( 390 ) 



\ 

 t\x)— idaiy\}{x,y,a)f{y)dii (3) 



o a 



From (2) follows that the smallest value of p is two. If we take 

 e.g. q=zO, 0:= ^, g{a)=::sin^ It, then J/=:'l, iV<^l/jr and 



00 



ƒ 8171- (( 







If therefore /{x) is a continuous function in .v, for \vhich the 

 integral equation of the first kind 



b 

 ƒ(..) = fK{^x,y) h{y) dy 



has a continuous solution h{ij), then (3) holds if we put 



l/:7rv = ] ah 



If however we put q^=0,6=\, g {u) =isi}V a, then J/=:^V=1 

 and 



00 



da = 



a^ 



and therefore 



2 * 1 



i|^ = — ^ ^-v' 7v i'V) (f. {y) -T --^in {a I A-,''), 



whilst p = 3. 



It is easy to see that after a choice of q and rf we can always 

 suffice by taking 



it! 

 g [a) z= (.sm'«) 2 , 



or also, if r is an integer and 



1 + cf^r<2 + rf, 



{j(a) = sin'' «. 



Another example is 



^ = ~ ^ ~V~ , (/) =r 2) 



rr V— 1 a ' w' ; 



~ V H r 



which is found for q = and r) = 1, if we take 



