1888.] A. Mukhopadli j.iy — On Fois.'inn'fs Integral. 105 



. e /TTe , u 



Taking the logarithmic differential, we have 



dO du 



sin 6 sin to 

 Therefore 



n-1 

 du 



(13) 



(sin u\^ /sin u\' 



(1 — ecos^A^'' 7/j_ /I — e cos n\^~^ 

 / \/ir^^ V ,n /l-ecos7yA^-l 



I = r 1(1 — 6 cos n, ) 



du. 

 Again, since iTom (14) we have 



(1 + e cos ^r - e^)"" 

 we have from (12), 



^t(1 "" 6 cos uY^ du, 



r sin^ 6 dO 1 f sin^ u du 



J(H-e cos ^r" (v/r=l2)2^-2^-lJ (l-ecos7.)2'-^ + l 



whicli is really equivalent to Poisson's transformation; and it is clear 

 from (12) that for 6 = Tr, 6 = 0, we have u = tt, u = 0. Thus, putting 



_ -2a 



l-a2\2 



1-f 

 we have 



J'^'" sinP 6 dO 1 r^ sin^ n du 



o (l-2acosf? + aT~ (1 - a?^^'^"^"^ Jo (H-2acos r^. + a^j^'-^ + l 



or putting u — 'r: — z 



this may be at once written 



J^"" sinP X doc 1 C^ S"^^ ^ ^^ 



o (l-2acos^+aT'"(l-«2)2^-^^-'Jo (1 -2acos a; + a2)P-^+l 

 as the variable is of no consequence in a definite integral. 

 14 



