590 Proceedings of the Royal Society of Edinburgh. [Sess. 
we write n for 1/ J4>kt, and so get, omitting for the present the factor Jir, 
frAu/n)*-*du, 
as the expression whose limit when n increases indefinitely is sought. 
Integrating by parts, and denoting \^f(u)du by F(u), this becomes 
[ nF{ujn)e “ 2 ]^ + 2 J | -(F (u/n) | u 2 e u2 du 
(3) 
Now the quantity in square brackets is zero at the inferior limit of 
integration, and at the superior limit is nF(p)e~ p2n2 , which approaches zero 
as limit when n increases indefinitely. Thus we have only to retain the 
integral in (3), the first factor of the integrand of which has by hypothesis 
the limit /( 0), and, moreover, approaches this limit boundedly, since F(x)/x 
is a continuous function with an unique limit at x = 0. Hence we can 
again use the last theorem referred to in the proof of the preceding 
theorem, since 
[ 2 u 2 e~ ll2 du — f ( - u)d(e~ u2 ) — [we~ M2 ]“ + [ e~ u2 du = ^ 7r . 
Jo Jo Jo 2 
We thus get for (3) the limit 0), which, replacing the omitted factor, 
proves the required result. 
§ 4. Theorem. — If f(u)/u is summable in the interval (p, q), where 
0 L p<q, then 
Ft 
t = 0 
/>/: 
e kv 2 t f(u) sin uv du = I ALA 
rm du . 
Jp u 
For since e~ kvH u sin nv is a bounded function of ( u , v), and f(u)/u is 
summable, 
dv\f e~ kv 2 t f(u) sin uv du = J® du fiu)^ e~ kvH sin uv dv. 
Also, since | e~ kv2t u sin nv dv \ Lqffe~ kvH dv is abounded function of (B, n) 
in the rectangle 0 Z B Z oo , p Ln Lq, and f(u)/u is a summable function, we 
have as in § 2, since \”e~ kv2t u sin uv dv exists, 
Now 
[ dv f e kv 2 t f(u) sin uv du = f du^^ I e kvH u sin uv dv . . ( 1 ) 
J 0 J p J p u J 0 
J* e~ kvH u sin uv dv = - [e kv2t cos uv ]“ + J^ 2 ktv e kvH cos uv dv 
U/U 7 
v e v cos- dv, 
v rtt 
= 1 + 
= 1+2 \^ve ~ v2 cos B'r dv. 
