Existance of Fourier transform

by kzhu
Tags: fourier transform, parseval theorem
kzhu is offline
Jan7-10, 10:04 PM
P: 12
1. The problem statement, all variables and given/known data
The sufficient condition of the existence of the Fourier transform of a function is that the function is absolutely integrable. I have identified a function that is absolutely integrable, but not square-integrable

[tex]f(t) = \frac{1}{\pi}\frac{1}{1+t^2}|t|^{\frac{-1}{2}}[/tex]

see page 177 of Fourier and Laplace Transforms by Beerends et al. What is the fourier transform of this function? Could this be evaluated?

2. Relevant equations

Since this function is not square-integrable, does it mean it has infinite energy and the Parseval's Theorem will not hold for it?

3. The attempt at a solution
I attempted to use MATLAB symbolic toolbox to separately find the Fourier transform of the function
[tex]g(t) = \frac{1}{\pi}\frac{1}{1+t^2}[/tex]

both of which exist but in a very complex form. I assume the Fourier transform of f(t) will be even more complex as Matlab failed to give an analytic form.

The implication of this little experiment shows that there are functions whose Fourier transform exists, but do not have finite energy.

Would appreciate if someone could confirm this with me.

Thank you.

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kzhu is offline
Jan11-10, 11:47 AM
P: 12
The definition of Fourier transform that I refer to is

[tex]X(j\omega) = \int_{-\infty}^{+\infty}x(t) e^{-j\omega t}dt[/tex]

and the inverse
[tex] x(t)=\frac{1}{2\pi} \int_{-\infty}^{+\infty} X(j\omega)e^{j\omega t} d\omega[/tex]

On a second thought, my question can read: Could someone provide a couple of functions that have fourier transforms by these definition but do not have finite energy?

Thank you.


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