At x=p which is a point of discontinuity for f (x) the value of the
integral is
(0)(0)10
22224
ff−+++
⎛⎞⎛⎞
⋅==
⎜⎟⎜⎟
⎝⎠⎝⎠
ppppp
Example 7.8.14 Find the Fourier transform of
2
1,1
()
0,1
xx
fx
x
−≤
⎧
=
⎨
>
⎩
Hence, evaluate
3
0
cossin
cos.
2
xxx
x
dx
x
∞
−
∫
[JNTU 2002, 2004S]
Solution The Fourier transform of f (x) is given by
11
2
11
1
2
23
1
23
233
{()}()()
0(1)0
(1)22
()()
1(1)
022
444
cossin(cossin)
isx
isxisxisx
isxisxisx
isisisis
FfxFsfxedx
ed ...
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