565 
1910-11.] On Fourier’s Repeated Integral. 
is zero when u is negative, and when u is positive it is a monotone decreas- 
ing function of u. 
Now 
f dv f f(u) cos uv du= f du f f(u) cos uv dv = ( sin B u du 
J o J o* J o J 6 J o u 
(i) 
But f(u/B) is a monotone function of u, and converges boundedly to/( + 0), 
as B increases indefinitely, and sin has a Harnack integral in the 
interval (0, oo ). 
Hence * 
fy(u\ S inu du 
J o' \ B / u 
has, as Q and B approach infinity in any manner whatever, an unique 
double limit, which is, therefore, that got by letting first B and afterwards 
Q increase without limit. Thus 
T , f Bp sf u \ sin u j r ,, , AN sinw, tt , v ax 
U /(t> du= /( + °) du=-f( + 0). 
B=oo J 0 \ B / u Jo u 2 
This is therefore also the limit of the integral on the extreme left of (1), 
which proves the theorem. 
§ 6. Theorem 4. — If for some value of the constant C 
f^ u )^ _ 9 9 where ffu) = J {/(w) +/( - u ) } , 
has a Lebesgue integral in the interval ( ~ p, p), then 
Jo do \ P - p f( u ) cos uv du = 7tC. 
Evidently we may replace f(u) by ffu) without altering the value of 
the integral. Also, since {ffu) — C)/u is summable, so is ffu), in the 
interval considered, and therefore since cos uv is bounded, we may, as usual, 
change the order of integration over a finite rectangle. Thus, as before, 
we have 
\ [ dv f fo(u) cos uv du = L t {flu) - * n du 
Jo J —p B—coJ 0 “ u 
n T . [ p sin B u -j T , f p fJu) - C . -d , 
= OL t du + Id / ^L2W gin g U ' 
B=a ,J 0 . U B=xJ 0 u 
Now, the first of the two integrals on the right is equal to / 
J 
b p sm u 
du, 
and has therefore ^ for unique limit when B increases indefinitely. Also, 
* For the general form of the theorem here used see W. H. Young, “ On a Theorem in 
the Harnack Integration of Series” (1910), § 3, Mess. Math., p. 105. 
