571 
1910-11.] On Fourier’s Repeated Integral. 
which occurs in both the extreme members of this inequality has an unique 
limit, say J. We thus get, since g( Q) has the unique limit zero, 
I = J. 
But this is true whichever of the possible limits J is. Therefore there is 
only one such limit — that is, JJdufo F{u, v)dv exists, and is equal to I. 
Since this is true whichever of the possible limits I is, this proves that 
there is only one such limit — that is, Jo dv\™¥(u, v)du exists, and is equal 
to JJduJo F(u, v)dv. 
Thus, as a preliminary step, we have proved that we can reverse the 
order of integration in JJ 5 dvj 3 F(u, v)du, both the repeated integrals in 
question in fact existing and being equal. 
Now 
r / b xr/ w r -nr \i-cosBw, r f( u )j f°/w 
I du] F(u,v)dv= f(u) du = 'L\-Ldu— I ^-^cosBudu, 
J q JO J q U J q U J a U 
and by the Second Theorem of the Mean, the last integral is numerically 
L 2f(q)/J$q. Hence, letting B increase without limit, the required result 
follows. 
§ 13. Theorem 9. — If in the interval {q, oo ) 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 
/' =c °'(u) 
form cos ku, then, provided I Mu exists, 
J q U 
[ dv I f(u) sin uv du = [ t^ldu, 
J o J q J q u 
both integrals necessarily existing. 
For, by the preceding theorem, 
poo poo poo p oo 
J dvj g(u) sin u(y + k)du — j dvj g(u ) sin uv du 
— J 9^du - j dvj g(u) si 
sin uv du , 
all these integrals existing. Changing the order of integration in the last 
integral (which we may do, as shown in the course of the preceding proof), 
and then performing the integration with respect to v, the last integral 
becomes 
- cos ku 
du. 
Hence we have 
u 
