Knowing that $f \in L^2(\mathbb{R^n})$, then $$\lim_{\epsilon\to0+}\int_{\mathbb{R^n}}e^{-i\langle x,\xi\rangle -\epsilon|x|}f (x)dx=\mathcal{F}_2f(\xi)$$ in $L^2(\mathbb{R^n})$, where $\mathcal{F}_2f$ denotes the Fourier transform in $L^2(\mathbb{R^n})$.
Meanwhile, at the first term after the limit we have a Fourier transform in $L^1(\mathbb{R^n})$, i.e. $\mathcal{F}_1(e^{-\epsilon|x|}f(x))(\xi) $ and therefore in theory to be well posed one should have $e^{-\epsilon|x|}f(x)\in L^1(\mathbb{R^n})$, but I don't think it can be deduced in any way... Furthermore, for the resolution I had thought of considering the sequence $\{e^{-\frac{|x|}{n}}f(x)\}_{n \in \mathbb{N}}$ which converges in $ L^2(\mathbb{R^n})$ to $f(x )$ and furthermore using which is a Cauchy sequence and Parseval's theorem we obtain the convergence in $ L^2(\mathbb{R^n})$ of the sequence $\{\mathcal{F}_1(e^{-\frac{|x|}{n}}f(x))\}_{n \in \mathbb{N}}$ and therefore it must necessarily converge to $\mathcal{F}_2f$ (note that $\epsilon=1/n->0^+$ for $n->+\infty$ ). I don't know if everything is right though, so if anyone can give me a hand, thank you very much.