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.