Relativistic Breit-Wigner Distribution

There is a single shape parameter \(\rho\) which takes values in \((0, \infty)\). The support is \(0 \leq x < \infty\).

\begin{eqnarray*} f\left(x, \rho\right) & = & \frac{k}{\left(x^2 - \rho^2\right)^2 + \rho^2}\\ F\left(x, \rho\right) & = & -\frac{i k\left(\frac{\tan^{-1}\left(\frac{x}{c}\right)}{c} - \frac{\tan^{-1}\left(\frac{x}{\bar{c}}\right)}{\bar{c}}\right)}{2\rho} \end{eqnarray*}

\begin{eqnarray*} \mu & = & \frac{k}{2\rho} \left[\frac{\pi}{2} + \tan^{-1}\left(\rho\right)\right]\\ \mu_2 & = & \frac{k\pi}{4} \left[\frac{1 - \rho i}{\sqrt{-1 - \rho i}} + \frac{1 + \rho i}{\sqrt{-1 + \rho i}}\right]\\ \mu_3 & = & \infty\\ \mu_4 & = & \infty\\ \end{eqnarray*}

where

\begin{eqnarray*} c & = & \sqrt{-\rho (\rho + i)}\\ \bar{c} & = & \sqrt{-\rho (\rho - i)}\text{ is its complex conjugate}\\ k & = & \frac{2\sqrt{2}\rho^2\sqrt{\rho^2 + 1}}{\pi\sqrt{\rho^2 + \rho\sqrt{\rho^2 + 1}}} \end{eqnarray*}

Implementation scipy.stats.rel_breitwigner