572 
Proceedings of the Royal Society of Edinburgh. [Sess. 
Changing k into — k, the right-hand side remains unaltered. Hence also 
[ dv [ g(u) sin u(y - k)du = I ftW cos k u du . 
Jo it J q u 
Adding these two results, we get the above theorem ; subtracting them we 
get a result to be stated in its proper place (Section 4, § 24). 
Con. — When the upper limit of integration with respect to v is finite , 
not infinite, we may reverse the order of integration. 
For, as above, 
j>J>) sin u{y + k)du = JJ +k dv\ q g(u) sin uv du - dv\ q g(u) sin uv du, 
and in each of the integrals on the right we may change the order of 
integration, as shown in the course of the proof of Theorem 8. Similarly 
we may change the order of integration, in dv\ q g(u) sin u(v — k)du, and 
therefore also in dv\ q g(u) cos ku sin uv du, which is half the sum of 
these two integrals. 
§ 14. Theorem 10 . — If in the interval (q, go ) the function f(u) can be 
expressed as the product of two functions, one of which, g(u), is monotone 
decreasing with zero as limit at infinity, and the other, h(u), is of the form 
sin ku, then, 
[ dv f f(u) sin uv du = ( — — ^ du, 
J 0 J q ‘ J q U 
both integrals necessarily existing. 
For, by a theorem proved in the next section (Section 4, § 23), 
J 0 dv\ q g(u) cos uv du = 0, 
and we may change the order of integration, writing 
lo d v \*g(u) cos uv du = \*du^g(u) cos uv du, 
these integrals existing. Hence, proceeding as in the preceding proof, we 
have 
POO roo r 30 POO P£ POO 
J 0 dv\ q g(u) cos u(y + k)du = J k dv J g(u) cos uvdu= - J 0 dv j q g(u) cos uv du 
r j r / \ 7 rgM s i n ku, 
= - I du q(u) cos uv dv= - ± du. 
J q J 0 ' W J q U 
Changing k into — k, we have, 
r / \ , 7W ["gM sin ku, 
/ dv I g(u) cos u(v - k)du = | du. 
J 0 J q J q U 
Subtracting these two results, and dividing by 2, our theorem follows ; 
adding them, we get the corresponding result to be stated later (Section 4, 
§24). 
