COMPUTATION OF RELATIVE CLASS NUMBERS OF CM-FIELDS 1 ...
MATHEMATICS OF COMPUTATION Volume 66, Number 219, July 1997, Pages 1185–1194 S 0025-5718(97)00863-6
COMPUTATION OF RELATIVE CLASS NUMBERS OF CM-FIELDS ´ STEPHANE LOUBOUTIN
Abstract. It was well known that it is easy to compute relative class numbers of abelian CM-fields by using generalized Bernoulli numbers (see Theorem 4.17 in Introduction to cyclotomic fields by L. C. Washington, Grad. Texts in Math., vol. 83, Springer-Verlag, 1982). Here, we provide a technique for computing the relative class number of any CM-field.
1. Statement of the results Proposition 1. Let n ≥ 1 be an integer and α > 1 be real. Set Pn (x) = 1 1 (1) + fn (s) = Γn (s)A−2s 2s − 1 2s − 2 and (2)
Kn (A) =
A2 iπ
Z
Pn−1
1 k k=0 k! x ,
α+i∞
fn (s)ds. α−i∞
Then, it holds (3)
0 ≤ Kn (A) ≤ 2Pn (nA2/n )e−nA
2/n
≤ 2n exp(−A2/n ).
Theorem 2. Let N be a totally imaginary number field of degree 2n which is a quadratic extension of a totally real number field N+ of degree n, i.e. N is a CMfield. Let wN be the number of roots of unity in N, QN ∈ {1, 2} be the Hasse unit index of N, and dN , ζN and dN+ , ζN+ be the absolute values of the discriminants and the Dedekind zeta functions of N and N+ , respectively. Let χN/N+ be the quadratic character assocciated with the quadratic extension N/N+ and P let φk be the coefficients of the Dirichlet series (ζN /ζN+ )(s) = L(s, χN/N+ ) = k≥1 φk k −s , p 1. Set AN/N+ = dN /π n dN+ . We have s Q w dN X φk N N Kn k/AN/N+ , (4) h− N = (2π)n dN+ k k≥1
and according to (3) this series (4) is absolutely convergent. Moreover, set n/2 λ def log AN/N+ (5) . B(N) = AN/N+ n Received by the editor December 5, 1995 and, in revised form, April 12, 1996. 1991 Mathematics Subject Classification. Primary 11R29, 11Y35; Secondary 11R42. c
1997 American Mathematical Society
1185
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´ STEPHANE LOUBOUTIN
1186
− Then, if λ > 1 and n are given, then the limit of |h− N −hN (M )| as AN/N+ approaches infinity is equal to 0, where h− N (M ) is the approximation of the relative class number obtained by disregarding in the series occurring in (4) the indices k > M ≥ B(N).
For example, if N of degree m = 2n is the narrow Hilbert class field of a real quadratic number field L of discriminant dL , we have m/4 λ m/8 dL logm/4 (dL /π 2 ). B(N) = 4π The following Proposition 3 explains how we compute the numerical values of the function A 7→ Kn (A) according to its series expansion: Proposition 3. Take A > 0. It holds
X
Kn (A) = 1 + π n/2 A + 2A2
(6)
Ress=−m (fn ).
m≥0
This series is absolutely convergent and for any integer M ≥ 0 we have π n/2 A2M+3 2 X (7) Ress=−m (fn ) ≤ . 2A (M + 1)(M !/2)n m>M
Finally, the following Proposition 4 explains how to compute recursively the values of the residues Ress=−m (fn ) occurring in (6): Proposition 4. We have (8)
1
Ress=−m (fn ) = −(−1)
nm
−1 A2m X −1−i 2 hi (m)((2m + 1)i + (2m + 2)i ) (m!)n i=−n
where the hi (m)’s are computed recursively from the hi (0)’s by using (9) i X
bi−j hj (m) hi (m + 1) = (m + 1)i−j j=−n
−1 X
and
hj (0)sj + O(1) = Γn (s)A−2s ,
j=−n
n−1 = ((k + n − 1)!/k!(n − 1)!). Thus, if where bk = Cn+k−1
(10)
Γn (s + 1) =
n−1 X
hi si + O(sn ),
i=0
then (11)
hj−n (0) =
j X (−2 log A)i i=0
i!
hj−i
(0 ≤ j ≤ n − 1).
For proving these results, obvious questions of convergence of series and integrals, and questions of inversions of integrals and summations will not be gone into. 1 Note
the misprint in the formula given in [Lou 2].
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COMPUTATION OF RELATIVE CLASS NUMBERS OF CM-FIELDS
1187
2. Introduction Prior to the method we have developed here, the only general method for computing the relative class number of any CM-field was that developed by T. Shintani (see [Oka 1] and [Oka 2] for examples of actual relative class number computations using Shintani’s ideas). However, his method requires the knowledge of a great deal of information on the maximal totally real subfield N+ . In particular, it requires the knowledge of a system of fundamental units of the group of totally positive units of N+ . However, what makes the concept of CM-field an attractive one is that the relative class number formula s s ! Q Q w d Res (ζ ) w d N N N s=1 N N N N h− = L(1, χN/N+ ) (12) N = (2π)n dN+ Ress=1 (ζN+ ) (2π)n dN+ enables us to get lower bounds on relative class numbers and solve class number and class group problems for CM-fields precisely because (12) does not involve any regulator (see [Lou-Oka] and [LOO]). Thus, the reader may possibly feel dissatisfied that he should have to know beforehand a good grasp of the unit group of N+ before − he can compute h− N , whereas (12) gives an expression for hN which does not involve units. The reader may now possibly feel satisfied that this paper shows how using (12) he indeed gets an efficient method for computing h− N provided that he only knows how to compute the decomposition of any rational prime into a product of prime ideals of N. The key point of our method is to establish the holomorphic continuation of s 7→ (ζN /ζN+ )(s) = L(s, χN/N+ ) in the same way Riemann did in the case of the Riemann zeta function (by using Mellin transformation) and to evaluate the resulting series at s = 1 (see section 4). Finally, we note that the results of this paper are better than those of [Lou 3]. Indeed, B(N) in (5) is nn/2 -fold better than the one we gave in [Lou 3]. Moreover, our proof of (3) (in section 3) is more satisfactory and elegant than the one we gave in [Lou 3]. 3. Proof of Proposition 1 We use: Lemma 5. Let α > 1 be real. We have Z
α+i∞
ds = u 2s − 1 s
α−i∞
(
0 √ iπ u
if u < 1, if u > 1;
Now, using
Z Z Γn (s) =
Z
α+i∞
( ds 0 = u 2s − 2 iπu s
and α−i∞
e−Tr(y) y s
if u < 1, if u > 1.
dy y
where the multiple integral ranges over (y1 , · · · , yn ) ∈ (R∗+ )n and where we set y = y1 y2 · · · yn and Tr(y) = y1 + y2 + · · · + yn , leads to Z α+i∞ Z Z 1 A2 dy 1 −Tr(y) 2 e + . Kn (A) = (y/A ) iπ 2s − 1 2s − 2 y α−i∞
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´ STEPHANE LOUBOUTIN
1188
Using Lemma 5 yields Z Z 2 Kn (A) = A
y≥A2
Z Z p −Tr(y) dy 2 2 ≤2 y/A + (y/A ) e e−Tr(y) dy. y y≥A2 2
For example, we get K1 (A) ≤ 2e−A . Now, using the arithmetic-geometric mean inequality yields that {(y1 , · · · , yn ); y ≥ A2 } is included in {(y1 , · · · , yn ); Tr(y) ≥ nA2/n }, which yields Z Z Kn (A) ≤ 2 e−Tr(y) dy. Tr(y)≥nA2/n
Then, the following easily proved Lemma 6 provides us with the desired result. We finally notice that we get a shorter and more satisfactory proof of [Lou 3, Proposition 1]: Pn−1 Lemma 6. Set Pn (x) = k=0 xk /k!. Then Z Z Z +∞ −α −Tr(y) Pn (α)e = e dy ≤ n e−y dy = ne−α/n . (y ,··· ,y )∈R∗ 1
n
Tr(y)≥α
+
α/n
Proof. Use {(y1 , · · · , yn ) ∈ R∗+ , Tr(y) ≥ α} n n o [ α (y1 , · · · , yn ), yi ≥ and yj ≥ 0 for j 6= i . ⊆ n i=1 4. Proof of Theorem 2 Let K be a number field of degree n = r1 + 2r2 , where r1 is the number of real places of K and r2 the number of complex places of K. Let ζK and RegK be the Dedekind zeta function and regulator of K. We set 1/2 AK = 2−r2 dK π − r1 +2r2 /2 , 2r1 hK RegK where wK is the number of roots of unity in K, wK s r1 FK (s) = AsK Γ Γ(s)r2 ζK (s). 2 Hence, FK a a simple pole at s = 1 with residue λK , and FK (s) = FK (1 − s). From now on, we let N be a CM-field of degree 2n, i.e. N is a totally imaginary number field of degree 2n which is a quadratic extension of a totally real number field N+ of degree n. Define the φk ’s by : X ζN (s) = φk k −s ( 1). ΦN/N+ (s) = ζN+ k≥1 Then, (ζN /ζN+ )(s) = L s, χN/N+ yields X (14) χN/N+ (I) φk = (13)
λK =
NN+ /Q (I)=k
where I ranges over the integral ideals of N+ of norm k. Now, ΦN/N+ = ζN /ζN+
and ΨN/N+ = FN /FN+
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COMPUTATION OF RELATIVE CLASS NUMBERS OF CM-FIELDS
1189
are entire and ΨN/N+ (s) = ΨN/N+ (1 − s). Notice that (15)
ΨN/N+ (1) =
h− λN N = λN+ QN wN
where QN ∈ {1, 2} is the Hasse unit index of N (see [Wa, Th. 4.16]). Since 2s−1 s s+1 Γ , Γ(s) = √ Γ π 2 2 using (13) for N and N+ leads to (16)
ΨN/N+ (s) = cN/N+ AsN/N+ Γn
s+1 2
where cN/N+ = 1/(4π)n/2
and AN/N+ =
Note that (17)
ΦN/N+ (s)
p dN /π n dN+ .
s cN/N+ AN/N+
Set (18)
1 ˆ Ψ N/N+ (x) = 2iπ
Z
α+i∞
α−i∞
1 = (2π)n
dN . dN+
ΨN/N+ (s)x−s ds
(α > 1),
ˆ i.e., Ψ N/N+ is the Mellin transform of the function ΨN/N+ . Using (18) and (16) yields X ˆ φk Hn kx/AN/N+ (19) (x > 0), Ψ N/N+ (x) = cN/N+ k≥1
with
(20)
Z α+i∞ 1 s+1 n x−s ds Γ Hn (x) = 2iπ α−i∞ 2 Z α+i∞ 1 Γn (S)x1−2S dS (x > 0 and α > 0). = iπ α−i∞
Now, we move the integral (18) to the line M
We want an upper bound on RM . We note that (14) yields |φk | ≤ dn (k). Moreover, !n k k X X dn (i) 1 ≤ ≤ logn (ek). Sn (k) = i i i=1 i=1 Thus, we have X dn (k) Kn (k/A) k k>M X ≤ 2enn−1 Sn (k) − Sn (k − 1) g(k/A) (if M ≥ A)
|RM | ≤
k>M
(by using (26)) X ≤ 2enn−1 Sn (k) g(k/A) − g((k + 1)/A) k>M
≤
2enn−1 X Sn (k)g 0 (k/A) A
(if M ≥ 2n/2 A)
k>M
(by using (27)) 2/n 4enn−1 X k logn (ek)e−n(k/A) ≤ 2 A k>M
(by using (25)) Z 4enn−1 X 4enn−1 ∞ = h(k) ≤ h(x)dx A2 A2 M
( if M ≥
k>M
(by using (28)).
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n+1 2
n/2 A ≥ 1)
´ STEPHANE LOUBOUTIN
1192
Now, we set B = (eA)2/n and we change the variable by setting x = Ay n/2 . We get Z ∞ dy 2 n |RM | ≤ 2e(n /2) y n logn (By)e−ny . y (M/A)2/n n/2 Since H(y) = y n+1 logn (By)e−ny decreases on [(M/A)2/n , +∞[ if M ≥ 2n+2 A n ≥ e(n/2)−1 (since its derivative H 0 (y) = (n + 1 − ny) log(By) + n y n logn−1 (By)e−ny satisfies H 0 (y) ≤ 0 if y ≥ (2n + 2)/n and B 2n+2 ≥ e), we get 2 Z ∞ dy H(y) 2 |RM | ≤ 2e(n2 /2)n y 2/n (M/A) Z ∞ dy ≤ 2e(n2 /2)n H((M/A)2/n ) y2 2/n (M/A) = 2e(n2 /2)n H((M/A)2/n )/(M/A)2/n , n/2 A ≥ e(n/2)−1 , then we have the following explicit upper i.e., if M ≥ 2n+2 n bound : (29) 2 n n |RM | ≤ 2e G((M/A)2/n ) where G(y) = y log(By)e−y and B = (eA)2/n . 2 n/2 and note that Now, we choose M ≈ B(N) = A nλ log A G(
λ log A) = On (A−λ/n log2 A) n
yields the desired result : (30)
|h− N
−
h− N (M )|
QN wN = n n/2 A|RM | = On 2 π
log2n A Aλ−1
.
5. Proof of Proposition 3 Let M ≥ 0 be a given integer. Shifting the integral (2) to the left to the line
COMPUTATION OF RELATIVE CLASS NUMBERS OF CM-FIELDS ´ STEPHANE LOUBOUTIN
Abstract. It was well known that it is easy to compute relative class numbers of abelian CM-fields by using generalized Bernoulli numbers (see Theorem 4.17 in Introduction to cyclotomic fields by L. C. Washington, Grad. Texts in Math., vol. 83, Springer-Verlag, 1982). Here, we provide a technique for computing the relative class number of any CM-field.
1. Statement of the results Proposition 1. Let n ≥ 1 be an integer and α > 1 be real. Set Pn (x) = 1 1 (1) + fn (s) = Γn (s)A−2s 2s − 1 2s − 2 and (2)
Kn (A) =
A2 iπ
Z
Pn−1
1 k k=0 k! x ,
α+i∞
fn (s)ds. α−i∞
Then, it holds (3)
0 ≤ Kn (A) ≤ 2Pn (nA2/n )e−nA
2/n
≤ 2n exp(−A2/n ).
Theorem 2. Let N be a totally imaginary number field of degree 2n which is a quadratic extension of a totally real number field N+ of degree n, i.e. N is a CMfield. Let wN be the number of roots of unity in N, QN ∈ {1, 2} be the Hasse unit index of N, and dN , ζN and dN+ , ζN+ be the absolute values of the discriminants and the Dedekind zeta functions of N and N+ , respectively. Let χN/N+ be the quadratic character assocciated with the quadratic extension N/N+ and P let φk be the coefficients of the Dirichlet series (ζN /ζN+ )(s) = L(s, χN/N+ ) = k≥1 φk k −s , p 1. Set AN/N+ = dN /π n dN+ . We have s Q w dN X φk N N Kn k/AN/N+ , (4) h− N = (2π)n dN+ k k≥1
and according to (3) this series (4) is absolutely convergent. Moreover, set n/2 λ def log AN/N+ (5) . B(N) = AN/N+ n Received by the editor December 5, 1995 and, in revised form, April 12, 1996. 1991 Mathematics Subject Classification. Primary 11R29, 11Y35; Secondary 11R42. c
1997 American Mathematical Society
1185
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´ STEPHANE LOUBOUTIN
1186
− Then, if λ > 1 and n are given, then the limit of |h− N −hN (M )| as AN/N+ approaches infinity is equal to 0, where h− N (M ) is the approximation of the relative class number obtained by disregarding in the series occurring in (4) the indices k > M ≥ B(N).
For example, if N of degree m = 2n is the narrow Hilbert class field of a real quadratic number field L of discriminant dL , we have m/4 λ m/8 dL logm/4 (dL /π 2 ). B(N) = 4π The following Proposition 3 explains how we compute the numerical values of the function A 7→ Kn (A) according to its series expansion: Proposition 3. Take A > 0. It holds
X
Kn (A) = 1 + π n/2 A + 2A2
(6)
Ress=−m (fn ).
m≥0
This series is absolutely convergent and for any integer M ≥ 0 we have π n/2 A2M+3 2 X (7) Ress=−m (fn ) ≤ . 2A (M + 1)(M !/2)n m>M
Finally, the following Proposition 4 explains how to compute recursively the values of the residues Ress=−m (fn ) occurring in (6): Proposition 4. We have (8)
1
Ress=−m (fn ) = −(−1)
nm
−1 A2m X −1−i 2 hi (m)((2m + 1)i + (2m + 2)i ) (m!)n i=−n
where the hi (m)’s are computed recursively from the hi (0)’s by using (9) i X
bi−j hj (m) hi (m + 1) = (m + 1)i−j j=−n
−1 X
and
hj (0)sj + O(1) = Γn (s)A−2s ,
j=−n
n−1 = ((k + n − 1)!/k!(n − 1)!). Thus, if where bk = Cn+k−1
(10)
Γn (s + 1) =
n−1 X
hi si + O(sn ),
i=0
then (11)
hj−n (0) =
j X (−2 log A)i i=0
i!
hj−i
(0 ≤ j ≤ n − 1).
For proving these results, obvious questions of convergence of series and integrals, and questions of inversions of integrals and summations will not be gone into. 1 Note
the misprint in the formula given in [Lou 2].
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
COMPUTATION OF RELATIVE CLASS NUMBERS OF CM-FIELDS
1187
2. Introduction Prior to the method we have developed here, the only general method for computing the relative class number of any CM-field was that developed by T. Shintani (see [Oka 1] and [Oka 2] for examples of actual relative class number computations using Shintani’s ideas). However, his method requires the knowledge of a great deal of information on the maximal totally real subfield N+ . In particular, it requires the knowledge of a system of fundamental units of the group of totally positive units of N+ . However, what makes the concept of CM-field an attractive one is that the relative class number formula s s ! Q Q w d Res (ζ ) w d N N N s=1 N N N N h− = L(1, χN/N+ ) (12) N = (2π)n dN+ Ress=1 (ζN+ ) (2π)n dN+ enables us to get lower bounds on relative class numbers and solve class number and class group problems for CM-fields precisely because (12) does not involve any regulator (see [Lou-Oka] and [LOO]). Thus, the reader may possibly feel dissatisfied that he should have to know beforehand a good grasp of the unit group of N+ before − he can compute h− N , whereas (12) gives an expression for hN which does not involve units. The reader may now possibly feel satisfied that this paper shows how using (12) he indeed gets an efficient method for computing h− N provided that he only knows how to compute the decomposition of any rational prime into a product of prime ideals of N. The key point of our method is to establish the holomorphic continuation of s 7→ (ζN /ζN+ )(s) = L(s, χN/N+ ) in the same way Riemann did in the case of the Riemann zeta function (by using Mellin transformation) and to evaluate the resulting series at s = 1 (see section 4). Finally, we note that the results of this paper are better than those of [Lou 3]. Indeed, B(N) in (5) is nn/2 -fold better than the one we gave in [Lou 3]. Moreover, our proof of (3) (in section 3) is more satisfactory and elegant than the one we gave in [Lou 3]. 3. Proof of Proposition 1 We use: Lemma 5. Let α > 1 be real. We have Z
α+i∞
ds = u 2s − 1 s
α−i∞
(
0 √ iπ u
if u < 1, if u > 1;
Now, using
Z Z Γn (s) =
Z
α+i∞
( ds 0 = u 2s − 2 iπu s
and α−i∞
e−Tr(y) y s
if u < 1, if u > 1.
dy y
where the multiple integral ranges over (y1 , · · · , yn ) ∈ (R∗+ )n and where we set y = y1 y2 · · · yn and Tr(y) = y1 + y2 + · · · + yn , leads to Z α+i∞ Z Z 1 A2 dy 1 −Tr(y) 2 e + . Kn (A) = (y/A ) iπ 2s − 1 2s − 2 y α−i∞
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´ STEPHANE LOUBOUTIN
1188
Using Lemma 5 yields Z Z 2 Kn (A) = A
y≥A2
Z Z p −Tr(y) dy 2 2 ≤2 y/A + (y/A ) e e−Tr(y) dy. y y≥A2 2
For example, we get K1 (A) ≤ 2e−A . Now, using the arithmetic-geometric mean inequality yields that {(y1 , · · · , yn ); y ≥ A2 } is included in {(y1 , · · · , yn ); Tr(y) ≥ nA2/n }, which yields Z Z Kn (A) ≤ 2 e−Tr(y) dy. Tr(y)≥nA2/n
Then, the following easily proved Lemma 6 provides us with the desired result. We finally notice that we get a shorter and more satisfactory proof of [Lou 3, Proposition 1]: Pn−1 Lemma 6. Set Pn (x) = k=0 xk /k!. Then Z Z Z +∞ −α −Tr(y) Pn (α)e = e dy ≤ n e−y dy = ne−α/n . (y ,··· ,y )∈R∗ 1
n
Tr(y)≥α
+
α/n
Proof. Use {(y1 , · · · , yn ) ∈ R∗+ , Tr(y) ≥ α} n n o [ α (y1 , · · · , yn ), yi ≥ and yj ≥ 0 for j 6= i . ⊆ n i=1 4. Proof of Theorem 2 Let K be a number field of degree n = r1 + 2r2 , where r1 is the number of real places of K and r2 the number of complex places of K. Let ζK and RegK be the Dedekind zeta function and regulator of K. We set 1/2 AK = 2−r2 dK π − r1 +2r2 /2 , 2r1 hK RegK where wK is the number of roots of unity in K, wK s r1 FK (s) = AsK Γ Γ(s)r2 ζK (s). 2 Hence, FK a a simple pole at s = 1 with residue λK , and FK (s) = FK (1 − s). From now on, we let N be a CM-field of degree 2n, i.e. N is a totally imaginary number field of degree 2n which is a quadratic extension of a totally real number field N+ of degree n. Define the φk ’s by : X ζN (s) = φk k −s ( 1). ΦN/N+ (s) = ζN+ k≥1 Then, (ζN /ζN+ )(s) = L s, χN/N+ yields X (14) χN/N+ (I) φk = (13)
λK =
NN+ /Q (I)=k
where I ranges over the integral ideals of N+ of norm k. Now, ΦN/N+ = ζN /ζN+
and ΨN/N+ = FN /FN+
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COMPUTATION OF RELATIVE CLASS NUMBERS OF CM-FIELDS
1189
are entire and ΨN/N+ (s) = ΨN/N+ (1 − s). Notice that (15)
ΨN/N+ (1) =
h− λN N = λN+ QN wN
where QN ∈ {1, 2} is the Hasse unit index of N (see [Wa, Th. 4.16]). Since 2s−1 s s+1 Γ , Γ(s) = √ Γ π 2 2 using (13) for N and N+ leads to (16)
ΨN/N+ (s) = cN/N+ AsN/N+ Γn
s+1 2
where cN/N+ = 1/(4π)n/2
and AN/N+ =
Note that (17)
ΦN/N+ (s)
p dN /π n dN+ .
s cN/N+ AN/N+
Set (18)
1 ˆ Ψ N/N+ (x) = 2iπ
Z
α+i∞
α−i∞
1 = (2π)n
dN . dN+
ΨN/N+ (s)x−s ds
(α > 1),
ˆ i.e., Ψ N/N+ is the Mellin transform of the function ΨN/N+ . Using (18) and (16) yields X ˆ φk Hn kx/AN/N+ (19) (x > 0), Ψ N/N+ (x) = cN/N+ k≥1
with
(20)
Z α+i∞ 1 s+1 n x−s ds Γ Hn (x) = 2iπ α−i∞ 2 Z α+i∞ 1 Γn (S)x1−2S dS (x > 0 and α > 0). = iπ α−i∞
Now, we move the integral (18) to the line M
We want an upper bound on RM . We note that (14) yields |φk | ≤ dn (k). Moreover, !n k k X X dn (i) 1 ≤ ≤ logn (ek). Sn (k) = i i i=1 i=1 Thus, we have X dn (k) Kn (k/A) k k>M X ≤ 2enn−1 Sn (k) − Sn (k − 1) g(k/A) (if M ≥ A)
|RM | ≤
k>M
(by using (26)) X ≤ 2enn−1 Sn (k) g(k/A) − g((k + 1)/A) k>M
≤
2enn−1 X Sn (k)g 0 (k/A) A
(if M ≥ 2n/2 A)
k>M
(by using (27)) 2/n 4enn−1 X k logn (ek)e−n(k/A) ≤ 2 A k>M
(by using (25)) Z 4enn−1 X 4enn−1 ∞ = h(k) ≤ h(x)dx A2 A2 M
( if M ≥
k>M
(by using (28)).
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n+1 2
n/2 A ≥ 1)
´ STEPHANE LOUBOUTIN
1192
Now, we set B = (eA)2/n and we change the variable by setting x = Ay n/2 . We get Z ∞ dy 2 n |RM | ≤ 2e(n /2) y n logn (By)e−ny . y (M/A)2/n n/2 Since H(y) = y n+1 logn (By)e−ny decreases on [(M/A)2/n , +∞[ if M ≥ 2n+2 A n ≥ e(n/2)−1 (since its derivative H 0 (y) = (n + 1 − ny) log(By) + n y n logn−1 (By)e−ny satisfies H 0 (y) ≤ 0 if y ≥ (2n + 2)/n and B 2n+2 ≥ e), we get 2 Z ∞ dy H(y) 2 |RM | ≤ 2e(n2 /2)n y 2/n (M/A) Z ∞ dy ≤ 2e(n2 /2)n H((M/A)2/n ) y2 2/n (M/A) = 2e(n2 /2)n H((M/A)2/n )/(M/A)2/n , n/2 A ≥ e(n/2)−1 , then we have the following explicit upper i.e., if M ≥ 2n+2 n bound : (29) 2 n n |RM | ≤ 2e G((M/A)2/n ) where G(y) = y log(By)e−y and B = (eA)2/n . 2 n/2 and note that Now, we choose M ≈ B(N) = A nλ log A G(
λ log A) = On (A−λ/n log2 A) n
yields the desired result : (30)
|h− N
−
h− N (M )|
QN wN = n n/2 A|RM | = On 2 π
log2n A Aλ−1
.
5. Proof of Proposition 3 Let M ≥ 0 be a given integer. Shifting the integral (2) to the left to the line
