1.5 The Incomplete Gamma Functions
9
Γ (z, a + 1) = aΓ (z, a) + z
a exp(−z).
(1.17)
The derivatives are given by
d
dz
γ (z, a) = −
d
dz
Γ (z, a) = z
a−1 exp(−z), and
(1.18a)
d 2
dz 2 y(z, a) = −
1 +
1 − a
z
d
dz
y(z, a) with y = γ (z, a) or y = Γ (z, a).
(1.18b)
The incomplete gamma function is related to the confluent hypergeometric
function 1 F 1 (a; b; z), Eq. (6.5), via
γ (z, a) = a
−1 z
a exp(−z) 1 F 1 (1, 1 + a, z) and
(1.19)
γ (z, a) = a
−1 z
a
1 F 1 (a, 1 + a, −z).
(1.20)
These equations will be useful for computing the incomplete gamma function by a
series expansion.
1.5.2 Computational Aspects
The series expansion of Eq. (6.5) converges for all allowed values, but in some
cases very slow. Therefore the computation will make use of the series expansion
Eqs. (1.19) and (1.20), on a continued fraction ansatz, and on the recurrence
formula.
From Eq. (6.5) we get
1 F 1 (1; 1 + a; z)
= 1 +
1
1 + a
z +
1
(1 + a)(2 + a)
z
2
+
1
(1 + a)(2 + a)(3 + a)
z
3
+ · · ·
(1.21)
which will diverge for a negative integers. The single terms of the sum can be
efficiently computed by
t 0 = 1 , t n =
1
n + a
t n−1
n = 1, 2, 3, · · · .
(1.22)
9
Γ (z, a + 1) = aΓ (z, a) + z
a exp(−z).
(1.17)
The derivatives are given by
d
dz
γ (z, a) = −
d
dz
Γ (z, a) = z
a−1 exp(−z), and
(1.18a)
d 2
dz 2 y(z, a) = −
1 +
1 − a
z
d
dz
y(z, a) with y = γ (z, a) or y = Γ (z, a).
(1.18b)
The incomplete gamma function is related to the confluent hypergeometric
function 1 F 1 (a; b; z), Eq. (6.5), via
γ (z, a) = a
−1 z
a exp(−z) 1 F 1 (1, 1 + a, z) and
(1.19)
γ (z, a) = a
−1 z
a
1 F 1 (a, 1 + a, −z).
(1.20)
These equations will be useful for computing the incomplete gamma function by a
series expansion.
1.5.2 Computational Aspects
The series expansion of Eq. (6.5) converges for all allowed values, but in some
cases very slow. Therefore the computation will make use of the series expansion
Eqs. (1.19) and (1.20), on a continued fraction ansatz, and on the recurrence
formula.
From Eq. (6.5) we get
1 F 1 (1; 1 + a; z)
= 1 +
1
1 + a
z +
1
(1 + a)(2 + a)
z
2
+
1
(1 + a)(2 + a)(3 + a)
z
3
+ · · ·
(1.21)
which will diverge for a negative integers. The single terms of the sum can be
efficiently computed by
t 0 = 1 , t n =
1
n + a
t n−1
n = 1, 2, 3, · · · .
(1.22)
