5. Regularizations of nonintegrable functions 19
Then (5.18) can be rewritten in the form
(5.21)
1
|x|2 k2 ± i0
= p.v.
1
|x|2 k2

iπδ(|x|2

k2).
Example 5.5 (Distributions
1
S±i0
, p.v.
1
S
, δ(S), where S(x) = 0 is a smooth
surface). A surface S(x) = 0 is called smooth if S(x)
C∞(Rn),
S(x) is
real-valued, and ∇S(x0) = (
∂S(x0)
∂x1
, . . . ,
∂S(x0)
∂xn
) = 0 when S(x0) = 0. We
have
1
S ±
(ϕ) =
Rn
ϕ(x)
S(x) ±
dx.
Let {αj(x)} be a partition of unity in
Rn,
i.e., αj(x) C0
∞(Rn)
and

j
αj(x)
1. We choose a partition of unity such that if supp αj {S(x) = 0} = ∅,
then
∂S
∂xnj
= 0 on supp αj for some nj, and the change of variables
(5.22) S(x) = t, xnj = xnj
has a
C∞
inverse on supp αj. Here xnj are n−1 remaining coordinates after
xnj is deleted.
Denote by
(5.23) xnj = h(t, xnj ), xnj = xnj
the inverse map to (5.22) . Making the change of variables (5.23), we get
Rn
αj(x)ϕ(x)dx
S(x) ±
=
Rn
αj(h(t, xnj ),xnj )ϕ(h(t, xnj ),xnj )
∂h(t,xnj )
∂t
t ±
dtdxnj .
Taking the limit as ε 0 we obtain, as in (5.18):
(5.24) lim
ε→0
Rn
αj(x)ϕ(x)dx
S(x) ±
= p.v.
Rn
αjϕ
∂h(t,xnj )
∂t
t
dtdxnj

Rn−1
αj(h(0,xnj ),xnj )ϕ(h(0,xnj ),xnj )
∂h(0,xnj )
∂t
dxnj .
Denote by the surface area element of the surface S(x) = 0. On supp αj
we have:
(5.25) = 1 + |∇xnj h(0,xnj )|2dxnj .
Since S(xnj , h(t, xnj )) t, we have:
(5.26)
∂S
∂xnj
∂h
∂t
= 1,
∂S
∂xk
+
∂S
∂xnj
∂h
∂xk
= 0 for k = nj.
Therefore
(5.27) = 1 +
∂S
∂xnj
−2
|∇xnj S|2dxnj = |∇xS|
∂h
∂t
dxnj .
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