66 
Proceedings of the Royal Society of Edinburgh. [Sess. 
The general method may he applied without any difficulty ; we shall 
introduce the polynomial B/(h) ; and in this particular case formula (6) 
will be written 
3V(1) B n y ( - 1 ) 0 
C/(l) CV(-l) 
In order to evaluate B« v (1)j we shall establish the following recurrence- 
fonnul^ * 
(n + 2v- \)G n _^ v {z)B n v {z) - nC*{z)B n _-f(z) + z 2 -l = 0. 
The demonstration is easy : we have a polynomial of degree 2n—2; and 
we shall prove that it has 2n—l roots, viz. the (n— 1) roots a 1? a 2 , ■ • • an-\ 
of C and the n roots of /3i ... /3 n of C n v (z). For a root ai, it 
becomes, since B n _ 1 v (ai)Cn_ 1 ' v (ai)= 1, 
— n ^n( a i) + ( a i 2 — l)C n _ 1 /v (aj), 
which is zero by recurrence-formula I ; and for a root /3i, it becomes 
(n + 2r- 1 )C n _F(/4 + (ft 2 - 1 )C " (ft), 
which is zero by II. Then, since 
(V(1) 
r (n + 2v) 
our formula gives us 
T(n + l)T(2i/) 
B/(l) = B n _/(l). 
But, C {(z) being 2vz, B *(z) is ~ ; so 
Ay 
B„hi) = 
\ 
2v 
and 
B n v ( 1 ) T(n+l)r(2v) 
(V(l) 2vT(?l+2v) 
The general method gives us then the following result : 
H 
B „»(*) (2v+i)r(»+i)r(2K) n dt 
x{Z) ~ ~ (1 -**)"+* ' 2vT(n + 2v) { > l(T^W- 
Two forms can then be proposed : we can, as we have said, express the 
above integral in terms of H 1 V , and write 
H n v (z) 
1 
(1 -z 2 y+i 
r(»+ nr^) M2) 
2vzT(n -i- 2v) 
+ r( ”t .-wiw 
zV(n + 2v) 
We can also observe that this integral is nothing else than the function 
H 0 v+1 (z), and write 
H„-(z)= - . 'Vffi + (2v + 1 + 1 )r(2l ') c„-(z)H/+ 1 (z). 
w (1 -z 2 )- +i 2vT(n + 2v ) w 0 w 
These are two reduction-formulae for H/. 
