Fourier transform of $\frac{1}{f(t)}$

Suppose we know Fourier transform of $f(t)$ is $F(\omega)$.
Can we find Fourier transform of $\frac{1}{f(t)}$.

I was thinking we can write $\frac{1}{f(t)}=( f(t))^{-{1}}$. So I guess a more general question is can we find Fourier transform of $(f(t))^k$ for any k?

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For $ k \geq 1$, the answer is as above. For $k < 0$, in general $F(\omega) = \int_{-\infty}^\infty \frac{1}{f(x)}e^{-2\pi i \omega x} dx$ need not be even defined. For instance, if
$$f = x^{-2}\chi_{|x| > 1} + \chi_{|\chi| \leq 1},\quad \chi_A(x) = \begin{cases} 1 & x \in A \\ 0 & x \notin A\end{cases}$$
then $F(\omega)$ is defined since the integrand is finite, but $1/f$ is not even integrable and I do not know if one can make sense of this without using tempered distributions. I do not know if one can salvage anything useful without using properties specific to the function $f$.

In general, $1/f$ may not have a Fourier transform, even in the sense of tempered distributions. Here’s one case where it does.

Suppose $f(t) = p(t)/q(t)$ where $p$ and $q$ are polynomials with no real roots.
The Fourier transform of $f$ can be obtained by expanding $f$ in partial fractions: if
$$ f(t) = r(t) + \sum_{j=1}^m \dfrac{a_j t + b_j}{(t – c_j)^2 + d_j^2}$$
where $r$ is a polynomial of degree $m$ and $a_j, b_j, c_j, d_j$ are real, $d_j > 0$, then $F(\omega)$ is
a linear combination of the Dirac $\delta$ and its derivatives up to the $m$’th derivative, $e^{-\omega(ic_j + d_j)} H(\omega)$ and $e^{\omega(-ic_j + d_j)} H(-\omega)$ ($H$ = Heaviside). And similarly for $1/f(t)$ where we use the partial fraction expansion of $1/f$.