WITH APPLICATION TO SPHEPJCAL HARMONIC ANALYSIS. 
197 
(•+1 
I Qlsin2^6 djji 
^ VP ^ __ I y '2^ (ft + O')! 1 («- + A — 1)!! 
1 
Ass O’ 
+ 1 
=:cN^->(- 1) 
(n — a — 1) !! (u — X)!! (A — c)!! (A A a) 0 
^+^>,-^,1 (>i + (t) : ! + A - 1) : ! (A + 1) ! 
shd 0 sill i:>0 dji 
1 
A = cr 
(?i - (T - 1) ! ! {n - Aj ’ ! (A - (t) :! (A A cr) ! ! • (X+ 1 +p)! I (A+ 1 -??) !! ‘ 
When p and n — cr are both even the integral vanislies. 
If(n “ cr) be odd and even, 
A -- cr 
r+i 
Ql sin 2)0 dp 
• -1 
X~i} — 1 
= s<=> (- 1) 
A = cr 
A = *;-l 
= cv N'"' (“ 1) 
1 
+ 1 
(a + <r — 1) 1 ! (?z + A)! I 
(u — a) '.'.{a — A — 111! (A — cr) 1! (X + o') ! ! j 
+ 0- _ 1) ' ; (>i 4- A) ;' (X + 1) ! 
cos 6 siiH 9 sin p9 dp 
1 
A=: (7 
(?i - cr)!! (a - X - 1)! ' (X-(t) : : (X + fx):! (X + 2 +p) l\{X + 2-p)ll' 
It will be noticed that the integral is zero whenever (a + o" + v) is even. Hence if 
sinpd is expressed in terms of tesseral functions, p ])eing even, only odd values of ji 
occur when cr is even, and only even values when cr is an odd iiuml)cr. The reverse 
will be the case when is odd. 
We find similarly, if (n — cr) is odd and p odd. 
r + l 
I cos pd dp 
1 “ 1) Etft + X)!_! 
A —cr 
•+1 
1 r- 
-;-cos 0 sin^ 0 COS p0 dp 
(a-cr)’:(/i - X - (X- cr)::(X + cr)!! ^ ‘ 
A = -<-I 
-cp (-1) 
A = (7 
+;>-<r-i 4 o-_ I ) ; ; 4 x)!! 
+ 1 )! 
(u — cr) ! !(7i — X — I) ! ! (A — cr)!!(A + cr):.(A42+p):.(A+2 —20 ^ 
and when n — cr and p are both even. 
+ 1 
-1 
Ql cosp0 dp 
" (/i-(r-l)!!(ft->)!!(A-(r)!!(A4cr)!!(A + l+y)!!(A + l-70!!’ 
M^hen n -- cr + y) is odd the integrals vanish for all values of p. 
We may now prove the theorem which has been mentioned in the first paragraph. 
If cr and p are both even, | Q^sin^dc/p, is zero, unless n is odd. In that case the 
quantity (A fi- 2 — ji) ! I which occurs in the denominator will be even, as A takes up 
