Fourier transform and singularities

[diraq]

Consider the Fourier transform of a complex function f(t):
f(t)=\int_{-\infty}^\infty F(\omega)e^{-i\omega t}
Here t and \omega 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 \text{limit}_{|\omega|\rightarrow\infty}F(\omega)=0. Now, if F(\omega) is analytically extended onto the complex plane, i.e., now \omega\in\mathbb C, then does the property \text{limit}_{|\omega|\rightarrow\infty}F(\omega)=0 still hold at infinity on the complex plane?

2) The width of f(t) and F(\omega) in real t and \omega domain respectively satisfy the uncertainty relation. When calculating f(t)=\int_{-\infty}^\infty F(\omega)e^{-i\omega t}, 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 \omega. To evaluate f(t), we only need to know the information about the singularity points of F(\omega) 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 F(\omega) and the corresponding residues?

Thanks!

 

[jasonRF]

Consider the Fourier transform of a complex function f(t):
f(t)=\int_{-\infty}^\infty F(\omega)e^{-i\omega t}
Here t and \omega 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 \text{limit}_{|\omega|\rightarrow\infty}F(\omega)=0. Now, if F(\omega) is analytically extended onto the complex plane, i.e., now \omega\in\mathbb C, then does the property \text{limit}_{|\omega|\rightarrow\infty}F(\omega)=0 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 $\omega \rightarrow \pm i \infty$.