1255 
(ee) 1 gnl 
0 
fu. (ae) J, (a da) = io 
0 
(NIELSEN p. 200), and for z=0O 
ri 
f: J, (az) J, (a) da = | J,(a) da =1. 
0 0 
8. Now we will apply our expansion to the problem of the 
momenta. In this problem the question is to determine the function 
fy) from the integral equation 
Gn = | f(y) y" dy. 
J 
where a, is a funetion which is given for all positive integral 
values of n. 
Putting 
[We VO (y) 
we obtain 
Cn = J ey" G (y) dy. 
0 
Supposing @(y) to be a function which satisfies the conditions of 
DirIicHLeT, we have 
6 (y) = b, == by fy (y) a b, Ps (4) sie NS 
SO 
ke) 
On 2 by fe yr pe (y) dy. 
0 
Now, this integral has the value zero, when p > n, therefore 
wD 
Ga Pl fen y” Pp (y) dy 
Moreover, according to the equation (7) 
n 
an =n! & (—1)r by at 
0 
so, with (10) 
ob 
a D ; oO b, pers = 
SW) = 4 2b, 9, (y= =~ fo aw Ly (2 Vary) de. 
uv Pp: 
82* 
