I'm trying to solve the following equation to find an approximation for $W(s,\alpha)$ given that I know $f(x,y)$.
\begin{equation} f(x,y)=\int_{-\infty}^{\infty}\int_{0}^{\pi}W(s,\alpha)\delta(g)d\alpha ds \end{equation}
Where $\delta$ is the Dirac delta function and $g(s,\alpha,x,y)$ is given by:
\begin{equation} g=x\cos(\alpha)+y\sin(\alpha)-s \end{equation}
--------------------------------------------------My attempt-----------------------------------------------
I have a strong suspicion that this can be solved numerically quickly using the Radon Transform, but I'm not sure how. I'm aware that the radon transform of a function $q(x,y)$ can be written as:
\begin{equation} R[q(x,y)]=p(s,\alpha)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}q(x,y)\delta(g)dxdy \end{equation} Using polar coordinates $x=r\cos(\theta)$, and $y=r\sin(\theta)$ this integral can be written as \begin{equation} p(s,\alpha)=\int_{-\infty}^{\infty}\int_{0}^{\pi}|r|q(r\cos\theta,r\sin\theta)\delta(r\cos\theta\cos\alpha+r\sin\theta\sin\alpha-s)d\theta dr \end{equation} This looks similar to my question above. Here I'm stuck.