1.4. HADAMARD’S “FINITE PART” METHOD 7
Exercise 1.17. Show that for any real numbers λ = 0 and μ and any holomor-
phic function f(z) = λ z + μ
z2
+
o(z2)
in a neighborhood of zero, with the notation
of (1.13) and for any Schwartz function φ on
R+
we have
fpz=0
˜
F a+z(φ) =
∞,Riesz
0
xa
φ(x) dx δa+k μ
φ(k−1)(0)
(k 1)!
.
Hint: Notice that
1
f(z)
=
1
λ z
μ z + o(z).
Applying the previous proposition to φ(x) =
e−x
leads to the following Riesz
extension of the Gamma function.
Given a complex number b, the map z
˜
Γ( b + z) is meromorphic on the plane
with simple poles in −b N {0}. The finite part of the integral
˜
Γ( b + z), which
is defined by the constant term in the Laurent expansion
(1.14)
ΓRiesz(b)
:= lim
z→0
˜
Γ( b + z)
1
z
Resz=0
˜
Γ( b + z) ,
coincides with the ordinary Gamma function Γ(b) whenever Re(b) 0.
Exercise 1.18. Show the following:
(1) If b / −N {0}, then
ΓRiesz(b)
=
˜
Γ( b) =
Γ(b + k)
b (b + 1) · · · (b + k 1)
for any k such that Re(b) + k 0.
(2) For any nonpositive integer b = −k show that
ΓRiesz(0)
= −γ;
ΓRiesz(−k)
= Resb=−k
˜
Γ( b)
k
j=1
1
j

(−1)k
k!
γ if k 0.
Hint: Apply Proposition 1.16 to a = −(k + 1) and φ(t) =
e−t.
1.4. Hadamard’s “finite part” method
As before, φ denotes a Schwartz function on
R+.
Clearly, for Re(a) −1,
Fa(φ) :=

0
xa
φ(x) dx = lim
→0

xa
φ(x) dx.
In particular, for Re(b) 0, we have
Γ(b) = lim
→0

xb−1 e−x
dx.
Following Hadamard (see e.g. [Sch]) we want to extend Fa(φ) to all complex values
of a in picking out the constant term denoted by
∞,Had
0
xa
φ(x) dx in the asymp-
totic expansion of the map

xa
φ(x) dx as 0, a procedure we are about
to describe.
Applying this to φ(x) =
e−x
yields an alternative extension
ΓHad(b)
:=
∞,Had
0
xb−1 e−x
dx
of the Gamma function to the whole complex plane.
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