6 Gauge Freedom of Entropies on q-Gaussian Measures
147
(1, n, k, 0; ξ) =
R
p 1 (x; ξ)
(n−1)(1−1)+1
x − μ
σ
2k
dx
= 2
∞
0
1
√
2πσ
exp
−
1
2
x − μ
σ
2
x − μ
σ
2k
dx
=
2
k
√ π
∞
0
e
−y y
k−
1
2 dy
=
2
k
√ π
k +
1
2
=
2
k
√
π
(2k − 1)!!
2 k
√ π
= (2k − 1)!!,
where (·) stands for the Gamma function, that is
(s) :=
∞
0
e
−x x
s−1 dx
for s > 0.
For q > 1, it tuns out that
(q, n, k, 0; ξ)
=
R
p q (x; ξ)
(n−1)(q−1)+q
x − μ
σ
2k
dx
= 2
∞
0
1
(Z q σ ) (n−1)(q−1)+q
1 +
q − 1
3 − q
x − μ
σ
2
(n−1)(q−1)+q
1−q
x − μ
σ
2k
dx
=
σ
(Z q σ ) (n−1)(q−1)+q
3 − q
q − 1
k+
1
2
∞
0
y
k−
1
2
(1 + y)
n−1+
q
q−1
dy
=
σ
(Z q σ ) (n−1)(q−1)+q
3 − q
q − 1
k+
1
2
B
3 − q
2(q − 1)
+ n − k,
1
2
+ k
.
Proposition 2 For a = 1 and ξ = (μ, σ ) ∈ R × q,a , we have that
g
(q,1)
μμ (ξ ) =
1
σ 2 ,
g
(q,1)
σ σ (ξ ) =
3 − q
σ 2 .
Proof It follows from Lemma 6 that
(1, 2, 0, 0; ξ) = 1,
,(1, 2, 1, 0; ξ) = 1,
,(1, 2, 2, 0; ξ) = 3,
implying
g
(1,1)
μμ (ξ ) = b
2
0 (1, 1)
1
σ 2 =
1
σ 2 ,
g
(1,1)
σ σ (ξ ) = b
2
0 (1, 1)
1
j=0
1
σ 2 (1 − 2 + 3) =
2
σ 2 .