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252 WEI WANG
where dV (ζ) denote the volume element of R 2n−3 . Repeating this procedure,
we can integrate all variables ζi with i ∈ S2 \ {1}. Then
(4.33)
I q
β,i2···in Πj /∈S2\{1}[A j
ij (z)]
βj +βj ij
·
R2n−2n2 +1
⎛
⎛
⎝2 q |r(z)| + |ζ1| +
⎞
|ζk| ⎠
· ⎝2 q |r(z)| + |ζ1| +
m
k /∈S2 t=2
k /∈S2
A k t (z)|ζk| t
Now integrate all variables ζi with i ∈ S0 by
(4.34)
for k > 2, we get
(4.35)
I q
Πj∈S1 [Aj
β,i2···in ij (z)]
βj +βj ij
·
R2n−2n0−2n2 +1
⎛
C
· ⎝2 q |r(z)| + |ζ1| +
dζdζ 1
(|ζ| + C) k Ck−2 ⎞
⎠
⎛
⎝2 q |r(z)| +
⎞
|ζk| + |ζ1| ⎠
m
k∈S1 t=2
k∈S1
A k t (z)|ζk| t
By condition C, S3 = ∅ and 1 ∈ S2, we see that
(4.36)
Therefore
(4.37)
2n0 + 2n1 + 2n2 = 2n,
2n2 + n1 = 2j + 2.
⎞
⎠
2n − 2j − 3 − 2n0 = n1 − 1,
2n − 2n2 − 2n0 + 1 = 2n1 + 1.
−(2n−2j−3+κ)
−(2+ P
j /∈S 2
βj +βj +κ)
ij dV (ζ).
−(2n−2j−2n0−3+κ)
„
− 2+ P
j∈S1 «
βj +βj +κ ij dV (ζ).