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 $f$.


1 Answer 1


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., $$\int \hat f(\omega ) \ \omega^{2 k} e^{-i \omega x} d \omega = (-1)^k \left(\frac{d}{dx}\right)^{2k} \ \int \hat f(\omega ) \ e^{-i \omega x} d \omega $$

  • 1
    $\begingroup$ You're uisng $k$ and $n$ interchangeably. $\endgroup$
    – Mark Viola
    Commented Jan 3 at 15:01
  • $\begingroup$ @Ronald Thank you. This is a nice trick. As Mark pointed out, $n$ in your answer should be $k$. $\endgroup$
    – Arindam
    Commented Jan 3 at 15:35
  • $\begingroup$ --and the name is Roland. Directly typing TeX with text of 5 point resolution is a random process. But ist not a trick. the correspondence of differential equations and algebraic equatiion by Fourier transfom was the kick start of modern mathematics and mathematical physics parallel to $\endgroup$
    – Roland F
    Commented Jan 3 at 19:28

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