Figure 37 - A large semicircular path. 

 Then, from [31a], around the large semicircle, approximately, 



/ — dt = K\t dt = iKR I e dO 



J dt J Jq 



l-2a 



[31d] 



Since by assumption 2 a > 1, this last expression goes to as ff -♦ ~. 



The remainder of the contour integral, therefore, must likewise become zero a.s R -> oa. 

 But if the small semicircles are allowed to shrinl< down onto their centers, the remainder 

 becomes in the limit the desired integral. Hence it must be that this latter integral itself 

 equals zero. It follows that Az = between the ends of the broken line, so that they join and 

 complete a finite polygon, at A^ in Figure 34. 



The last segment of the broken line makes an angle (let) n with the first, or the first 



makes an angle 



a„^j ff = (2-2a) ;7 ^ [2 -(Oj + 02 . . . a„)]7T 



with the last; in Figure 34 this is the exterior angle at /!,. The number of actual vertices then 

 depends upon the value of £« . 



If 2 a = a, + a, . . . a„ < 2, the polygon has an actual vertex corresponding to t = <x, 

 with an exterior angle a^_^.n. The two adjacent sides A^ A^ and A^ A^ in Figure 34, have 

 lengths 



f [real] ^ dt, [ [real] ^ dt 

 J-oo ^^ ''a ^^ 



62 



