I'm looking to understand how to compute this distributional Fourier transform:
$$\mathcal{F}(x \delta(x)) = \int x \delta(x) e^{-ikx} \, dx$$
Were it $x f(x)$ I would use the common rule that $\mathcal{F}(x f(x)) = i \partial_k \mathcal{F}(f(x) = i \partial_k \delta(k)$. But I'm not sure it applies here.
Apologies if this has been asked before. Searching for questions related to $\delta(x)$ doesn't seem to work very well since it gives a lot of questions about $\delta(x)$ itself.
You can either compute the Fourier transformation directly, since $$ \mathscr{F}[x\delta(x)](k) = \left.\int_\mathbb{R} x\delta(x)e^{-ikx} \,\mathrm{d}x = xe^{-ikx}\right|_{x=0} = 0, $$ or, if you really want to use the relation $\mathscr{F}[xf(x)](k) = i\partial_k\hat{f}(k)$, one obtains $\mathscr{F}[x\delta(x)](k) = i\partial_k\hat{\delta}(k) = 0$, given that $\hat{\delta}(k) = 1$.