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∫
∞
⋅⋅⋅⋅==
0
)sin()(
2
)()}({ dtttfFtf ω
π
ω
s
F
Inverse sine transform
∫
∞
−
⋅⋅⋅==
0
1
)sin()()()}({ dttFtfF
s
ωωωF
Fourier cosine transform
∫
∞
⋅⋅⋅⋅==
0
)cos()(
2
)()}({ dtttfFtf ω
π
ω
c
F
Inverse cosine transform
∫
∞
−
⋅⋅⋅==
0
1
)cos()()()}({ dttFtfF
c
ωωωF
Fourier transform (proper)
∫
∞
∞−
−
⋅⋅⋅== dtetfFtf
tiω
π
ω )(
2
1
)()}({F
Inverse Fourier transform (proper)
∫
∞
∞−
−−
⋅⋅== dteFtfF
tiω
ωω )()()}({
1
F
Example 1
– Determine the Fourier transform of the function f(t) = exp(- t), for t
>0, and f(t) = 0, for t<0.
The continuous spectrum, F(ω), is calculated with the integral:
∫∫
+−
∞
∞→
+−
=
ε
ω
ε
ω
ππ
0
)1(
0
)1(
2
1
lim
2
1
dtedte
titi
.
1
1
2
1
1
))1(exp(1
2
1
lim
ωπω
εω
π
ε
ii
i
+
⋅=
+
+−−
=
∞→
This result can be rationalized by multiplying numerator and denominator by
the conjugate of the denominator, namely, 1-iω. The result is now:
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