1.5. DISCREPANCIES 13
As a result, since
φtk) (
= t−k−1φ(k)(t−1·), for any a / ∈ −N we have
˜
F
a
(φt) :=
(−1)k
(a + 1) · · · (a + k)
Fa+k
φtk)(
=
ta
˜
F
a
(φ),
i.e.,
˜
F
a
is still a homogeneous distribution. However, for a = −k with k ∈ N we
have
˜
F a(φt) =
1
(k − 1)!
⎛
⎝t−k
k
j=1
φ(k−1)(0)
j
−
t−k−1
∞
0
log(x)
φ(k)(t−1
x)
dx⎠
⎞
for
=
1
(k − 1)!
⎛
⎝t−k
k
j=1
φ(k−1)(0)
j
−
t−k
∞
0
log(tx)
φ(k)(x)
dx⎠
⎞
=
t−k
˜
F a(φ) +
φ(k−1)(0)
(k − 1)!
log t
=
t−k
˜
F a(φ) + Resa=−k
˜
F a(φ) log t ,
so that the extended distribution is no longer homogeneous. A discrepancy arises
with the loss of homogeneity of the extended homogeneous distribution at negative
integers. Consequently,
˜
F
a
is homogeneous whenever z →
˜
F
a+z
(φ) is holomorphic
at zero.
1.5.2. The extended Gamma function: obstruction to the functional
equation. The extended Gamma function
˜
Γ obeys the following property for
Re(b) 0:
(1.18)
˜
Γ( b + 1) = b
˜
Γ( b),
but a discrepancy arises since property (1.18) breaks down at nonpositive integers.
Exercise 1.25. For any complex value b, show that
˜
Γ( b + 1) = b
˜
Γ( b) + Resz=0
˜
Γ( b + z).
Consequently,
˜
Γ obeys the functional equation
˜
Γ( b + 1) = b
˜
Γ( b) outside the poles,
but at a pole −k in Z≤0 we have
˜
Γ( −k + 1) = −k
˜
Γ( −k) +
(−1)k
k!
.
In both cases investigated in sections 1.5.1 and 1.5.2, the presence of a residue
is responsible for an obstruction to the expected property.
