Mathematical Methods for Physicists: A concise introduction - Site Map

FOURIER INTEGRALS AND FOURIER TRANSFORMS
Figure 4.14.
The box function.
Solution: The Fourier transform of f …x† is, as shown in Fig. 4.14,
g…!† ˆ 1 Z 1
p f …x 0 †e i!x 0 dx 0 ˆ 1 Z a
p …1†e i!x 0 dx 0 ˆ 1
0
p ei!x 2
2
2 i!
1
r
2 sin !a
ˆ ; ! 6ˆ 0:
!
p
For ! ˆ 0, we obtain g…!† ˆ
2= a.
The Fourier integral representation of f …x† is
Now
Z 1
1
f …x† ˆp
1
2
Z 1
1
Z
sin !a
1
! ei!x d! ˆ
1
a
g…!†e i!x d! ˆ 1 Z 1
2 1
Z
sin !a cos !x
1
d! ‡ i
!
1
2 sin !a
e i!x d!:
!
sin !a sin !x
d!:
!
The integrand in the second integral is odd and so the integral is zero. Thus we
have
f …x† ˆp
1
2
Z 1
1
g…!†e i!x d! ˆ 1 Z 1
1
sin !a cos !x
!
d! ˆ 2
Z 1
0
a
a
sin !a cos !x
d!;
!
the last step follows since the integrand is an even function of !.
It is very dicult to evaluate the last integral. But a known property of f …x† will
help us. We know that f …x† is equal to 1 for jxj a, and equal to 0 for jxj > a.
Thus we can write
2
Z 1
0
sin !a cos !x
d! ˆ 1 jxja
!
0 jxj > a
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