590 



u{ 



sinXTt f ƒ'(§) 



sin Xjz C 



x)r= 7 



jr J(.' 



1— / 



dl^ 





Z ^2n 

 2 



n/r ;< 





l 



of which the last integral can be brought bj^ partial integrating into 

 the form : 



V 



^(-1) 



©•■ 



/ r (n + 



;. + 1 



(2n+ A_l)(.^_|)2«+/-2 

 2n + Ï— 2 





1 



2;! 



-) (...-|)2«+>-2 



nl r\n -\- 



A+1 



+ 





Ar\f{^dl. \^{-\Y 



(.^. _ ^)2n+;.-2 



?// r n + 



A + 1\ ■ 2n + A — 2 ' 



b}' partial integrating the last part we find for it, if we put : 



/r-0 (..) = /ƒ (^)./5: 



ƒ' 



f— n (^\ ,1' ) 



/(-0(5)./5. ^(-1)'' 



^\2« 2«+; 3 



2 (-^) 



w / r n + 



A+ 1 



Summarizing we arrive at the following form for the solution 



of ila) : 



sin Xjtr f (5) 



sini.^r fa) 



u Lv) = I — c/c — 



sinXjr{2 — X)rr /;. + ! 



•T 



■:/■[ 



3—/ 



{,V-^) 2 J i_;|.(.t— 1)1 



(76) 



(^-§)2- 



'/(^J + 



/(-')(!), 





