While calculating a Fourier inverse, I get the following expression $$\int_{-\infty}^{\infty} \omega^{2k} \hat{f}(\omega) e^{-iwx}dw$$
where $f \in L^2(\mathbb{R})$.
I though of taking the inverse the inverse Fourier transform of $w^{2k}$, call it $g$, and then the expression becomes
\begin{align} \int_{-\infty}^{\infty} \hat{g}(\omega) \hat{f}(\omega) e^{-iwx}dw &=\int_{-\infty}^{\infty} \widehat{g*f} (\omega) e^{-iwx}dw\\ \end{align}
\begin{align} &=g*f \end{align}
But the inverse Fourier transform of $w^{2k}$ turns out to be unbounded.
Is there some other way to calculate the integral? I am expecting it in terms of expression of $f$.
The factor $\omega^{2 k}$ in
$$ \omega^{2 k} e^{-i \omega x} = (i \frac{d}{dx})^{2k} e^{-i \omega x }$$
can be interchanged with integration, if $\hat f$ is smooth with compact support, eg., $$\int \hat f(\omega ) \ \omega^{2 k} e^{-i \omega x} d \omega = (-1)^k \left(\frac{d}{dx}\right)^{2k} \ \int \hat f(\omega ) \ e^{-i \omega x} d \omega $$