Page 487 - Advanced engineering mathematics

14.1 The Fourier Integral 467

1 1 2sin(ω)
A ω = cos(ωξ)dξ =
π −1 πω
and
1 1
B ω = sin(ωξ)dξ = 0.
π −1
The Fourier integral of f is
∞

2sin(ω)
cos(ωx)dω.
πω
0
Because f is continuous for x =±1, the integral converges to f (x) for x =±1. At x = 1the
integral converges to
1 1 1
( f (1+) + f (1−)) = (1 + 0) = .
2 2 2
Similarly, the integral converges to 1/2at x =−1. This Fourier integral is a faithful representation
of the function except at 1 and −1, where it averages the ends of the jump discontinuities there.
In view of this convergence, we have
⎧
⎪1 for −1 < x < 1
1 ∞ 2sin(ω) ⎨
cos(ωx)dω = 1/2 for x =±1
π 0 ω ⎪
0 for |x| > 1.
⎩
There is another expression for the Fourier integral of a function that is sometimes
convenient to use. Insert the coefficients into the Fourier integral:
∞

[A ω cos(ωx) + B ω sin(ωx)]dω =
0

∞ ∞ ∞
1 1
f (ξ)cos(ωξ)dξ cos(ωx) + f (ξ)sin(ωξ)dξ sin(ωx) dω
π π
0 −∞ −∞
1 ∞ ∞
= f (ξ)[cos(ωξ)cos(ωx) + sin(ωξ)sin(ωx)]dξ dω
π 0 −∞
1 ∞ ∞
= f (ξ)cos(ω(ξ − x))dξ dω.
π 0 −∞
This gives us the equivalent Fourier integral representation
1 ∞ ∞
f (ξ)cos(ω(ξ − x))dξ dω (14.4)
π 0 −∞
of f (x) on the real line.

SECTION 14.1 PROBLEMS

In each of Problems 1 through 10, write the Fourier integral k for −10 ≤ x ≤ 10
2. f (x) =
representation (14.1) of the function and determine what 0 for |x| > 10
this integral converges to.
⎧
⎪−1for −π ≤ x ≤ 0
⎨

x for −π ≤ x ≤ π 3. f (x) = 1 for 0 < x ≤ π
1. f (x) = ⎪
⎩
0 for |x| >π 0 for |x| >π
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October 14, 2010 16:43 THM/NEIL Page-467 27410_14_ch14_p465-504