Normalizable, but singular distribution I have obtained a probability distribution for the observable $l$ which takes the form:
$$ \frac{dP}{dl}=\frac{(1-\sqrt{1-3l^{2}})^{2}}{l^{3}\sqrt{1-3l^{2}}}\exp\left[-\frac{4\pi}{9l^{2}}(1-3l^{2})^{3/2}+\frac{4\pi}{9l^{2}}\right]$$
where $P(l)$ is the probability that $l$ is between $0$ and $l$. This distribution is valid for $l\in\{0,\frac{1}{\sqrt{3}}\}$. It is normalizable because:
$$\int_{0}^{\frac{1}{\sqrt{3}}}\frac{dP}{dl}dl=constant$$
However, it is singular at the value $l=\frac{1}{\sqrt{3}}$. This is counterintuitive because a normalizable distribution means that the sum of probabilities can be normalized to $1$. However, the value $l=\frac{1}{\sqrt{3}}$ has infinite probability.
How is this consistent? Is this observed in any well known physical system? What can it mean about the behavior of this system?
Thank you
 A: There is a difference between the probability distribution (or: probability density) and the propability. The former is your function $p(l) = \textrm{d} P / \textrm{d} l$, we obtain by integrating over some range of $l$,
\begin{equation}
 P([a,b]) = \int_a^b \textrm{d}l\ p(l).
\end{equation}
I.e. it is only meaningful to talk about the probability associated with an interval and if the volume of that interval goes to zero, $\vert a - b \vert \to 0$, so does $P([a,b])$. And as you already noted, your $p(l)$ integrated over any finite region always gives a finite number (i.e. it is an integrable function).
To see why there is no divergence for $\vert a - b \vert \to 0$ with $b = 1/\sqrt{3}$ (in fact we are showing that $p(l)$ is integrable despite the divergence), we can investigate $p(l)\textrm{d}l$. Changing variables to $x = \sqrt{3}l \in [0,1]$ and then to $x = \sin \theta$ with $\theta \in [0,\pi/2]$, we obtain
\begin{equation}
  p(l)\textrm{d}l = 3 \frac{(1-\cos\theta)^2}{\sin^3 \theta} \exp\left[ - \frac{4\pi}{3 \sin^2\theta}(\cos^{3/2}\theta - 1) \right] \textrm{d} \theta.
\end{equation}
This is manifestly finite as $\theta \to \pi/2$ (corresponding to $l = 1/\sqrt{3}$). Note that under the change of variables $\textrm{d}l = \textrm{d}x / \sqrt{3} = \cos\theta \textrm{d}\theta / \sqrt{3}$. This factor of $\cos\theta$ cancels the $\sqrt{1-3l^2} = \cos\theta$ in the denominator.
This shows that when considering intervals, the divergences of your probability distribution are unproplematic.
Appendix:
To see that the limit $\theta \to 0$ is also unproblematic, expand in the numerator, $(1- \cos\theta)^2 \to \theta^4/4$, and the denominator, $\sin^3\theta \to \theta^3$. This leaves a power of $\theta \to 0$. The exponential goes to $\exp(- \textrm{large}) \to 0$.
