1.1. HOMOGENEOUS DISTRIBUTIONS 3
(2) The inverse of the Gamma function
1
Γ(z)
defined on the half-plane Re(z)
0 extends to a holomorphic map at z = 0 and
(1.4)
1
Γ(z)
= z +
γz2
+
o(z2).
In particular,
(
1
Γ
)
(0) = 1.
Proof. (1) The derivative of Γ at 1 reads
Γ (1) = ∂zΓ(1 + z)|z=0
= ∂z
∞
0
tz e−t
dt
|z=0
=
∞
0
log t
e−t
dx
= −γ.
The derivative at k ∈ N − {1} reads
Γ (k) = lim
z→0
Γ (k + z) − Γ (k)
z
= lim
z→0
(k + z − 1) · · · (z + 1) · Γ (z + 1) − (k − 1)!
z
= (k − 1)!
⎛
⎝
k−1
j=1
1
j
−
γ⎠
⎞
= Γ(k)
⎛
⎝
k−1
j=1
1
j
−
γ⎠
⎞
,
so that for k ≥ 2,
Γ (k)
Γ(k)
=
k−1
j=1
1
j
− γ.
(2) It follows from (1.2) that
1
Γ(z)
=
z
Γ(z + 1)
=
z
Γ(1) + Γ (1)z + o(z)
= z +
γz2
+
o(z2).
Exercise 1.8. Show that
(1.5) Γ(z) ∼0
1
z
− γ,
where by f(z) ∼0 g(z) we mean that limz→0
f(z)
g(z)
= 1. Hint: Formula (1.5) follows
from (1.2) setting b = z, k = 1 or equivalently from (1.4).
Exercise 1.9. Show that2
Γ(z) Γ z +
1
2
=
21−2z
√
π Γ(2z).
Exercise 1.10. Show that for two complex numbers a and b with positive real
parts, we have
Γ(a)Γ(b) = B(a, b)Γ(a + b),
where B(a, b) :=
1
0
ta−1(1
−
t)b−1dt
is the Beta function.
2I
thank Stephan Weinzierl for pointing out this formula to me.
