X-ray transform

Remarks

The same line is given by $(\theta, x)$ and $(-\theta, x)$, and $P f(-\theta, x) = P f(\theta, x)$. For $f \in L^1(\R^n)$ the transform is defined for almost every $x \in \theta^\perp$, for each $\theta$. By Fubini's theorem, $\int_{\theta^\perp} P f(\theta, x)\, e^{-i x \cdot \xi}\, dx = \hat f(\xi)$ for $\xi \in \theta^\perp$, where $\hat f(\xi) = \int f(y)\, e^{-i y \cdot \xi}\, dy$; this is the projection-slice theorem for lines.

Example: $f(y) = e^{-\abs{y}^2}$. For $x \perp \theta$, $\abs{x + t\theta}^2 = \abs{x}^2 + t^2$, so $P f(\theta, x) = \sqrt{\pi}\, e^{-\abs{x}^2}$.

Non-example: $f(y) = (1 + \abs{y})^{-1}$ is continuous and bounded but has no X-ray transform, since $\int_\R (1 + \abs{x + t\theta})^{-1}\, dt$ diverges like $\int dt / \abs{t}$.

For $n = 2$, let $\theta'$ be $\theta$ rotated by a right angle. Then $P f(\theta, s\theta')$ equals the Radon transform $R f(\theta', s)$, since both integrate $f$ over the line $y \cdot \theta' = s$.