1879.] Definite Integrals involving Elliptic Functions. 



we find, by means of the formula 



2 sn x cn x dn x 

 sn zx= , 



l_#>sn 4 a> 



th at j* log (1 - ft 2 sn 4 aj) da = - J a-K' + K log 



Similarly 



335 



(16). 



log cn 



J-logcn 2 2xdx— 



) 



Ck Ck 



log dn 2xdx = log dn xdx, 

 Jo Jo 



whence, from the formulae 



1-2 sn 2 a3 + & 2 sn 4 a5 

 cn za? = , 



■sn^x 



, l-2& 2 sn 2 a; + k 2 sn 4 a; 

 dn 2a?= , 



l-^SIL^X 



by the use of (16), we find 



J K J log (1 - 2 sn 2 a; + W sn 4 a3) 2 (fo = - JfrK' + K log . 



j K log(l-2^ 2 sn 2 ^ + ^ 2 sn 4 aj)^ =-j7rK / + Klog^^!^ . 



These two integrals may also be put respectively in the forms 

 f K 



Jlog (cn 2 a? — sn 2 « dn 2 «) 2 ^a3 (19) 



Jo 



(17) 



(18), 



Ck 



log (dn 2 x — W sn 2 x cn 2 x) dx 



o 



(20) 



the former expression is written 1 log ( ) 2 , as the quantity in brackets 

 is negative from x—^K. to $=K. 

 § 7. Using the formulae 



1 — cn 2a?=2sn 2 x dn 2 x — & 2 sn 4 «), 



1 + cn 2a?= 2cn 2 a? (1 -ft 2 sn 4 a?), 



1 - dn 2x = %W sn 2 x cn 2 a? ( 1 - W- sn 4 a?) , 



1 + dn 2a; = 2dn 2 x -~(l—k % sn 4 x), 



cn 2 a3=(dn2a3 + cn2a3) H-(l + dn2af), 



dn 2 a?= (A/ 2 + dn 2a3 + k 2 cn 2a?) + (1 + dn 2x) , 

 and, observing that 



0(2a3)c?a3=i (p(x)dx, 

 o Jo 



