72 Proceedings of the Royal Irish Academy. 



where (p [x] = f(x) dx, and G (1) =j= 0. From this we obtain 



* {T) = ^i 



xav(a)/G(l) x 



da + - — ■ 



H* 



G\t) 

 1) 



dt 



ilf% 



1 , yg(y) , 



row y 



i-f'*.£i* 



rfa. 



o ©(1) 



Here, on account of the nature of <j> (%), G and D must lie to the right of the 

 axis of imaginary quantities, and from consideration of the second integral 

 we see that D must also lie to the left of the line ij = 1 + y, supposing that 

 g{x) is of the form x y p(x), p(x) being always finite for 0<x<a. (We put 

 as before a = >j + id.) Integrating by parts in the denominator, we get 



/(*) 



d(j> 



dx 27rijc 



a;a-l v („) 1 



f i K -^—da + — 



■ t«-W{t)dt 



J 



d 



2tti dx 



r * a \y- a ff (y) dy 



D a ^t«-i G(t) dt 



da. 



In this expression we may write a + 1 instead of a, supposing at the same 

 time that C lies to the right of the line t\ = — 1, and .D between the lines 

 tj = - 1 and ij = y ; then we have 



a,«+l fV«-l'3f(y) dy 



J 





rfa. 



In general, if we have G (1) = G'(l) = . . . = £<--'> (1) = 0, and 0W(1) + 



after repeated integrations by parts our equation becomes 



<?(") (1) (x) = (- 1)" x™g (if) + f 1 (?(" +l ) (0 4. (to) <ft, 

 where 4) (*) = J ... J f (?) *c> 



the integrations being /i + 1 in number. From this we obtain 



*w-is 



v (a) 



ct?('0(l)- t«G(»+i)it)dt 

 1 



(-l)»ar« j'y-«+»0(y)(fy 



; Jtf£(»)(l) -f 1 *«<?(«+!)(*)<& 



J 



rfa 



rf«, 



