My question rather counts to physics, but it still involves statistical mathematics.
Given the equation of motion for a pendulum: $$x(t) = A\,\cos(\omega\,t)$$
This is an object with an $x$-coordinate varying from $x \in [-A,A]$
Now for calculating the Probability distribution, that predicts the probability of finding the Pendulum in a certain interval I stumbled on the following:
$$ P = \dfrac{2}{T}\, \dfrac{1}{|v|}$$
I'm not sure why this is true but I'm also not familiar with statistics yet.
Anyway, the Probability can be explicitly calculated:
$$P = \dfrac{2}{T}\dfrac{1}{\omega\,A\,|\sin(\omega\,t)|} = \dfrac{1}{\pi\,A\,\sqrt{1-\cos^2(\omega\,t)}}$$
Now by geometry: $\omega\,t = \varphi = \arccos(x/A)$, so finally:
$$P(x) = \dfrac{1}{\pi\,A\,\sqrt{1-(x/A)^2}}$$
The only problem arising: $P(\pm A) \to \infty$. How can it be interpreted the probability distribution goes to infinity? This would imply that the probability for finding the Pendulum at $x \in [-A, x_i]$ with $x_i \in (-A,A]$ always is to be infinit, isn't it?
Where is $$ P = \dfrac{2}{T}\, \dfrac{1}{|v|}$$ from ?
If we assume that time $t$ is uniformly distributed on the interval $[0,2\pi/\omega]$. The CDF of $x(t)$ is then \begin{align} F(x)&=\frac{\omega}{2\pi}\operatorname{Leb}\Big\{t\in[0,2\pi/\omega]:A\cos(\omega t)\le x\Big\}=\frac{\omega}{2\pi}\Big\{\frac{2\pi}{\omega}-\frac{1}{\omega}\arccos\frac{x}{A}\Big\}\\ &=1-\frac{1}{2\pi}\arccos\frac{x}{A}\,. \end{align} Where Leb denotes Lebesgue's measure. The PDF of $x(t)$ is therefore $$ P(x)=\frac{d}{dx}F(x)=\frac{1}{2\pi}\frac{1}{\sqrt{1-x^2/A^2}}\,. $$ It is not surprising that this PDF has infinite peaks at $x=\pm A$ because the speed of the pendulum around the maximum amplitudes is slowest.
There is nothing wrong with an infinite probability density. The probability of finding the pendulum in an interval $[x,y]$ is $F(y)-F(x)$ and this is finite for every interval.