I'm having a trouble understanding a proof in regards to the inversion of the Radon transform in $\mathbb{R}^3$.
The statement is as follows: if $f \in \mathcal{S}(\mathbb{R^3})$, then $\Delta(\mathcal{R}^{*}\mathcal{R}(f)) = -8\pi^2f$ (where $\mathcal{S}$ denotes the Schwatrz space and $\Delta$ is the Laplacian).
Proof: Using the inversere Fourier transfrom we have
$\mathcal{R}(f)(t,\gamma) = \int_{-\infty}^{\infty} \hat{f}(s\gamma)e^{2\pi its} ds$ ($\hat{f}$ is the Fourier transform of $f$)
Hence $\mathcal{R}^{*}\mathcal{R}(f(x)) = \int_{S^2}\int_{-\infty}^{\infty} \hat{f}(s\gamma)\exp(2\pi ix \cdot \gamma s) ds d\sigma(\gamma)$
$\Delta(\mathcal{R}^{*}\mathcal{R}(f(x))) = \int_{S^2}\int_{-\infty}^{\infty} \hat{f}(s\gamma){\color{magenta}{(-4\pi^2s^2)}}\exp(2\pi ix \cdot \gamma s) ds d\sigma(\gamma)$
$=-4\pi^2\int_{S^2}\int_{-\infty}^{0} \hat{f}(s\gamma)\exp(2\pi ix \cdot \gamma s) s^2 ds d\sigma(\gamma) -4\pi^2\int_{S^2}\int_{0}^{\infty} \hat{f}(s\gamma)\exp(2\pi ix \cdot \gamma s) s^2 ds d\sigma(\gamma) $
$=-8\pi^2\int_{S^2}\int_{0}^{\infty} \hat{f}(s\gamma)\exp(2\pi ix \cdot \gamma s) s^2 ds d\sigma(\gamma)$
$=-8\pi^2f(x)$
After the Laplacian has been applied, I have no idea where the ${\color{magenta}{(-4\pi^2s^2)}}$ comes from.
As $\gamma \in S^{2}$, $\Delta\left( \exp(2\pi ix \cdot \gamma s) \right) = (2\pi i s )^{2} \gamma \cdot \gamma \, \exp(2\pi ix \cdot \gamma s) =-4\pi^{2}s^{2} \exp(2\pi ix \cdot \gamma s)$.