ARISING FROM THE LOAD OF NEIGHBOURING OCEANIC TIDES. 
39 
Making use of these formulae we obtain 
P e _/ 'AJ 0 (kr)Ji (ha) dh = - f 
JO 7T J 
f (fa-).J, (ha) 'A = 1 I 
J o A * 7T J 
To find these integrals, put 
a = 
a 
i r 
cos 0 dO 
7 r Jo 
(R 3 +z 3 )*' 
1 r 
a — r cos 0 
7T Jo 
R 2 
1 f 
a—r cos 0 
IT Jo 
R 2 
a 
(3 = 
d6>-- 
7T J 
(R 2 + 2 2 )^0- 
,,r a —cos 0 7 ^ 
0 R 2 (R 2 + z 2 )* 5 
a—r cos 0 
R 2 
dO. 
a~ + 7- + z- 
6ar 
1-/3 
e* =-—• 
1+/3 
e s = - 
a 
a 
so that 
e 3 < e 2 < e l5 
6i + e 2 + e 3 = 0, 
and change the integration variable from 0 to s by 
cos 0 = as + /3, 
( 2 s+ 6 ^ ds 
then 
2 
{W + z 2 Y 
Put again 
,.{4 (s-c,) (s-e.) (s-e,)} 1 
s = * («). 
• ( 11 ) 
• ( 12 ) 
then, since s or 9 (u) is real and lies between e 3 and e 2 , s = e 3 and s = e 2 correspond to 
u = ft) 3 and u = w 2 respectively, if we take 9 f ( u ) to be positive* ; where «] and w 3 
denote the real and imaginary half-period and o > 2 = «i + w 3 . Thus 
f cos 6 d6 , f “ 2 /1 , „ ( \\ 7 
JO \Xi liS J' Jw 3 
= «" — >h)- • • 
For the evaluation of the other integrals, write 
a 2 +r 2 —2arfi 
9 (v) 
2ara 
* If we assume 9' (u) to be negative, then s = e 3 and s =e 2 correspond to u = w 2 and u 
respectively. But the same result will, as a matter of course, be obtained after integration. 
. (13) 
■ (14) 
= 2 OJj -j- (Uj 
