OCCURRING IN SPHERICAL HARMONIC ANALYSIS. 
391 
f; 
Jo 2 
Cl(j> 7T 1 
p cos 3 (f> + q sin 2 0 2 fpc[ 
(a) 
This result will still be true if for p, q we substitute the unreal quantities c+re 0 * 7 , 
c+re-V, as may indeed be shown by direct integration. 
Let us differentiate both sides of the equation (a), cr times with regard to p and a 
times with regard to q. We find 
2<x! 
f 2 “ — 
J o (p 
sin 2cr 0 cos 2<r 0 
cos 3 0 + q sin 3 0) 2o " fl 
Making the above substitutions for p, q we get 
sin 3a 20 
1.3 .. . (2er —1) 
2 <r 
I 2 1 
J CmY +h 
2<r ' ! gfc 
1.3 . . . 2cr — 1 7 rj o 
{c + r(A’ cos 3 0 + e # sin 3 0) } s 
2 ( 7+1 
cl(f)= 
c 3 + 2 crjjb + P 
2(7+1 
2 
L '^+-+<-r£-r$*- 
1.3 .. . (2(7 — 1) c 3<r+1 \ r?/x' 
the last line being derived from its predecessor by cr differentiations with regard to /x. 
And now expanding the expression under the integral sign in powers of r, and 
equating the coefficients of r'~ a , we find 
^ a ' a ' 2°'-( 2 (e 9j cos 3 0-j-e -0 - 7 sin 3 0V _<r sin 3 ' 7 ' 20cZ0 
cl/x 17 2a! l — cr! 7rJ 0 v r r 
On putting 20=0, and substituting the expression just found in equation (a), we 
finally obtain for (i, cr) one of the forms corresponding to C, viz. 
2 + cr 
9,rr1 
a! i\ 
! o-! 1 p . . 
—2 cos o-0i/ r -1 (/x+J*'Cos00 <r sin 3<r 0c/0 . 
There is obviously another expression, corresponding to the second form of C, which 
is easily written. It is to be found by expanding, as above, in powers of the reciprocal 
of r. 
The expansion of the binomial inside of the integral (c) and subsequent integration 
with regard to 0 give the form 'corresponding to D, viz. :— 
i + a! ^ , f • (i — a)(i—a — 1) . 0 , , 
MS 2 cos A + 1) V --V+ 
<rf) 
Integration of the Product of Two Harmonics and Proof of Laplace’s Coefficient. 
§ 8. Let us take two points A and B at distances a and b from the origin, A being- 
on the axis of 2 and B at an angular distance a from it. 
