OF UNRESTRICTED DEGREE, ORDER, AND ARGUMENT. 
465 
the points 1, — 1 being indefinitely small, and the semi-circles round the point 0 being 
so also ; the parts of the integral taken along the loops and semi-circles in the limit 
vanish, and we have only to consider integrals taken along the real axis. The integral 
consists of four parts, the following :— 
(1.) From 0 to — 1, phase of t equal to — n, and the phases of t — 1, t + 1 
equal to v, — 2tt respectively. 
(2.) From — 1 to 0, phase of t equal to — n, and the phases of t — 1, t + 1 
equal to u and 0 respectively. 
(3.) From 0 to 1, phases of t, t — 1, t -J- I equal to 0, tt, 0 respectively. 
(4.) From 1 to 0, phases of t, t — 1, t -J- 1 equal to 0, — u, 0 respectively. 
Taking v for the modulus of t, we have in the first two parts of the integral 
t = ve~ in , and in the other two parts t = v ; hence, the integral is 
or 
f (l — v 2 )"' v p (e' un . e _2,U7r . e~ p+Un — e nin . 1. e~ p+Un + e' un .1.1— e~' ,un .1.1 )dv, 
Jo v 
(e nn — e~ nni ) (1 + e~ pm ) I* (1 — v 2 ) n v p civ, 
Jo 
(p - F 
which is equal to 
2i sin mr . (1 + e pm ). Jr 
n ( n) n 
n (n + 
p + 1 
To show that the result is the same, when the real parts of n + 1, p + 1 are not 
both positive, we find by integration by parts 
/•(-1 + , 1-) i iv) I Q + , 1-) 
j (t 2 — 1)" tP clt = — j ( t 2 — 1 Y t p+2 clt, 
and also 
f 
(-i+,i-) 
(t 2 — \ytp dt = — 
2n -h p + 3 f (-1+ > 1-) 
2 n + 2 
r 
(t 2 — i y +1 t p dt. 
By successive use of these two formulm, we find 
r(-i+, i-) 
(; t 2 — 1)” tP clt = { — 1 Y 
n t^^+n+s+X 
n ( 1 ~—+n) n l^—^+s 
2 
f(-i + , 1-) 
n fF±i+7t+s 
X 
(£2 _ 1)«+A t p+2 ' dt, 
3 o 
n (n) 
n ( 7i -f- xj 
MDCCCXCVI.-A. 
