Nonainusoidal Osailtatione of a Cylinder in a Free Surface 



If this expression is substituted into (C-5) and the transform is 

 converted we find that 



"'"-'^'"— J.„ (p'-K-^)( | p|-K) ° dp 



2AK' r°° eP^cos px 

 ^ Jq <P^- K'2)(P - K) P 



A f py / 1 



= \ e cos px s : - 



^ Jo t {K+K')(p+K') 



+ — i J:^ jdp. (C.7) 



1 2K' 



(K*- K)(p - K') ■ (K'^-K^(p-K) 



Apparently there are poles at p = K' and p = K. However if the 

 inverse transfornm of the right-hand side of Eq. (C-6) is taken, it is 

 readily seen that the integral must be integrated as a principal-value 

 integral in order to recover the original function A sin K'|x|. This 

 means that the integral in Eq. (C-7) associated with the second and 

 third terms in the square bracket should be taken as P. V. integrals. 



If we let 



r eP^cospx r e 



' Jo P+K' ^ 'Jo p+K' ^' 



and make the change of variable t = i(p +K')x, we can show that 



I, = Reie"^' \ 4-^*= ^«i[«"^'^i<iK'z)]. 



Again the change of variable t = i(p- K')z enables us to show that 

 ,00 py '-'Oo 



- I e cos px J T> 1 e *^ , 

 ^2=1 ^^P=^®?1 TITTUP 



^ -Jq p - k' Jo p - k' 



= Re. le E, (-iK'z)t i"""© J 



where ± signs correspond to the case of x ^ 0. Similarly the change 



/oo 

 (e" cos px)/(p - K) dp leads to 



13= Re. [e'"^'Ej{-lKz)T l^re''*^'] 



949 



