338 
NOTE ON LANDEN’S THEOREM. 
[768 
leading to 
/1 - k 2 x 2 7 C . /(1 - k') 2 - t 2 /(1 + kj - 
V 1 - x 2 dx ~ 1 2 + 4 V (1 + kj - t 2 + V (1 - k') 2 - 
-t 2 
(1 - ky - t 2 
dt. 
The form in which the transformation is usually employed (see my 
Functions, pp. 177, 178) is 
leading to 
where 
y = (1 + k')x 
(1 + k') dx 
j 1 — x 2 
v 1 —k 2 cc 
k 2 a? * 
dy 
Vl — a?. 1 — k 2 x 2 V1 — y 2 .1 — \ 2 y 2 ’ 
1 -k' 
X = 
1 + AT 
If, to identify the two forms, we write y = ——p and in the last 
introduce t in place of y, the last equation becomes 
dx dt 
Vl — ¿r 2 .1 — k 2 x 2 V{(1 — ky — t 2 } {(1 + k') 2 — t 2 ) 
Comparing with Landen s form, in order that the two may be identical, 
have 
1 — k 2 x 
« — k +1 A A 1 -- - r, lA /(1 + vy - n 
r 4 V (1 + ky — t 2 4 V (1 -ky-t?) 
viz., this is 
that is, 
x V(1 — k') 2 — t 2 V(1 + k') 2 — t 2 , 
1 _ k 2 x 2 = i {V(l - k'f - t 2 + V(1 + k') 2 -1 2 } 2 , 
1 - k 2 x 2 = i [1 + k' 2 -t 2 + V{(1 - A;') 2 - t 2 ) {(1 + k') 2 -1 2 }], 
where the function under the radical sign is 
(1 — k' 2 ) 2 — 2(1 + k' 2 ) t 2 + t l (=T suppose); 
and this must consequently be a form of the original integral equation 
t=fe v / rr^- 
In fact, squaring and solving in regard to x 2 with 
we have 
,_k 2 + t 2 — V2 7 
Elliptic 
equation 
we must 
the assumed sign of the radical,