Page 378 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 14] FOURIER INTEGRALS 369

This is equivalent to
a 1 2 sin a
ð ð 2
2
ð1Þ dx ¼ 2 d
a 1
ð 2 ð 2
1 sin a 1 sin a
or d ¼ 2 d ¼ a
2 2
1 0
ð 2
1 sin a a
i.e., 2 d ¼
0 2
By letting a ¼ u and using Problem 14.5, it is seen that this is correct. The method can also be used to
ð 2
1 sin u
find 2 du directly.
0 u
PROOF OF THE FOURIER INTEGRAL THEOREM
14.8. Present a heuristic demonstration of Fourier’s integral theorem by use of a limiting form of
Fourier series.
Let
1
X n x n x
a 0
a n cos þ b n sin
f ðxÞ¼ þ ð1Þ
2 L L
n¼1
1 ð L n u 1 ð L n u
f ðuÞ cos du and f ðuÞ sin du:
where a n ¼ b n ¼
L L L L L L
Then by substitution (see Problem 13.21, Chapter 13),
1 ð L 1 X L n
1 ð
f ðuÞ cos ðu xÞ du
f ðxÞ¼ f ðuÞ du þ L L ð2Þ
2L L L
n¼1
ð
1
If we assume that j f ðuÞj du converges, the first term on the right of (2) approaches zero as L !1,
1
while the remaining part appears to approach
1 X 1 n
1 ð
lim f ðuÞ cos ðu xÞ du
L!1 L L ð3Þ
n¼1 1
This last step is not rigorous and makes the demonstration heuristic.
Calling ¼ =L,(3)can be written
1
X
f ðxÞ¼ lim Fðn Þ ð4Þ
!0
n¼1
where we have written
1 ð 1
f ðuÞ cos ðu xÞ du
Fð Þ¼ 1 ð5Þ
But the limit (4)isequal to
ð ð ð
1 1 1 1
d f ðuÞ cos ðu xÞ du
f ðxÞ¼ Fð Þ d ¼
0 0 1
which is Fourier’s integral formula.
This demonstration serves only to provide a possible result. To be rigorous, we start with the integral
1 ð 1 ð 1
d f ðuÞ cos ðu xÞ dx
0
1
and examine the convergence. This method is considered in Problems 14.9 through 14.12.