= Moment differentiation of the Fourier transform
{title2=$\widehat f^{(k)}(\xi)=\int(-ix)^kf(x)e^{-ix\xi}\,dx$}
If $f\in L^1(\mathbb R)$ and $x^kf\in L^1(\mathbb R)$ for $0\le k\le r$, then its <Fourier transform> is $r$ times <continuously differentiable> and the displayed formula holds. The <dominated convergence theorem> permits each <derivative> under the <integral>. Thus $r+1$ <vanishing moments> give a zero <Taylor polynomial> through degree $r$ at the origin. Unweighted $L^1$ membership only guarantees a <continuous function> as the <Fourier transform>.
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