575 
1910-11.] On Fourier’s Repeated Integral. 
§ 17. Theorem 11. — If in the interval (q, cc ) 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 any periodic 
function whose square is summable, then, provided 
f vJ^ldu and [ 
J q U J q 
f(u) 1 
u 
du 
both exist, we have 
[ dv ( f(u) sin uv du= I •t^-du. 
J 0 J q J q U 
We shall, in the first instance, assume that the Fourier series of h(u) is 
free of constant term. 
Since the square of h{u) is summable, the Fourier coefficients a n and b n 
of the function h(u) are such that the series, whose general terms are 
respectively \a n \/n and \ b n \jn, converge absolutely, by the lemma of § 16. 
Now, in any finite interval ( q , Q) of values of u, the function g(u) sin uv 
has bounded variation, and h(u) is summable, since (h(u)) 2 is summable. 
Therefore * we may integrate the product term-by-term, using the not 
necessarily convergent Fourier series of h(u). Thus 
Jg g(u)h(u) sin uv du = g{u) sin uv(a n cos nu + b n sin nu)du . . (1 ) 
»i== i 
Let m be the first integer greater than 2B, so that, for all integers n \ m, 
n - B > \n, 
and therefore, for all values of v in the closed interval (0, B), 
n + v A n — v A n - B > \n. 
Hence, using the Second Theorem of the Mean to bring g(q) outside the 
sign of integration, and expressing the two products of sines and cosines 
which appear in the integrand as the sum or difference of sines and cosines 
in the usual way, we get for the absolute value of the integral under the 
summation sign in (1) the following inequality, supposing n\m : — 
=g(i)I\°-n\ + \K\) .... ( 2 ) 
Now, denoting the sum of the first 2(m— 1) terms of the Fourier series 
of h(u) by s m , so that 
m— 1 
s m = cos nu + b H sin nu), 
71 = 1 
* W. H. Young, “On the Integration of Fourier Series” (1910), Theorem 2, presented 
to the London Mathematical Society. 
