1.3. RIESZ REGULARISATION 5
The following proposition gives the asymptotics at negative integers.
Proposition 1.14. At negative integers −k ∈ −N, we have
(1.12)
˜
Γ( −k + z) ∼0
(−1)k
k!
⎛
⎝
1
z
− γ +
k
j=1
1
j
⎞
⎠
.
Proof. By formula (1.9) we have
˜
Γ( −k + z) =
˜
Γ( z)
(z − k) · · · (z − 1)
∼0
1
z
− γ
(z − k) · · · (z − 1)
∼0
(−1)k
k!
1
z
− γ 1 +
z
k
· · · (1 + z),
from which (1.12) follows.
Remark 1.15. Formula (1.12) yields back the residue at z = −k given by
formula (1.10).
1.3. Riesz regularisation
We now want to make sense of
˜
F −k(φ) in spite of the fact that a = −k arises
as a pole in (1.7). Following Riesz (see e.g. [Sch]), we pick the constant term at
z = 0 in the Laurent expansion of the map z →
˜
F
−k+z
(φ).
Proposition 1.16. Given a Schwartz function φ on
R+ and a complex number
a, the map z → Fa+z(φ) is meromorphic on the plane with simple poles in −a − N.
The constant term in the Laurent expansion3
∞,Riesz
0
xa
φ(x) dx := fpz=0
˜
F a+z(φ)
:= lim
z→0
˜
F
a+z
(φ) −
1
z
Resz=0
˜
F
a+z
(φ)
(1.13)
coincides with the ordinary integral
∞
0
xa φ(x) dx whenever Re(a) −1.
(1) If a / ∈ −N, then
∞,Riesz
0
xa
φ(x) dx =
˜
F a(φ) =
(−1)k
(a + 1) · · · (a + k)
Fa+k(φ(k)),
where k is any integer such that Re(a) + k −1.
(2) If a = −k ∈ −N, then
∞,Riesz
0
x−k
φ(x) dx = −
1
(k − 1)!
∞
0
log x
φ(k)(x)dx
+ Resa=−k
˜
F a(φ)
k−1
j=1
1
j
,
setting the sum over j equal to zero if k = 1.
Proof. The case a / ∈ −N follows from the previous discussion. We therefore
prove the proposition when a = −k for some integer k ∈ N.
3Here
fp stands for the finite part.
