Integral Transformations - Heat Conduction: Mathematical Models and Analytical Solutions - page 450

which is a complex-valued function of ω . By Eq. (B.5), we obtain an integral ex-

pression of f ( t )

+ ∞

F − 1 f

( ω ) =

1
2

β − i ω

2 e iω t d

( )=

f t ω

π β

2

+ ω

− ∞

+ ∞

1
2

β cos ω t + ω sin ω t

= d ω

2 2

π β + ω

− ∞

+ ∞

1
π

β cos ω t + ω sin ω t

= d ω .

2 2

β + ω 0

Thus we can also obtain the integral

⎧
⎨

0 , t < 0 ,

+ ∞

β cos ω t + ω sin ω t

2 ,

d ω =

= ,

t 0

β

2

+ ω

2

⎩ 0

e − β t

π , t > 0 , β > 0 .

Example 2. Find the image function and the integral expression of the impulse
function

A

, | t | < a ,

f ( t )=

0 , | t |≥ a .

Solution . By the definition of an image function, we have

⎧
⎨

2 A sin ω a

a

, ω = 0 ,

f

A e − iω t d t

ω

( ω )= F [ f ( t )] = =

⎩

− a

2 aA , ω = 0 .

Since 2 A sin

ω a

→ 2 aA as ω → 0, ω = 0 is a removable discontinuous point. Thus

ω

we may write

2 A sin ω a

f

( ω )= F [ f ( t )] = .

ω

The integral expression of f ( t ) can be obtained by taking the inverse Fourier trans-

formation

+ ∞

F − 1 f ( ω ) =

1
2

2 A sin ω a

e iω t d

f ( t )= ω

π ω − ∞

+ ∞

A
π

sin ω a cos ω t

= d ω .

ω

− ∞