﻿Problems in Diffraction. 65 



From (4) by multiplication by J n (V) 



K.(.)J.(*>J.<-0 • fiElSFi'-W-^-W] 



, e WfV. __* I.'. . (8, 



With substitution of the value o£ jj->--. this reduces to 



\2/ \2 J 2 sin mr( ^2iT,ie nlo s n - n J 

 Hence when the real part of n is positive this becomes 



K„(.,)J„(.0 = ^g')", • • • (9) 



«ince the second part vanishes at infinity. 

 The expression 



cos)?{ir -(fl-fl')} 

 sin nir 



vanishes at infinity if 6 — 6' lies between and 2tt. 

 3. Consider the integral 



COS n(7T-6—b') t / /\ XT / \J / 1A n 



-A — — - J (mr ) K,(mr)dn, . (10) 



where r>r' and 6>&\ 



the integral being taken over the path A in the rc-plane. . 



The integral is equal to 2iri XB, where SB is the sum of 

 the residues of this function. 



Eur. 1. 



L 



recti axis iiv n jjtane 



The path A (fig. 1) has a small semicircle at the origin. 

 Phil. Mag. b. b. Vol. 12. No. 67. July 1906. F 



