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.