Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 3, Article 38, 2002
EXPLICIT UPPER BOUNDS FOR THE AVERAGE ORDER OF dn (m) AND
APPLICATION TO CLASS NUMBER
OLIVIER BORDELLÈS
22, RUE J EAN BARTHÉLEMY,
43000 LE PUY- EN -VELAY, FRANCE.
borde43@wanadoo.fr
Received 29 June, 2001; accepted 13 March, 2002
Communicated by J. Sándor
A BSTRACT. In this paper, we prove some explicit upper bounds for the average order of the
generalized divisor function, and, according to an idea of Lenstra, we use them to obtain bounds
for the class number of number fields.
Key words and phrases: Multiplicative number theory, Average order, Class number.
2000 Mathematics Subject Classification. 11N99, 11R29.
1. I NTRODUCTION
Let K be a number field of degree n, signature (r1 , r2 ) , discriminant d (K) , Minkowski
r
1
bound b (K) := b = nn!n π4 2 |d (K)| 2 and class number h (K) . We denote by OK the ring of
algebraic integers of K. We are interested here in finding explicit upper bounds for h (K) of the
type
1
h (K) ≤ ε (n) |d (K)| 2 (log |d (K)|)n−1 ,
where ε (n) is a positive constant depending on n, and log is the natural logarithm.
There are several methods to get such bounds for h (K) : Roland Quême in [8] used the
geometry of numbers to prove that if b > 17,
r 2
1
2
R (K) h (K) ≤ w (K)
|d (K)| 2 (2 log b)n ,
π
where R (K) is the regulator of K, and w (K) is the number of roots of unity in K.
In [5], Stéphane Louboutin proved, by using analytic methods, that
r
n−1
1
w (K) 2 2
e
log
|d
(K)|
R (K) h (K) ≤
|d (K)| 2
,
2
π
4 (n − 1)
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
We would like to thank Professor Joszef Sándor for his helpful comments. We also are indebted to Professor Patrick Sargos for the proof
of the Erdös-Turán inequality in the form used here.
053-01
2
O LIVIER B ORDELLÈS
and, if K is a totally real abelian extension of Q,
n−1
1
log
d
(K)
R (K) h (K) ≤ d (K) 2
.
+ 0.025
4 (n − 1)
The methods used to get these bounds are very deep, but it is necessary to compute the
regulator (which is usually not easy), or use the Zimmert’s lower bound for R (K) (see [11]):
R (K) ≥ 0.02 w (K) e0.46r1 +0.1r2 .
We want to prove some inequalities involving h (K) in an elementary way: we have
h (K) ≤ |{a : integral ideal of OK , N (a) ≤ b}| ,
where N (a) denotes the absolute norm of a, and, using an idea of H.W. Lenstra (see [4]), we
can see, by considering how prime numbers can split in K, that, for each positive integer m, the
number of integral ideals a of absolute norm m is bounded by the number of solutions of the
equation
a1 a2 · · · an = m (ai ∈ N∗ ) .
Lenstra deduced that
(1.1)
h (K) ≤ |{(a1 , . . . , an ) ∈ (N∗ )n , a1 a2 · · · an ≤ b}| .
Now the idea is to work with the generalized divisor function dn , since (1.1) is equivalent to:
Lemma 1.1. Let K be a number field of degree n ≥ 2, and b be the Minkowski bound of K.
Then:
X
h (K) ≤
dn (m) .
m≤b
In an oral communication, J.L. Nicolas and G. Tenenbaum proved that, for any integer n ≥ 1
and any real number x ≥ 1,
X
x
(log x + n − 1)n−1 .
(1.2)
dn (m) ≤
(n − 1)!
m≤x
(one can prove this inequality by induction).
Hence, by Lemma 1.1 and (1.2), we get Lenstra’s result, namely:
b
(log b + n − 1)n−1 .
h (K) ≤
(n − 1)!
2. N OTATION
We mention here some notation that will be used throughout the paper:
General. m, n, r, s will always denote positive integers, x a real number ≥ 1, and [x] denote
the integral part of x, the unique integer satisfying x − 1 < [x] ≤ x.
• ψ (x) := x − [x] − 21 , and e (x) := e2iπx . ψ is 1-periodic and |ψ (x)| ≤ 12 .
• γ ≈ 0.5772156649015328606065120900... is the Euler constant.
• For any finite set E, |E| denotes the number of elements in E.
On number fields. K is a number field of degree n ≥ 2, signature (r1 , r2 ) , discriminant d (K) ,
r
1
Minkowski bound b = nn!n π4 2 |d (K)| 2 , class number h (K).
On arithmetical functions. By 1, we mean the arithmetical function defined by 1 (m) = 1 for
any positive integer m.
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E XPLICIT U PPER B OUNDS FOR THE AVERAGE O RDER OF dn
3
The generalized divisor function dn is defined by
X
d1 (m) = 1, dn (m) :=
1 (n ≥ 2) ,
a1 a2 ···an =m
and, if n = 2, we simply denote it by d (m) .
If f and g are two arithmetical functions, the Dirichlet convolution product of f and g is
defined by
m
X
(f ∗ g) (m) :=
f (δ) g
.
δ
δ|m
3. BASIC P ROPERTIES OF THE G ENERALIZED D IVISOR F UNCTION
The properties of the generalized divisor function can be found in [5], [9] and [10]. For our
purpose, we only need to know that dn is multiplicative (i.e. dn (rs) = dn (r) dn (s) whenever
gcd (r, s) = 1) and, for any prime number p and any non-negative integer l, we have :
n+l−1
l
dn p =
,
l
where ab denotes a binomial coefficient ([9], equality (4)).
It’s important to note that we have
(3.1)
dn = 1
∗ ... ∗ 1}
| ∗ 1 {z
(n ≥ 1) .
n times
One knows that the average order of dn (m) is ∼ (log m)n−1 / (n − 1)! : to see this, one can
use the following result ([9], equality (18)):
X
1
1
n−1
+O
(x > 1, n ≥ 2) .
dn (m) = x (log x)
(n − 1)!
log x
m≤x
Our aim is to compute several constants κ (n) depending (or not) on n such that
X
dn (m) ≤ κ (n) x (log x)n−1 .
m≤x
We will need the following lemma:
Lemma 3.1. Let x ≥ 1. Then:
X 1
ψ (x)
ε
= log x + γ −
+ 2
m
x
x
m≤x
with
1
|ε| ≤ .
4
This result is well-known, and a proof can be found in [2].
4. R ESULTS
Theorem 4.1. Let n ≥ 1 be an integer and x ≥ 1 a real number. Then:
n−1
X
1
dn (m) ≤ x log x + γ +
.
x
m≤x
Theorem 4.2. Let n ≥ 1 be an integer and x ≥ 6 a real number. Then:
X
dn (m) ≤ 2x (log x)n−1 .
m≤x
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4
O LIVIER B ORDELLÈS
5. A PPLICATION TO C LASS N UMBER
Theorem 5.1. Let K be a number field of degree n, Minkowski bound b and class number h (K) .
Then:
h (K) ≤ b log b + γ + b−1
n−1
.
Theorem 5.2. Let K be a number field of degree n ≥ 2, Minkowski bound b and class number
h (K) . Then, if b ≥ 6,
h (K) ≤ 2b (log b)n−1 .
Theorem 5.3. Let K be a number field of degree n, discriminant d (K) and class number h (K) .
Then :
h (K) ≤
1
2n−1
|d (K)| 2 (log |d (K)|)n−1 .
(n − 1)!
More generally, if a > 0 is satisfying a ≥ 2 (n − 1) / (log |d (K)|) , then
h (K) ≤
a+1
2
n−1
1
|d (K)| 2
(log |d (K)|)n−1 .
(n − 1)!
6. P ROOFS OF THE T HEOREMS
In the following proofs, we set
Sn (x) :=
X
dn (m) .
m≤x
Proof of Theorem 4.1.
Sn (x) =
X
X
1
m≤x a1 ....an =m
=
XX
a1 ≤x a2 ≤x
X
...
1
an ≤x/(a1 ...an−1 )
x
a ...an−1
a1 ≤x
an−1 ≤x 1
!n−1
X1
=x
,
a
a≤x
≤
X
...
and we use Lemma 3.1 to conclude the proof.
J. Inequal. Pure and Appl. Math., 3(3) Art. 38, 2002
X
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E XPLICIT U PPER B OUNDS FOR THE AVERAGE O RDER OF dn
Proof of Theorem 4.2.
then
Z
(1) We first note that, since

1,








n + 1,







 2n + 1,








(n2 + 5n + 2)


,

2
Sn (t) =



(n2 + 7n + 2)


,


2





 (3n2 + 7n + 2)



,


2







(3n2 + 9n + 2)


,
2
e2
−2
t Sn (t) dt =
1
7
3e−2
−
24
2
2
n +
5
if 1 ≤ t < 2,
if 2 ≤ t < 3,
if 3 ≤ t < 4,
if 4 ≤ t < 5,
if 5 ≤ t < 6,
if 6 ≤ t < 7,
if 7 ≤ t < 8,
1093 9e−2
−
840
2
n + 1 − e−2 ,
and then, if n ≥ 2,
e2
Z
t−2 Sn (t) dt <
(6.1)
1
2n2
.
3
(2) Let x ≥ 6, n ≥ 1. The theorem is true if n = 1, since S1 (x) = [x] ≤ x, so we prove
the result for n ≥ 2.
We first check that the theorem is true when 6 ≤ x < e2 . Indeed, in this case, we
have
2x (log x)n−1 ≥ 12 (log 6)n−1 > 4n2 ≥
3n2 + 9n + 2
= Sn e2 ≥ Sn (x) .
2
so we can suppose that x ≥ e2 and n ≥ 2.
We prove the inequality by induction : if n = 2,
X X
S2 (x) =
1 ≤ x log x + x ≤ 2x log x.
r≤x s≤x/r
Assume it is true for some n ≥ 2. By (3.1), we have:
X
Sn+1 (x) =
(dn ∗ 1) (m)
m≤x
=
XX
dn (δ)
m≤x δ|m
=
X
dn (δ)
δ≤x
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δ
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6
O LIVIER B ORDELLÈS
≤x
X dn (δ)
δ
δ≤x
x
Z
t−1 d (Sn (t))
1−
Z x
= Sn (x) + x
t−2 Sn (t) dt
=x
1
Z
e2
−2
= Sn (x) + x
Z
x
t Sn (t) dt + x
t−2 Sn (t) dt.
e2
1
Using (6.1) and induction hypothesis, we get
Z x
2n2 x
Sn+1 (x) ≤ 2x (log x)
+
+ 2x
t−1 (log t)n−1 dt
3
e2
2n2 2n+1
2x
n
n−1
=
(log x) + x 2 (log x)
+
−
n
3
n
n
= 2x (log x) − xfn (x) ,
n−1
where
fn (x) :=
2
2−
n
2n2 2n+1
n−1
(log x) − 2 (log x)
+
−
.
3
n
n
Now we have
2n2
fn (x) ≥ fn e2 = 2n −
≥ 0,
3
hence
Sn+1 (x) ≤ 2x (log x)n .
This concludes the proof of Theorem 4.2.
Proof of Theorems 5.1 & 5.2. Direct applications of Theorems 4.1 and 4.2.
Proof of Theorem 5.3. Let a > 0, and suppose x ≥ e(n−1)/a . Then n − 1 ≤ a log x, and, using
(1.2),
Sn (x) ≤
(6.2)
(a + 1)n−1
x (log x)n−1 .
(n − 1)!
1
Now, Since b < |d (K)| 2 , we have, by Lemma 1.1,
X
h (K) ≤
dn (m) .
m≤|d(K)|1/2
We then use the inequality ([6], Lemma 10)
|d (K)| ≥ e2(n−1)/3
and (6.2) with a = 3 to get the first part of Theorem 5.3.
The 2nd part comes directly from (6.2). This concludes the proof of Theorem 5.3.
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E XPLICIT U PPER B OUNDS FOR THE AVERAGE O RDER OF dn
7. U SING
THE
7
C ONVOLUTION R ELATION IN A D IFFERENT WAY
We now want to prove another bound, using the Dirichlet hyperbola principle:
Theorem 7.1. Let K be a number field of degree n, Minkowski bound b and class number h (K) .
Then, if b ≥ 36,
(i) n = 2p (p ≥ 1) ,
b
h (K) ≤ p−2
(log b)p (log b + p − 1)p−1 ,
2 (p − 1)!
(ii) n = 2p + 1 (p ≥ 1) ,
b
2
p
p−1
p
h (K) ≤ p
(log b) log b (log b + p − 1)
+
(log b + p) .
2 (p − 1)!
p
We first need the following result:
Lemma 7.2. Let x ≥ 6 be a real number and k ≥ 1 an integer. Then:
X dk (m)
≤ 2 (log x)k .
m
m≤x
Proof. The result is true if k = 1, so we suppose k ≥ 2. Suppose first that x ≥ e2 . By partial
summation, we can write, using Theorem 4.2,
Z x
X dk (m)
−1
= x Sk (x) +
t−2 Sk (t) dt
m
1
m≤x
Z e2
Z x
k−1
−2
≤ 2 (log x)
+
t Sk (t) dt + 2
t−1 (log t)k−1 dt
e2
k+1
1
2
<
2k
2
2
(log x)k + 2 (log x)k−1 +
−
,
k
3
k
and one can check that
k−1
2 (log x)
2k 2 2k+1
+
−
≤
3
k
3 1
−
2 k
(log x)k
if x ≥ e2 and k ≥ 2, hence
X dk (m) 3 1 ≤
+
(log x)k ≤ 2 (log x)k .
m
2 k
m≤x
Now, if 6 ≤ x < e2 and k ≥ 2, we get
2 (log x)k ≥ 2 (log 6)k >
X dk (m) X dk (m)
6k 2
1
>
245k 2 + 1093k + 840 =
≥
,
5
840
m
m
2
m≤x
m≤e
which concludes the proof of Lemma 7.2.
Proof of Theorem 7.1. Let x ≥ 36 be a real number. If n = 2p is even, using (3.1) again, we
can write:
X
X
X
dn (m) =
dn/2 ∗ dn/2 (m) =
(dp ∗ dp ) (m) ,
m≤x
m≤x
m≤x
and, by the Dirichlet hyperbola principle, we get, for any real number T satisfying 1 ≤ T ≤ x,
X
X
X
X
X
dn (m) ≤
dp (m)
dp (r) +
dp (m)
dp (r) ,
m≤x
m≤T
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m≤x/T
r≤x/m
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8
O LIVIER B ORDELLÈS
and then, using (1.2),
x
dn (m) ≤
(p − 1)!
m≤x
X
(
p−1
X dp (m) x
log + p − 1
m
m
m≤T


X dp (m) p−1
x
+
log + p − 1
,

m
m
m≤x/T
x
and, with Lemma 7.2, if min T, T ≥ 6, we get
x p o
2x (log x + p − 1)p−1 n
(log T )p + log
,
(p
−
1)!
T
m≤x
1
1
and we choose T = x 2 (so min T, Tx = x 2 ≥ 6) to conclude the proof.
If n = 2p + 1 is odd, then we write:
X
X
X
dn (m) =
d(n−1)/2 ∗ d(n+1)/2 (m) =
(dp ∗ dp+1 ) (m) .
X
m≤x
dn (m) ≤
m≤x
m≤x
8. C ASE
Q UADRATIC F IELDS
√ We suppose in this section that K = Q
d , where d ∈ Z \ {0, 1} is supposed to be
squarefree. We denote here ∆ the discriminant and h (d) the class number. We recall that:

 d, if d ≡ 1 (mod 4) ,
∆=

4d, if d ≡ 2 or 3 (mod 4) .
OF
The problem of the class number is in this case utterly resolved: for example, if d < −4, we
have (see [1], Corollary 5.3.13)
−1 X d
d
h (d) = 2 −
,
2
k
16k<|d|/2
where kd represents the Kronecker-Jacobi symbol. Nevertheless, we think it would be interesting to have upper bounds for h (d).
We also note that, by [3], we can replace, in Lemma 1.1, the Minkowski bound b by the
bound β defined by:
 p
if ∆ ≥ 8,
 ∆/8,
β :=
 p
−∆/3, if ∆ < 0.
We can see that the problem of the class number of a quadratic field is then connected with
that of having good estimations of the error-term
X
R (x) :=
d (m) − x (log x + 2γ − 1)
m≤x
(Dirichlet divisor problem).
1
One can prove in an elementary way that R (x) = O x 2 (see below). Voronoï proved
1
3
that R (x) = O x log x . If we use the technique of exponent pairs (see [2]), we can have
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E XPLICIT U PPER B OUNDS FOR THE AVERAGE O RDER OF dn
9
27 R (x) = O x 82 . By using very sophisticated technics, Huxley succeeded in proving that
23
461
146
73
R (x) = O x (log x)
.
The following result is well-known, but, to make our exposition self-contained, we include
the proof:
Lemma 8.1. Let x ≥ 1. Then :
x 3
X
+ .
d (m) ≤ x (log x + 2γ − 1) + 2 ψ
4
m
m≤x1/2
m≤x
X
Proof. By the Dirichlet hyperbola principle, we have:
X
X
d (m) =
1
m≤x
rs≤x
X X
=
r≤x1/2
= 2
1+
s≤x/r
X X
X X
s≤x1/2
X X
1−
r≤x1/2
r≤x/s
1
s≤x1/2
√ 2
1− x
r≤x1/2 s≤x/r
√ 1 2
= 2
[x/r] −
x−ψ x −
2
r≤x1/2
X √ 1
− x − ψ2
= 2
x/r − ψ (x/r) −
x
2
1/2
X
√
r≤x
√ √
√
√ 1
− + 2ψ x x + x − ψ x ,
4
and, by using Lemma 3.1, we get
X x √
X
√ √ 1
−1
− 12
−2
ψ
− x+ψ x
d (m) = 2x
log x + γ − x ψ x + εx
r
2
1/2
m≤x
r≤x
√ 1
√ √
√
√ 1
+ − x − ψ2
x − + 2ψ x x + x − ψ x
2
4
X x
√ 1
= x (log x + 2γ − 1) + 2ε + − ψ 2
x −2
ψ
,
4
r
1/2
r≤x
and we conclude by noting that |ε| ≤
1
4
√ and 14 − ψ 2 ( x) ≤
1
4
if x ≥ 1.
Corollary 8.2. Let x ≥ 1. Then:
X
d (m) ≤ x (log x + 2γ − 1) +
m≤x
Proof. Use |ψ (t)| ≤
1
2
in Lemma 8.1.
√
3
x+ .
4
We get, using Lemma 1.1:
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10
O LIVIER B ORDELLÈS
√ d be a quadratic field of discriminant ∆. Then :
Corollary 8.3. Let K = Q
 r 14

∆
∆ 1
3
3


log ∆ + 2γ − 1 − log 2 +
+ ,
if ∆ ≥ 8,



8 2
2
8
4
h (d) ≤
r

14


1
∆
1
∆
3


log (−∆) + 2γ − 1 − log 3 + −
+ , if ∆ < 0.
 −
4
3 2
2
3
Example 8.1. If d = 13693, then, using PARI system (see [1]), we get h (d) = 15. The bound
of Corollary 8.3 gives
h (d) < 166.
Example 8.2. If d = −300119, then we have h (d) = 781, and Corollary 8.3 gives
h (d) < 1889.
For bigger discriminants, it could be interesting to have a lower exponent on the error-term.
We want to prove this explicit version of Voronoï’s theorem:
Lemma 8.4. Let x ≥ 3. Then:
X
x
< 6 x 13 log x.
ψ
m m≤x1/2
We first need an effective version of Van Der Corput inequality:
Lemma 8.5. Let f ∈ C 2 ((N ; 2N ] 7→ R) . If there exist real numbers c ≥ 1 and λ2 > 0
satisfying
λ2 ≤ f 00 (x) ≤ cλ2 (N < x ≤ 2N ) ,
then:
X
n
o
1
− 12
− 21
2
e
(±f
(m))
≤
4π
cN
λ
+
2λ
.
2
2
N <m≤2N
Proof. We first prove the following result:
Let f ∈ C 2 ([N ; 2N ] 7→ R) satisfying
(i) f 0 (x) ∈
/ Z if N < x < 2N,
(ii) there exists λ2 ∈ 0; π1 verifying f 00 (x) ≥ λ2 (N ≤ x ≤ 2N ) .
Then:
X
1
−1
(8.1)
e (±f (m)) ≤ 4π − 2 λ2 2 .
N ≤m≤2N
Since
X
N ≤m≤2N
X
e (−f (m)) = N ≤m≤2N
e (f (m)) ,
00
we shall prove (8.1) just for f, and since f (x) > 0 for x ∈ [N ; 2N ] , f 0 is a strictly increasing
function.
Let x be a real number satisfying 0 < x < 12 . By (i) , we can define real numbers u, v, N1 , N2
such that u := f 0 (N ) , v := f 0 (2N ) , and f 0 (N1 ) = [u] + x, f 0 (N2 ) = [u] + 1 − x. We have :
X
X
X
X
e (f (m)) =
e (f (m)) +
e (f (m)) +
e (f (m)) ,
N ≤m≤2N
N ≤m<N1
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N1 ≤m≤N2
N2 <m≤2N
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E XPLICIT U PPER B OUNDS FOR THE AVERAGE O RDER OF dn
11
with
X
N ≤m<N1
0
f (N1 ) − f 0 (N )
e (f (m)) ≤ max {N1 − N, 1} = max
,1
f 00 (ξ)
for some real number ξ ∈ (N ; N1 ) , then, by (ii) ,
X
x
[u] + x − u
e (f (m)) ≤ max
, 1 ≤ max
,1 ,
λ2
λ2
N ≤m<N1
and we have the same for
X
x
v + x − [u] − 1
, 1 ≤ max
,1 ,
e (f (m)) ≤ max
λ2
λ2
N2 <m≤2N
and we use Kusmin-Landau inequality (see [7]) to get
X
πx 2
e (f (m)) ≤ cot
≤
.
2
πx
N ≤m≤N
1
2
We then have:
X
x
2
,1 +
.
e (f (m)) ≤ 2 max
λ
πx
2
N ≤m≤2N
We then choose x =
λ2
π
12
x
λ2
, so
1
= (πλ2 )− 2 ≥ 1 if λ2 ≤ π −1 , and we get
X
1 −1
e (f (m)) ≤ 4π − 2 λ2 2 .
N ≤m≤2N
We are now ready to prove Lemma 8.5:
1
1
If λ2 > π1 , then 4π − 2 cN λ22 > 4π −1 N > N , so we suppose λ2 ≤ π1 .
We take u, v as above, and we define
[u; v] ∩ Z := {l + 1, ..., l + K}
for some integer l and positive integer K, and define
Jk := {m ∈ Z, l + k − 1 < m ≤ l + k} ∩ [u; v]
We have, by (8.1),
X
N <m≤2N
(1 ≤ k ≤ K + 1) .
K+1
1
−1
XX
e (f (m)) ≤
e (f (m)) ≤ 4π − 2 (K + 1) λ2 2 ,
k=1 m∈Jk
and, by the mean value theorem,
K − 1 ≤ v − u = f 0 (2N ) − f 0 (N ) ≤ cN λ2 ,
thus
X
N <m≤2N
1
−1
e (f (m)) ≤ 4π − 2 (cN λ2 + 2) λ2 2 .
This concludes the proof of Lemma 8.5.
J. Inequal. Pure and Appl. Math., 3(3) Art. 38, 2002
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12
O LIVIER B ORDELLÈS
Proof of Lemma 8.4. We write
X x X
x x
X
1
ψ
ψ
+
ψ
=
≤ x 3 + |Σ| .
m m
m 1
m≤x 12
m≤2x1/3
2x1/3 <m≤x 2
1 1i
1
We then split the interval 2x 3 ; x 2 into sub-intervals of the form (N ; 2N ] with 2x 3 < N ≤
1
x 2 : the number J of such intervals satisfies
1
2J−1 N ≤ x 2 < 2J N,
1
and since N > 2x 3 , we have

1
2
log x /N
J =
log 2

log x
+ 1 <
.
6 log 2
We then have :
X
|Σ| ≤
max
2x1/3 <N ≤x1/2 N <m≤2N
x log x
ψ
.
m 6 log 2
Moreover, using Erdös-Turán inequality (see Appendix A), we get, for any positive integer H,
( H
)
X
x X 1 X
N
1 X 1 X
hx hx ψ
+
e
e
≤
+H
,
2
m 2H π
h
m h m N <m≤2N
N <m≤2N
h=1
h>H
N <m≤2N
so, by Lemma 8.5, with λ2 = hx/ (4N 3 ) and c = 8, we get
( H
X
x X 1
3
1
3
1
N
ψ
+ 16π − 2
x 2 (N h)− 2 + N h−1 2 x− 2
≤
m 2H
N <m≤2N
h=1
)
X 1
− 12
3
1
3
−
−5/2
+H
x2 Nh
+ N 2x 2h
h>H
1
1
3
3
N
− 32
−1 2
+ζ
N 2 x− 2
≤
+ 16π
2 xHN
2
2H
! )
3 12
Z ∞ 1
x 2 −3
N
+H
t 2+
t−5/2 dt
N
x
H
1
1
3
3
N
−
− 32
−1 2
+ ζ
+ 2/3 N 2 x 2 ,
≤
+ 16π
4 xHN
2
2H
P
− 32
where ζ 32 := ∞
. The well-known bound ζ (σ) ≤ σ/ (σ − 1) (σ > 1) gives ζ
k=1 k
2
11
≤
<
4,
hence
3
3
X
o
x n
1
1
3
3
N
ψ
+ 64π − 2 xHN −1 2 + N 2 x− 2 .
≤
m 2H
N <m≤2N
3
2
+
We then choose
h
−1
H = 2 Nx
J. Inequal. Pure and Appl. Math., 3(3) Art. 38, 2002
− 31
i
.
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E XPLICIT U PPER B OUNDS FOR THE AVERAGE O RDER OF dn
13
Considering the inequality 1/ [y] ≤ 2/y (y ≥ 1) , we get
X
x 1
3
1
3
3
1
ψ
≤ 64π − 2 2− 2 + 2 x 3 + 64π − 2 N 2 x− 2 ,
m N <m≤2N
and
|Σ| ≤
n
1
o log x
3
1
3
1
64π − 2 2− 2 + 2 x 3 + 64π − 2 x 4
,
6 log 2
1
1
and since x ≥ 3, x 4 ≤ 3−1/12 x 3 , then
n
1
o x 13 log x
1
1
− 32
−2
− 12
< 5 x 3 log x.
|Σ| ≤ 64π
2 +3
+2
6 log 2
We obtain with Lemma 1.1:
√ d be a quadratic field of discriminant ∆. Then:
Corollary 8.6. Let K = Q
 p
1
3
∆/8
log
∆
+
2γ
−
1
−
log
2
+ 6 (∆/8)1/6 log (∆/8) + 34 , if ∆ ≥ 72,

2
2


p
h (d) ≤
−∆/3 12 log (−∆) + 2γ − 1 − 21 log 3



+6 (−∆/3)1/6 log (−∆/3) + 34 , if ∆ < −27.
A PPENDIX A.
We want to show here this special form of the Erdös-Turán inequality used in this paper:
Theorem A.1. Let H, N be positive integers, and f : (N ; 2N ] 7→ R be any function. Then:
X
ψ
(f
(m))
N <m≤2N
)
( H
X 1 X
N
1 X 1 X
+
e (hf (m)) + H
e
(hf
(m))
≤
.
2
2H π
h
h h=1
N <m≤2N
N <m≤2N
h>H
Proof. For any positive integers h and H, we set
Z 1/H
H
c (h, H) :=
e (−ht) dt.
2πih 0
(1) We first note that
1
|c (h, H)| ≤
min
2π
(A.1)
1 H
,
h h2
.
Indeed, if h ≤ H, then
H
|c (h, H)| ≤
2πh
Z
1/H
|e (−ht)| dt =
0
1
,
2πh
and if h > H, then the first derivative test gives
Z
H 1/H
H
2
H
H
|c (h, H)| ≤
e (−ht) dt ≤
·
=
<
.
2
2πh πh
2πh 0
2πh2
(πh)
J. Inequal. Pure and Appl. Math., 3(3) Art. 38, 2002
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14
O LIVIER B ORDELLÈS
(2) Let x, t be any real numbers. Since ψ (x) ≤ ψ (x − t) + t, we get
Z 1/H
Z 1/H
ψ (x) dt ≤
(ψ (x − t) + t) dt,
0
0
and then
Z
1/H
ψ (x) ≤ H
(A.2)
ψ (x − t) dt +
0
1
.
2H
P
The partial sums of the series h≥1 {− sin (2πhx) / (hπ)} are uniformly bounded,
hence
Z 1/H
Z
∞
1 X 1 1/H
ψ (x − t) dt = −
sin (2πh (x − t)) dt
π h=1 h 0
0
Z
∞
1 X 1 1/H
= −
{e (hx) e (−ht) − e (−hx) e (ht)} dt
2πi h=1 h 0
Z
Z
∞
∞
1 X e (−hx) 1/H
1 X e (hx) 1/H
= −
e (−ht) dt −
e (ht) dt
2πi h=1 h
2πi h=1 −h
0
0
X e (hx) Z 1/H
1 X
= −
e (−ht) dt = −
c (h, H) e (hx) ,
2πih 0
H h∈Z, h6=0
h∈Z, h6=0
hence, using (A.2),
ψ (x) ≤
X
1
−
c (h, H) e (hx) ,
2H h∈Z, h6=0
and
X
N <m≤2N
X
X
N
−
c (h, H)
e (hf (m))
2H h∈Z, h6=0
N <m≤2N
∞
X
X
N
≤
+ 2
c (h, H)
e (hf (m)) ,
2H
ψ (f (m)) ≤
h=1
N <m≤2N
hence
(A.3)
∞
X
X
N
+2
|c (h, H)| e (hf (m)) .
ψ (f (m)) ≤
2H
N <m≤2N
N <m≤2N
h=1
X
(3) Since we also have ψ (x) ≥ ψ (x + t) − t, we get in the same way
X
X
X
N
ψ (f (m)) ≥ −
+
c (h, H)
e (−hf (m))
2H h∈Z, h6=0
N <m≤2N
N <m≤2N
∞
X
X
N
≥ −
− 2
c (h, H)
e (−hf (m))
2H
N <m≤2N
h=1
∞
X
X
N
≥ −
−2
|c (h, H)| e (−hf (m)) ,
2H
h=1
J. Inequal. Pure and Appl. Math., 3(3) Art. 38, 2002
N <m≤2N
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E XPLICIT U PPER B OUNDS FOR THE AVERAGE O RDER OF dn
15
and since e (−hf (m)) = e (hf (m)), we obtain
(A.4)
∞
X
X
N
ψ (f (m)) ≥ −
−2
|c (h, H)| e (hf (m)) .
2H
N <m≤2N
N <m≤2N
h=1
X
The inequalities (A.3) and (A.4) give
∞
X
X
X
N
+2
|c (h, H)| e (hf (m))
ψ (f (m)) ≤
2H
N <m≤2N
N <m≤2N
h=1
( H
X
X
N
=
+2
|c (h, H)| e (hf (m))
2H
N <m≤2N
h=1
)
X
X
+
|c (h, H)| e (hf (m)) ,
h>H
N <m≤2N
and we use (A.1).
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J. Inequal. Pure and Appl. Math., 3(3) Art. 38, 2002
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