Page 498 - Advanced engineering mathematics

478 CHAPTER 14 The Fourier Integral and Transforms

it is enough to derive the operational formula when n = 1. Integrate by parts:
∞

−iω 0 t
[ f (t)](ω) = f (t)e dt
−∞
∞

∞
−iωt −iωt
= f (t)e − f (t)(−iω)e dt
∞
−∞
∞

= iω e −iωt f (t)dt
−∞
ˆ
= iω f (ω),
where we have used the fact that f (t) has limit 0 at ∞ and at −∞ to conclude that
∞
−iωt
f (t)e = 0.
−∞
Now an inductive argument leads to the conclusion for the nth derivative.
EXAMPLE 14.9
We will solve the differential equation
y − 4y = H(t)e −4t .

Apply the Fourier transform to the differential equation to get
F[y (ω)]− 4 ˆy(ω) = F[H(t)e −4t ](ω).

From the operational rule with n = 1,

F[y ](ω) = iω ˆy(ω).
Further, from Example 14.4, with 4 in place of 5,
1
F[H(t)e −4t ](ω) = .
4 + iω
Therefore
1
iω ˆy − 4 ˆy = .
4 + iω
Solve for ˆy to get
−1
ˆ y(ω) = .
16 + ω 2
From Example 14.3,
−1 1

−1 −4|t|
y(t) = F =− e .
16 + ω 2 8
The operational formula can be adjusted to accommodate a finite number of jump disconti-
nuities of f . If these occur at t 1 ,··· ,t M and if
lim f (t) = lim f (t) = 0,
t→−∞ t→∞
then
M

ˆ −it j ω
F[ f (t)](ω) = iω f (ω) − ( f (t j +) − f (t j −))e .
j=1
Each term f (t j +) − f (t j −) is the magnitude of the jump discontinuity at t j .

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October 14, 2010 16:43 THM/NEIL Page-478 27410_14_ch14_p465-504