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.