Therefore, the third integral of Equation (115) is given by 



/ 



( ) du = - ZiTKi 



exp(iu,x cos 9) 



[l+4(niK)-'"^^ cos 0] 



1/2 



+ 2iTKi 



exp(iu„x cos 6) 



[l+4(mK)-^^^ cos 6] 



1/2 



c 



I 



iK exp(-ux cos 9) , 



1/7 1/2 ^ 

 o iv-[(K) ' -Km) ' V cos 9] 



(D.4) 



By adding Equations (D.2) through (D.4), g_ is expressed 

 tt/2 



't^i-\ 



K exp(-vx cos 9) dv 

 o iv+[(K)-'-^^-i(iii)-'-^^v cos 9] 



Tr/2 



2tt 



d9 







J 



K exp(-vx cos 9) dv 



o iv-[(K)-'-^^--i(in)^^^v cos 9] 



-1 



K exp(-iu x cos 9) 



o [4(mK)-^^^ cos 9-1] 



1/2 



d9 



. r"/' K exp 

 ^ 1 [1-4 C 



K exp(-iu X cos 9) 



1/2 

 mK) ' cos 9] 



tt/2 



d9 - - 



= 1 



K exp(iu,x cos 9) 

 [1+4 (mK)-^^^ cos 9] 



d9 



tt/2 



•' 



K exp(iu„x cos 9) 



o [l+4(mK)"^''^ cos 9] 



1/2 



(D.5) 



2. x < 



We can follow the same method to transform the integral with respect to u. But, 

 in this case, we take different integral paths: for the first and second integrals, 

 we evaluate the complex integrals given in Case 1 in the first quadrant, and for the 



61 



