?-5r-^£?^<<83°>'Zt**l*HlkiSK***0*>m 



contour with a large semicircle in R , along which the Integral vanishes 

 because of exponential smallnees of exp(-itJt) there. To Interpret the 

 meaning of K (t) in R_ we write it as X(t)/K (t), so that 



„ , . 1 f ie~ ltZ K.(t) ,„ t - ,,. 



H + (S) -2^I J (t + k) K(t)(t-a) dt (7 ' 37) 



The pole at t ■ s lies outside the contour, K (t) is analytic within 

 the contour, and 



m 



(t + k Q ) K(t) - [ t _ k ) (t + k,) 



so that the only singularity within the closed contour is at t ■ -k , 

 and therefore 



ie ik »* K <-k ) .Ik £ 



V 8) " 2^1 ( ~ 27rl) ^ - .. r " 77TTT • 



7- k i- k i\ (s+k.) * 



WK) ^' s) 



u / ^ ie" 1 ^ ie lk > & 1 -is£ ik.Jl, ., ... 



H - (8) " (s + k ) K(8) _ (s + k) " 7Tir (e " « 1 ) • (7.39) 

 o + * * 



This instructive example shows the importance of removing 

 exponential growth at infinity; although the original H(s) is analytic 

 In R it is not algebraically mall at infinity there, and use of the 

 Cauchy integrals shows how it can be split into H (s) + H («0 , with 

 H + (s) - 0(s _1 ) at infinity in R + as in (7.38) and (7.39). 



Coming now to the integral terms in the integral equations (7.33), 

 (7.34), we can this time complete the contour with a large semicircle 

 in R+, because along that semicircle the factor exp(itt) will ba 



46 



ilUMMinim M „^ , { n ^M t^H' i i-----''^'~---^ : -"^- 



