Fourier transform of x(t)=1

**asmani** · May 24th 2011, 07:46 AM

Hi all

I know that the Fourier transform of x(t)=1 is X(jω)=2πδ(ω) by using the duality property.
This implies:
$\int_{-\infty }^{+\infty }e^{-j\omega t}dt=2\pi\delta(\omega)$
Consequently, for ω≠0:
$\int_{-\infty }^{+\infty }e^{-j\omega t}dt=0$
And as a result:
$\int_{-\infty }^{+\infty }\cos t\: dt=0$

Is this result true?!

Thanks in advance

**ojones** · July 17th 2011, 06:38 PM

No; $\delta(\omega)\ne 0$ .

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