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I have to find inverse Fourier transform of \begin{align*} F(\omega)=e^{-\frac{\omega^2}{2}}\frac{1}{\mathsf{sinc}( \omega )} \end{align*} I was wondering whether it exists maybe in a sense of distributions.

Using duality between Fourier transform and inverse Fourier transform. We can ask what is Fourier transform of \begin{align*} f(x)=e^{-\frac{x^2}{2}}\frac{1}{\mathsf{sinc}( x )} \end{align*} and whether it exits?

Note, that the function $f(t)$ has vertical asymptots at for $x=n \pi$ for $n \in \mathbb{Z}$ \begin{align} \mathsf{sinc}( x )=\frac{\sin(x)}{x} \end{align} equals $0$ at integer multiplies of $\pi$. Is this a problem?

One way to look at the problem is find convolution of $\mathcal{F}(e^{-\frac{x^2}{2}})$ and $\mathcal{F}(\frac{1}{\mathsf{sinc}( x )})$. However, this direction can be problematic since $\mathcal{F}(\frac{1}{\mathsf{sinc}( x )})$ might not exist.

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  • $\begingroup$ The Fourier transform of a convolution is a product. So what you are looking for is a convolution of the inverse Fourier transform of the first function (which will be a Gaussian also) and the inverse Fourier Transform of the reciprocal sinc. $\endgroup$ – Paul Jan 28 '15 at 21:43
  • $\begingroup$ Yes, good. But what is the inverse Fourier of $\frac{1}{sinc}$? $\endgroup$ – Boby Jan 28 '15 at 21:45
  • $\begingroup$ Since 1/sinc is unbounded, I can't imagine it has one. Is there some bound to address this issue? $\endgroup$ – Paul Jan 28 '15 at 21:51

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