Page 486 - Advanced engineering mathematics

466 CHAPTER 14 The Fourier Integral and Transforms

Now the Fourier series on [−L, L] can be written
1 L 1 L
f (ξ)dξ ω + f (ξ)cos(ω n ξ)dξ cos(ω n x)
2π −L π −L
L

+ f (ξ)sin(ω n ξ)dξ sin(ω n x) ω.
−L
Let L →∞,so [−L, L] expands to cover the entire real line. Then ω → 0. Examine what
happens in the terms of the last equation. First,
1 L
f (ξ)dξ ω → 0
2π −L
because the integral converges (hence is bounded). The other terms in this equation resemble a
Riemann sum for a definite integral. As L →∞ and ω → 0, this expression approaches the
limit
1 ∞
f (ξ)cos(ωξ)dξ cos(ωx)
π
−∞
∞

+ (ξ)sin(ωξ)dξ sin(ωx) dω.
−∞
This is the Fourier integral of f (x) on the real line, and it has the form
∞

(A ω cos(ωx) + B ω sin(ωx))dω (14.1)
0
in which the Fourier integral coefficients of f are
1 ∞
A ω = f (ξ)cos(ωξ)dξ (14.2)
π −∞
and
1 ∞
B ω = f (ξ)sin(ωξ)dξ. (14.3)
π
−∞
The integration variable ω replaces the summation index n in this integral representation.
As with Fourier series, the relationship between the integral (14.1) and f (x) must be
clarified. This is done in the following theorem.

THEOREM 14.1 Convergence of the Fourier Integral
∞
Suppose f (x) is defined for all real x and that | f (x)|dx converges. Suppose f is piecewise
−∞
smooth on every interval [−L, L] for L > 0. Then at any x the Fourier integral (14.1) of f
converges to
1
( f (x+) + f (x−)).
2
In particular, if f is continuous at x, then the Fourier integral converges at x to f (x).

EXAMPLE 14.1
Let

1for −1 ≤ x ≤ 1
f (x) =
0for |x| > 1.
Certainly f is absolutely integrable. The Fourier integral coefficients of f are

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October 14, 2010 16:43 THM/NEIL Page-466 27410_14_ch14_p465-504