112 
MK. L. N. G. FILUN ON AX ArPKOXLMATE SOLUTION FOK LENDING A 
1 
+ h,{- ly-* 2-'-' + . . . + zhu 
(\/ — l2 4" — l) (\/ — l2 + — o) . . . 
(\/ — Iz — At — 3)(\/ — Iz — 4? + l) 
4- (v/^ITlz — 4^—1) iV — Iz — 4i — 5) . . . 
[\/ — \z At — 3)(v/ — l2 4"'i^~l"l)_ 
?il(— l)‘2'^ 4- ^3{~ 1)‘"‘2"*““ 4“ • • • + '^2(+l 
(x/ — Iz4"'^^ — Oi'/ — i2 4"-i^ — 5) • • • 
(y _ - 4i-3) (v/ - 12 - 4^4-1) 
— (\/— 12 ~ 4^^1) is/ — \z — At —b) .. . 
(x/ — l 2 4" 4^ — 3) (x/ — 1^ 4~ + 1)_ 
_ i 
- ‘?1 
we obtain 
-Ki2! + 1 - /. 
(_iy 77- 
{2t + 1 )! 4 
(-ly TT 
TTZ 
‘7 
^21+1 i'z) st;ch 4- X-2(+i (~) 
- ^2t+i {z) tanh 4- X2(-fi (2) sech ^ 
^2i+i — (2^ + 1)! 4 dz-^+- 
aud since J, = K, 4- L,, we find that the required integral 
_ ^ y-i-iy 
A tZo (20! 
I xfjot (2) sech y 4- X2t (2) tanh 
TTZ 
Y 
4 - 
jL 
Yz 
TF V 77 
ifj2t{z) tanh “ — X2« (2) sech 7 
77Z 
0 
J 
i=ii 
+ ^ 
(_iy d-‘+i . 
A tZ {"At + ij! 
I '/'•«+1 (2) sech 4 - X2J+1 (2) tanh - 
d 
TTZ 
‘7 
dz 
'2£ + l 
(2) tanh .y — X2£-n (2) sech 
TTZ 
4~ il.2,j+-, 
§ 25 . Justijication of the Procedure employed, in the last Section. 
We have to show that, in the case of the integral 
Jn = 
J( 
(2£6)“ 
1 0 sinh'*’''^ 226 
(sinh u 4“ u cosh u) cos uz du 
(sinh u 4- u cosh u) cos uz du 
)o(l - 
we were justitied in expanding (1 — ascending powers ol e 
