Fourier transform of a polynomial multiple ($\mathcal{F}[x^{n}f](k)=i^n{\partial }_k^n\hat{f}(k)$)

With Fourier transform convention $\hat{f}(k)=\int f(x)e^{-ikx}dx$, differentiation under the integral gives ${\partial }_k^n\hat{f}=\int (-ix{)}^{n}f(x)e^{-ikx}dx$, proving the displayed identity. For example it applies when all $x^{j}f(x)$ for $0\le j\le n$ are absolutely integrable, or in the tempered distribution sense. It is the dual counterpart of the Fourier transform of a derivative.