584 Proceedings of the Royal Society of Edinburgh. [Sess. 
meaning and remains true when we write infinity for Q. Replacing 
sin Bu 
u 
by its integral form, this gives us 
^du\_g(u){h(u) - s m } cos uv dv\ — % \^du\_g{u){a n cos nu -f b n sin nu }J® cos uv dv\ (4) 
both sides accordingly existing and having equal values. 
f x o(uJ 
Now, since j ^-f^du exists, the corresponding terms of the infinite 
series on the right-hand sides of (3) and (4) have been already proved (§ 24) 
to be equal, and therefore the same is true of the left-hand sides. 
But, again, for any integral value of n less than m it is equally true, 
and has been showm in the same theorems, that 
cos nu 
r 
J duj g(u) 
cos , [ B 7 r / x 
cos uv . dv = dv g{u) 
0 Jo 
cos uv . du, 
q jo sin nu Jo J q ' sin nu 
so that both sides exist and are equal. 
Adding all such equations for integral values of n less than m, after 
multiplying by suitable coefficients, we have 
I<z du^g(u)s m cos uv dv = J*^ dv^ g(u)s m cos uv du. 
Hence, expressing the fact, already proved, that the left-hand sides of 
(3) and (4) are equal, and adding the equation so obtained to that last 
written down, we have 
J dv j f(u ) cos uv du = J du J f(u) sin uv dv = j 
sin B u du. 
q u 
Proceeding to the limit with B, we get, in this case, the required result, 
using the theorem of Riemann-Lebesgue (§ 10). 
Next, let \ a 0 be the constant term in the Fourier series of h(u). Then, 
by what has just been proved, 
Jo dv J g(u){h(u) — ^a 0 } cos uv du = 0. 
But, by Theorem 14, 
|-a 0 | 0 dv\ q g(u) cos uv du = 0. 
Adding these two equations, the required result follows. 
/ °°£r(h) 
Cor. 1. — If h(u) is an even function , the condition that / ^-Aiu 
J q U 
should exist, may be omitted. 
For the coefficients b n are then identically zero, therefore Theorem 15 
is then not used, while Theorem 16 does not require the above condition. 
Cor. 2. — If h(u) is a bounded function, the condition that 
J q 
f(u) 
Idu 
exists need not be mentioned in the enunciation. 
