Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 4 / iii / Solution

Let $A(k)$ be the Fourier transform of $e^{-x^2}$. Since $(e^{-x^2}{)}^{\prime }=-2x{e}^{-x^2}$, integration by parts gives $\mathcal{F}[x{e}^{-x^2}]=-(ik/2)A(k)$. Differentiation under the integral then yields $A^{\prime }(k)=-(k/2)A(k)$. The Gaussian integral supplies $A(0)=\sqrt{\pi }$, soApplying part (ii) with $n=1$ givesIt is imaginary and odd, as expected for the transform of a real odd function.