I was wondering whether there is a way to show this identity

$$\pi \int_{\mathbb{R}^3} \frac{f(x)}{|x|} dx = \int_{\mathbb{R}^3} \frac{\widehat{f(x)}}{|x|^2} dx $$ without using distributions for $f \in \mathcal{S}(\mathbb{R}^n)$.

The reason I ask is the following: Obviously this identity makes perfect sense in the classical way, but the most obvious proof would use the fourier transform of $\frac{1}{|x|^2}$ which is clearly not defined in $L^1$ or $L^2$.

Thus, I would be interested, whether one can proof this identity with classical methods, i.e. no distribution they, too?

I was wondering whether there is a way to show this identity

$$\pi \int_{\mathbb{R}^3} \frac{f(x)}{|x|} dx = \int_{\mathbb{R}^3} \frac{\widehat{f(x)}}{|x|^2} dx $$ without using distributions for $f \in \mathcal{S}(\mathbb{R}^n)$.

The reason I ask is the following: Obviously this identity makes perfect sense in the classical way, but the most obvious proof would use the fourier transform of $\frac{1}{|x|^2}$ which is clearly not defined in $L^1$ or $L^2$.

Thus, I would be interested, whether one can proof this identity with classical methods, i.e. no distribution theory, too?