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 $\omega \rightarrow \pm i \infty$.