# Inverse Fourier transform of $\omega^{2k}\hat{f}(\omega)$

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., $$\int \hat f(\omega ) \ \omega^{2 k} e^{-i \omega x} d \omega = (-1)^k \left(\frac{d}{dx}\right)^{2k} \ \int \hat f(\omega ) \ e^{-i \omega x} d \omega$$
• You're uisng $k$ and $n$ interchangeably. Commented Jan 3 at 15:01
• @Ronald Thank you. This is a nice trick. As Mark pointed out, $n$ in your answer should be $k$. Commented Jan 3 at 15:35