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472 CHAPTER 14 The Fourier Integral and Transforms

ˆ
We also denote F[ f ] as f :
ˆ
F[ f ](ω) = f (ω).

EXAMPLE 14.3
We will determine the transform of e −c|t| , with c a positive number. First, write

e −ct for t ≥ 0
f (t) = e −c|t| =
e ct for t < 0.
Then
∞

e
F[ f ](ω) = e −c|t| −iωt dt
−∞
0 ∞
ct −iωt
e
= e e dt + e −ct −iωt dt
−∞ 0
0 ∞
= e (c−iω)t dt + e −(c+iω)t dt
−∞ 0
0 ∞
1 −1

= e (c−iω)t + e −(c+iω)t
c − iω c + iω
−∞ 0
1 1 2c

= + = .
c + iω c − iω c + ω 2
2
We can also write
2c
ˆ
f (ω) = .
2
c + ω 2
EXAMPLE 14.4
Let H(t) be the Heaviside function, defined by

1for t ≥ 0
H(t) =
0for t < 0.
We will compute the Fourier transform of f (t) = H(t)e −5t . This is the function

e −5t for t ≥ 0
f (t) =
0 for t < 0.
From the definition of F,
∞ ∞ ∞

e
e
ˆ
f (ω) = H(t)e −5t −iωt dt e −5t −iωt dt = e −(5+iω)t dt
−∞ 0 0
1 ∞ 1
=− e −(5+iω)t = .
5 + iω 0 5 + iω
EXAMPLE 14.5
ˆ
Let a and k be positive numbers. We will determine f (t), where

k for −a ≤ t < a
f (t) =
0 for t < −a and t ≥ a.

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October 14, 2010 16:43 THM/NEIL Page-472 27410_14_ch14_p465-504