WITH APPLICATION TO SPHEPJCAL HARMONIC ANALYSIS. 
189 
The integral on the right-hand side has been found in the last paragraph. Writing 
p + e = cr and e — p — X, we note that [n — X — 2) = {n — e + p — 2) is necessarily 
positive as e <,n — p, and that cr — \ is |)ositive and even. If, further, n — cr he 
even n — X must be even also, because the difference between these quantities is 
a — X — 2p. We may now write the value of the integral 
_, rn-^e + p-l)!!0^^e + p-2)!! 1 
J_l * duP ^ (h — e — p) !! (« + e — p -f- 1) !! (2p — 2) !! 
if e ^ — p and + p + e is even, 
= 0, in all other cases. 
If p = 1, the above reduces to 
§ 0 . f|Q:Q^,dp, {^[Q'^QUrlp. 
'+1 
'-1 
+1 
•'-1 
Before discussing the general integral of the product of two tesseral functions, it is 
convenient to obtain the solution in a few special cases. When the type of the two 
tesserals differs hy 2, we may transform the integrals as follows ; 
f + l 0’+2 7(r + 2 p a- /crp 
(I _ ^2) , dP _ 0^,, lii 7 
+ 1 
-1 
1 r+i 
<r + 2 
f + l 
I ( 1 - P') 
afjL' ' ■ dfjL. 
2/ + lJ_]'" \ did 
cr+ J 
=2] dp. 
dll \ o II 
By the application of PvODRiguez’s theorem the right-hand side reduces to 
'2i% -f" 
dp. 
Integrating each term partially cr ff 1 times we arrive at the equation 
f*‘QrQrc/M 
'-1 
- 2i + 1 (fv^ [(*■ + ^7 + 2) (i + <r + 1) P. ,, - (i - (t) (l' - o- - 1) P._,] if.. 
If in be odd, or if i > n, the Integrals on the rightdiand side vanish. If t<w 
and i -\- n even, we must substitute. 
+ idP« 
-1 dji 
r+ipp 
" P;+i = I Pjff = 2, as found above. 
