SSSsfeifJOMUoasmas* 



Hv(s) - *$- y(W-) 



(a + k } (-2k ) K (-k ) 

 o o - o 



(5.5) 



i v "i^ 2 /fvi-^ i r i _ i ~\ 



■ W " T V ^ } (s + k ) (8 - k ) K (8) " (-2k ) K. (-k ) 

 o|_ o- o-oj 



The entire function E(s) is again zero and thus we find 



K + (s) Y + (s) - 7 _ r - i yin - rT - ^L y(Df) - ± - (5.6) 



0-0 o o - o 



Ifrom this equation we can now determine the value of y(0+), for 

 we recall from $3 that the behavior of Y (s) as s ■*■ °° in R is related 

 to the behavior of y(x) as x ■* Of — i.e., to y(0+). As a + » in R, 



we have 



* + (s) •- J y(0+) exp(isx) dx + ... 

 ^o 



- lyC**) + (provided Ims > 0) 



(5.7) 



while K (s) ■* 1 as s ■*■ » in R . Therefore the leading terns of (5.6) 

 state that 



M2ti. 



s 

 which requires 



y((H-) 



mo z y(0+) 1_ 



sK (-k ) 2k T s K (-k ) 

 - o o - o 



2k T K (-k ) K (-k ) 

 — o 



The solution can now be completed in precisely the same manner 



as before. This method — of examining the detailed behavior, possibly 



to several terms, Jn the expavision of Y,(s) as a ■*■ °> in R. —is 



+ + 



24 



(5.8) 



(5.9) 



