Fourier transform and singularities

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Consider the Fourier transform of a complex function f(t):
[tex]f(t)=\int_{-\infty}^\infty F(\omega)e^{-i\omega t}[/tex]
Here t and [itex]\omega[/itex] are on real axis. Let's suppose f(t) is square integrable. Here are my questions:

1) Since f(t) is square integrable, so we have [itex]\text{limit}_{|\omega|\rightarrow\infty}F(\omega)=0[/itex]. Now, if [itex]F(\omega)[/itex] is analytically extended onto the complex plane, i.e., now [itex]\omega\in\mathbb C[/itex], then does the property [itex]\text{limit}_{|\omega|\rightarrow\infty}F(\omega)=0[/itex] still hold at infinity on the complex plane?

2) The width of f(t) and [itex]F(\omega)[/itex] in real t and [itex]\omega[/itex] domain respectively satisfy the uncertainty relation. When calculating [itex]f(t)=\int_{-\infty}^\infty F(\omega)e^{-i\omega t}[/itex], if the answer to the first question is "yes," then for different sign of t, the contour can be closed by half circles on the complex plane for [itex]\omega[/itex]. To evaluate f(t), we only need to know the information about the singularity points of [itex]F(\omega)[/itex] on complex plane. So, the question is: is there any simple rule of thumb relation between the width of f(t) and the position of the singularity of [itex]F(\omega)[/itex] and the corresponding residues?

Thanks!
 

jasonRF

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Consider the Fourier transform of a complex function f(t):
[tex]f(t)=\int_{-\infty}^\infty F(\omega)e^{-i\omega t}[/tex]
Here t and [itex]\omega[/itex] are on real axis. Let's suppose f(t) is square integrable. Here are my questions:

1) Since f(t) is square integrable, so we have [itex]\text{limit}_{|\omega|\rightarrow\infty}F(\omega)=0[/itex]. Now, if [itex]F(\omega)[/itex] is analytically extended onto the complex plane, i.e., now [itex]\omega\in\mathbb C[/itex], then does the property [itex]\text{limit}_{|\omega|\rightarrow\infty}F(\omega)=0[/itex] still hold at infinity on the complex plane?
No. The counterexample is the function ##f(t)=e^{-t^2/2}##. It has the Fourier transform ##F(\omega)=\sqrt{2\pi}e^{-\omega^2/2}##, which blows up as ##\omega \rightarrow \pm i \infty##.
 

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