1910-11.] 
On Fourier’s Repeated Integral. 
569 
interval of integration is infinite. It will be at once seen that this 
extension is allowable, provided the conditions which Lebesgue postulates 
in a Unite interval hold in the infinite interval — that is, provided the 
absolute value of the integrand may be integrated over the infinite 
interval. 
In fact, if | g(x) | has an integral from q to infinity, so has g(x) sin Bx, 
since it lies between -f- 1 g(x) | and — | g(x) j ; thus we may write 
| \ q 9( x ) s i n Bxdx\ = \ g(x) sin Bx dx + \^g(x) sin B (x)dx | 
L I $g( x ) sin Bxdx\+ j*| g(x) | dx. 
Therefore, by the theorem of Riemann-Lebesgue, the interval ( q , Q) 
Lit | r q g( x ) sin Bxdx \ Z J" j g(x) | dx, 
being finite 
which, b}^ choosing Q sufficiently large, is smaller than any assigned 
positive quantity. Therefore 
L t g{x ) sin Bx dx — 0. 
B= 00 
Similarly, 
LI J g(x) cos Bx dx = 0. 
B=cc 
§ 11. Theorem 7. — If J™ |f(u)| du exists, then 
fdvj *f(u) sin uvdu = J ~du. 
Since f(u) sin uv lies between + \f(u)\ and — \f{u)\, it may be integrated 
between the limits q and infinity. Also 
sin uv du L \f(u) | du L J" \f(u) | du, 
which shows that the integral on the left is a bounded function of the 
ensemble (v, Q) in the whole region 
q L Q = 20 ? OZrZoo. 
Hence 
\filv Bt f(u) sin uvdu = L t J® dvf^fiu) sin uv du — L t du ^ f(u) sin uvdv 
Q= 
Q=c 
= r-Mdu- f" S&cosBu du, 
J q U J q U 
change of order of integration being allowable by the theorem already 
quoted in § 2. But, since i Z ^ , | f(u)/u | has an integral from q to oo . 
Hence, by the theorem of Riemann-Lebesgue, as pointed out in § 10, this last 
integral has zero as unique limit when B is indefinitely increased, whence 
the required result at once follows. 
