Take as an example $f(t)=1/\sqrt{1+t^2}$, which is a function in $L^2(\mathbb{R})$ that is not in $L^1(\mathbb{R})$. So the Fourier transform of $f$ is not absolutely convergent. However, the following limit does exist as a function in $L^2$:
$$
\hat{f}(s)=\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\frac{1}{\sqrt{1+t^2}}e^{-ist}dt \\
= \lim_{R\rightarrow\infty}\sqrt{\frac{2}{\pi}}\int_{0}^{R}\frac{1}{\sqrt{1+t^2}}\cos(st)dt.
$$
The above limit not only converges in the $L^2$ norm as $R\rightarrow\infty$, but it also converges pointwise everywhere except at $s=0$, which can be seen by looking at this integral in light of the alternating series test for convergence, keeping in mind that $1/\sqrt{1+t^2}$ is strictly monotone and converges to $0$ as $t\rightarrow\infty$. However, the integral does not converge absolutely for any $s\in\mathbb{R}$.
The above integral converges pointwise using the same reasoning if $1/\sqrt{1+t^2}$ is replaced by $1/(1+t^2)^{p}$ for any $p > 0$. However, the result is a Fourier transform that is not in $L^2(\mathbb{R})$ if $0 < p < 1/4$ because $1/(1+t^2)^p \notin L^2(\mathbb{R})$ for such $p$. So it cannot converge in $L^2$ for $0 < p < 1/4$.
I don't believe that the dominated convergence theorem says much about these simple cases.
First, a few quick things:
$1$. The uncertainty principle typically either has no $\pi$ in it or there is a $\pi$ in the denominator, not the numerator. I will provide it with no $\pi$, as it implies the version with it in the denominator. Also, if you’re working with the square of the $L^2$ norms, then the factor is $1/4$. At least, this is always how I've seen it stated.
$2$. There are a lot of scaling conventions for the Fourier transform. I'm going to use the one that makes it unitary, but I'll leave some extra steps in case you do not.
Onto the proof, observe that $$\xi\cdot \mathcal{F} f(\xi)=\mathcal{F}\left(\frac{1}{i}\frac{d}{dx}f\right)(\xi),$$ and $$\left[x,\frac{1}{i}\frac{d}{dx}\right] f=if.$$ Using these relations, Plancherel, and Cauchy-Schwarz, we see that
\begin{align*}
\left\|xf\right\|_{L^2}\|\xi\cdot \mathcal{F}f\|_{L^2}&=\left\|xf\right\|_{L^2}\left\|\mathcal{F}\left(\frac{1}{i}\frac{d}{dx}f\right)\right\|_{L^2}=\left\|xf\right\|_{L^2}\left\|\left(\frac{1}{i}\frac{d}{dx}f\right)\right\|_{L^2}\\
&\geq \left|\left\langle\frac{1}{i}\frac{d}{dx}f, xf\right\rangle\right|\geq \left|\text{Im}\left\langle\frac{1}{i}\frac{d}{dx}f, xf\right\rangle\right|\\
&=\frac{1}{2}\left|\left\langle\left[x,\frac{1}{i}\frac{d}{dx}\right]f,f\right\rangle\right|\\
&=\frac{1}{2}\left\|f\right\|_{L^2}^2=\frac{1}{2}\left\|f\right\|_{L^2}\left\|\mathcal{F}f\right\|_{L^2}\\
&=\frac{1}{2}.
\end{align*}
If one, instead, uses the semiclassical Fourier transform, then we pick up a factor of $h$ (or $\hbar$, if you like), which is consistent with the traditional version that you'll see.
Best Answer
You can define $r_+^{-m}$ as $$(r_+^{-m}, \phi) = \int_0^\infty r^{-m} \left( \phi(r) - \sum_{k = 0}^{m - 2} \frac {\phi^{(k)}(0)} {k!} r^k - \frac {\phi^{(m - 1)}(0)} {(m - 1)!} r^{m - 1} H(1 - r) \right) dr$$ and then define $r^{-m}$ in $\mathbb R^n$ as the integral in spherical coordinates: $$(r^{-m}, \phi) = \left( r_+^{-m + n - 1}, \int_{\mathcal S_r} \phi dS \right),$$ where the inner integral is taken over the surface of an $(n - 1)$-sphere. Then the formula $$\mathcal F[r^\lambda] = \int_{\mathbb R^n} r^\lambda e^{i \boldsymbol x \cdot \boldsymbol \xi} d \boldsymbol x = \frac {2^{\lambda + n} \pi^{n/2} \Gamma \!\left( \frac {\lambda + n} 2 \right)} {\Gamma \!\left( -\frac \lambda 2 \right)} \rho^{-\lambda - n}$$ still holds for $\lambda = -n -2 k - 1$, while for $\lambda = -n - 2k$ the result will be the regular part of $\mathcal F[r^\lambda]$, which isn't a homogeneous function of $\rho$: $$\mathcal F[r^{-n - 2k}] = [(\lambda + n + 2k)^0] \mathcal F[r^\lambda] = \\ \frac {\pi^{n/2}} {\Gamma(k + 1) \Gamma \!\left( k + \frac n 2 \right)} \left( -\frac {\rho^2} 4 \right)^k \left( -\ln \frac {\rho^2} 4 + \psi \!\left( k + \frac n 2 \right) + \psi(k + 1) \right), \\ k \in \mathbb N^0.$$