) =
2
n≤x
X
2
Z
1+1/ log x+iT
1+1/ log x−iT
ζ 4 (s) xs
ds + O
ζ(2s) s
x1+
T
.
2.
1
τ (n) =
2πi
n≤x
X
where the
H
2
I
x
ζ 4 (s) xs
ds + O x
+ x1/2 T 2 ,
ζ(2s) s
T
is over the rectangle with vertices 1 + 1/ log x + iT and 1/2 + ± iT .
99
3.
X
2
τ (n) = xP (log x) + O x
n≤x
x
T
+x
1/2
T
2
,
where P (u) is a cubic polynomial with leading coefficient
1
1 1
= 2 . (2 ≤ T ≤ x1/2 )
3! ζ(s)
π
We choose T so that
x
= x1/2 T 2 ,
T
that is, T ≤ x1/6 . So
X
τ (n)2 = xP (log x) + O(x5/6+ ),
n≤x
or
X
τ (n)2 =
n≤x
1
x log2 x + O(x log2 x).
2
π
Notation ::
• G is a finite abelian group
• Hom(G, C× ) is the set of (group) homomorphim from G to C - itself a group.
• Zq = Z/qZ for q ∈ N.
×
• Z×
q = (Z/qZ) the multiplicative group of unites in Z/qZ / reduced residue classes
(mod q).
×
Given h : Z×
q → C , define an arithmetic function

 h (n mod q) if (n, q) = 1,
χ(n) =
0
if (n, q) > 1.
100
Such χ have the property ::
• χ is periodic with period q.
• χ(n) 6= 0 ⇐⇒ (n, q) = 1.
• χ(mn) = χ(m)χ(n) for all m, n ∈ Z.
Such an arithmetic function χ is called a Dirichlet Character (mod q).
Example :: q = 12.
1 2 3 4
5
6
7
8 9 10 11 12
χ0
1 0 0 0
1
0
1
0 0
0
1
0
χ1
1 0 0 0 -1
0 -1 0 0
0
1
0
χ2
1 0 0 0 -1
0
0 0
0
-1
0
1 0 0 0
0 -1 0 0
0
-1
0
χ1 χ2
1
1
In general, we let χ0 denote the principal character (mod q) ::

 1 if (n, q) = 1,
χ0 (n) =
 0 if (n, q) > 1.
101
Examples ::
• χ(1) = 1, and χ(n) is a root of unity for (n, q) = 1.
• The set of { Dirichlet Characters (mod q) } is a group for all q ∈ N (for z ∈ {|z| = 1},
z −1 = z).
In fact, this group is isomorphic to Z×
q itself.
Fundamental Theorem for Finitely Generated Abelian Groups :: Any G is
“isomorphic” to exactly one group of the form
Zr1 × Zr2 × · · · × Zrk ,
where r1 |r2 | · · · |rk .
Alternatively, of the form
Zbp11 × · · · × Zbpll ,
where each pbi i is a prime power, not necessarily distinct.
102
Example :: q = 7.
1
2
3
4
5
6
7
χ0
1
1
1
1
1
1
0
χ
1
α2
α
α2
α
-1 0
χ2
1 α4 = α2
α2
α2
α2
1
χ3
1
1
α3 = −1
1
-1
-1 0
χ4
1
α2
α2
α2
α2
1
χ5 = χ = χ−1
1
α2
α
α2
α
-1 0
0
0
3 is a primitive root (mod 7), its powers are 3, 2, 6, 4, 5, 1 (mod 7). So we let α = e2πi/6 ...
103
Generality ::
- χ(−1) is always -1, +1, we call χ odd or even, respectively.
- If χ2 = χ0 (that is, χ(n) ∈ {−1, 0, 1} for all n), then χ is called real or quadratic.
(Greg :: I guess we don’t call χ0 quadratic)
- Let q be prime, then
n
χ(n) =
q
(Legendre Symbol)
is a quadratic character (mod q).
(Lecture 31)
Warm-up Question :: Describe the structure of Z×
108 .
- 1008 = 24 × 32 × 7.
So
×
×
∼ ×
Z×
1008 = Z16 × Z9 × Z7
∼
= (Z2 × Z4 ) × Z6 × Z6 .
Standard Form 1 ::
∼
= Z2 × Z4 × (Z2 × Z3 ) × (Z2 × Z3 ).
Standard Form 2 ::
∼
= Z2 × Z2 × Z6 × Z12 .
Orthogonality Relations for Characters
For any finite Abelian group G, any χ ∈ Hom(G, C× ),

 |G| if χ is trivial,
X
χ(g) =
 0 otherwise.
g∈G
104
Proof. If χ is nontrivial, choose h ∈ G with χ(h) 6= 1. Then
X
X
X
χ(h)
χ(g) =
χ(hg) =
χ(g).
g∈G
g∈G
g∈G
So since χ(h) 6= 1, we get
X
χ(g) = 0.
g∈G
More generaly, for any χ, ψ ∈ Hom(G, C× ),

 |G| if χ = ψ,
X
χ(g)χ(g) =
 0 if χ =
6 ψ.
g∈G
For any h, g ∈ G,
X
χ∈Hom(G,C× )

 |G| if g = h,
χ(g)χ(h) =
 0 otherwise.
“Proof.” Choose χ ∈ Hom(G, C× ) such that
χ(g) 6= χ(h) · · ·
(Reality Check Problem...why does such a ψ exist?)
Given g ∈ N, this means ::
• For any a, b, ∈ N,
X
χ (mod q)

 φ(q) if a ≡ b (mod q) and (a, q) = 1,
χ(b)χ(a) =
0
otherwise.
• For any χ (mod q),

 φ(q) if χ = χ ,
0
χ(n) =
0
if χ =
6 χ0 .
(mod q)
X
n
Note :: The last fact can be restated if χ 6= χ0 . Then for any k ∈ N,
k+q−1
X
χ(n) = 0.
n=k
So what can we say about
∞
X
χ(n)
n=1
nβ
105
?
(Homework #2 implies converges for β > 0)
Therefore, if we define the Dirichlet L-function
L(s, χ) =
∞
X
χ(n)
ns
n=1
,
then L(s, χ) converges for Re(s) > 0 (χ 6= χ0 ). In fact, σc = 0 (for Re(s) < 0, χ(n)
9 0),
ns we still have σa = 1.
Remark :: If χ0 is the principal character (mod q), then
L(s, χ0 ) =
∞
X
∞
X
χ0 (n)n−s =
n−s = ζ(s)
n=1
(n,q)=1
n=1
Y
(1 − p−s ).
p|q
Note that, for any (a, q) = 1,
X
X
χ(a)L(s, χ) =
χ (mod q)
=
χ(a)
χ (mod q)
∞
X
−s
n
n=1
= φ(q)
∞
X
χ(n)n−s
n=1
X
χ(a)χ(n)
χ (mod q)
X
n−s . (Re(s) > 1)
n=1
n≡a (mod q)
In fact, for Re(s) > 1,
L(s, χ) =
Y
(1 − χ(p)p−s )−1 .
p
Thus, taking logarithmic derivatives,
∞
X
L0
(s, χ) = −
Λ(n)χ(n)n−s .
L
n=1
Therefore,
X
χ (mod q)
L0
χ(a) − (s, χ) = φ(q)
L
106
∞
X
n=1
n≡a (mod q)
Λ(n)n−s .
(Lecture 32)
1
2
3 4
5
6 7
8
9 10 11 12 13 14 15 16 17 18 19 20 21
χ
1 -1 0 1 -1 0 0 -1 0
1
-1
0
1
0
0
1
-1
0
1
-1
0
χ∗
1 -1 0 1 -1 0 1 -1 0
1
-1
0
1
-1
0
1
-1
0
1
-1
0
General Fact :: If χ1 and χ2 are characters modulo q1 and q2 , χ1 χ2 is a character modulo
LCM(q1 , q2 ).
Special Case :: q ∗ = q, any χ∗ (mod q), χ0 (mod q). Then χ∗ χ0 is a character modulo
LCM(q ∗ , q) = q. This new character χ∗ χ0 has the same values as the old character χ∗ ,
except its zero when it has to be.
107
Example :: q = 12.
1 2 3 4
5
6
7
8 9 10 11 12
χ0
1 0 0 0
1
0
1
0 0
0
1
0
χ1
1 0 0 0 -1
0 -1 0 0
0
1
0
χ2
1 0 0 0 -1
0
0 0
0
-1
0
1 0 0 0
0 -1 0 0
0
-1
0
χ1 χ2
1
1
χ0 is induced by χ (mod 1),
χ2 is induced by χ (mod 4),
χ1 χ2 is induced by χ 6= χ0 (mod 3).
Definition :: Given a character χ (mod q), let q ∗ denote the smallest modulus for which
there exists a character χ∗ satisfying
χ∗ χ0 = χ .
|{z} |{z}
|{z}
q∗
q
We say ::
• χ∗ induces χ.
• χ∗ is a primitive character modulo q ∗ .
108
q
In particular, χ is primitive if and only if q ∗ = q, q ∗ is called the conductor of χ∗ .
Important Fact ::
{character modulo q} ↔
•
[
{primitive characters modulo q ∗ }.
q ∗ |q
Plan :: For
L(s, χ) =
∞
X
χ(n)n−s ::
n=1
• Analytic Continuation / Functional Equation.
• Number of Zeros in Critical Strip.
• Zero-free Region.
• Asymptotic / Explicit formula for ::
ψ(χ; q, a) =
X
Λ(n),
n≤x
n≡a (mod q)
θ(χ; q, a) =
X
log p,
p≤x
p≡a (mod q)
π(χ, q, a) =
X
1 = #{p ≤ x : p ≡ a (mod q)}.
p≤x
p≡a (mod q)
- Functional Equation :: Let


 0 if χ is even, 
a=
 1 if χ is odd. 
so that χ(−1) = (−1)a .
Define, for a primitive character χ(mod q) ::
q (s+a)/2 s + a Γ
L(s, χ).
ξ(s, χ) =
π
2
109
Then
ξ(1 − s, χ) =
ia q 1/2
ξ(s, χ),
τ (χ)
where τ (χ) is the Gauss sum
τ (χ) =
q
X
χ(m)e2πim/q .
m=1
It turns out that |τ (n)| = q 1/2 , so that
a 1/2 i q τ (n) = 1.
One example of a Gauss sum :: Take q = prime,
χ(m) =
m
q
.
Then
τ (χ) =
q X
m
q
m=1
q
=
X
e2πim/q
#{b (mod q) : b2 ≡ m (mod q) − 1 e2πim/q
m=1
q
=
X
2πib2 /q
e
.
b=1
From the functional equation (together with the Euler product)
L(s, χ) =
Y
(1 − χ(p)p−s )−1
for σ > 1,
p
we see that the only zeros of L(s, χ) outside the critical strip {s : 0 ≤ σ ≤ 1} are
when
–
s+a
2
is a non-positive integer.
– 0, -2, -4, . . . for χ even.
– -1, -3, -5, . . . for χ odd.
– Number of zeros
110
Define
N (T, χ) = #{ρ : 0 ≤ Re(ρ) ≤ 1, |Im(ρ)| ≤ T, L(s, χ) = 0},
counted with multiplicity. Then
1
T
N (T, χ) =
log
2
2π
qT
2πe
+ O(log(qT ))
if χ is primitive modulo q.
Note :: If χ∗ (mod q ∗ ) induces χ (mod q), then
L(s, χ) = L(s, χ∗ )
Y
(1 − χ∗ (p)p−s ).
p|q
(p-q ∗ )
(Lecture 32)
If |z| = 1, then 0 ≤ Re(3 + 4z + z 2 ) ≥ 8.
1. Zero-free region for L(s, χ)
We start from
L(s, χ) =
Y
(1 − χ(p)p−s )−1 ,
p
log L(s, χ) =
X
κ(n)χ(n)n−s ,
σ ≥ 1.
n
Then
0 ≤ Re
∞
X
κ(n)n−σ (3 + 4χ(n)n−it + χ2 (n)n−2it )
n=1
= Re(3 log ζ(σ) + 4 log L(σ + it, χ) + log L(σ + 2it, χ2 )).
Exponentiating,
1 ≤ ζ(σ)3 L(σ + it, χ)4 L(σ + 2it, χ2 ) .
Since L(σ + 2it, χ2 ) isn’t approaching a pole as σ → 1+ , unless χ2 = χ0 and t = 0,
the same argument as for ζ(s) yields L(1 + it, χ) 6= 0 and in fact.
111
Let χ be a character (mod q). Then L(s, χ) has no zeros in the region
σ >1−
c
,
log(qτ )
(some c > 0)
with the possible exception of a single real zero σ for a quadratic character χ.
A real number
c
log q
1>β >1−
for which L(β, χ) = 0 is called an exceptional zero, or Siegel zero the (quadratic)
character χ is called an exceptional character.
We do know :: If L(σ, χ) = 0, then
c
σ ≤1− √ .
q
2. Formulas for
X
χ(x; q, a) =
Λ(n)
n≤x
n≡a (mod q)
(Lets assume (a, q) = 1).
- Explicit Formula
ψ(x; q, a) =
1
φ(q)
1
=
φ(q)
X
χ (mod q)
X
χ (mod q)
χ(a)
X
Λ(n)χ(n)
n≤x
1
χ(a)
2πi
Z
c+i∞
−
c−i∞
L0
xs
(s, χ) ds.
L
s
Pulling the conour way to the left ::
ψ(x; q, a) =
x
1
−
φ(q) φ(q)
X
χ(a)
X
χ (mod q) non-trivial zeros ρ of L(s, χ)
112
xρ
+ O(x1/4+ ).
ρ
If we use the zero-free region, we get ::
Prime Number Theorem for Arithmetic Progressions
√
Let q ≤ exp(c log x). If there’s no exceptional character (mod q), then
ψ(x; q, a) =
p
x
+ O(x · exp(−c log x)).
φ(q)
If, however, there’s an exceptional zero ρ corresponding to an exceptional character χ1 ,
then
ψ(x; q, a) =
p
χ1 (a) xβ
x
−
+ O(x · exp(−c log x)).
φ(q)
φ(q) β
(χ1 (a) = χ1 (a) since χ1 is real)
Note :: If q ≤ log x, then
c
c
ρ>1− √ >1− √
.
q
log x
Corollary :: If q ≤ log x, then
ψ(x; q, a) =
p
x
+ O(x · exp(−c log x)).
φ(q)
By comparison, if Generalized Riemann Hypothesis (GRH) is true, then
x
+ O(x1/2 log2 x) for all q ≤ x.
φ(q)
√
Note that even this isn’t good when q > x.
ψ(x; q, a) =
Facts to tick off
1.
X
p≤x
p≡a (mod q)
1
log log x
=
+ O(1),
p
φ(q)
for q ≤ x.
(partial summation from Corollary)
113
2. Brun-Titchmarsh Inequality
π(x; q, a) = #{p ≤ x : p ≡ a (mod q)} <
2x
,
φ(q) log(x/q)
for 1 ≤ q ≤ x.
3. Bombieri-Vinogradov Inequality
“GRH is true on average, for q ≤ x1/2− .”
4. Siegel-Walfisz Theorem
There exists an ineffective constant C() for any > 0 such that L(β, χ1 ) = 0, then
β ≤1−
C()
.
q
5. There exists an ineffective constant H such that if ζ(s) has non-trivial zeros up to
height H, then RH is true.
Proof. If RH is false, let ζ(σ + it) = 0 with σ > 1/2, and take H = t. If RH is true,
take H = 1.
The End.... =)
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