598 
Proceedings of the Royal Society of Edinburgh. [Sess. 
and therefore has, as b increases, the limit zero, so that 
lo dv^e~ kv2f g(u) sin u ( v ~ Y)du 
exists. Hence 
L t f dv f e~ kv2t g(u) sin u(v - Y)du = L t Lt [ di$^ [ e~ kvn u sin u(v - Y )dv 
t=oJ 0 J q t = 0 B=oo J q U J 0 
— Ltf duf^- f e~ kv2t u sin u{y - V)dv, 
t=0j q U J 0 
g(u)/u being positive and summable in ( q , oo ). But this last expression 
may be written 
f du^^\d[ cos uV +2 [ ve~ v2 cos u(—=- - Y W-iA 
J q U <=0 \ J 0 \Jkt ) 1 5 
by the same reasoning, using the value of the integral already found, and 
putting B = oo . In fact is summable in (g, oo ), and the remaining 
factor is a bounded function of ( u , t). 
Now, the integral with respect to v on the right of the preceding 
equation is a bounded function of (u, t), and, except for the value zero of u , 
has the unique limit zero for each fixed value of n when t diminishes to 
zero, by the extended theorem of Riemann-Lebesgue. Hence, when we 
multiply this limit by the bounded function g(u)/u , and integrate from q 
to infinity, the result is zero. Thus the last equation becomes 
L£ [ dv [ e m g(u) sin u(v - Y )du = f cos uV du 
t=oJ 0 J q J q U 
Similarly, changing Y into — Y, 
( 1 ) 
Lt f dv f e kvn g(u) sin u(v + Y )du = f cos uY du . (2) 
t=zoJ 0 J q J q U 
Subtracting (1) and (2), the required result follows. 
§ 11. Theorem. — If f(u) is expressible as the product of two factors , one 
of which, g(u), is monotone decreasing with zero as limit at infinity , while 
the other is of the form sin uY, then 
Ltf dv f e~ kv2t f(u ) sin uv du= f du , 
both these integrals necessarily existing. 
For, subtracting the equations (3) and (4) of § 9, this result follows. 
§ 12. Theorem. — If f(u) is expressible as the product of two factors, one 
of which, g(u), is monotone decreasing with zero as limit at infinity, while 
