6 Gauge Freedom of Entropies on q-Gaussian Measures
149
φ q,n,k, j;σ (z) :=
z
σ
2k p q (z; 0, σ )
(n−1)(q−1)+q
− q (z; 0, σ )
j
=
z
σ
2k
p q (z; 0, σ )
(n−1)(q−1)+q
z
2
+ r (q, σ )
2
(Z q σ ) 1−q (3 − q)σ 2
− j
,
where we set
r (q, σ ) :=
− ln q
1
Z q σ
· (Z q σ ) 1−q (3 − q)σ 2 .
The function φ q,n,k, j;σ has poles of order j at ±ir (q, σ ). For R > r (q, σ ), let L R
and C R be smooth curves in C defined by
L R := {z : [−R, R] → C | z(θ ) = θ },
C R := {z : [0, π] → C | z(θ ) = Re
iθ
},
respectively. The residue theorem yields that
L R ∪C R
φ q,n,k, j;σ (z)dz = 2π i · Res(φ q,n,k, j;σ ; ir (q, σ )),
(6.9)
where Res(φ q,n,k, j;σ ; ir (q, σ )) stands for the residue of φ q,n,k, j;σ at z = ir (q, σ ).
Lemma 7 For n, j ∈ N, k ∈ {0, 1, . . . , n} and (μ, σ ) ∈ R × q,a , then
(q, n, k, j; μ, σ ) = 2π i · Res(φ q,n,k, j;σ , ir (q, σ )).
Proof If we show that
lim
R→∞
C R
φ q,n,k, j;σ (z)dz = 0,
then we have the desired result by letting R → ∞ in (6.9).
Take R > r (q, σ ) large enough. We calculate that
C R
φ q,n,k, j;σ (z)dz
≤ R
π
0
φ q,n,k, j;σ (Re
iθ
)
dθ
= R
π
0
R
σ
2k
p q (Re
iθ
; 0, σ )
(n−1)(q−1)+q
R
2 e
2iθ
+ r (q, σ )
2
(Z q σ ) 1−q (3 − q)σ 2
− j
dθ
≤ C R
2(k− j)+1
π
0
exp q
−
R
2 e
2iθ
(3 − q)σ 2
(n−1)(q−1)+q
dθ,
where the constant C depends on q and σ .