Potentials,  JEolotropie  and  Isotropic. 
/LL  =  0  to  1,  i.e.  we  require 
I    PL+i/'2'+1  dfi=n(2n  f  3)-2y,     for  p=l, 
J  0 
I    (f^'L+2  ~  Pk+2K9+1  **=»(2n  +  5)  -2q,  for  ;;  =  2, 
J  o 
and 
jV»(I»-..  P^)^=  »(2p  +  2n.  + 1)  - 2q 
for  the  general  case. 
583 
K30) 
No^ 
and 
fV<P2„+i^=P^+i(i)-%  fV'-'P^+i^  ; 
Jo  Jo 
the  last  term  vanishes  for  £  =  0,  1,  2,...rc,  but  no  further,  and 
therefore 
J    0 
Again 
ri(^p;!+2-p;,i+2v2»+i^=p;,!+2(i)-(2?+3)rV2«+1PL+2'^ 
Jo  Jo 
while 
fV<+1  P:,„+2rf/4  =  P2,!+2  (1)  -  (2q  +  l)  f  Vp*.+s 
Jo  Jo 
<7/<t. 
The  last  integral  vanishes  for  q  =  0,  1,  2,...w,  but  no  further, 
and  therefore 
f1('tFk+2-:pk+>3?+1^=(»+1)(2^+3)-(2<?+3)=»(2»+5)-2g. 
Jo 
The  solution  is  therefore  fully  verified  for  the  cases/?  =  1  and 
2,  which  belong  to  the  two  problems  stated,  the  cases  of 
potentials  P  and  Q.  For  the  general  case  a  mode  of  proof 
will  be  briefly  indicated.  Carry  out  the  operation  I  by  using 
P^+1  —  /LfcP/=(s-r-l)P«  differentiated  any  number  of  times, 
hen  in  the  result  I*-1.  ?££,  yields 
PL+2,-1  -(p-l)(2^  +  ^  +  l)PL+2u_2+a  residue, 
