316 PROCEEDINGS OF THE AMERICAN ACADEMY. 



we easily obtain the solution 



^_ p g-zVX^=fc'xj'(A?-)(/A . 



~T~Jo v^^^^' * 



This result was given by Sommerfeld, Ann. der Phys. 1909, Bd. 28, 

 p. 683. 



From this I obtain for a ring of sources of radius a, 



M= / \ \ J(Xr) J(Xa)dX, (8) 



J A /A 2 _ „2 



which reduces to (7) when a = 0, and when k = 0, to 



u= I ^'^J{\r) J(Xa) dX, (9) 



the well-known expression for the Newtonian potential of a ring ; and 

 when r = 0, to 



uiz)= ^-===Xj{Xa)d\ (10) 



J V A^ — K^ 



which can be shown to be equal to 



the correct expression for points on the axis of symmetry. 

 Now since 



J(\r) = - / <i'^"°s«(^(o, (11) 



ttJo 



we have 



^^==X/(Xr)/(A«)a 



1 /^=0 fn rn , XflX 



= I I I d9d(ae^^ ('■ COS 0. +o cos b)-2 Va^k2 ^ (12) 



"■Jo Jo Jo -^/A"^ — K^ 



We will divide this into two parts, the integrals from to k, and k to x . 

 Putting now 



A = K sin (fi, Vx^ — K^ = IK cos <^, dX = k cos <f) d<f>, 



