198 Mr. A. Cayley on an Analytical Theorem connected with 

 and the integral is 



l-/ti L v/i2 + 2(l-/ii)(l + 6) bJ 

 For the second integral we have 



G=b^ + 2b, L=b^ + 2{l-fi){l + b), 



H = 26(1 + 6)0, M = {l-fji){2 + b){l + b)0, 

 N=2(l-/i)(l+i)202; 

 and thence 



LN-M2=(l-yLi)(l +^)ft«(l +6)90^ 



MG-LH = -(l+/i) 63(1 + ^)0, 

 L + 2M + N = 62 + 2(1-/a)(1+6){(1+0)2 + 6(0 + 02)}, 



L =62 + 2(1-/16) (1 + 6); 



and the value of the integral is 



?^ r 1 



(l + 6)(l-^)0 L Vj2+3(l_^)(l + 4)((l + 0f +S(0 + 0')) 



' \. 



'v/6H2(l-/i)(l+6)J 

 For the third integral, 



G=6^ L=6^ 



H=26(l+^)0, M = (l-/t)6(l + 6)0, 

 N = 2(1 -/i) (1 + 6)^02, 

 and thence 



LN-M2= (l-ft)(l +fi)b%l + b)W, 



MG-LH=-(l+/*)63(l + 6)0, 

 L+2M + N = 62 + 2(1 -/*)(! + 6)(02 + 6(0 + 02)), 



L = 62; 



and the value of the integral is 



^ I 1 11 



(l + 6)(l-/i)0 L^52^2(l-;i)(l+6)(02 + 6(0 + 02)) *J* 



Hence the expression for y is 



r 1 n 



L \/62 + 2(l+uUl + 6) AJ' 



y^ -A62 log(]+A) /- 1 



(i-/t)(l+6) A L\/62 + 2(l+yu,)(l + 6) 

 _ hb^ log(l+A) 



-{l-f,){l + b) A 

 1 



v/62 + 2(l-/ti)(l + 6)((l+0)«+6(0 + 02)) ^/62+2(l 



-M(1+6)J' 



