Single-Particle Electrodynamics - Assassination Science

6.7.1 The bare inverse-cube integral
We first consider the inverse-cube integral with f(n d ) = 1:
η 0
( 4
3 πε3 ) −2 ∫
∫
d 3 r
r≤ε
d 3 r ′ ˜r −3
The r s -integral is trivial, as before; we are left with
( ) 4 −1 ∫ {
2ε
η 0
3 πε3 4πrd 2 dr d 1 − 3 ( ) rd
+ 1 ( ) }
rd
3
1
0
2 2ε 2 2ε (rd 2 + . (6.102)
w2 d )3/2
Now, to avoid wasted effort, we shall, from the outset, keep in mind that,
ultimately, we shall be taking the limit w d → 0; we shall, for this reason,
retain the quantity w d only so far as required to regularise the expressions
involved. Thus, the integrals of the second and third terms of (6.102)—which
are perfectly well-behaved at both limits of integration—can be performed
with w d = 0 immediately:
r ′ ≤ε
d .
( ) 4 −1 ∫ {
2ε
η 0
3 πε3 4πrd 2 dr d − 3 ( ) rd
+ 1 ( ) }
rd
3
1
0
2 2ε 2 2ε (rd 2 + w2 d )3/2
= − 9η ∫ 2ε
0
dr
4ε 4 d + 3η ∫ 2ε
0
dr
0 16ε 6 d rd 2 + O(w d )
0
= − 9η 0
2ε 3 + η 0
2ε 3 + O(w d)
= − 4η 0
ε 3 + O(w d). (6.103)
The integral of the first term in (6.102), however, must be computed with
w d finite: to compute
( ) 4 −1 ∫ 2ε
η 0
3 πε3 4πr 2 1
d dr d
0 (rd 2 + w2 d )3/2
we must employ the change of variable (6.87) (with r and ˜r replaced by r d
and ˜r d , of course); noting further that
r 2 d = w2 du 2
1 − u 2 ,
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