338 Mr. J. W. L. Glaislier on [Nov. 20, 



putting for 00 its value, viz. ( — - — J , this gives 



logexdx=^K , +iKlo^^^ .... (38). 



K 



10 



Also Ha?= ^/h . sn xQx, 



and integrating the logarithm of this equation, using (5) and (38), 

 we have 



£ log U. /.,= -i^K' + iK log (^5). 



Replace x, y in (36) by mx, nx and the formula becomes 



Q(m-\-n)xQ(m—n)x — Q yl, O Q ( nx l q _£2 sn 2 ma , sn 2 wa .) 



whence, since (p being any integer) 



f K f K 



log (px)dx = log Qxdx, 



Jo Jo 



we have 



j K log(l-/ l : 2 sn 2 ??z^sn 2 7^)^ = 2^Klog0O- K \ogOxdx^ . (39), 

 if m and n be different integers, and 



= 3^Klog0O— ^logexdx^ . (40), 



if m and n be equal integers. 

 Thus we find 



| K log (l-^sn^m^sn 3 nx)dx= - ^K'+fKlog^) . (41). 

 if m and n be different integers, and 



= -irfC' + Klog(^') . (42), 



if m and » be equal integers. 

 The relation 



log (1 — h 2 sn^mx sn 2 nx)dx=%\ log (1 — Jc 2 sn 4: mx)dx . . (43) 



Jo . Jo 



is curious. The value of the latter integral is of course immediately 

 derivable from (16). 



§ 10. (Lemma.) If 0(#) be an even function of x, and if 0(aj + 2K) 

 = 0(a;), then 



fK (*K fK 



0(# +v)cfcy-|- 0(# — y}dx=2\ (/>(x)dx t 

 Jo Jo Jo 



