ACTA ARITHMETICA

Olivier Ramare

Description

ACTA ARITHMETICA * (200*) Approximate formulae for L(1, ?), II by Olivier Ramare (Lille) 1. Introduction and results. Upper bounds of |L(1, ?)| are mainly useful in number theory to study class numbers of algebraic extensions. In [1]–[3] Louboutin establishes bounds for |L(1, ?)| that take into account the behavior of ? at small primes. His method uses special representations of L(1, ?) and does not extend to odd characters. For instance in [2] he uses L(1, ?) = 2∑n ∑ l≤n ?(l)/(n(n + 1)(n + 2)) which comes from an integra-tion by parts; such a formula fails in the odd case. But the effect of this integration by parts is in fact similar to the introduction of a smoothing, something we did in [5], the only difficulty being to handle properly the Fourier transform of functions behaving like 1/t near ∞. This method gives good numerical results in a uniform way. In this note we improve on the results given in [2] and [3] and extend them to the odd character case. Let us mention that we take this opportunity to correct several typos occurring in [5].

• euler ?-function


• louboutin has


• then


• character modulo


• upper bounds


• combine louboutin's


• while generalizing both


• p2 ?


• bound comes simply

Sujets

• Upper and lower bounds

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 14
• Langue: English

Extrait

ACTA ARITHMETICA
* (200*)
Approximate formulae for L(1;), II
by
Olivier Ramare (Lille)
1. Introduction and results. Upper bounds ofjL(1;)j are mainly
useful in number theory to study class numbers of algebraic extensions.
In [1]{[3] Louboutin establishes bounds forjL(1;)j that take into account
the behavior of at small primes. His method uses special representations
ofL(1;) and does not extend to odd characters. For instance in [2] he usesP P
L(1;) = 2 (l)=(n(n + 1)(n + 2)) which comes from an integra-
n ln
tion by parts; such a formula fails in the odd case. But the e ect of this
integration by parts is in fact similar to the introduction of a smoothing,
something we did in [5], the only di cult y being to handle properly the
Fourier transform of functions behaving like 1=t near1. This method gives
good numerical results in a uniform way.
In this note we improve on the results given in [2] and [3] and extend
them to the odd character case. Let us mention that we take this opportunity
to correct several typos occurring in [5].
We rst state a general formula.
Theorem. Let be a primitive Dirichlet character modulo q and let h
be an integer prime to q. Let F :R!R be such that f(t) = F (t)=t is in
2 0 00 1C (R) (also at 0), vanishes at 1 and f and f are inL (R). Assume
also that F is even if is odd, and odd if is even. Then, for every > 0,
we have
Y X(p) 1 F (n)
1 L(1;) = (n)
p n
n1pjh
(n;h)=1
1X( h)() F (t)
+ c (m)(m) e(mt=(qh))dt:h
qh t
m1 1
2000 Mathematics Subject Classi cation: Primary 11M20.
[1]2 O. Ramare
Here the Gauss sum() is de ned by
X
(1) () = (a)e(a=q)
a modq
and the Ramanujan sums c (m) byh
X
(2) c (m) = e(ma=q):h
a mod h
2i Of course e() = e , and a mod h denotes summation over all invertible
residue classes moduloh. We further restrict our attention to square-freeh.
Here are two interesting choices forF which we take directly from Propo-
sition 2 of [5]. Set
2 Xsint 2 sgn(m)
(3) F (t) = + ;3 2 t (t m)
m2Z
1 1
F (t)3
(4) j(u) = e(ut)dt = (u) ((1 t) cott + 1)dt;[ 1;1]t
1 juj
2sint
(5) F (t) = 14
t
which satis es
1
F (t)4 2(6) e(ut)dt = i (1 juj) (u):[ 1;1]
t
1
Notice furthermore that F and F take their values in [0; 1].3 4
In order to compute e cien tly the resulting sums we select several levels
of hypotheses, starting by the most general ones. We use the Euler-function
and the number!(t) of distinct prime factors of t.
Corollary 1. Let be a primitive Dirichlet character modulo q and
h an integer prime to q. Assume q is divisible by a square-free k and set
= 0 if is even, and = 5 2 log 6 = 1:41648::: if is odd. Then
Y X (p) (hk) logp 1 L(1;) logq + 2 +!(h) log 4 + p 2hk p 1
pjh pjhk
2 !(h)is bounded from above if is even and qk 4 by

!(k) 1 !(h)+1 2 !(h)(h)2 log(q4 ) if qk 4 ,
p
h q 1:81 +!(h) log 4 logq if k = 1,