= Solution
Let $A(k)$ be the <Fourier transform> of $e^{-x^2}$. Since $(e^{-x^2})'=-2xe^{-x^2}$, <integration by parts> gives $\mathcal F[xe^{-x^2}]=-(ik/2)A(k)$. Differentiation under the <integral> then yields $A'(k)=-(k/2)A(k)$. The Gaussian <integral> supplies $A(0)=\sqrt\pi$, so
$$
A(k)=\sqrt\pi e^{-k^2/4}.
$$
Applying part (ii) with $n=1$ gives
$$
\boxed{\mathcal F[xe^{-x^2}](k)=iA'(k)=-\frac{i\sqrt\pi}{2}k e^{-k^2/4}.}
$$
It is imaginary and odd, as expected for the transform of a real odd function.
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