The point of using half-range transforms is that conditions are 

 different according as x < 0, so that a two sided-transform cannot 

 immediately be applied. For x > we have the equatio of motion 



&■ + k 2 y - t s . (2.16) 



dx 2 l 



Multiply by exp(lsx) and integrate from to ». Then the last term 



produces 



k* Y + (s) 



provided seR , while the first term can be integrated by parts to give 



rd 2 y isx , dy isx . . isx » 2„ , \ 

 — \ e dx « ■£ e -is [ye 1 -s 2 Y + (s) 

 „ ox o 



From the anticipated behavior, y^expi' ik x) as x •*■ +<*> we set that 

 j2- e isx and ye isx both vanish as x ■+ +«° provided seR + . If we write 



y(o+) - l ~% & y(x), y'(CH-) <* ^ k w y ; (x) (2.17) 



then we have 



(s 2 - k\) Y + (s) - -y*(0+) + isy(0+) (2.18) 



as a statement, for ssR., of the equation of motion in x > 0. 

 + 



Applying the tame procedure, this time for s<cR_, to the equation 



&.+ k 2 y - o (2.19) 



dx 2 ° 



which holds in x < 0, produces 



(s 2 - k 2 ) Y (s) - + y' (0--) - isy(O-) (2.20) 



11 



^^^-^ '-^fecy^ 



