1910-11.] Sommerfeld’s Form of Fourier’s Repeated Integrals. 593 
Now in the case when we have cosu(v — V), the last integral is 
numerically 
L f(q 4 - 0)2tt . 
using again the Second Theorem of the Mean, and remembering the fact 
that 
I f Q sin ku 
I J q U 
du L 7 r. 
Thus in this case we may, changing q to Q, write 
\l+c dv \ l ¥ ( u r v > t ) du = z > 
where 0 has zero as unique limit when the ensemble (c, Q) approaches 
(0, oo ) in any manner. 
But this is also true when we have sin u(v — Y), since in this case 
j exists by hypothesis, and we have 
f du[ f(u) sin u(v - Y )dv j = I / (o 
J q J V+C \ \ J q u 
Thus we have only to take 
2 — g- W-c)2#2 
os uc.— cos u( B' - Y)du I L 2 [ &^du. 
J a U 
zffMdu. 
J q U 
Hence, in either case, we have the inequality 
Jv+cH"? F ( M ’ V > t ) du - Z l!v+e dv lq' P (^ V ’ t) du \v+c dv lq A U > V ’ *) d U + Z. 
Now let c approach zero in such a way as to give an unique limit I for 
the central integral, and subsequently let Q move off to infinity in such a 
way that v, t)du, which has taken the place of the integral 
appearing in the extremes of the last inequality, has an unique limit J. 
Since at the end of this process 0 has vanished, we have 
I = J. 
But this is true whichever of the possible limits J was ; therefore all 
such limits coincide, and we have 
y^dv^Yiu, v, t)du — I, 
the integral on the left existing. This is, however, true whichever of the 
possible limits I was, so that all such limits coincide with I, and we have 
\™du\^vF{u, v , t)dv = \\dv\ q F(m, v, t)du, 
both these integrals existing. 
VOL. XXXI. 
38 
