I was trying to answer this question: Identity for all uniformly bdd functions follows "by basic Fourier (transform) analysis"? which involved: $$\mathcal{F}^{-1}(1/\varphi(-\cdot)),$$ where $\varphi = \mathcal{F}(P)$, where $P$ is a probability measure on $\mathbb{R}$.
I noticed that using a Fourier analysis trick that I might write it as $\mathcal{F}(1/\varphi(\cdot))$ (which might not be a good idea... if it is wrong, then please tell me). So, I tried to search "Fourier transform of $1/f$" in Google, which then autocorrected to "inverse Fourier transform of $1/f$", which I thought was so strange. I then searched Google but because of formatting I did not find anything. So, I tried Wolfram Alpha did not understand what I was asking.
So, I was wondering if there is something up with this expression or what? Thanks.
Update: What, if any, smoothness properties of this do we know? We know by http://math.mit.edu/~jerison/103/handouts/fourierint2.13.pdf that $\varphi$ is continuous and bounded, but that doesn't look good for the function in question since it might not even be in $L^2$?