OF UNRESTRICTED DEGREE. ORDER, AND ARGUMENT. 
481 
on making the substitution u = hz, this becomes 
fri + m 
(i - 2 ph + h*y» +i dh% 
or 
, r(, + ,4-) 
hn | \ 2 / 
h n+n 
(1 - 2 nh + 7d)« +i 
clli . (50). 
In this integral the phases of 1 — hz, 1 — — are to be zero at the points M, L in 
which the lines joining the points —, 2 to the origin cut the path. 
In the particular case m = 0, we have 
which is reducible to 
p» M 
AfGF)_£_ dh 
2 t rt J (1 - 2 fill + hj ’ 
h n _ 
(1 - 2 fill + hy 
dh 
(51), 
the integral being taken along a closed path which includes both the points z, — > 
and excludes the point 0. 
By changing n into — (n + 1), we obtain the formulae 
-L 2 m n _?) / 2 _ i um f( 2+i = ) _ lyyyy _ „ / 52 ) 
— 2t rt n(-i) ^ • J (1-2 nh, + m 
(1 - 2 fth + hy 
P »W = 2hk- 
h~ n ~ l 
(1 — 2 fjJi + hy 
dh 
(53), 
the integral (53) being taken as in (51). 
25. Next consider the expression (B), the path being the same as in Art. 22; we 
find 
3 Q 
MDCCCXCVI. —A. 
