574 Proceedings of the Koyal Society of Edinburgh. [Sess. 
Therefore, taking the repeated limit of both sides, first with respect to 0 , 
and then with respect to n, we get 
H - \ z JA x ) dx - \‘,M x ) dx - • • • • - = 0. 
But, by (ii.), already proved, U ~Lt )dx+ \‘ q f 2 (x)dx+ .... +HfJx)dx] 
n=x> z=B 
exists, therefore, by the last equation, (i.) and (ii.) follow simultaneously. 
Similarly the lemma is true when the upper limit of integration is q 
and the lower limit is p. Hence by addition the required result follows. 
Example. — If the series whose general term is n h a n converges, and 
co 00 
f(x) — sin nx , g(x) = 2a n (l - cos nx), 
«= 1 • n=i 
then 
J" 0 x ~ §f(x)dx and J 0 x ~ $ g(x)dx 
exist, and we have their values, since, by a known result,* 
| 0 « _ 5sin nxdx— J'lTrn—.^x~i{\ - cos nx)dx, 
given as follows, 
\*x -| f(x)dx = J2TT^ria n m ~ I g(x)dx. 
n — 1 
§ 16. We shall also require the following lemma, which is an immediate 
consequence of a theorem of Harnack’s, recently generalised by Fatou f : — 
Lemma. — If a n , b n are the Fourier constants of a function f(x) 
whose square is summable, then the series 2a n /rd +e , 2b a /m +e converge 
absolutely , e>0. 
For , by Harnack’s theorem, 2(a n 1 2 + b n 2 ) converges , since it is equal to 
Hence eaph of the series 2<x n 2 , Uof converges, and therefore so do also the 
series whose general terms are respectively 
a 2 + -J—r and b n 2 + ^ 
n i+-n 
1 + 2e 
But 
1 / 1 ( a 2 4- ^ 
d+ e ~ 2 \ n n 1+2e )’ 
therefore the series 2 | a n | /m +e . converges. Similarly the series 2 | b, t \ /n* +e 
converges. This proves the lemma.J 
* See Pringsheim, loc. cit., pp. 375-6. 
+ Fatou, loc. cit. Harnack, Math. Ann., xix. (1882), p. 225. 
X See A. Pringsheim, Munch. Ber., 30 (1900), p. 63. 
