13 Luglio 2004 alle 9:50
#3602
::
Io farei così
`
\begin{multline}\label{eq:densitaquattro}
f(\varphi)=\frac{1}{2\pi}\exp
\bigg(-\frac{r^2_{los}}{2\sigma^2}\bigg)\\
\cdot\bigg[1+\sqrt{\frac{\pi}{2}}\frac{r_{los}\cos
\varphi}{\sigma}\exp\bigg(\frac{r^2\cos^2\varphi}{2\sigma^2}\bigg)\bigg]\\
\cdot\bigg[1+\textrm{erf}\bigg(\frac{r_{los}\cos\varphi}{\sigma\sqrt{2}}\bigg)\bigg]
\end{multline}
`