1.4. HADAMARD’S “FINITE PART” METHOD 11
Iterating the integration by parts procedure gives

x−k
φ(x) dx =
1
(k 1)!

x−1 φ(k−1)(x)
dx +
−k+1
k 1
φ( ) + · · ·
+
−k+j
(k 1) · · · (k j)
φ(j−1)(
) + · · · +
−1
(k 1)!
φ(k−2)(
)
=
1
(k 1)!

log x
φ(k)(x)
dx
log
(k 1)!
φ(k−1)(
)
+
−k+1
k 1
φ( ) + · · · +
−k+j
(k 1) · · · (k j)
φ(j−1)(
)
+ · · · +
−1
(k 1)!
φ(k−2)(
). (1.16)
A Taylor expansion around = 0 of φ and its derivatives then gives rise
to the finite part
∞,Had
0
x−k
φ(x) dx
=
1
(k 1)!

0
log x
φ(k)(x)
dx +
φ(k−1)(0)
(k 1)!(k 1)
+ · · ·
+
φ(k−1)(0)
(k 1) · · · (k j) (k j)!
+ · · · +
φ(k−1)(0)
(k 1)!
=
1
(k 1)!

0
log x
φ(k)(x)
dx +
φ(k−1)(0)
(k 1)!
k−1
j=1
1
k j
.
Exercise 1.22. Show that for any positive real number λ =

we have
fp
=0

λ
xa
φ(x) dx =
∞,Had
0
xa
φ(x) dx δa+k μ
φ(k−1)(0)
(k 1)!
.
Compare with Exercise 1.17.
Applying Proposition 1.21 to φ(x) =
e−x
leads to the following Hadamard
extension of the Gamma function (which we also call cut-off Gamma function):
ΓHad(b)
:= fp
=0

xb−1 e−x
dx .
Exercise 1.23. (1) Show that if Re(b) 0, the map

xb−1 e−x dx
lies in A[ ]. The corresponding asymptotic expansion has constant term
given by
ΓHad(b)
= lim
→0

xb−1 e−x
dx = Γ(b).
(2) Show that for b / Z≤0, the map

xb−1 e−x dx lies in C A[ ] b.
The corresponding asymptotic expansion has constant term given by
ΓHad(b)
=
Γ(b + k)
b(b + 1) · · · (b + k 1)
.
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