s '3^'*3$««? , t«B>«-jeaws*»r,» 



that Y + (s) are analytic in R , respectively, and that Y (s) is analytic 

 everywhere (an entire function). For x < 0, deform the contour into the 

 upper half-plane R. . The contributions from the large semicircle in R, to 

 the Y and Y integrals are zero, because exp(-isx) is exponentially small 

 and Y and Y are at least as small as s 1 at infinity in R . Further, 



Y and Y are analytic within the closed contour £_ _^ , and so their 



contour integrals vanish. Thus for x < 



-M*- 



y(s) -^1 Y_(s) exp(-isx) ds (7.44) 



which we may try to evaluate by completing the contour in 3 — thcugh 

 note that we know nothing in advance about the behavior of Y (s) at 

 infinity in R . 



If x > t we complete the contour with a large semicircle in R , 

 along which, although Yj(s) is exponentially large, the product 

 Y (s) exp(-isx) is exponentially small when x > £. Similar arguments 

 then give 



W"iJ Y + 



y(x) - ~ J Y + (s) exp(-isx) ds (7.45) 



Finally, if < x < *., close the contour for Y (s) in R , that 

 for Y (s) in R + , and we find that the integrals of Y_(s), Y (s) 

 vanish, leaving 



y(x) '—I Y (s) exp(-isx) ds (7.46) 



/IT J 1 



Because of (7.43) it is only possible to close the contour here in R , 

 Y (s) exp(-isx) being exponentially large in R when < x < £. 



These arguments apply generally in three-part boundary value 

 problems. Here the integrals (7.44-7.46) can all be evaluated by 



48 



^-^r^Vin,-..--.,.,:^:^. 



