566 
Proceedings of the Royal Society of Edinburgh. [Sess. 
by the theorem of Riemann - Lebesgue, since 
f 2 (u)-C 
u 
is a summable 
function, by hypothesis, the second integral on the right has zero for 
unique limit, which proves the theorem. 
Note. — If as u approaches zero ,f 2 (u) has an unique limit — in other 
words, if / 2 ( + 0) exists, — it is evident that the only possible value for 
C is/ 2 ( + 0). 
That the expression may still have a Lebesgue integral without this 
occurring is evident, if we reflect that the integrability of (ffuj — ty/u, 
when f 2 (u) has C as unique limit, will not be affected, if, for example, at a 
countable set of points having u = 0 as limiting point, we ascribe any value 
whatever to f 2 (u). 
§ 7. Theorem 5. — If in the interval (0, p), f(u) can be expressed as 
the product of two functions, one of which, g(u), has bounded variation, 
and the other, h(u), is such that 
h(u) - C 
has a Lebesgue integral in that interval, then 
j dvj f{ii) cos uvdu— ^-C{< 7 ( + 0)}. 
For, as before, we may without loss of generality suppose g(u) to be 
monotone decreasing. Also, since f(u) is summable, and therefore change of 
order of integration between finite limits is allowable, 
[ dv [ g(u)h(u) cos uv du — It [ h{u s * n ^ u du 
Jo Jo' B=°oJ 0 U 
= C U P. g (") smB “ tfa+ U G ;/(a) sin B u du. 
B=ooJ 0 U B=00 J 0 U 
But since (h(u) — C )/u is summable and g(u) is bounded, their product is 
summable, whence, by the theorem of Riemann-Lebesgue, the last integral 
has the unique limit zero. Further, since g(u) is monotone decreasing, it 
follows, as in the proof of Theorem 3, by the Second Theorem of the Mean, 
that the first of the two integrals on the right of the preceding equality has 
the unique limit ^g( + 0). 
Cor. — If a similar statement is also true for the interval ( — p, 0), 
then, C' being the corresponding constant, 
f dv I f{u) cos uv du — — {C g{ + 0) - C'g( - 0)}. 
J o J -p 2 
