1.4. HADAMARD’S “FINITE PART” METHOD 9
Exercise 1.20. Deduce from the previous lemma that the map

log x
e−x
dx lies in A[ ] A[ ] log , and that we have
lim
→0

log x
e−x
dx = −γ.
Proposition 1.21. Let φ denote a Schwartz function on
R+.
(1) If Re(a) −1, the map

xa φ(x) dx lies in C A[ ] a+1, and we
have
lim
→0

xa
φ(x) dx =

0
xa
φ(x) dx = Fa(φ).
(2) If a / −N, the map

xa
φ(x) dx lies in C A[ ]
a+1
with constant
term given by
∞,Had
0
xa
φ(x) dx =
(−1)k
(a + 1) · · · (a + k)

0
xa+k φ(k)(x)
dx =
˜
F a(φ)
for any integer k such that a + k −1.
(3) If a = −k −N, the map

xa
φ(x) dx lies in A[ ]
−k+1
A[ ] log
with constant term given by
∞,Had
0
x−k
φ(x) dx =
1
(k 1)!

0
log x
φ(k)(x)
dx
+ Resa=−kFa(φ)
k−1
j=1
1
j
=
˜
F
−k
(φ),
where the sum is set to zero if k = 1.
Proof. (1) If Re(a) −1, we have

xa
φ(x) dx =

0
xa
φ(x) dx
0
xa
φ(x) dx
=

0
xa
φ(x) dx
a+1
1
0
xa
φ( x) dx.
The map
1
0
xa
φ( x) dx lies in A[ ] so that the map

xa
φ(x) dx
lies in C A[ ]
a+1
and
lim
→0

xa
φ(x) dx =

0
xa
φ(x)dx.
(2) If a / −N, we proceed by induction on k with Re(a) ] (k + 1), −k]
using integration by parts to show that

xa
φ(x) dx lies in C⊕A[ ]
a+1.
The step k = 0 holds by the previous item. One integration by parts
provides the induction step. Indeed, if the map

xa+1
φ(x) dx lies
in C⊕A[ ]
a+2,
then the map

xaφ(x)
dx lies in C⊕A[ ]
a+1
since

xa
φ(x) dx =
xa+1
a + 1
φ (x) dx +
xa+1
a + 1
φ(x)

=
1
a + 1

xa+1
φ (x) dx
a+1
a + 1
φ( ).
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