OF UNRESTRICTED DEGREE, ORDER, AND ARGUMENT. 
479 
Transform the integral by means of the equation u — z 2 — ( z~ — l)v, so that v is 
the independent variable, we have then, 
- ~ 1)*' 
m^ — n — r i—1 
d + ,0-) 
y 2 {'ll + Til I 
1 ] n+ 
(z*- 1) 
/ 1 \-l-m 
(v — 1-- j v~*~ m (z 2 — 1) dv, 
or, 
r(i+> 0—) 
(^ 2 — _ j )- 2m <; i 
1 \ "I n+m 
l — — \v \ (v — Y)~ h ~ m v~ h ~ m do ; 
in this integral v — 1 has the phase zero at that point of the path in which v is 
positive and greater than unity ; this expression may be written in the form 
1 / 1 \r ]T (n 4- rrT\ f( 1 + ’°-) 
1)'( 1 - A v-*-~r (v - 1)-*-’dv ; 
s 2 / II (n + m — r) II (r) 
now 
rU+, o-) 
V 
l m + r (y — 1 )-^ m civ = 2 L cos nnr 
II (- \ — m ) n (- -L — m + r) 
IT ( — 2m + r) 
hence the expression becomes 
“ “ 1 ) ^ ,l z n+m —— 2 —~ 2t C0S mnY (. “ n - m, i- - m, 1 - 2m, 1 - ^ j, 
or 
JL_ 
2 2m 
(^ 2 
I \-bn „n+m ^ (2m 1) 
. 47 tl sin m7rF 
— m, \ — m, 1 — 2m, 1 
or 
/ o i . . n (m — i) n (— 4) n / . 
— (u. 2 — 1) h*z n+m 2i sm niTr. — ■—-:- - h — n — m, 4 — m, 1 — 2m, 1 
vr ’ II (m - 4) 2 
