1910-11.] Sommerfelcl’s Form of Fourier’s Repeated Integrals. 595 
But the left-hand side is independent of Q and, letting Q move off to 
infinity, the right-hand side vanishes. Therefore all the limits on the left 
coincide, and we get the required equation, 
£o!o v > fy* M =o* 
§ 8. Theorem. — If in the interval (q, cc ) the function f(u) is monotone 
decreasing with zero as limit at infinity , then 
e kv2 f(u ) sin uv du 
If 
f(u) 
du, 
r oo -co 
Lt dv\ 
t=0 J 0 J q 
provided J ~“du exists * 
By the lemma of § 6, putting V = 0, we have 
JT dv\ q e~ kvH f(u) sin uv du — f* du\* e~ kvH f(u) sin uv dv 
CduMf 
J q U J I 
e~ kvH u sin uv dv. 
Now 
Jo e kv2t u sin uv < 
iv = - [e 
_ 1 _ cog j>, u + 2 
cos + 2 kt^v cos e kv2t dv 
r 
UV 
v cos — v dv 
Jht 
(1) 
and is therefore numerically less than 2 + 2 ^ve~ v 2 dv, i.e. less than 3 
always. 
Similarly 
J^e~ kv2t u sin uv dv< 3e~ km , 
so that \™e~ kvH u sin uv dv exists, and \*e~ kvH u sin uv dv converges boundedly 
to it as limit when B increases indefinitely. 
Hence also J 0 dv\ q e~ kv 2 f(u) sin uv du also exists, for by the same reason- 
ing as the above \ldvf* q e~ kvn f(u) sin uv du is numerically less than 
3 e~ km J &^-du, and therefore vanishes when B increases indefinitely. 
Thus we may write 
f dv f e~ kv2 f(u) sin uv du—^ ^ f du^^—i f e kv2t u sin uvdv 
'=°7o J q u Jo 
= Ltf du M f e~ kvH u sin uv dv, 
t=<>J q U Jo 
since f(u)/u has an integral from ( q , oo ), and the integral multiplying it 
* This theorem, and that of § 15, can also be proved even more simply by the method 
of § 14. 
