280 Methods for the Solution of Special Problems [CH. vni 



Again, since the function / is periodic in 6 with a period 2?r, it follows 

 that, when 6 = 2?r, expression (260) may be written in. the form 

 f(r, 2-7T, z t a, a, 0) -f(r, - 2?r, z, a, - a, 0), 



and this clearly vanishes. Thus expression (260) vanishes when 6 = and 

 when 6 = 2?r. That is to say, it vanishes on both sides of the semi-infinite 

 conducting plane. 



It is now clear that expression (260) satisfies all the conditions which 

 have to be satisfied by the potential. The problem is accordingly reduced 

 to that of the determination of the function f(r, 6, z, a, a, 0). 



336. Let us write 



then the distance R from r, 0, z to a, a, is given by 

 fi? ^2 _ 2ar cos (6 a) + a? + z* 



= 2ar {cos i (p - p) - cos (0 a)} + z*. 

 Take new functions R and/(w) given by 



R' 2 = 2ar (cos i (p - p) - cos (d - u)} + z 2 , 



The function /(w) has infinities when u = a, a + 2w, a 4-Tr. ..., its residue 

 being unity at each infinity. Also, when u a, the value of R becomes R. 

 Hence the integral 



f(u)du (261), 



where the integral is taken round any closed contour in the w-plane which 

 surrounds the value u = a, but no other of the infinities of f(u), will have as 



its value 2i7r x -= . We accordingly have 



The integral just found gives a form for the potential function in ordinary 

 space which, as we shall now see, can easily be modified so as to give the 

 potential function in the Riemann's space which we are now considering. 



We notice first that -^, , regarded as a function of r, 0, and z, is a solution 



of Laplace's equation, whatever value u may have. Hence the integral (261) 

 will be a solution of Laplace's equation for all values of f(u), for each term 

 of the integrand will satisfy the equation separately. 

 If we take 



