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

(λ + m)(λ + m − 1) · · · (λ + 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

(λ + m) · · · (λ + 1)

dm

dx1 m

x+(ϕ)μ

=

(−1)m

(λ + m) · · · (λ + 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 dω is the

surface area element of the unit sphere in

Rn.