where 
1 
a, == 4h (i) at 
0 
Ano Ors == DS. le, 
1 1 
shan 2 | h (t) cos 2 nt dt, n= 2 fn (4) sin Zar nt dt 
0 0 
Now the functions cos 2% nt and sin 2a nt are for any value of n 
expansible in power-series 
oo oo 
sae ' zis 
cos 2 nt = N pap, OF, sit avant — XN u Bu t*, 
0 0 
converging uniformly in the interval (0,1). Sinee A(t) is limited 
in that interval we may use the following reduction 
1 1 
| h(t) cos 2 nt dt = [h(t) Np A, te dt = Se? A, 4 h (t) tr dt 
Pd Pd 
0 0 0 0 0 
and in a similar manner we find 
1 1 
t) sin 20 nt dt = Nz B, A(t) te de 
[resin nn Di „no C 
0 
0 0 
© co 1 
Hence by (13) all coetficients in the expansion of Fourier are 
zero, and therefore h(t) is identically zero in the interval (0,1). 
Since, further, g (1) = h'(@, the same thing holds for g (4), and since 
(jt =g'( (except at t= 0), the generating function f(t) itself is 
zero in the interval (0,1). This is the second part of Lrrcn’s theorem. 
Since the first part follows immediately from the second, the theorem 
has been proved in the particular case that the arithmetical progres- 
sion of zeros of a(w) has 1 for its common difference. 
If this difference is equal to the positive number 7 and if, therefore, 
the zeros are given by formula (2), we substitute 
ii =o) C= yy. ¢ = He 
by which the integral passes into 
1 5 5 Et é 
„fre is ds = [roe Mites sting 41-401) 
yf 
0 0 
1 Sia Vg ta 
The function — „(5 Jer “— p(s) has the properties 1 and 2 
q 
of § 2, so that the foregoing arguments may be applied to it. The 
