ARISING FROM THE LOAD OF NEIGHBOURING OCEANIC TIDES. 
41 
the value of v given in (i.) of (18), observe that the integral on the left-hand side of 
(15) and the function Vr] X —w^{y) change their values continuously as 6 varies from 
0 to 1, while m remains unchanged in this variation. In the limit as 6 -> 0, the value 
of the integral is nil and 
2'Vt] l — 2oo 1 £(v) — 2't I'-iri, 
and therefore we have 
(i.) m — — v!. 
Similarly for the value of v given in (ii.) of (18), proceeding to the limit 0 -> 0, we find 
(ii.) m — —(n'+ l). 
The value of 9' (v) will be obtained from 
P' 2 ( v ) = 4 [> M-eJ • 0 (v)-e a ] . \_9 {v)-e z \ 
In the present case we have 
„/ / \ , -z(a 2 — r 2 ) 
V (v) = +i — - L 
x ~ 2 ar 
It may easily be shown that the value of 9' (v) is a positive imaginary quantity for 
the value of v given in (i.) of (18) and a negative imaginary quantity for (ii.). 
Therefore we have to take 
for 
and 
for 
where 
9' (v ) = +i 
(a 2 ~r 2 ) 
2 ar 
m = —n , 
v — (2 n T 1 ) Wi T (2 ft! + @) cog \ 
(19) 
9'(v) =-i 
K-f 2 )! 
9 
ar 
m = — (n'+ l), 
v — (2n T1 ) co x T (2 n !T 2 — 6 ) wg j 
0 < 0 < 1. 
( 20 ) 
Lastly, one more integral which can be evaluated without the knowledge of elliptic 
functions is 
r a—r cos 0 
d6 = - for r < a, 
K 
2 • 
a 
= 0 „ r > a. 
( 21 ) 
