z=0 



1_ 



2fT 



S L. 



>0<Q) 1 * rm 3 A 



d S Q (36) 



z=0 



because of the symmetry. Now we employ the integral representation 



77 oo 



PQ 



2tt 



ZZ = -^r I d9 I exp[-k{|z-z' |+i(x-x')cos 9 + i(y-y')sin 6}] dk (37) 



-TT 



in Equation (34) . Then the Fourier transform of <J> ) becomes 



z=0 



exp[ik(x cos + y sin 6)] dxdy 



z=0 



I 



9* (Q) 



3n Q 9n Q 



exp[kz'+-ik(x' cos + y* sin 0)] d S 



(38) 



Combining Equations (35) and (38), we find 



" ^ J J 



9 

 ~ — -((>_ ~k~ ) exp[kz+ik(x cos + y sin 0)] ds 



^0 J \ 



~ lk cos ~ 



3x 



exp[ik(x cos + y sin 0)] dy 



z=0 





 1 k_ f x _ 



4tt Yo 3x 2 4tt 



Y / T 



(39) 



20 



