where an approximation sin2 0 = 2(1 cos 0) + O[(x xo)4] is used to obtain the
last term inside the first bracket. Here we may drop off the term O [(x xo)4/p] ,
which is essentially 0(e1), and its 01 part will be incorporated into the calculations
of Da-terms later. Then, using the same techniques as used to find Aa-terms, we
can reduce Eq. (4-213) to
Q2 + q2rE3 3 1
Qt() ME q2 E X- A2 2 (r2 + j) X (2 +1 co 0)] -3/2
2 f2[r 2 f2 (r2 + J2 o
q2 (1 ) EJA -3/2 Cos 1
(62 + t COS 1/2
v2fro (r + J2)3/2 o A
q2E iroX-1-1
2 (r2 + J2Po (4-24)
As we saw before, by Eq. (4-170) (62 + 1 cos )-3/2 A`1 in the limit A -i 0
and the first term on the right hand side will vanish. The second term will also
give no contribution to the regularization parameters because (X-3/2 cos } = 0.
Only the last term, which is ~ po0 will give non-zero contribution according to the
argument in the analysis presented above (see Eq. (4-202)). Using Eq. (4-171) in
the limit A 0 and taking "()" process, Eq. (4-214) becomes
t\ 1 q2 Et (- 3/2 ) 5
lim Qe(1) [c 2 P, (cos 0). (4-215)
r^2 r- (1 + J2/.2)3/2
The identity (x- P) (1 a sin2 )-P 21 (p, ( ; 1, a) = Fp, with a
2/ (r2 + J2) is taken from Appendix C of Ref. [18], and we take the limit 0 -- 0
lim Qti) [-'] 2 E1F3/ 1. (4-216)
Ao/eo 2 r2 (1 + J2/r2)3/2 0
Now the remaining part is
t[] 3q2 [0t (2)] pl -
Qt(2)[ -1 t (2)] t to (4-217)