When calculating the Fourier transform of a function of the form $f(\vec{r}) = \frac{1}{4 \pi \left|\vec{r}\right|}$, one encounters the problem that the resulting integral does not converge, i.e. $$\mathcal{F}\left[\frac{1}{4 \pi |\vec{r}|}\right]\!\!\left(\vec{p}\right) = \int_{\mathbb{R}^n} \frac{1}{4\pi |\vec{r}|} e^{-i \vec{p} \cdot \vec{r}}\mathrm{d}^3\! r = \frac{1}{\left|\vec{p}\right|} \int_0^\infty \sin\!\left(\left|\vec{p}\right| \cdot r\right) \mathrm{d}r$$ (see e.g. this arXiv paper, appendix A). This problem comes up when trying to express the Coulomb potential in momentum space.
In the link above, this is circumvented by introducing an attenuating factor $e^{- \lambda \left|\vec{r}\right|}$ and subsequently taking the limit $\lambda \to 0$, i.e. $$\mathcal{F}\left[\frac{1}{4 \pi |\vec{r}|}e^{- \lambda \left|\vec{r}\right|}\right]\!\!\left(\vec{p}\right) = \int_{\mathbb{R}^n} \frac{1}{4\pi |\vec{r}|} e^{- \left(\lambda\left|\vec{r}\right| + i \vec{p} \cdot \vec{r}\right)}\mathrm{d}^3\! r = \frac{1}{\left|\vec{p}\right|^2 + \lambda^2}$$ $$\stackrel{\lambda\to 0}{\Rightarrow} \mathcal{F}\left[\frac{1}{4 \pi |\vec{r}|}\right] = \frac{1}{\left|\vec{p}\right|^2}$$
My question is whether this is mathematically justified. Essentially, as I see it, this operation comes down to exchanging the order of taking the limit and the integration: $$\mathcal{F}\left[\lim_{\lambda\to 0}\frac{1}{4 \pi |\vec{r}|}e^{- \lambda \left|\vec{r}\right|}\right]\!\!\left(\vec{p}\right) \stackrel{?}{=} \lim_{\lambda\to 0}\mathcal{F}\left[\frac{1}{4 \pi |\vec{r}|}e^{- \lambda \left|\vec{r}\right|}\right]\!\!\left(\vec{p}\right)$$
I'm guessing the question can somehow be answered with the concept of uniform convergence, but mathematical details like these are not often part of my physics courses, so I'm not very familiar with its application.