148
H. Matsuzoe and A. Takatsu
Assume q > 1. By the property that
B(s + 1, t) =
s
s + t
B(s, t)
for s, t > 0,
we have that
(q, 2, k, 0; ξ) =
σ
(Z q σ ) (2−1)(q−1)+q
3 − q
q − 1
k+
1
2
B
3 − q
2(q − 1)
+ 2 − k,
1
2
+ k
=
σ
(Z q σ ) (q−1)+q
3 − q
q − 1
k+
1
2
f 2 (k)
(
1
q−1
+ 1) ·
1
q−1
B
3 − q
2(q − 1)
,
1
2
=
1
(Z q σ ) 2(q−1)
3 − q
q − 1
k (q − 1)
2 f 2 (k)
q
,
where we set
f 2 (0) : =
3 − q
2(q − 1)
+ 1
·
3 − q
2(q − 1)
=
(q + 1)(3 − q)
4(q − 1) 2 ,
f 2 (1) : =
3 − q
2(q − 1)
·
1
2
=
3 − q
4(q − 1)
,
f 2 (2) : =
3
2
·
1
2
=
3
4
.
This leads to that
g
(q,1)
μμ (ξ ) =
4
(3 − q) 2
b
2
0 (q, 1)
(Z q σ ) 2(1−q) σ 2 (q, 2, 1, 0; ξ) =
1
σ 2 ,
g
(q,1)
σ σ (ξ ) =
b
2
0 (q, 1)
(Z q σ ) 2(1−q) σ 2
2
k=0
2
k
(−1)
k
(q, 2, k, 0; ξ)
=
1
σ 2
2
k=0
2
k
(−1)
k
3 − q
q − 1
k
(q − 1)
2 f 2 (k)
=
3 − q
σ 2 .
Fix n, j ∈ N, k ∈ {0, 1, . . . , n} and ξ = (μ, σ ) ∈ R × q,a . Let us compute
(q, n, k, j; ξ) with the use of the residue theorem. Note that
(q, n, k, j; μ, σ ) = (q, n, k, j; 0, σ ).
Define a complex valued function φ q,n,k, j;σ on C by