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14.3 The Fourier Transform 471

to obtain
1 1 ∞ ∞ 1
( f (x+) + f (x−)) = f (ξ) e iω(ξ−x) + e −iω(ξ−x) dξ dω
2 π 0 −∞ 2
1 ∞ ∞
= f (ξ)e iω(ξ−x) dξ dω
2π 0 −∞
1 ∞ ∞ −iω(ξ−x)
+ f (ξ)e dξ dω.
2π 0 −∞
In the next-to-last integral, replace ω with −ω and compensate for this change by replacing
∞
0
··· dω with ··· dω. This enables us to write
0 −∞
1
( f (x+) + f (x−))
2
1 0 ∞ −iω(ξ−x) 1 ∞ ∞ −iω(ξ−x)
= f (ξ)e dξ dω + f (ξ)e dξ dω.
2π −∞ −∞ 2π 0 −∞
Combine these integrals to obtain
1 1 ∞ ∞ −iωξ iωx
( f (x+) + f (x−)) = f (ξ)e e dξ dω. (14.9)
2 2π
−∞ −∞
This is the complex Fourier integral representation of f (x) on the real line.Ifwelet
∞

C ω = f (ξ)e −iωξ dξ,
−∞
then this integral representation is
1 1 ∞
( f (x+) + f (x−)) = C ω e iωx dω.
2 2π
−∞
We call C ω the complex Fourier integral coefficient of f .
We may use this complex Fourier integral as a springboard to the Fourier transform, the idea
of which is contained in equation (14.9). For emphasis in how we want to think of this equation,
write it as
1 1 ∞ ∞
( f (x+) + f (x−)) = f (ξ)e −iωξ dξ e iωx dω. (14.10)
2 2π
−∞ −∞
The term in large parentheses on the right in equation (14.10) is the Fourier transform of f .We
summarize this discussion as follows.

If f is absolutely integrable on the real line, then the Fourier transform F[ f ] of f is the
function defined by
∞

F[ f ](ω) = f (t)e −iωt dt.
−∞

Thus, the Fourier transform of f is the coefficient C ω in the complex Fourier integral
representation of f .
Because of the use of the Fourier transform in applications such as signal analysis, we
usually use t (for time) as the variable in the defining integral, and ω as the variable of the
transformed function F[ f ]. Engineers refer to ω in the transformed function as the frequency of
the signal f .

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October 14, 2010 16:43 THM/NEIL Page-471 27410_14_ch14_p465-504