1910-11.] Sommerfeld’s Form of Fourier’s Repeated Integrals. 599 
the other is of the form cos uY, then 
L tj dv J e~ 7cv2t f(u) sin uv du = J (i^ldu, 
'provided J du exists. 
For, adding the equations (1) and (2) of § 10, we have at once the 
required result. 
§ 13. To prove the next two theorems we require the following lemma : — 
Lemma. — If g(u) is a monotone decreasing function of u with zero as, 
limit at infinity, and h(u) has its square summable in every finite 
interval, then we may reverse the order of integration and write 
^dvf° q G{u, v, t)h(u)du = ^du^G(u, v, t)h(u)dv , 
where 
G (u, v, t) = e kv g{u) _ . 
sin uv 
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 h n 
of the function h(u) are such that the series whose general terms are 
respectively \a n \ /n and \b n \ /n converge absolutely.* 
Now, in any finite interval (q, Q) of values of u, h(u) is summable, since 
its square is summable, and G {u, v, t) has bounded variation ; therefore f 
we may integrate the product term-by-term, using the not necessarily con- 
vergent Fourier series of h(u). Thus 
v, t)h(u)du = % J^G (u, v , t)(a n cos nu + b n sin nu)du . . (1) 
Let m be the first integer greater than 2B. Then, for all integers 
n A m, and for all values of v in the closed interval (0, B), 
n + v \ n — v \ n — B> \n. 
Hence, using the Second Theorem of the Mean to bring g(q) outside the sign 
of integration, and breaking up the products cos nu cos uv, cos nu sin uv, 
sin nu cos uv, and sin nu sin uv into the sum or difference of two sines or 
cosines in the usual way, we get for the absolute value of the integral 
* In fact, the series whose general terms are n~ l -~ e a n and n~ l -~ e h n where 0 < e, converge 
absolutely. See my recent paper on Fourier integrals already cited, § 16. 
t W. H. Young, “On the Integration of Fourier Series” (1910), Theorem 2, presented 
to the London Mathematical Society. 
