BOUNDARY PROBLEMS AND DEVELOPMENTS. 115 



/. s2?(*)- 



Writing 



X 



Jv= - Jy(x) (8*j) Z(t) R(t) F(t)-dt 



a 



we have from (98) 



S£?(*)- =i J/i-dX. 



From this it is immediately seen that 



(105) S ( i } (a)-=0. 



Consequently it is necessary to consider further only the case x =£ a. 

 We have 



X 



\ ^ fe,m,p,g=l 



{5 mp + l/X[^ mp «)]}5 Pfl 7 fl (0 /«(0 dtyj, 

 i.e. ' 



(106) Jv=~(f e^W-W )+*&>-*& S * m (0 /,(*) <ft) + 



a 



a; 



L/\( f £ ^|r^)-r^)| + «^)-^(O4 7p(0/pWfop ^ (x) + 



W A.p-l 



1/ 



8iktkp(t)]dt). 



For X on any arc C M „, however, each term under the sign of summation 

 in the last matrix on the right of this expression is seen to be of the 

 type <ps(t, X). Their sum therefore is of the same type, and by lemma 

 2 the integral is of type ^(X). Consequently we have, upon integrat- 

 ing by parts the elements in the first matrix on the right of (106), 15 



Jv - £(- 4 | ~fi(x -0)+e "W&Ma + 0) + 



X 



I eAVi(x) ~ Vi(t) { \u t)eB,{x) ~ Biil) \ dt \ ) +X- 



a 



15 Recall that r<(a) = £,-(a) = 0. 



