' ^^>'*;H^-s , v^> 





y(X «) e i8(X + *> dX 



S i8 f y(x 



+«,) e lsX dX 



^ i8£ j ly(^+) + X y»(*+) + ...] e lsX dX 



+ -- e y(U) + 



(■?) 



as s ■* °° in 8 . Thus Y (s) has erponential behavior, exp is&, at 

 infinity in R . From the differ* tial equation we have 



(s 2 - k 2 ) Y (s) = [-y'(£+) + is y(£+)] e isl 



o + 



(7.6) 



(7.7) 



from which (7.6) is obvious. Again, there is a pole at s « k eR unless 



o + 



-y*(«.+) + ik y(i+) = 



(7.8) 



For the middle portion of the string we define 



r 



Yj(s) - I y(x) exp isx dx 



» n 



(7.9) 



Since the integration is over a finite range only and the integrand is 

 bounded and continuous, Y (s) is analytic in the entire complex pl.ne 

 (the Fourier Transform of a function with compact support — i.e., 

 vanishing outside some finite range, is an ENTIRE function). As 8 +» 

 with Im s > 0, 



Y,(s ) ~ (y(Oi-) + xy'(CH-) + ...) e isx dx - 0(s _1 ) 

 Jo 



(7.10) 



because when Im s > only small values of x contribute to the integral, 

 so that we can expand y(x) about r, ■ 0+ and also extend the integration 



36 



