70 



Proceedings of the Royal Irish Academy. 



Hence | R(x) - R n {x) | may be made less than t"x*, t" being arbitrarily 

 small. But we have 



R n {x) 



2iri 



D 



X a\ y-l-a + p Vn (y) - \ i K{t)r, n {ty)dt dy 



J L ^_0 J _ 



1 - [ t«K(t)dt 



J 



da 



1 



2wi 



W: 



t K(t)dt 

 "*o(l-f' t°K(t)dt 



p— a ( i'p + 1 \ p-a + l-i 



&, ( 1 - t K(t)dt )a 



+ 



a + 1 



b n 1 1 - t K(t)dt)a 



+ . . . 4 



l p - a + n 



Now it is easy to show that the integral 



da I 



da. 



taken on that part of the infinite circle which lies to the right of D is equal 

 to zero, and hence that the integrals 



(p-a) 



-[' t"K{t)dt 



J 



I da | , . . . 



(p - a + n) 



t a K{t)dt 



da, 



taken on the same part, are also equal to zero for x < a. 



Hence in the expression for R„(x) we may substitute for D the closed 

 contour consisting of I) and this part of the infinite circle. The function to 

 be integrated has poles in the interior of this contour, namely, the points 



a = p, p + 1, . . . p + n, 



and the roots of the equation 



1 - 



t a K(t)dt = 0. 



We can take p such that none of the first set of poles coincides with the 



