486 Lord Rayleigh on the Influence of Obstacles 



much positive as negative, S„ vanishes for every odd value 

 of ft. This holds even when a and ft are unequal. 

 Again, 



S 2n = t(m l + im)- 2n = ;- 2n 2(— m' + m)- 2 " 



= (-l)»2(-*V + m)- 2 »=(-l) w S 2w . 



Whenever n is odd, S 2 » = — S 2 », or S 2 » vanishes. Thus for 

 square order, 



S 6 = S 10 = S 14 = . . . = (20) 



This argument does not, without reservation, apply to S 2 . 

 In that case the sum is not convergent ; and the symmetry 

 between m and m f 9 essential to the proof of evanescence, only 

 holds under the restriction that the infinite region over which 

 the summation takes place is symmetrical with respect to the 

 two directions a and ft — is, for example, square or circular. 

 On the contrary, we have supposed, and must of course continue 

 to suppose, that the region in question is infinitely elongated 

 in the direction of a. 



The question of convergency may be tested by replacing 

 the parts of the sum relating to a great distance by the corre- 

 sponding integral. This is 



CC dxdy CC cos 2nd r dr d6 



JJ {* + iyf n ~' = JJ ^~ 



and herein 



J 



r -2n+l dr = r -2»+2/(_2?l + 2) ; 



so that if n > 1 there is convergency, but if n = l the integral 

 contains an infinite logarithm. 



We have now to investigate the value of S 2 appropriate to 

 our purpose ; that is, when the summation extends over the 

 region bounded by x= +u, y= +v, where u and v are both 

 infinite, but so that v/u—0. If we suppose that the region 

 of summation is that bounded by x= +v, y= +v } the sum 



Fig. 2. 



vanishes bj' symmetry. We may therefore regard the sum- 

 mation as extending over the region bounded externally by 

 a= +oo, y= +v, and internally by #= +v (fig. 2). When 

 v is very great, the sum may be replaced by the corresponding 



