Moment differentiation of the Fourier transform (${\hat{f}}^{(k)}(\xi )=\int (-ix{)}^{k}f(x)e^{-ix\xi }dx$)

If $f\in L^1(\mathbb{R})$ and $x^{k}f\in L^1(\mathbb{R})$ 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.