exponentially small and the integral along the semicircle will contribute 

 nothing. The functions D (t), S (t) are analytic within the closed 

 contour, and the function 1/K_(t) » K (t)/K(t) has a pole at t ■ k only. 

 Therefore (7.33) gives 



D.(s) (k. - It ) e*V D.OO 

 [G.(-s) - H + <8>] + ^ + ^__ ±_ . 



and (7.34) gives 



S + (s) (k, -k o ) e lk i ft S+(ki ) 



W-> '■♦^l + .$ff (T^T 



(7.40) 



(7.41) 



The unknown constants D (kj), S + (kj) are found by putting e - k in 



(7.40), (7.41) and then we have completely determined the functions 



D jl( 8 )i S,(s), from which Z.(e), Y (-s) can be found, and hence the 

 v *r t — 



whole solution is determined. While there are no difficulties of 

 principle, the algebra is very tedious and there is no point in giving 

 it here. 



The only remaining point of interest concerns the inversion of 

 the Fourier integral for this three-part problem. We have 



y(*) " 2» J l Y _< 8 > + V 8 > + V 8)1 e*^- 18 *) *» 

 with the integral along a path from -°°to +«in R, and we recall that 



(7.42) 



Y_(s) - 0(8-') 



Y(s) - 0(8 _1 ) 



e i8 * 

 - 0(^ ) 



B 



isi 

 Y + (s) - 0(^-) 



as Is \ •*■ <*> in R 



as | s | ■* 



in R. 



as Is -»• » in R 



as | s ->■ 



in R. 



(7.43) 



47 



-.■^.., , .■..^ J ^-,...w-^,,,.: J .^^.»r. : ..^_^.,.^ fa L;.-^^-,^^^-,-,.-. 



jftai&flfeBfca aflfc t -■• viit- .tut- antw -^t^- -■ 



