428 Mr Basset, On finding the Potentials of [May 10, 



The case of in = has been solved by Mr Gallop, who finds that 

 U =- dfi e - ^ 3 cos Xv cos fivJ (fip) dv (8). 



T Jo JO 



If therefore we put m = in (7) and substitute the above value 

 of U , we find that the integral 



q /»GO /'CO /»CO 



Y o — — I dfi I dxl du 



7T Jo Jo Jo 



e - '" 3 Xit(/> ^) cos Xv cos /W (Xw) /„ (fip) cZv (9) 



o 

 satisfies the conditions that 



V = <p(p), P <c, z=0\ 



£-* *»<■ -« (10) ' 



5. In order to find U 1 , consider the integral 



I dfi \ e - ** 3 sin Xv sin fivJ l (ftp) dv. 



o Jo 



Now 



1 IT 



( n r e / as ■ 2 /u/w /* ?sin 2 6W0 



e~^ cos (up cos 0) sin' Udvdfi = / ^f- — ^ ^ 



Jo Jo ^ Jof +p'cos-0 



Differentiating with respect to £, we obtain 



r 



j^e-^J 1 (fip)dfi = -^l- 



r 



Let £= ^ — iv where t = >/— 1, then 



/•OO 



t I e - ^ 3 sin /xv/j (fip) dfi = imaginary part of 

 Jo 



z — tv 



p ^_ ttf )« +p »Ji 



Transform the left-hand side into polar co-ordinates, and put 



J? cos 2% = r 2 - a 2 , 



i2 sin 2^ =2 rv cos 0, 

 so that R = (r 4 + v 4 + 2rV cos 26 f, 



and we shall obtain 



e-^ 3 sin fivJ t (jip) dfi = ^ — K {v (R + r - v 2 )* - ^ (-B - r 2 + v 2 )^} 

 Jo ' -^P V^ 



