= Solution
For the <Gaussian function>, direct differentiation gives $f'=-tf$. Applying the <Fourier transform> identities from part (a),
$$
i\omega F=-iF',\qquad F'=-\omega F.
$$
Solving this first-order <ordinary differential equation> gives
$$
\boxed{h(\omega)=-\omega,\qquad F(\omega)=C e^{-\omega^2/2}.}
$$
In fact, the normalization can also be fixed: the <Gaussian integral> gives $F(0)=\int_{\mathbb R}e^{-t^2/2}\,dt=\sqrt{2\pi}$, so $C=\sqrt{2\pi}$ with the convention used here. The transform has the same Gaussian shape.
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