Complex Analysis - Maths KU

382 Chapter 6 Series and Residues
1
y
Figure 6.25 Graph of f in Example 4
x
EXAMPLE 4 Fourier Transform
Find the Fourier transform of f(x) =e −|x| .
Solution The graph of f,
⎧
⎨ e
f(x) =
⎩
x , x < 0
e−x , (21)
, x ≥ 0
is given in Figure 6.25. From the expanded definition of f in (21), it follows
from (19) that the Fourier transform of f is
� 0
F{f(x)} = e x e iαx � ∞
dx + e −x e iαx dx = I1 + I2. (22)
−∞
We shall begin by evaluating the improper integral I2.One of several ways of
proceeding is to write:
I2 = lim
b→∞
=
� b
0
1
αi − 1 lim
b→∞
e −x(1−αi) dx = lim
b→∞
0
e −x(1−αi)
αi − 1
�
�
�
�
� e −b cos bα + ie −b sin bα − 1 � =
b
0
e
= lim
b→∞
−b(1−αi) − 1
1
1 − αi .
αi − 1
Here we have used limb→∞ e −b cos bα = 0 and limb→∞ e −b sin bα = 0 for b>0.
The integral I1 can be evaluate in the same manner to obtain
I1 =
1
1+αi .
Adding I1 and I2 gives the value of the Fourier transform (22):
F{f(x)} =
1 1
+
1 − αi 1+αi
or F (α) = 2
.
1+α2 EXAMPLE 5 Inverse Fourier Transform
Find the inverse Fourier transform of F (α) =
2
.
1+α2 Solution The idea here is to recover the function f in Example 4 from the
inverse transform (20),
F −1 {F (α)} = 1
� ∞
2π
2
1+α2 e−iαx dα = f(x). (23)
−∞
To evaluate � (23), we let z be a complex variable and introduce the contour
1
integral
π(1 + z2 ) e−izxdz.Note that the integrand has simple poles at
C