74 



Proceedings of the Royal Irish Academy. 



dividing by G (1), which we suppose not equal to zero, we obtain an equation 

 of the form 



<t> (x) -\<t> (fix) = 4> (x) + 1 ' K(t) <j> (tx) dt, 

 where A is a constant. A solution of this equation is given by the formula 



1 ' x a \° y l - a ip(y)dy 



* « = 2.1 



d 1 - A M * - j ' t'K(t)dt 



This is evident by the method we have formerly used, if the convergence of 

 the integral can be proved. Putting 



U(a) 



t«K{t)dt 



we get 

 where 



and 



1 - Aju" 



4>(x) = r (x) + Zl(%) + ■ • • + Zm-l(z) + $m{x), 



W = 9 



!7rt 



*«[t7<«)] r 



D 



/-I — a 



#y>fc 



* m(li:) _ 2Vt 



1 - Aju" 

 *■ [27(a)]" f' r !-««# 



fl!a, 



1 - V 



i) 





^O. 



We must see that D does not pass through any root of the equations 

 1 - A/«« = 0, 1 - V - I t a K(t)dt = 0. 



J M 



We can prove exactly as before that for any m ^ a certain N, <j> m (x) has 

 a meaning. 



The integrals £ r (.r) demand some discussion. Let us begin with the 

 case r = 0, i.e., with the integral 



Ux) = 



Lt 



x a 



y 



•l-a 



Ky) d y 



2ni D n = x J9„ 



1 -Aft" 



da. 



We have | \P(x) \ < Mx y ; let us suppose that | X/x y \ < 1 ; then we can 

 take the line D left of the line t) = y, so as to have on D the inequality 



| A/Le- | < 1. 



