Single-Particle Electrodynamics - Assassination Science

we find
( ) 4 −1 ∫ 2ε
η 0
3 πε3 4πr 2 1
d dr d
0 (rd 2 + = 3η ∫ 2ε/(4ε 2 +wd 2)1/2
0
u 2
du
w2 d )3/2 ε 3 1 − u . (6.104)
2
Finally, by noting that
∫
u 2 ( ) 1 + u 1/2
du
1 − u = ln − u
2 1 − u
(which may be verified by direct differentiation), we find
( ) 4 −1 ∫ 2ε
η 0
3 πε3 4πrd 2 dr d ˜r d −3 = 3η 0 4ε
ln − 3η 0
0
ε 3 w d ε + O(w d). (6.105)
3
Coupling the result (6.105) with that of (6.103), we thus find that
η 0
( 4
3 πε3 ) −2 ∫
∫
d 3 r
r≤ε
d 3 r ′ ˜r d −3 = 3η 0
r ′ ≤ε
ε 3
0
4ε
ln − 7η 0
w d ε + O(w d). (6.106)
3
Clearly, the first term in (6.106) encapsulates the (logarithmic) divergence
of the integral in the limit w d → 0; the second term, on the other hand,
represents a finite contribution.
To write this result in a somewhat more
shorthand notation, we define two new integral constants, η ′ 3 and η ′′
3 (the
unit coëfficients here are arbitarily chosen):
we then have
η 0
( 4
3 πε3 ) −2 ∫
r≤ε
η ′ 3 ≡ η 0
ε 3 , (6.107)
η ′′
3 ≡ η ′ 3 ln 4ε
w d
; (6.108)
∫
d 3 r
d 3 r ′ ˜r d −3 = −7η 3 ′ + 3η 3, ′′ (6.109)
r ′ ≤ε
where we take it as understood that there are terms of order w d present (that
will of course have no bearing on the final results). It is the presence of η ′′
3,
of course, that renders the unappended integral η 3 infinite.
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