This might be a result of a lack of properly understanding the Radon transform or maybe a lack of integration techniques. I do understand that one integrates a certain function over determined hyperplane. But when it comes to actual computations i struggle a bit. For example, how could one calculate $$\int_{\mathbb{R}^n} e^{-|x|^2}\delta(\langle\omega,x\rangle-t)\:dx$$ i.e, the Radon transform of the bump function $e^{-|x|^2}$ at the hyperplane $\langle\omega,x\rangle=t$.
Thank you all in advance.
I will assume $\omega \in \mathbb{S}^{n-1}$ and $t \in \mathbb{R}$. The key here is to rotate the coordinate system in such a way that the hyperplane in question is perpendicular to one of the axes. Let $R \in \mathrm{SO}(n)$ be a rotation matrix that maps $\omega$ to $\mathrm{e}_n$ (of course, any other standard basis vector also works). Then we can write $$ \mathcal{R} \left(x \mapsto\mathrm{e}^{-\left|x\right|^2} \right) (\omega,t) \equiv \int \limits_{\mathbb{R}^n} \mathrm{e}^{-\left|x\right|^2} \operatorname{\delta}(\langle\omega,x\rangle - t) \, \mathrm{d} x = \int \limits_{\mathbb{R}^n} \mathrm{e}^{-\left|R x\right|^2} \operatorname{\delta}(\langle R\omega, R x\rangle - t) \, \mathrm{d} x \, , $$ since $\langle \cdot,\cdot \rangle$ is rotationally invariant. The same is true for the integration measure, so changing variables to $y = R x$ we find $$ \mathcal{R} \left(x \mapsto\mathrm{e}^{-\left|x\right|^2} \right) (\omega,t) = \int \limits_{\mathbb{R}^n} \mathrm{e}^{-\left|y\right|^2} \operatorname{\delta}(\langle \mathrm{e}_n,y\rangle - t) \, \mathrm{d} y = \int \limits_{\mathbb{R}^n} \mathrm{e}^{-\left|y\right|^2} \operatorname{\delta}(y_n - t) \, \mathrm{d} y \, . $$ The integral over $y_n$ is now trivial and those over the perpendicular directions are just Gaussian integrals, so $$ \mathcal{R} \left(x \mapsto\mathrm{e}^{-\left|x\right|^2} \right) (\omega,t) = \pi^{(n-1)/2} \mathrm{e}^{-t^2} \, . $$ Clearly, this idea works for the Radon transform of any rotationally symmetric function, as long as the integral converges.