= Fourier transform of a polynomial multiple
{c}
{title2=$\mathcal F[x^nf](k)=i^n\partial_k^n\widehat f(k)$}
With <Fourier transform> convention $\widehat f(k)=\int f(x)e^{-ikx}dx$, differentiation under the <integral> gives $\partial_k^n\widehat f=\int(-ix)^nf(x)e^{-ikx}dx$, proving the displayed identity. For example it applies when all $x^jf(x)$ for $0\leq j\leq n$ are absolutely integrable, or in the <tempered distribution> sense. It is the dual counterpart of the <Fourier transform of a derivative>.
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