5. Regularizations of nonintegrable functions 17
Assuming that Re λ −1, we have for an arbitrary positive integer m:
(5.12)

0
x1
λϕ(x1)dx1
=
1
0
x1
λ
ϕ(x1)
m−1
k=0
ϕ(k)(0)
k!
x1
k
dx1
+
1
0
m−1
k=0
ϕ(k)(0)
k!
x1+kdx1 λ
+

1
x1
λϕ(x1)dx1
=
1
0
x1
λ
ϕ(x1)
m−1
k=0
ϕ(k)(0)
k!
x1
k
dx1
+

1
x1
λϕ(x1)dx1
+
m−1
k=0
ϕ(k)(0)
k!
1
λ + k + 1
.
The right hand side of (5.12) is meromorphic in λ for Re λ −m 1
with simple poles at the points −1, −2,..., −m, with residues
ϕ(k)(0)
k!
, k =
0,...,m 1, and it defines a linear continuous functional on D for Re λ
−m 1. This functional is an analytic continuation of x+(ϕ)
λ
defined origi-
nally for Re λ −1. We denote this functional also by x+.
λ
Since m in (5.12)
is arbitrary, the distribution x+
λ
is defined for any complex λ except negative
integers. Therefore x+(ϕ)
λ
is a meromorphic function of λ for each ϕ D
with simple poles at −1, −2,..., −n,... and x+
λ
is a regularization of a
nonintegrable function x+
λ
for Re λ −1. For λ C such that Re λ −1
and for any integer m 0, we get using integration by parts
(5.13)

0
x+ϕ(x1)dx1
λ
=
1
+ m)(λ + m 1) · · · + 1)
dm
dx1
m
x++m(ϕ).λ
Since the left and the right hand sides (5.13) have analytic extension to
C\N−, where N− are negative integers, we get (by the uniqueness of analytic
continuation) that (5.13) holds for all λ C\N−. In particular, if −m−1
Re λ −m, let μ = λ + m. Then x+
μ
is a regular functional: x+(ϕ)
μ
=

0
xμϕ(x)dx
and we obtain from (5.13):
(5.14) x+(ϕ)
λ
=
1
+ m) · · · + 1)
dm
dx1 m
x+(ϕ)μ
=
(−1)m
+ m) · · · + 1)

0
x1+m λ
dmϕ(x1)dx1,mdx1
where −m 1 Re λ −m.
5.2. Regularization in
Rn.
Example 5.4 (Distributions p.v.
1
|x|2−k2
,
1
|x|2−(k±i0)2
, k 0). Introduce
polar coordinates in
Rn
: x = rω, where r = |x|, |ω| = 1 and is the
surface area element of the unit sphere in
Rn.
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