1910-11.] 
On Fourier’s Repeated Integral. 
567 
§ 8. Theorem 6. — If f(u) is a summable function in the interval 
( — p, p), and J M f(u)du has at the origin a differential coefficient 
whose value is f(0), then 
|c *dv\ p _ p f(u) cos uv du = tt/(0), 
provided only this repeated integral exists. 
Let 
g(u) — f(u) sin \u!u y 
then 
\if w)du = \^F)ff u)du -/' 
u cos \u — sin \u 
ir 
f(u)du 
du. 
But, since the differential coefficient of \ u f(u) du at u = 0 is /( 0), the first 
term on the right has, when h approaches zero, the unique limit hf(0). 
Hence * 
Lit- f‘g(u)du = if(0) + Utl r ™t" 2 MCOg W r f{u)du]du. 
h —0 h J 0 h = 0 h J 0 u 2 \J o' J 
But 
sin hu - lu cos \u 
and 
J f(u)du 
are both continuous functions of u ; hence the integrand of the integral on 
the right, being the product of these two functions, is itself a continuous 
function of u, and consequently the limits on the right are the same as 
Ut g in ( h f(u)du, 
h=0 h 2 Jo 
which are uniquely zero. 
Hence, finally, 
L t~ I g(u)du = if(0) = g(0), 
h=0/l J 0 
that is, at the point u — 0 , the function g(u) is the differential coefficient of 
its integral. 
Denote the repeated integral, whose existence was postulated, by I. 
Then 
sin B u 
= Lt [ f(u ) 
B=qo J —p 
J -du. 
Since this limit is unique, we may replace B by J(2m+1), so that 
I = Lt r g ( M ) si - n J( ^ +1 ) M rf M= U f w g(M )Sin|(2m + l)y^, 
m—coj —p Sin 2 U m=<x.J —ir Sin -ph 
where, if need be, we regard g(u) as being zero outside the interval (—p,pf 
This last integral is 2i r times the sum of the first (2m + 1) terms of the 
* I use the symbol L££ to denote a possible plurality of limits, and L t to denote an 
h~0 h=0 
unique limit, known to exist. 
