564 
Proceedings of the Royal Society of Edinburgh. [Sess. 
Theorem 1, bis .— The repeated integral 
Jo dv\ p _ p f(u) sin uv du , 
where p is any finite positive quantity , necessarily exists, and is equal to 
r -tMdu, 
J -p u 
where f-^u) = J(f(u) — f( — u)), provided only the latter integral exists. 
The proof is immediate, since the substitution of fi(u) for f(u) does 
not alter the repeated integral. In fact, 
s i n uvdu= - \ P _ p f( - u ) sin uv du, 
which proves the truth of the statement. 
Section 2. 
On the Fourier Cosine-integral, JJdvJpf(u) cos uvdu, in which the 
Limits of Integration p and q with respect to u are finite. 
§ 4. Theorem 2 . — The repeated integral 
\1 dv \lA u ) cos uv du, 
where p and q are finite and have the same sign, necessarily exists and 
is zero, provided only that f(u) is summable. 
To prove this theorem, we have merely to repeat the argument used 
in Section 1 to prove that change of order of integration between finite 
limits is allowable. We hence get for the value of our repeated integral 
L t I sin B u du. 
B — oo J V U 
But, since p and q have the same sign, f(u)/u is summable in the 
interval (p, q). Hence, by the theorem of Riemann-Lebesgue, the required 
result follows. 
Cor . — If p and q have opposite signs, the value or values of the limits 
represented by the repeated integral in question are independent of the 
particular fixed values we may choose to assign to p and q. 
§ 5. Theorem 3 . — If f(u) is a function of bounded variation in the 
interval ( — p, p), then 
[ dv f f(u) cos uv du=~ {/(•+ 0) +/( - 0) } = tt/ 2 ( + 0), 
J o J -p 2 
where f 2 (u) is defined to be J(f(u)-f f( — u)). 
It is clearly sufficient to prove the theorem on the hypothesis that f(u) 
