= Endpoint concentration in the arcsine distribution
An unbounded <probability density function> does not imply an atom. Here
$$
\mathbb P(X=0)=\mathbb P(X=1)=0,\qquad
\mathbb P(0<X<\varepsilon)=\frac2\pi\arcsin\sqrt\varepsilon
\sim\frac2\pi\sqrt\varepsilon\quad(\varepsilon\downarrow0).
$$
The last equivalence follows from $\arcsin u/u\to1$. The endpoint interval has much more <probability> than the same interval under a <uniform distribution>, whose value is $\varepsilon$, but its <probability> still tends to zero. This is an <integrable singularity>.
For example, the first 1% of the unit interval contains about $6.377\%$ of the <probability>; the last 1% contains the same amount because the <arcsine distribution> is a <symmetric probability distribution>. Each exact endpoint still has zero <probability>. Density and interval <probability> answer different questions.
\Image[/arcsine-cdf-and-equal-bin-probabilities.png]
{title=Endpoint density diverges, but every finite bin has finite probability}
{description=Original exact curves and bin probabilities. Left: the arcsine <cumulative distribution function> compared with the uniform one. Right: ten equal-width bins have probabilities $F((j+1)/10)-F(j/10)$, not heights sampled from the divergent density. Their total is one.}
A numerical integration routine must respect these endpoint <integrable singularities>. The substitution $x=\sin^2\theta$ gives
$$
f(x)\,dx=\frac2\pi\,d\theta,\qquad0<\theta<\pi/2.
$$
It turns a difficult-looking density integral into a constant measure in angle. Integrating finite bins is also a faithful way to draw the distribution near its endpoints; sampling $f(0)$ to make a bar height is not.
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