Page 379 - Schaum's Outline of Theory and Problems of Advanced Calculus

370 FOURIER INTEGRALS [CHAP. 14

L sin v 0 sin v
ð ð
14.9. Prove that: (a) lim dv ¼ ; ðbÞ lim dv ¼ .
!1 0 v 2 !1 L v 2
ð L sin v ð L sin y ð 1 sin y
(a)Let v ¼ y. Then lim dv ¼ lim dy ¼ dy ¼ by Problem 12.29, Chap. 12.
!1 0 v !1 0 y 0 y 2
0 sin v L sin y
ð ð
Let v ¼ y. Then lim dv ¼ lim :
ðbÞ dy ¼
!1 L v !1 0 y 2
14.10. Riemann’s theorem states that if FðxÞ is piecewise continuous in ða; bÞ, then
b
ð
lim FðxÞ sin xdx ¼ 0
!1 a
with a similar result for the cosine (see Problem 14.32). Use this to prove that
L sin v
ð
ðaÞ lim f ðx þ vÞ dv ¼ f ðx þ 0Þ
!1 0 v 2
0 sin v
ð
ðbÞ lim f ðx þ vÞ dv ¼ f ðx 0Þ
!1 L v 2
where f ðxÞ and f ðxÞ are assumed piecewise continuous in ð0; LÞ and ð L; 0Þ respectively.
0
(a)Using Problem 9(a), it is seen that a proof of the given result amounts to proving that
L sin v
ð
lim f f ðx þ vÞ f ðx þ 0Þg dv ¼ 0
!1 0 v
f ðx þ vÞ f ðx þ 0Þ
is piecewise contin-
v
This follows at once from Riemann’s theorem, because FðvÞ¼
uous in ð0; LÞ since lim FðvÞ exists and f ðxÞ is piecewise continuous.
n!0þ
(b)A proof of this is analogous to that in part (a)ifwe make use of Problem 14.9(b).
ð
1
14.11. If f ðxÞ satisfies the additional condition that j f ðxÞj dx converges, prove that
1
ð ð 0
1 sin v sin v
lim f ðx þ 0Þ; ðbÞ lim f ðx 0Þ:
ðaÞ f ðx þ vÞ dv ¼ f ðx þ vÞ dv ¼
!1 0 v 2 !1 1 v 2
We have
sin v sin v sin v
ð ð L ð
1 1
dv
f ðx þ vÞ dv ¼ f ðx þ vÞ dv þ f ðx þ vÞ ð1Þ
0 v 0 v L v
ð ð L ð
sin v sin v sin v
1 1
dv
f ðx þ 0Þ dv ¼ f ðx þ 0Þ dv þ f ðx þ 0Þ ð2Þ
0 v 0 v L v
Subtracting,
ð ð L
1 sin v sin v
dv
f f ðx þ vÞ f ðx þ 0Þg dv ¼ f f ðx þ vÞ f ðx þ 0Þg
0 v 0 v
ð ð
1 sin v 1 sin v
dv
þ f ðx þ vÞ v dv f ðx þ 0Þ v
L L
Denoting the integrals in (3)by I; I 1 ; I 2 , and I 3 , respectively, we have I ¼ I 1 þ I 2 þ I 3 so that
jIj @ jI 1 jþjI 2 jþjI 3 j ð4Þ
ð ð
1 1 1
sin v
Now jI 2 j @ dv @ j f ðx þ vÞj dv
f ðx þ vÞ v
L L L