1879.] Definite Integrals involving Elliptic Functions. 339 

 for, if u denote the left hand member of this equation, 



^=i E ^(, +2/ ) C? ,+[ E -f 9( ,-,)^ 



an Jo ay Jo ay 



dy J o dy J o cfy 



j o cfo r v ' J o dx 



=p(K + ^-0Q/)-0(K-^ + 0(-^) ; 

 = 0, 



so that u is independent of ?/ and the lemma is proved. Now from 

 (36), 



[ K log Q(x+y)dx + [ K log S(x-y)dx=2 [* log exdx+ZKlog By 

 Jo Jo Jo 



fK 



— 2K log 90 + log (1 — ft 2 sn 2 x sn 2 y) dx, 



and since Ox satisfies the conditions snpposed in the lemma, the left- 

 hand member of this eqnation is equal to the first term on the right- 

 hand side, so that we obtain the formula 



| K log (1-/j 2 sn 2 ,!' sn 2 w>7.i' = 2Klog— =Klog(^P)- 2Klo g" e U (M)> 

 Jo Qy \ w J 



write in the form 



(l_ a 2 sn 2 x )dx = Klog(^^-2Kloge(^n-^. 



which we may write in the form 

 "'k 



log 







In virtne of the formula 



sn + ?/) sn (x — y) 



sn- x — sn 4 y 

 1 — A; 2 sn 2 ro sn 2 y 



, . x f si — sn 2 x — sn 2 y + & 2 sn 2 x sn 2 y 



cn (x+y) cn (x-y) = — 



I — # sn- x sn- z/ 



a , , n t / n 1 — J$ sn 2 re — 7j 2 sn 2 7/ + A' 2 sn 2 d 1 sn 2 ?/ 



dn (a; + y) dn (x-y)= — _ J 



1 — Jcr sn- a; sn~ y 



we can deduce by means of (44), since sn 2 #, cn 2 », and duo; each 

 satisfy the conditions of the lemma, that 



j *J log (sn 2 x - snhjY~dx = — JttK' + K log - 2K log 9y . (45) , 



log (1 — sn 2 x — sn 2 y + 7j 2 sn 2 x sn 2 ?/) 2 d?j 



= -i 7 rK / + Klog^?^-2Klog0^/ . (46), 



