Swiss Federal Institute of Technology Zurich
Department of Mathematics
Bachelor Thesis
Fall 2010
Troy Koltes
Soundararajan’s Upper Bound on
Moments of the Riemann Zeta Function
Submission Date:
Advisor:
December 2010
Dr. Paul-Olivier Dehaye
CONTENTS
iii
Contents
1 Introduction
1
2 The
2.1
2.2
2.3
2.4
3
3
4
5
6
Main Theorem and its Corollary
Preliminaries . . . . . . . . . . . . . .
A conjectured bound for µ(S(T, V )) .
Soundararajan’s main theorem . . . .
An upper bound on Mk (T ) . . . . . .
3 Proof of the Main Theorem
3.1 Outline . . . . . . . . . . .
3.2 An upper bound on log(|ζ( 21
3.3 Two lemmas . . . . . . . .
3.4 Proof of Theorem 2.3 . . . .
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9
. 9
. 9
. 19
. 23
4 Extension to the Family of Quadratic Dirichlet L-functions
29
5 Summary
31
Bibliography
32
iv
CONTENTS
Chapter 1
Introduction
The Riemann zeta function is defined as
ζ(s) =
X 1
n≥1
ns
,
Re s > 1.
By analytic continuation, it becomes a meromorphic function on the entire complex plane,
with a simple pole at s = 1. It vanishes at even negative integers, the trivial zeros. The
famous Riemann Hypothesis states that all non-trivial zeros of ζ(s) lie on the critical line
Re s = 12 .
We inspect moments of the Riemann zeta function. For T > 0 and k ∈ C, these are
defined as
Z T
|ζ( 12 + it)|2k dt.
Mk (T ) =
(1.1)
0
Additionally assuming the Riemann Hypothesis, the integrand in Mk (T ) is thus integrated
along the critical line of zeros of the Riemann zeta function. We will present a theorem
by Soundararajan [8] which, eventually, implies the following asymptotic upper bound for
Mk (T ) for real k > 0, conditional on the Riemann Hypothesis:
Mk (T ) k,ε T (log T )k
2 +ε
,
∀ε > 0.
(1.2)
Rather than deriving any new results, we try to make Soundararajan’s proof more explicit
and include as many details as possible. In particular, an attempt is made at directly
verifying all computations that are only sketched in the original paper.
Finding bounds of Mk (T ) is a hard problem, even more so when one removes the assumption of the Riemann Hypothesis. For real k > 0, it has been speculated that, in
fact, a stronger relation than the one stated above holds, without assuming the Riemann
Hypothesis:
2
Mk (T ) ∼ Ck T (log T )k .
(1.3)
Keating and Snaith [4] have conjectured a precise value for Ck as a consequence of results
from random matrix theory. Up to now, this asymptotic equivalence has only been proved
for k = 1, 2 by Hardy/Littlewood and Ingham, respectively. For k = 1/m (with m a
positive integer), a result by Heath-Brown [2] states that
2
2
T (log T )k k Mk (T ) k T (log T )k .
1
2
Introduction
Jutila [3] improved on this result by showing that the dependence on k can be removed.
Assuming that the Riemann Hypothesis is true, the lower bound Mk (T ) k T (log T )k
has been proved by Ramachandra [6] for real k > 0.
2
In the following chapter, we present the main theorem, which gives us bounds on the
measure of large values of |ζ( 12 + it)|. We also derive the corollary yielding the upper
bound on Mk (T ) stated in Equation (1.2).
The third chapter will discuss the proof of the main theorem in full detail. This requires
a proposition which elegantly bounds the logarithm of the absolute value of ζ(s) along a
certain part of the critical line. Finally, a mean value estimate for certain Dirichlet series
helps us complete the proof of the main theorem.
In Chapter 4, we describe an extension of the previous results to families of L-functions
in the particular case of quadratic Dirichlet L-functions, under the assumption of the
Generalized Riemann Hypothesis.
Chapter 2
The Main Theorem and its
Corollary
2.1
Preliminaries
Throughout this presentation, we write logk (x) for log log . . . log(x) if k ≥ 3.
|
{z
k times
}
Also, for two functions f, g : C → C we will use the standard notation
f g ⇐⇒ f = O(g).
We denote by µ the Lebesgue measure on (R, B(R)).
The characterization of the moments of the Riemann zeta function will rely on the main
theorem. This theorem bounds, for large T , the measure of the set
S(T, V ) := {t ∈ [T, 2T ] | log |ζ( 12 + it)| ≥ V }.
(2.1)
The following restatement of Mk (T ) will be essential in proving the corollary that leads
to the desired upper bound on the moments of ζ(s).
For T > 0,
Z 2T
T
|ζ( 12
+ it)|
2k
Z 2T
dt =
1
e2k log |ζ( 2 +it)| dt
(2.2)
T
=
Z 2T Z log |ζ( 1 +it)|
2
T
Z
2ke2kV dV dt
−∞
2T Z ∞
= 2k
1l
T
Z ∞
= 2k
−∞
(t)
1
{t | log |ζ( 2 +it)|≥V }
e2kV µ(S(T, V )) dV.
· e2kV dV dt
(2.3)
(2.4)
−∞
We can exchange integrals in (2.3) since (2.2) exists; this is the case because ζ( 12 + it) is
bounded in the compact interval [T, 2T ].
3
4
2.2
The Main Theorem and its Corollary
A conjectured bound for µ(S(T, V ))
The main Theorem 2.3, due to Soundararajan [8], will prove a weaker version of a conjecture further below. This conjecture is derived from a result by Selberg [7],[5]; to be more
precise, Selberg proved a limit theorem (in the probabilistic sense) for the modulus of the
Riemann zeta function on the critical line. It states that, unconditionally, for any fixed
a ∈ R,
q
µ(S(T, a
lim
1
2
log log T ))
T
T →∞
where
= 1 − Φ(a),
(2.5)
x
t2
1
e− 2 dt
Φ(x) = √
2π −∞
is the standard normal distribution function. One of the central steps of Selberg’s proof is
to find a suitable approximation of ζ(s) on the critical line. In fact, the proof of the main
theorem in the following chapter also strongly relies on a proposition with a similar bound.
Z
Conjecture 2.1. For V ≥ 0 and T large,
!
√
log log T
V2
µ(S(T, V )) T
· exp −
,
V
log log T
(2.6)
where the implicit constant is absolute.
.q
This conjecture is natural because a “substitution” of a in Equation (2.5) by a
1
2
log log T
Φ0 (x)
x ,
and the approximation 1 − Φ(x) ∼
for x → ∞, would lead to the speculated upper
bound. Unfortunately,
this
substitution
is
not possible as Selberg’s theorem is only valid
√
for V of size log log T .
If one could prove Conjecture 2.1, this would have strong consequences for the upper
bound on Mk (T ). The following corollary displays that we would obtain the upper bound
2
T (log T )k , with the implicit constant solely depending on k.
2
Corollary 2.2. If Conjecture 2.1 is true, then Mk (T ) k T (log T )k for real k > 0.
Proof. Let Z be a random variable (on some probability space) such that Z ∼ N (0, 1).
Then,
Mk (2T ) − Mk (T ) =
Z 2T
T
|ζ( 21 + it)|2k dt
Z ∞
= 2k
= 2k
e2kV µ(S(T, V )) dV
−∞
Z k log log T
2
e
2kV
Z ∞
µ(S(T, V )) dV + 2k
−∞
≤ 2kT
Z k log log T
2
e
2kV
Z ∞
dV + 2k
−∞
2kT e
k2 log log T
Z ∞
+ 2k
k
2 log log T
k
2 log log T
√
2kV
e
T
k
2 log log T
e2kV µ(S(T, V )) dV
e2kV µ(S(T, V )) dV
log log T
V2
exp −
V
log log T
!
dV
2.3 Soundararajan’s main theorem
5
√
≤ 2kT e
k2 log log T
≤ 2kT ek
2
log log T
= 2kT ek
2
log log T
!
log log T ∞ 2kV
V2
e
exp −
+ 2kT k
dV
log log T
2 log log T −∞
√
q
2
2 √ ( log log T )
+ 4k T π k
· E[exp (2k 12 log log T · Z)]
2 log log T
√
2
+ 8kT π · ek log log T
Z
2
k T (log T )k .
(2.7)
By telescoping and using Mk (0) = 0 it holds by (2.7) that
Mk (T ) =
∞
X
Mk (2−j T ) − Mk (2−j−1 T )
j=0
∞
X
k
2−j−1 T (log(2−j−1 T ))k
2
j=0
k2
≤ T (log T )
∞
X
1
j+1
2
j=0
2
= T (log T )k ,
which proves the corollary.
2.3
Soundararajan’s main theorem
Theorem
2.3. Assume the Riemann Hypothesis. Let V ≥ 3 and T be large.
√
If 10 log log T ≤ V ≤ log log T , then
V
V2
µ(S(T, V )) T √
exp −
log log T
log log T
4
1−
log3 T
!
.
(2.8)
If log log T < V ≤ 21 (log log T ) log3 T , then
V
V2
µ(S(T, V )) T √
exp −
log log T
log log T
7V
1−
4(log log T ) log3 T
2 !
.
(2.9)
If 21 (log log T ) log3 T < V , then
µ(S(T, V )) T exp −
1
V log V
129
.
(2.10)
All implicit constants above are absolute.
Proof. See p. 23.
In the range 0 ≤ V ≤ log log T , Jutila [3] showed that a similar bound holds:
V2
µ(S(T, V )) T exp −
log log T
where the implicit constants are absolute.
V
1+O
log log T
!
,
(2.11)
6
The Main Theorem and its Corollary
Interestingly, to prove this approximation, Jutila showed that
2
T (log T )k Mk (T ) T (log T )k
2
1
is true in the case k = m
, m ∈ N, T ≥ 2, with absolute implicit constants. This was an
improvement on Heath-Brown’s [2] result (for the same k):
2
2
T (log T )k k Mk (T ) k T (log T )k .
However, Soundararajan’s bound on Mk (T ) will be a consequence of the bounds on
µ(S(T, V )) and not vice versa.
It is clear that Soundararajan’s and Jutila’s bounds are closely related to Conjecture 2.1.
Consequently, Soundararajan’s proof of the bound on Mk (T ) as a corollary of Theorem
2.3 will use an approach similar to the proof of Corollary 2.2.
2.4
An upper bound on Mk (T )
The following corollary, assuming the Riemann Hypothesis, gives the bound on Mk (T )
mentioned in the introduction; it is nearly of the conjectured magnitude described in
Equation (1.3).
Corollary 2.4. Assume the Riemann Hypothesis. Then for every real k > 0 and every
ε > 0, it holds that
Z T
Mk (T ) =
0
|ζ( 12 + it)|2k dt k,ε T (log T )k
2 +ε
.
(2.12)
Proof. See p. 8.
To prove Corollary 2.4, we will use the following crude combination of Theorem 2.3 and
Jutila’s bound in Equation (2.11).
Proposition 2.5. Assume the Riemann Hypothesis. For every real k > 0, we have
(
µ(S(T, V )) k
T (log T )o(1) exp(−V 2 / log log T ),
T (log T )o(1) exp(−4kV ),
if 3 ≤ V ≤ 4k log log T
if V > 4k log log T.
√
Proof of Proposition 2.5. Note that in the range 3 ≤ V ≤ 10 log log T , Jutila’s theorem
(see Equation (2.11)) yields
V2
µ(S(T, V )) T exp −
log log T
!
V2
exp(o(1)) T exp −
log log T
!
.
By the
√ conditions imposed on V , we can rewrite the r.h.s. of (2.8) in the following way:
If 10 log log T < V ≤ log log T , then
V
V2
µ(S(T, V )) T √
exp −
log log T
log log T
4
1−
log3 T
!
2.4 An upper bound on Mk (T )
≤T
p
7
log log T (log T )
o(1)
= T (log T )
4
log3 T
V2
exp −
log log T
V2
exp −
log log T
!
!
.
The last step follows from the fact that
1
2
log3 T = o(1) · log log T
and from taking exponentials, which yields
(2.13)
√
log log T = (log T )o(1) .
If 4k ≤ 1, then the case 3 ≤ V < 4k log log T is taken care of by the two bounds above.
If 4k > 1, we pick T large enough to assure that 4k log log T ≤ 21 (log log T ) log3 T . Then,
in the range log log T < V ≤ 4k log log T (≤ 21 (log log T ) log3 T ), the r.h.s. of (2.3) can be
bounded as follows:
V
V2
µ(S(T, V )) T √
exp −
log log T
log log T
7V
1−
4(log log T ) log3 T
2 !
V2
224k 3 log log T
49 log log T
log log T exp −
+
−
log log T
log3 T
16(log3 T )2
≤ 4kT
p
= 4kT
p
log log T · (log T )
224k3 (log3 T )−49/16
(log3 T )2
V2
= T (log T )o(1) exp −
log log T
V2
· exp −
log log T
!
!
!
.
The last step above is proved by means of a similar argument as in (2.13):
log(4k) +
1
2
· log3 T +
224k 3 (log3 T ) − 49/16
(log log T ) = o(1) log log T,
(log3 T )2
(2.14)
By taking exponentials, the desired equality follows.
For the range V > 4k log log T , we consider all three cases of Theorem 2.3 to find a simpler
bound. If 4k log log T < V ≤ log log T (i.e., if k < 14 ), then
µ(S(T, V )) T
4kV (log log T ) 4 log log T
log log T exp −
+
log log T
log3 T
p
= T (log T )o(1) exp(−4kV ).
If log log T < 4k log log T < V ≤ 12 (log log T ) log3 T , let T be large enough to assure that
256k log log T < 21 (log log T ) log3 T . We consider two subcases:
If log log T < 4k log log T < V ≤ 256k log log T ,
µ(S(T, V )) 256kT
= 256kT
p
p
7 · 256k log log T
1−
4(log log T ) log3 T
log log T exp −4kV
2 !
C(k) log log T
D(k) log log T
log log T exp −4kV +
−
log3 T
(log3 T )2
8
The Main Theorem and its Corollary
= T (log T )o(1) exp(−4kV ),
where C(k), D(k) ∈ R are constants dependent on k.
Considering the second subcase, if 256k log log T < V ≤ 12 (log log T ) log3 T , then we have
Tp
V2
µ(S(T, V )) log log T (log3 T ) exp −
2
log log T
256kV
≤ T (log T )o(1) exp −
64
o(1)
= T (log T )
exp(−4kV ).
7
1−
8
2 !
We now turn to the last case. If 12 (log log T ) log3 T < 4k log log T < V , let T be large
enough to ensure that V > e129·4k . Then,
1
µ(S(T, V )) T exp −
V log e129·4k
129
= T exp(−4kV ) T (log T )o(1) exp(−4kV ).
Hence, we have proved for V > 4k log log T that
µ(S(T, V )) T (log T )o(1) exp(−4kV ).
(2.15)
Combining the above bounds yields the proposition.
We can now use the weaker approximations from Proposition 2.5 to find the wanted upper
bound on Mk (2T ) − Mk (T ) and, consequently, also on Mk (T ).
Proof of Corollary 2.4, p.6. Let Z be a random variable (on some probability space) such
that Z ∼ N (0, 1). Then,
Mk (2T ) − Mk (T ) =
Z 2T
T
|ζ( 12 + it)|2k dt
Z ∞
= 2k
≤ 2kT
e2kV µ(S(T, V )) dV
−∞
Z 3
e2kV dV
−∞
+ 2kT (log T )
o(1)
Z 4k log log T
2kV
e
3
+ 2kT (log T )
o(1)
Z ∞
V2
exp −
log log T
!
dV
e2kV −4kV dV
4k log log T
k T e6k + T (log T )o(1) log log T · E[exp(2k
p
o(1)
+ T (log T )
q
1
2
log log T · Z)]
2
· exp(−8k log log T )
k T + T (log T )o(1) · ek
2
log log T
+ T (log T )o(1)
2
= T (log T )o(1)+k .
Therefore, for any k > 0, ε > 0, it holds that
Mk (2T ) − Mk (T ) k,ε T (log T )k
2 +ε
.
By telescoping in the same way as in Corollary 2.2, we can bound Mk (T ) by T (log T )k
as well. The claim follows.
2 +ε
Chapter 3
Proof of the Main Theorem
3.1
Outline
The diagram in Figure 3.1 describes the relations between different lemmas, propositions
and corollaries used in Soundararajan’s proof of Theorem 2.3 and of the bound on the
moments Mk (T ) (Corollary 2.4).
We have already seen that finding approximations for the measures of S(T, V ) is the crucial
step in finding this bound on Mk (T ).
The proof of Theorem 2.3 itself is the consequence of a bound on the function log |ζ( 12 + it)|
by a sum involving the von Mangoldt function Λ(n) (and another term of similar magnitude), which will be described in Proposition 3.1. Interestingly, the contribution of zeros
of ζ(s) near 12 + it to this bound is minor. To raise the effectiveness of the proposition,
we show in Lemma 3.5 that a restriction of the sum to primes is unproblematic, i.e., the
resulting error is, in an appropriate sense, negligible. Then, a second lemma (Lemma 3.6)
gives us a general bound for mean values, or “moments”, of Dirichlet sums over primes.
Applying this mean value estimate to the sum from the proposition and considering different ranges for the concerning variables eventually gives us the wanted upper bounds on
µ(S(T, V )).
The proof of the proposition bounding log |ζ( 12 + it)| will involve manipulating the product
0
expansion of ζζ (s) and Stirling’s approximation formula for Γ(s). During this computation,
a need for a bound on log |ζ(σ0 + it)|, for σ0 > 1/2, arises (Corollary 3.4). This will be
0
taken care of by bounding ζζ (s) (using residue calculus) and integrating.
3.2
An upper bound on log(|ζ( 21 + it)|)
As mentioned before, Soundararajan’s proof of Theorem 2.3 relies on the subsequent
“tailor-made” bound on the logarithm of |ζ(s)| along the critical line, conditional on the
Riemann Hypothesis. Denote by Λ the (arithmetical) von Mangoldt function:
(
Λ(n) :=
log p, if n = pk for some prime p and some k ≥ 1
0,
otherwise.
9
(3.1)
10
Proof of the Main Theorem
Bound on the moments
2
Mk (T )k,ε T (log T )k +ε
Corollary 2.4
Simplified bound on
µ(S(T, V )) (2 cases)
Proposition 2.5
General bound on
µ(S(T, V )) (3 cases)
Theorem 2.3
Reduction of the
sum in the bound
on log |ζ( 12 + it)| to
a sum over primes
Lemma 3.5
Bound on
log |ζ( 12 + it)| by a
sum involving Λ(n)
Proposition 3.1
Mean value approximation of a
short Dirichlet sum
Lemma 3.6
Bound on
log |ζ(σ0 + it)|
for σ0 > 1/2
Corollary 3.4
Reformulation
0
of − ζζ (s) by
residue calculus
Lemma 3.3
Figure 3.1: Relations between different results in view of Theorem 2.3. The red (thick)
boxes contain statements relying on the Riemann Hypothesis.
3.2 An upper bound on log(|ζ( 12 + it)|)
11
Proposition 3.1. Assume the truth of the Riemann Hypothesis. Let T be large, t ∈ [T, 2T ],
and 2 ≤ x ≤ T 2 . Also, let λ0 be the unique positive real number such that e−λ0 = λ0 + 12 λ20 .
Then the following upper bound holds for any λ ≥ λ0 :
log |ζ( 21 + it)| ≤ Re
log( nx ) (1 + λ) log T
1
. (3.2)
+
+O
1/2+λ/log x+it log n log x
2
log
x
log
x
n
n≤x
Λ(n)
X
Proof. See p. 16.
Remark. The existence and uniqueness of a λ0 ∈ (0, ∞) as defined in Proposition 3.1
follows because
f : [0, ∞) → R
x 7→ −e−x + x +
x2
2
(3.3)
is strictly increasing, with f (0) = −1 and f (1) > 1.
In fact, it holds that 0.491 < λ0 < 0.492.
The following corollary of Proposition 3.1 gives a nice bound on the absolute value of
ζ( 21 + it).
Corollary 3.2. Assume the truth of the Riemann Hypothesis. For all large t, it holds that
|ζ( 12
3 log t
+ it)| ≤ exp
.
8 log log t
Proof. We bound the logarithm in Equation (3.1) directly for x = (log T )2−ε , where ε > 0.
Let t ∈ [T, 2T ]:
log |ζ( 21
log n log( nx ) (1 + λ0 ) log T
1
+
+O
+ it)| ≤
1/2
2
log x
log log T
n log n log x
n≤x
X
≤
1
X
n≤(log T )2−ε
n1/2
+
(1 + λ0 ) log T
+ O(1)
4 − 2ε log log T
(1 + λ0 ) log T
+ O(1)
4 − 2ε log log T
log T
(1 + λ0 )
=
o(1) +
.
log log T
4 − 2ε
≤ (log T )
2−ε
2
+
For T large and after taking exponentials, choosing t = T yields
|ζ( 21 + iT )| ≤ exp
log T
log log T
The claim then follows by choosing ε <
o(1) +
(1 + λ0 )
4 − 2ε
.
2−4λ0
3 .
Before proving Proposition 3.1, we state a lemma which allows us to deduce a corollary
bounding log |ζ(σ0 + it)| for σ0 > 12 . This will eventually help us bound log |ζ(s)| on the
critical line.
Define R := {ρ 6= −2, −4, −6, . . . | ζ(ρ) = 0}, the nontrivial zeros of ζ(s). If the Riemann
Hypothesis is assumed, write R = {ρ = 21 + iγ | ζ(ρ) = 0}, for γ real.
12
Proof of the Main Theorem
Lemma 3.3. For any s 6= 1 with s ∈
/ R, i.e. not coinciding with a zero of ζ(s) on the
critical line, and for any x ≥ 2, it holds that
X Λ(n) log(x/n)
ζ0
1
− (s) =
+
s
ζ
n
log x
log x
n≤x
−
0
ζ
ζ
0
(s)
+
1 X xρ−s
log x ρ (ρ − s)2
∞
1 X
x−2k−s
x1−s
+
,
(1 − s)2 log x log x k=1 (2k + s)2
where the two sums converge absolutely. The sum over ρ is understood as
X
lim
T →∞
|Im ρ|<T
xρ−s
.
(ρ − s)2
Note that this lemma is unconditional on the truth of the Riemann Hypothesis.
Proof. Recall the series expansion of the logarithmic derivative of ζ on Re z > 1,
∞
X
Λ(n)
ζ0
.
− (z) =
ζ
nz
n=1
(3.4)
Write s = σ + it (with σ, t ∈ R) and set c := max(1, 2 − σ). If w ∈ [c − i∞, c + i∞], we
have
(
Re(s + w) = σ + c =
σ + 1,
2,
σ>1
σ ≤ 1.
Concluding that Re(s + w) > 1, we can apply the series expansion (3.4) to the following
integral, which yield:
1
2πi
Z c+i∞
c−i∞
ζ0
xw
1
− (s + w) 2 dw =
ζ
w
2πi
=
Z c+i∞ X
∞
Λ(n) xw
ns+w w2
c−i∞ n=1
dw
Z c+i∞ w log(x/n)
∞
X
Λ(n) 1
e
n=1
ns 2πi
c−i∞
w2
(3.5)
dw.
(3.6)
The exchange of the sum and the integral above is permissible since the sum converges
uniformly absolutely on Re w = c. To evaluate the integral in (3.6), we use residue
calculus:
First, let x < n; then log( nx ) < 0. If a, b ∈ C, denote the oriented line segment from a
to b by [a, b]. For any B > 0, let C(B)
be the right half-circleof radius B with center
c, oriented clockwise, i.e. C(B) := B · e−iaπ | a ∈ [−1/2, 1/2] . Also define M (B) :=
C(B) ∪ [c − iB, c + iB]. Then,
Z c+iB w log(x/n)
e
c−iB
w2
Z
dw =
M (B)
ew log(x/n)
dw −
w2
Z
C(B)
ew log(x/n)
dw.
w2
(3.7)
Since the integrand is holomorphic in a neighborhood of the simply connected set with
boundary M (B), an application of Cauchy’s integral theorem yields that the integral along
3.2 An upper bound on log(|ζ( 12 + it)|)
13
M (B) vanishes.
Additionally, since log( nx ) < 0, we have
Z
πB
ew log(x/n)
dw
sup
exp B log( nx ) cos(aπ)
≤ 2
2
C(B)
w
B a∈[−1/2,1/2]
π
π
x
π
≤
B
exp B log( n ) cos( 2 ) =
B
→ 0,
if B → ∞.
Letting B → ∞ on both sides of Equation (3.7), it follows that
Z c+i∞ w log(x/n)
e
w2
c−i∞
dw = 0,
if x < n.
Now, let x ≥ n; then log( nx ) ≥ 0. Similarly to the first case, for any B > c, let D(B)
denote
the left half-circleof radius B with center c, oriented counterclockwise, i.e. D(B) :=
B · eiaπ | a ∈ [1/2, 3/2] . Also set N (B) := D(B) ∪ [c − iB, c + iB]. Therefore,
Z c+iB w log(x/n)
e
w2
c−iB
Z
dw =
N (B)
ew log(x/n)
dw −
w2
Z
D(B)
ew log(x/n)
dw.
w2
(3.8)
The integrand has a pole at w = 0. Residue calculus yields
Z
N (B)
ew log(x/n)
ew log(x/n)
dw
=
2πi
Res
w=0
w2
w2
= 2πi Resw=0
1 + w log( nx ) + O(|w|2 )
w2
= 2πi log( nx ).
As in the first case,
Z
ew log(x/n)
πB
π
dw
→ 0,
≤ 2 exp B log( nx ) cos( π2 ) =
2
D(B)
w
B
B
if B → ∞.
Thus, letting B → ∞ in Equation (3.8), we have
Z c+i∞ w log(x/n)
e
c−i∞
w2
dw = 2πi log( nx ),
if x ≥ n.
We insert this result into (3.6) and find
1
2πi
Z c+i∞
c−i∞
X Λ(n)
ζ0
xw
− (s + w) 2 dw =
log(x/n).
ζ
w
ns
n≤x
(3.9)
For k, j ∈ Z>0 , set Bk = −Re(s) − 2k − 1. Also, select Tj in such a way that Tj → ∞ as
j → ∞ and for all ordinates γ = Im ρ, where ρ ∈ R, we have ||γ|−Tj −Im s| (log(Tj ))−1 ,
for j large (see Davenport and Montgomery [1], Section 17). By our choice of Bk , we can
apply some estimates by Davenport and Montgomery [1], Section 17.
For fixed j and w ∈ [Bk + iTj , Bk − iTj ] we have the bound
0
ζ
(s + w) log(2|s + w|) log(2|w|),
ζ
(3.10)
14
Proof of the Main Theorem
keeping in mind that s is fixed.
This estimate can be applied since Re(s + w) ≤ −1 for all k.
For fixed k and w ∈ [Bk + iTj , c + iTj ] or w ∈ [Bk − iTj , c − iTj ], whenever Re(s + w) ≤ −1,
a similar estimate is valid:
0
ζ
(s + w) log(2|Tj |).
ζ
For w ∈ [Bk + iTj , c + iTj ] with −1 ≤ Re(s + w) ≤ 2, we have, by our choice of Tj ,
0
ζ
(s + w) log2 (|Tj |).
ζ
Finally, if Re(s + w) ≥ 2,
ζ0
ζ (s
+ w) is bounded on these two segments.
We can now integrate along the rectangle with vertices c − iTj , c + iTj , Bk + iTj , Bk − iTj
and apply Cauchy’s theorem. Denote, in the same order, the four sides by γi , i = 1, 2, 3, 4.
Then we have for fixed k:
Z
ζ0
xw
Bk x−1
(c + 1)xc
− (s + w) 2 dw ≤ 2
O(log(2Tj )) +
O(log2 (Tj )),
γ2,4 ζ
w
Tj + 1
Tj2
(3.11)
where we split up the integration paths γ2 and γ4 at [1 + iTj ] and [1 − iTj ], respectively,
estimating the two resulting integrals separately with the bounds of Davenport and Montgomery [1]. Thus, this integral vanishes for j → ∞.
On γ3 for Tj → ∞, we rewrite
Z
ζ0
xw
− (s + w) 2 dw = ζ
w
γ3
Z
∞
−∞
ζ0
xBk +it
(s + Bk + it)
dt
.
ζ
(Bk + it)2 (3.12)
Letting k → ∞, the integrand vanishes. To be able to exchange the limit and the integral,
we need to show that the integrand is absolutely integrable, uniformly for large k. Choose
a positive integer k0 ≥ −Re s/2. This implies that Bk ≤ −1 when k ≥ k0 . For these k,
the integrand above is bounded in t ∈ [−1, 1]. Hence, using the fact that log(1 + y) ≤ y
for y ≥ 0, we find
Z ∞ 0
xBk +it ζ
(s + Bk + it)
dt ≤ O
(Bk + it)2 −∞ ζ
Z ∞
log(2|Bk + it|)
2
1
(Bk2 + t2 )
!
dt
=O
!
Z ∞
log(2) + 21 log(Bk2 + t2 )
dt
2
2
≤O
Z ∞
log t2 + (1 + Bk2 /t2 )
1
1
Z ∞
≤O
1
Bk + t
t2 (1 + Bk2 /t2 )
2 log t
1
+
dt
t2
t2
!
dt
= O(1),
uniformly in k ≥ k0 . Hence, we may apply the Dominated Convergence Theorem, which
yields that the integral along [Bk + i∞, Bk − i∞] vanishes for k → ∞.
Consequently, the two horizontal and the vertical integrals tend to 0 if we first let j → ∞,
then k → ∞.
3.2 An upper bound on log(|ζ( 12 + it)|)
15
We can now use residue calculus to rewrite the l.h.s. of (3.5).
0
w
Define f (w) := − ζζ (s + w) xw2 . Since we count residues in the region left of the line
0
Re w = c, the contributing residues of − ζζ (s+w) come from the region left of Re(w+s) =
c + σ > 1 by definition of c. Therefore, all (trivial and non-trivial) zeros of ζ(s) contribute.
1
2πi
Z c+i∞
f (w) dw = Resw=0 f (w) + Resw=1−s f (w) +
X
c−i∞
Resw=ρ−s f (w)
ρ
+
∞
X
Resw=−2k−s f (w)
k=1
ζ0
= − (s) log x −
ζ
0
ζ
ζ
0
(s)
+
∞
X xρ−s
X
x1−s
x−2k−s
−
−
,
2
(1 − s)2
(−2k − s)2
ρ (ρ − s)
k=1
where the sum over ρ is understood in the symmetric sense as
X
lim
T →∞
|Im ρ|<T
xρ−s
.
(ρ − s)2
Then this sum converges absolutely (see Davenport and Montgomery [1], Section 17), and
the absolute convergence of the sum on the right follows by geometric summation and
|x| > 1.
The claimed equality follows after combining this with (3.9) and dividing by log x.
1
2
Corollary 3.4. Let T be large, σ0 >
holds that
and 2 ≤ x ≤ T 2 . For any s 6= 1 s.t. s ∈
/ R, it

Λ(n)
log(x/n)
1 ζ0
−
(σ0 + it)
nσ0 +it log n log x
log x ζ
n≤x
log |ζ(σ0 + it)| = Re 
X
1 X
+
log x ρ
Z ∞
σ0
!
xρ−s
dσ + O((log x)−1 ) ,
(ρ − s)2
(3.13)
where the sum over ρ is understood as
X Z ∞
lim
T →∞
|γ|<T
σ0
xρ−s
dσ.
(ρ − s)2
Proof. We integrate the real parts of both sides of the equality in Lemma 3.3 for fixed
t ∈ [T, 2T ] from σ0 > 12 to ∞. For 2 ≤ x ≤ T 2 , we have
log |ζ(σ0 + it)| =
Z ∞
−Re
σ0
ζ0
(σ + it) dσ
ζ

Λ(n)
log(x/n)
1 ζ0
−
(σ0 + it)
nσ0 +it log n log x
log x ζ
n≤x
= Re 
+
+
X
1 X
log x ρ
1
log x
Z ∞
σ0
Z ∞X
∞
σ0
xρ−s
1
dσ −
2
(ρ − s)
log x
!
x2k−s
dσ .
(2k + s)2
k=1
Z ∞
σ0
x1−s
dσ
(1 − s)2
16
Proof of the Main Theorem
The sum over ρ can be exchanged with the integral since it converges absolutely (see
Davenport and Montgomery [1]), uniformly in 12 ≤ σ < ∞. In fact, the integrals of the
last two terms in the r.h.s. of Lemma 3.3 are O((log x)−1 ) for 2 ≤ x ≤ T 2 , t ∈ [T, 2T ].
This is true because looking at the integral of the former term for large x (and therefore
also T ) yields:
Z
Z ∞ 1−σ
∞
x1−s
1
x
dσ
dσ ≤
σ0 (1 − s)2 log x
log x σ0 t2
Z ∞
≤
1
T 2 log x
e(1−σ) log x dσ
σ0
x1−σ0
x(log x)2
1
.
log x
≤
Also, bounding the integral of the last term of the r.h.s. of Lemma 3.3, we get
∞ Z ∞
∞
1 Z ∞X
1 X
x−2k−σ
x2k−s
dσ
≤
dσ
log x σ0
(2k + s)2 log x k=1 σ0 |2k + σ0 + it|2
k=1
≤
1
∞
X
x−2k
σ02 xσ0 log x k=1 log x
1
,
log x
where the exchange of the sum and the integral is possible due to the uniform absolute
convergence of the sum on 12 ≤ σ < ∞.
The claim of the corollary follows.
We now present the proof of Proposition 3.1. Using Stirling’s formula, we will bound the
difference of log |ζ( 12 + it)| and log |ζ(σ0 + it)| for σ0 > 21 . Combining this with a version
of Corollary 3.4, we obtain the desired estimate of log |ζ( 12 + it)|.
Proof of Proposition 3.1, p.11. Since we are assuming the truth of the Riemann Hypothesis, we have R = {ρ = 21 + iγ | ζ(ρ) = 0}, with γ real.
When considering s = σ + it ∈ C, we pick t in such a way that t 6= γ for all ρ ∈ R. Define
F (s) = F (σ + it) :=
X
1
1
2σ − 1
+
=
s−ρ
s−ρ
(σ − 1/2)2 + (t − γ)2
Im(ρ)>0
Im(ρ)>0
X
=
σ − 1/2
.
(σ − 1/2)2 + (t − γ)2
ρ∈R
X
Since
X
1
1
1
+
≤
< ∞,
2
ρ
ρ
|ρ|
Im(ρ)>0
Im(ρ)>0
X
the series converges absolutely (see Davenport and Montgomery [1]). Clearly, F (s) ≥ 0
for σ ≥ 1/2.
3.2 An upper bound on log(|ζ( 12 + it)|)
17
By an application of the Hadamard product theorem, ζ(s) can be written as a product
over its nontrivial zeros (see Titchmarsh [9]). For all s ∈ C, it holds that
s
exp(s(log(2π) − 1 − γ0 /2)) Y
s
eρ ,
1−
1
ρ
2(s − 1)Γ(1 + 2 s)
ρ
ζ(s) =
where γ0 = limn→∞ − log n + nk=1 n1 is the Euler-Mascheroni constant.
Taking the logarithmic derivative, this yields
P
X
ζ0
γ0
1
1 Γ0
1
1
,
(s) = log(2π) − 1 −
−
−
(1 + 12 s) +
+
ζ
2
s−1 2 Γ
s
−
ρ
ρ
ρ
(3.14)
where the sum over ρ converges absolutely using
X
ρ
and the convergence of
we find
1
ρ |ρ|2 .
P
Considering the real parts of both sides of Equation (3.14),
ζ0
γ0
(s) = log(2π) − 1 −
ζ
2
γ0
= log(2π) − 1 −
2
Re
1
1 X
s
+ =
s−ρ ρ
ρ(s
− ρ)
ρ
− Re
X
1
1
Γ0
Re
− Re (1 + 21 s) +
s−1 2
Γ
ρ
− Re
X 1
1
1
Γ0
1
− Re (1 + 21 s) + F (s) +
+ .
s−1 2
Γ
ρ
ρ
Im(ρ)>0
1
1
+
s−ρ ρ
(3.15)
√
The sum on the right of (3.15) evaluates to 1 − γ20 − log(2 π) (see Davenport and Montgomery [1], Section 12). Hence we can simplify (3.15):
Re
ζ0
1
1
1
Γ0
(s) = log(π) − Re
− Re (1 + 12 s) + F (s).
ζ
2
s−1 2
Γ
(3.16)
Stirling’s formula for Γ says that, for σ > 0,
Γ(s) =
2π
s
1 s 2
s
e
1 + O(|s|−1 ) .
(3.17)
Again, we take the logarithmic derivative and get
Γ0
1
(s) = log s −
+ O(|s|−1 ) = log s + O(|s|−1 ).
Γ
2s
(3.18)
Inserting this approximation into (3.16), we find
Re
ζ0
1
1
1 (s) = Re
+ log π − Re log( 12 s + 1) + O(|s|−1 ) + F (s)
ζ
s−1 2
2
1
= − log( 12 s + 1) + F (s) + O(1).
2
(3.19)
If we fix σ and take t ∈ [T, 2T ], then, for T large,
−Re
ζ0
1
(s) = log T − F (s) + O(1).
ζ
2
(3.20)
18
Proof of the Main Theorem
We integrate the l.h.s. of (3.20) for fixed t along the real part from
Z σ0
1
2
−Re
1
2
to σ0 >
1
2
and get
ζ0
(σ + it) dσ = Re log ζ( 12 + it) − log ζ(σ0 + it)
ζ
= log |ζ( 12 + it)| − log |ζ(σ0 + it)|.
Doing the same for the r.h.s. of (3.20), we end up with
Z σ0
log T
1
2
2
+ O(1) − F (s) dσ =
log T
+ O(1) (σ0 − 12 ) −
2
= (σ0 −
Let h(x) = log(1 + x2 ) −
x2
,
1+x2
1
2)
Z σ0
F (σ + it) dσ
1/2
(σ0 − 12 )2 + (t − γ)2
log T
1X
.
+ O(1) −
log
2
2 ρ
(t − γ)2
x ∈ R. Then h : R → R is smooth with h(0) = 0. Since
h0 (x) =
2x3
,
(1 + x2 )2
we find that 0 is a critical point and also a maximum of h. It follows that
log(1 + x2 ) ≥
x2
.
1 + x2
We use this to find the bound
log |ζ( 12 + it)| − log |ζ(σ0 + it)| = (σ0 − 12 )
≤ (σ0 −
1
2)
(σ0 − 1/2)2
log T
1X
+ O(1) −
2
2 ρ (σ0 − 1/2)2 + (t − γ)2
log T
F (σ0 + it)
−
+ O(1) .
2
2
(3.21)
We now use Corollary 3.4 to remove the dependence on log |ζ(σ0 + it)| and conclude the
claim of the proposition.
Observe that the contribution to log |ζ(σ0 + it)| from the nontrivial zeros ρ ∈ R can be
bounded as follows:
XZ ∞
X Z ∞ xρ−s
X
x1/2−σ
x1/2−σ0
dσ
≤
dσ
=
2
2
2
σ0 (ρ − s)2
σ0 |σ0 + it − ρ|
ρ
ρ
ρ ((σ0 − 1/2) + (t − γ) ) · log x
=
F (σ0 + it)x1/2−σ0
.
(σ0 − 1/2) log x
Working this bound and (3.20) into the r.h.s. of Corollary 3.4, we obtain
log |ζ(σ0 + it)| ≤ Re
X
n≤x
Λ(n)
nσ0 +it log n
+ O((log x)−1 ) +
log(x/n)
log T
F (σ0 + it)
+
−
log x
2 log x
log x
F (σ0 + it)x1/2−σ0
.
(σ0 − 1/2)(log x)2
We combine this inequality with (3.21) by addition, which yields a bound for the logarithm
of |ζ(s)| on the critical line:
3.3 Two lemmas
log |ζ( 21
19
log( nx ) log T
Λ(n)
+ it)| ≤ Re
+
nσ0 +it log n log x
2
n≤x
X
1
1
σ0 − +
2 log x
x1/2−σ0
1
σ0 − 1/2
+ F (σ0 + it)
−
−
2
(σ0 − 1/2)(log x)
log x
2
!
+ O((log x)−1 ). (3.22)
Set σ0 = 12 + logλ x for any λ ≥ λ0 . Remember that for these λ, f as defined in (3.3) is
nonnegative, hence e−λ − λ − λ2 /2 ≥ 0. Observe that
x1/2−σ0
1
σ0 − 1/2
e−λ − λ − λ2 /2
−
−
=
≤ 0,
(σ0 − 1/2)(log x)2 log x
2
λ log x
since log x > 0. Therefore, the term including F (σ0 + it) in (3.22) is nonpositive and can
be disregarded in the computation of an upper bound. This is a crucial step since this
term includes the contribution from zeros ρ close to 12 + it. These zeros therefore have a
minor effect on the bound for log |ζ( 12 + it)| and we obtain the claimed approximation
log |ζ( 12 + it)| ≤ Re
3.3
log( nx ) (1 + λ) log T
1
+
+O
. (3.23)
1/2+λ/log
x+it
2
log x
log x
n
log n log x
n≤x
X
Λ(n)
Two lemmas
Before we continue with the proof of the main theorem, we will need two lemmas.
The first lemma shows that restricting the sum in Proposition 3.1 to primes gives only a
small error term of order log3 T . The second lemma is a mean value estimate for Dirichlet
series, which will be applied to Proposition 3.1 in the proof of the main theorem.
Lemma 3.5. Assume the truth of the Riemann Hypothesis.
Let T ≤ t ≤ 2T , 2 ≤ x ≤ T 2 and σ ≥ 12 . Then,
X
Λ(n) log(x/n) log3 T + O(1).
nσ+it log n log x n≤x
(3.24)
n6=p
Proof. First, consider the sum over n = pk with k ≥ 3. Then,
X
X 1 log(p) X 1
Λ(n) log(x/n) ≤
pkσ log(pk ) nσ+it log n log x p3/2
k
k
n≤x
n=pk
k≥3
p ≤x
k≥3
(3.25)
p ≤x
k≥3
≤
X
1
n≤x
n3/2
1,
(3.26)
for x (and therefore T ) large.
Now, we look at the contribution from n = p2 . Section 18 in Davenport and Montgomery [1] gives, under assumption of the Riemann Hypothesis, the following bound on
P
ψ(z) :=
n≤z Λ(n) for z ≥ 2 large:
√
ψ(z) = z + O( z log2 z).
20
Proof of the Main Theorem
Using Abel summation, we find
X Λ(n)
n≤z
ns
=
ψ(z)
+s
zs
Z z
ψ(u)u−s−1 du.
1
Set s = 2it for T ≤ t ≤ 2T . Then, for z large,
X log p
p≤z
p2it
√
√
O( u log2 u)
du + O( z log2 z)
2it
2it−1
2it
2it+1
n
z
u
1 u
n≤z
!
Z z√
√
1
u log2 u
2it
= 2it−1 + 2it−1
z log2 z)
+ 2itO
du
+
O(
2it+1
z
z
(1 − 2it)
u
1
z √
2
+ z log z
T
≤
X Λ(n)
=
1
+ 2it
Z z
1
+
where the implicit constant is absolute.
From this, we deduce a (conditional) upper bound on the sum over inverse primes to the
2it-th power, using Abel summation.
Assume z ≤ T :
X 1
p≤z
p2it
=
X log p
p≤z
1
p2it log p
!
!
√
√
Z z
u/T + u log2 (u)
z
z log2 z
+
+O
=O
du
T log z
log z
u log2 u
2
√
√
z
=O
+ z log z + O(1 + x)
T log z
√
= O( z log z),
To
we will split up the sum (over primes). Let
√ prove our upper bound for prime squares,
2 ≤ y ≤ T and first assume that y > log4 T . Then, for T large,
X
X
X
1
log(y/p)
1
log(y/p)
1
log(y/p)
≤
+
2σ+2it
2σ+2it
2σ+2it
log y p
log
y
p
log
y
p≤y p
p≤log4 T
log4 T <p≤y
X 1 log(y/p) X
1 log(y/p) +
≤
p log y 4
p2σ+2it log y p≤log4 T
log T <p≤y
X 1 X
1 log(y/p) ≤
+
p 4
p2σ+2it log y p≤log4 T
log T <p≤y
X
1
log(y/p)
≤ O(log3 T ) + (3.27)
2σ+2it
log y log4 T <p≤y p
Clearly, from these bounds we can deduce that for y ≤ log4 T , we have
X
1
log(y/p)
= O(log T ).
3
2σ+2it
log y p≤y p
3.3 Two lemmas
21
We turn back to the estimation of the right sum in Equation (3.27) with y > log4 T . Abel
1
summation with the arithmetical function a(n) := 1l{n prime} nlog(y/n)
2it log y and the C -function
√
φ(y) := 1/y 2σ yields, for 2 ≤ y ≤ T , T ≤ t ≤ 2T and T large:
X
X log(y/p)
X
log(y/p) 1
log(y/p)
1
1
=
−
4
2σ
2it
2it
2σ
2it
p log y (log T )2σ
log4 T <p≤y p p log y p≤y p log y y
p≤log4 T
Z y
X
log(y/p) 2σ
du
+
2σ+1
p2it log y log4 T u
p≤u


X
X
Z y
1 log p 
2σ 
y
≤ 2σ + O(1) +
du
+
2σ+1
y
log4 T u
p≤u p2it p≤u p2it log y 

√
√
Z y
u log u
u log(u)2 
1
u


= O(1) + O 
+
+
du
2σ+1
2σ+1
 log4 T u2σ+1
u
log y |{z}
T
u
log y 
≤1
log T
1
1
+
+
2
4
log T
log (T ) log y log y
≤O
!
= O(1),
where the implicit constant is absolute since we assumed
that σ is fixed. The bound on
R y log2 u
2
log
y
√
. The claim of the lemma follows
the rightmost integral follows by 1 u3/2 du = O
y
√
by replacing y with x.
We now state and prove the second lemma, which will give us an upper bound on means
of short Dirichlet series.
Lemma 3.6. Let T be large and 2 ≤ x ≤ T . Choose k ∈ Z>0 in such a way that xk ≤
For any complex sequence (ap ), indexed by primes, it holds that

k
Z 2T X
2
X
2k
ap |ap | 
.
dt T k! 
1/2+it
p
p
T
p≤x
p≤x
Proof. We rewrite the k-th power of the sum in the integrand as

X
ap
p≤x
p1/2+it

k
 =
X bk,x (n)
The coefficient bk,x is defined as follows: Let n =
(
bk,x (n) =
n1/2+it
n≤xk
αi
i=1 pi .
Qr
.
Then, it follows that
if ri=1 αi =
6 k
Pr
Q
k
r
α
i
i=1 api , if
i=1 αi = k.
α1 ,...,αr
P
0,
This representation is a consequence of the multinomial theorem.
Rewriting the integral in the claim yields
Z 2T X
ap 2k
dt =
p1/2+it
T
p≤x
X
m,n≤xk
bk,x (m)bk,x (n)
√
mn
Z 2T it
n
T
m
dt
T
log T .
(3.28)
22
Proof of the Main Theorem
X |bk,x (n)|2
=T
n
n≤xk
!
X
+O
m,n≤xk
|b (m)bk,x (n)|
√ k,x
,
mn| log(m/n)|
(3.29)
m6=n
by separating the terms with m = n and m 6= n. By the binomial formula, it holds that
√
2|bk,x (m)bk,x (n)/ mn| ≤ |bk,x (m)|2 /m + |bk,x (n)|2 /n.
Applying this to the rightmost term in (3.29), we find the bound
X
m,n≤x
m6=n
|b (m)bk,x (n)|
√ k,x
≤
mn| log(m/n)|
k
X
m,n≤xk
m6=n
X |bk,x (n)|2 X
=2
n≤xk
n
m≤xk
m6=n
X |bk,x (n)|2
=4
n≤xk
n
n≤xk
n≤xk
X
X |bk,x (n)|2
≤
n
X |bk,x (n)|2
n
n−m
n )|
1
≤ 4xk
n
−
m
1≤m<n
n≤xk
T
n≤x
1
| log(1 −
1≤m<n
n
xk log(xk )
X |bk,x (n)|2
1
=4
| log(m/n)|
n
k
X
X |bk,x (n)|2
xk
≤4
|bk,x (m)|2 /m + |bk,x (n)|2 /n
| log(m/n)|
≤4
1
|
log(m/n)|
1≤m<n
X
X |bk,x (n)|2
n≤xk
n
n
n−m
1≤m<n
X
1 X |bk,x (n)|2
m
n
k
X
1≤m<xk
n≤x
T log(T / log T ) X |bk,x (n)|2
log T
n≤xk
n
,
since xk ≤ T / log T .
Hence,
2k
X |bk,x (n)|2
.
dt T
Z 2T X
ap
1/2+it
p
T
p≤x
n≤xk
n
The sum on the r.h.s. can be bounded by
X |bk,x (n)|2
n≤xk
n
X
=
X
p1 <···<pr ≤x αP
1 ,...,αr ≥1
αi =k
k
α1 , . . . , αr
!2
|ap1 |2α1 · · · |apr |2αr
pα1 1 · · · pαr r
!
X
≤ k!
X
p1 <···<pr ≤x αP
1 ,...,αr ≥1
αi =k

= k! 
X |ap |2
p≤x
p
k
|ap1 |2α1 · · · |apr |2αr
α1 , . . . , αr
pα1 1 · · · pαr r
k
 ,
where the inequality holds since
k
α1 , . . . , α r
!
=
k!
≤ k!.
α1 ! · · · αr !
3.4 Proof of Theorem 2.3
3.4
23
Proof of Theorem 2.3
Using the previous lemmas, we are now able to prove Theorem 2.3, using a few more
bounds and suitable choices of parameters in Proposition 3.1.
√
Proof of Theorem 2.3, p.5. We assume 10 log log T ≤ V ≤
large and V > 38 log T / log log T , we have
3
8
log T / log log T since for T
µ(S(T, V )) = µ({t ∈ [T, 2T ] | log |ζ( 12 + it)| ≥ V }) = 0
by Corollary 3.2, so the range V >
Define
A :=
3
8
log T / log log T does not contribute to the measure.

1


 2 log3 T,
log log T
 2V
if V ≤ log log T
if log log T < V ≤ 21 (log log T ) log3 T
if V > 21 (log log T ) log3 T.
log3 T,

1,
(3.30)
Also set x = T A/V (≤ T ) and z = x1/ log log T . Inserting Lemma 3.5 into Proposition 3.1
for λ = λ0 and x, z as just defined, yields
X
x log(
)
Λ(n)
1 + λ0 log T
1
n +
+
O
log |ζ( 12 + it)| ≤ 2 log x
log x
n≤x n1/2+λ0 /log x+it log n log x log( xp ) 1 + λ0
X
log p
+
V + O(log3 T )
≤ 2A
p≤x p1/2+λ0 /log x+it log p log x ≤ S1 (t) + S2 (t) +
with
1 + λ0
V + O(log3 T ),
2A
X
log( xp ) 1
S1 (t) = p≤z p1/2+λ0 /log x+it log x and
X
x log(
)
1
p S2 (t) = .
z<p≤x p1/2+λ0 /log x+it log x For any t ∈ S(T, V ), we have either
S1 (t) ≥ V
1−
7
8A
or
S2 (t) ≥
V
.
8A
This statement is true since assuming the contrary, i.e., reversing the inequalities and
making them strict would yield, using λ0 < 299/400:
V ≤ log |ζ( 12 + it)| ≤ S1 (t) + S2 (t) +
But
V
400A
1 + λ0
V
V + O(log3 T ) < V −
+ O(log3 T ).
2A
400A
= O(log3 T ) does not hold for any choice of V and A (compare (3.30)).
Let k ≤ V /A − 1 be a positive integer. By our choice of A and since T is large, we have
A
T
V
xk ≤ T V ( A −1) ≤
T
2
log log T log3 T
≤
T
.
log T
24
Proof of the Main Theorem
This means that we can apply Lemma 3.6 with
ap =

0,
1

pλ0 / log x
p≤z
p > z.
log(x/p)
log x ,
This yields
Z 2T
2k
|S2 (t)|
T
Z 2T X
ap 2k
dt =
dt
p1/2+it
T
(3.31)
p≤x

k
log2 (x/p) 
T k! 
p1+2λ0 / log x log2 x
z<p≤x
1
X

≤ T k! 
X 1
p
z<p≤x
k

T (k(log log x − log log z + O(1)))k
T
k log
log T A/V
log T A/(V log log T )
!!k
!
+ O(1)
k
= T k(log3 T + O(1))
.
(3.32)
Fix k as the largest integer satisfying k ≤ V /A − 1. Observe that
Z 2T
2k
|S2 (t)|
T
V
dt ≥ µ({t ∈ [T, 2T ] | S2 (t) ≥ V /(8A)})
8A
2k
.
Using this together with (3.32) and absorbing the O(1) into log3 T , we deduce the bound
µ ({t ∈ [T, 2T ] | S2 (t) ≥ V /8A}) T
8A
V
2k
(2k log3 T )k .
(3.33)
2
Since T is large, observe
√ that for any choice of V , we have A ≤ log3 T ≤ log(V ) = 2 log(V )
by our assumption 10 log log T ≤ V . Also note that 2k > V /A > 1. This gives a more
concise bound in (3.33):
T
8A
V
2k
2V log(V 2 )
A
(2k log3 T )k T V −2k (8A)2V /A
!V /A
= T V −2V /A (256AV log V )V /A
T V −V /A (512 log2 V )V /A
V
T V − 2A
V
log V
= T exp −
2A
.
The last bound follows from the fact that (512 log2 V )V /A = O(V V /(2A) ) because
lim
T →∞
512 log2 V
√
V
!V /A
= 0,
(3.34)
3.4 Proof of Theorem 2.3
25
since T → ∞ implies that V → ∞ and A ≥ 1. In short, we find
V
µ ({t ∈ [T, 2T ] | S2 (t) ≥ V /8A}) T exp −
log V
2A
.
(3.35)
We proceed in a similar way to estimate µ ({t ∈ [T, 2T ] | S1 (t) ≥ V1 := V (1 − 7/(8A))}).
Assume 1 ≤ k ≤ log(T / log T )/ log z, so z k√≤ T / log T is satisfied. Applying Lemma 3.6
and the simplified Stirling formula k! = O( k(k/e)k ) with the complex sequence
ap =


log(x/p)
1
,
pλ0 / log x log x
p≤z
0,
p > z,
we find
Z 2T
T

|S1 (t)|2k dt T k! 
X1
p≤z
p
k

√ k log log z k
T k
e
√ k log log T k
T k
,
e
using z = T A/(V log log T ) ≤ T .
By the same method as for S2 (t), the measure can be bounded in the following way:
√ k log log T k
µ ({t ∈ [T, 2T ] | S1 (t) ≥ V1 }) T k
.
eV12
(3.36)
Choose
(
k=
bV12 / log log T c, V ≤ (log log T )2
b10V c,
V > (log log T )2 .
Selected in this way, for V ≤ log log T , we have A =
k≤
1
2
log3 T and
V12
V2
log(T / log T ) 2 log log T
log(T / log T )
≤
≤V ≤
V =
,
log log T
log log T
log T
log3 T
log z
log T
since T is large. For log log T < V ≤ (log log T )2 , we have A = log2V
log3 T and we find
that
V2
2 log(T / log T ) 2 log(T / log T )
V12
k≤
≤
≤
V =
,
log log T
log log T
log T (log3 T )
log z
since T is large. Further, in the second range for k, if V > (log log T )2 , we use A = 1. It
holds that
log(T / log T ) log log T
log(T / log T )
k ≤ 10V ≤
V =
log T
log z
since T is large.
We conclude that k ≤ log(T / log T )/ log z is satisfied.
Hence, with this choice of k we can further bound Equation (3.36):
26
Proof of the Main Theorem
µ ({t ∈ [T, 2T ] | S1 (t) ≥ V1 })
V1
T√
log log T
V 2 / log log T
1 1
e
√ +T V
V12
V
exp −
T√
log log T
log log T
!
V
V12
T√
exp −
log log T
log log T
!
V12
V
exp −
T√
log log T
log log T
!
10V log log T
eV 2 (1 − 7/(8A))2
10V
eV (1 − 7/(8A))2
√
10 V
log V
+ T exp
− 10V log
2
+ T exp
log V
+ 10V log(256) − 5V log V
2
!!
+ T exp (−4V log V ) .
(3.37)
Summarizing the above, we have
V12
V
exp −
µ ({t ∈ [T, 2T ] | S1 (t) ≥ V1 }) T √
log log T
log log T
!
+ T exp (−4V log V ) .
(3.38)
Combining (3.35) and (3.38) will yield the claim of the theorem. Since
µ(S(T, V )) ≤ µ ({t ∈ [T, 2T ] | S1 (t) ≥ V1 or S2 (t) ≥ V /(8A)})
≤ µ ({t ∈ [T, 2T ] | S1 (t) ≥ V1 }) + µ ({t ∈ [T, 2T ] | S2 (t) ≥ V /(8A)}) ,
we estimate these two measures for each of the three ranges in Theorem 2.3 separately.
Set B1 (T, V ) := {t ∈ [T, 2T ] | S1 (t) ≥ V1 } and B2 (T, V ) := {t ∈ [T, 2T ] | S2 (t) ≥ V /(8A)}.
√
√
If 10 log log T < V ≤ log log T , then A = 12 log3 T . Bounding (3.35) for 10 log log T <
V ≤ 12 log log T and using 2V / log log T ≤ 1 yields
V
µ(B2 (T, V )) T exp −
log V
log3 T
V
log3 T
T exp −
(log 10 +
)
log3 T
2
2V 2
T exp −
log log T
V2
T exp −
log log T
!
2
1
+
log3 T
2
V
V2
T√
exp −
log log T
log log T
In the second subcase,
1
2
!
4
1−
log3 T
!
.
log log T < V ≤ log log T , we find, using V / log log T ≤ 1,
V
log V
log3 T
V
T exp −
(− log 2 + log3 T )
log3 T
µ(B2 (T, V )) T exp −
V2
T exp −
log log T
4
1−
log3 T
!
3.4 Proof of Theorem 2.3
27
V2
V
exp −
T√
log log T
log log T
4
1−
log3 T
!
.
For√ the bound on (3.38), it is not necessary to consider these subcases separately. For
10 log log T < V ≤ log log T , we directly find
V
V2
µ(B1 (T, V )) T √
exp −
log log T
log log T
7
1−
4 log3 T
2 !
+ T exp(−4V log V )
!
14
V
V2
1−
T√
exp −
log log T
4 log3 T
log log T
log3 T
+ T exp −4V log 10 +
2
!
2
V
V
4
T√
exp −
1−
+ T exp (−8V )
log log T
log3 T
log log T
V
V2
T√
exp −
log log T
log log T
4
1−
log3 T
!
V2
V
exp −
T√
log log T
log log T
4
1−
log3 T
!
8V 2
+ T exp −
log log T
!
.
√
This proves Theorem 2.3 for the range 10 log log T < V ≤ log log T .
If log log T < V ≤ 12 (log log T ) log3 T , then A =
log log T
2V
V 2 log V
µ(B2 (T, V )) T exp −
(log log T ) log3 T
V2
T exp −
log log T
log3 T . Bounding (3.35) yields
!
!
V
V2
T√
exp −
log log T
log log T
7V
1−
4(log log T ) log3 T
2 !
.
The bound on (3.38) follows directly from this choice of range for V :
V
V2
µ(B1 (T, V )) T √
exp −
log log T
log log T
7V
1−
4(log log T ) log3 T
2 !
7V
1−
4(log log T ) log3 T
2 !
7V
1−
4(log log T ) log3 T
2 !
+ T exp(−4V log V )
V2
V
exp −
T√
log log T
log log T
V2
+ T exp −
log log T
!
V
V2
T√
exp −
log log T
log log T
This proves Theorem 2.3 for the range log log T < V ≤ 21 (log log T ) log3 T .
.
28
Proof of the Main Theorem
If 12 (log log T ) log3 T < V , then A = 1. Again, we first bound (3.35):
µ(B2 (T, V )) T exp −
V
log V
2
1
V log V
129
T exp −
.
Finally, we want to prove that
µ(B1 (T, V )) T exp −
1
V log V
129
.
(3.39)
The right term in the r.h.s. of Equation (3.38) clearly fulfills this bound, so we only need
to consider the left term. We want to show that
V
V2
1
√
exp −
+
V log V
64
log
log
T
129
log log T
!
= o(1).
(3.40)
Since V > 12 (log log T ) log3 T , write V = α(log log T ) log3 T with some α > 1/2 which may
depend on T . Keeping in mind that T is large, the l.h.s. of (3.40) is reformulated as:
exp log α +
log3 T
α2 (log log T )(log3 T )2
+ log4 T −
2
64
1
α log log T (log3 T )(log α + log3 T + log4 T )
129
!!
1
1 log α
1
2
2
+
+
= exp α log log T (log3 T ) o(1) −
64
129α 129 α log3 T
+
|
2
2
= exp α log log T (log3 T )
1
1
+
o(1) −
| 64 {z 129α}
{z
= o(1)
}
!!
(3.41)
<0
= o(1),
since α > 1/2 (uniformly) implies that the exponent in (3.41) tends to −∞ for T → ∞.
Hence, we have shown that the bound in (3.39) holds, which completes the proof of
Soundararajan’s Theorem 2.3.
Chapter 4
Extension to the Family of
Quadratic Dirichlet L-functions
Under the assumption of the Generalized Riemann Hypothesis, Soundararajan has extended Proposition 3.1, which is the crucial element of the proof of Theorem 2.3, to work
for certain families of L-functions as well (see [8]). In particular, he gives a modified
version of the bound in Proposition 3.1 for quadratic Dirichlet L-functions.
Recall that an integer d is a fundamental discriminant if and only if one of the following
conditions holds:
ˆ d ≡ 1 (mod 4) and d is square-free
ˆ d = 4m for some square-free integer m ≡ 2 or 3 (mod 4).
Proposition 4.1. Assume the Generalized Riemann Hypothesis. Let x ≥ 2, λ0 as defined
in Proposition 3.1 and d be a fundamental discriminant. Denote by χd the corresponding
d-th primitive quadratic character, defined as the Kronecker symbol: χd (n) = nd . Then,
log L( 12 , χd )
Λ(n)χd (n)
log(x/n) 1 + λ0 log |d|
1
+
+O
.
≤
1/2+λ
/
log
x
0
2
log x
log x
n
log n log x
2≤n≤x
X
Note that, due to the Generalized Riemann Hypothesis, we have L( 12 , χd ) ≥ 0: Assuming
L( 21 , χd ) < 0 would imply the existence of a real zero ρ with 12 < ρ < 1 since L(2, χd ) > 0
and L(σ, χd ) 6= 0 for real σ ≥ 1.
If L( 12 , χd ) = 0, we interpret log L( 21 , χd ) as −∞ to keep the inequality in the proposition
valid.
Comparing this with the original proposition for | log(ζ(s))|, we see that the real variable
t has been replaced by the fundamental discriminant d, for example in the numerator
(log |d|)(1 + λ0 ) of the second term. We also remark that the sum in the proposition
restricted to prime
squares now is of order 21 log log x instead of log3 x due to the fact that
χd (p2 ) = dp dp = 1 when p - d and that the “dampening” effect of 1/pit in the sum
is not present any more. After some more work which consists in finding a statement
similar to Lemma 3.6, Soundararajan showed that it is possible to use the modified proposition to find an analogue of Theorem 2.3. Since d is discrete, the measure applied in the
29
30
Extension to the Family of Quadratic Dirichlet L-functions
modified case is the counting measure, i.e., the modified theorem bounds the number of d
with absolute value ≤ X such that log L( 12 , χd ) ≥ V + 12 log log X. Notice the additional
1
2 log log X compared to the definition of the set S(T, V ) for the Riemann zeta function
due to the larger effect of the sum in Proposition 4.1 restricted to prime squares. More
precisely, Soundararajan stated the modified theorem for 2 different ranges of X:
Theorem
4.2. Assume the Generalized Riemann Hypothesis. For X ≥ 2 large and
√
log log X ≤ V = o((log log X) log3 X), it holds for fundamental discriminants d that
!
#{d | X ≥ |d| and
log L( 21 , χd )
V2
≥ V + log log X} X exp −
(1 + o(1)) .
2 log log X
1
2
If V ≥ (log log X) log3 X, we have
#{d | X ≥ |d| and log L( 21 , χd ) ≥ V + 21 log log X} X exp (−cV log V )
for a constant c > 0.
We see that the modified theorem states quite similar upper bounds to the ones for the
Riemann zeta function.
Finally, this theorem is used to find a bound on the moments of L( 12 , χ):
Corollary 4.3. Assume the Generalized Riemann Hypothesis. Let X ≥ 2 be large. Then
for every real k > 0 and every ε > 0, it holds that
X
1
L( 21 , χd )k k,ε X(log X) 2 k(k+1)+ε ,
|d|≤X
where the sum is over fundamental discriminants d.
Again, we see the similarity to the original bound apart from the additional term k/2
in the exponent, which arises because we are bounding the number of d with absolute
value ≤ X such that log L( 12 , χd ) ≥ V + 21 log log X instead of just log L( 12 , χd ) ≥ V as in
the original Corollary 2.4. To prove Corollary 4.3, the sum is interpreted as an integral
with respect to an appropriate counting measure, and a similar reformulation as seen in
Equation (2.4) is used. Implementing the bounds on the (counting) measure from the
modified Theorem 4.2 yields the stated result.
Chapter 5
Summary
Approximating moments of the Riemann zeta function is a difficult problem. In this
presentation we followed Soundararajan’s proof of the bound
Mk (T ) k,ε T (log T )k
2 +ε
for real k > 0 and ε > 0, under the assumption of the Riemann Hypothesis (Corollary 2.4).
An attempt was made to give as many details as possible to make the proof more explicit
than in the original paper.
To prove the approximation of the moments, a theorem bounding the measure of the set
S(T, V ) as defined in Chapter 1 was used (Theorem 2.3). Proving this theorem required
several auxiliary statements, as depicted in Figure 3.1. The crucial point was Proposition
3.1, which allowed us to bound |ζ( 12 + it)| by a certain sum involving the von Mangoldt
function. In addition to being helpful in proving Theorem 2.3, this proposition yielded
Corollary 3.2, an upper bound on |ζ( 21 + it)|. The actual proof of Theorem 2.3 relied
on the proposition, where the sum was restricted to primes (Lemma 3.5). A mean value
estimate (Lemma 3.6) applied to this sum over primes yielded, with some more work, the
upper bounds from Theorem 2.3.
Lastly, a slight modification of the proposition eventually was an important step in finding
an upper bound on moments for the family of quadratic Dirichlet L-functions, as mentioned
in Chapter 4.
In fact, a version of Soundararajan’s theorem and bound on moments can also be applied
to other families of L-functions, for example the family of quadratic twists of an elliptic
curve (see [8]).
Obviously, a future proof of the bound on Mk (T ) without using the Riemann Hypothesis,
confirming the conjecture mentioned in the introduction (see Equation (1.3)), would be a
vast improvement of the result covered in this paper. Still, Soundararajan’s bound might
give valuable hints to the “right direction” of a possible proof of the unconditional result.
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Summary
Bibliography
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