Integral Transformations - Heat Conduction: Mathematical Models and Analytical Solutions - page 461

Solution. By the Dirichlet integral + ∞
0

sinx
x

= 2 ,wehave d x

⎧
⎨

− 2 , < ,

x 0

+ ∞

sin ω x

0 , x = 0 ,

d ω =

ω ⎩ 0

2 ,

x > 0 .

Thus

⎧
⎨

1
2 ,

− x < 0 ,

F − 1 1
i

+ ∞ + ∞

1
2

1 1

π

sin ω x

e iω x d

0 , x = 0 ,

= ω = d ω =

ω π i ω ω ⎩

− ∞ 0

1
2 ,

x > 0 .

Since

F − 1 F − 1 1

i

1
i

F − 1 πδ ( ω )+ = [ πδ ( ω )] +

ω ω

⎧
⎨

1
2 ,

0

− x < 0

, x < 0

1
2 +

=

0 =

0 =

H ( x ) ,

⎩ 1

2 ,

1 , x >

x >

1

we finally obtain F [ H ( x )] = πδ ( ω )+ .

i ω

Remark. The Fourier transformations of some functions in the tables of Fourier
transformation are the generalized Fourier transformation.

B.1.5 The Multiple Fourier Transformation

Consider a function f of three variables x , y and z . By taking a Fourier trans-

formation with respect to x ,wehave

( x , y , z )

+ ∞

f

e − iω 1 x d x

( ω 1 , y , z )= f ( x , y , z ) ,

− ∞

+ ∞

1
2

f

e iω 1 x d f ( x , y , z )= ( ω 1 , y , z ) ω 1 .

π − ∞