Probability distribution with $p(x) = p(1/x)$ I am looking for a probability distribution defined on $x\in(0,\infty)$ that satisfies, for all $x$:
$$p(x) = p(1/x)$$
Naturally, we must have:
$$\int_0^\infty p(x) \, dx= 1$$
and thereore:
$$
\int_0^1 p(x) \, dx + \int_1^\infty p(x) \, dx = 1
$$
Does such a probability distribution exist?

It is not true that $\int^1_0 p(x)\, dx$ and $\int^\infty_1 p(x) dx$ necessarily both need to be equal to $0.5$, just that their sum needs to be equal to $1$.
As easy example could be $$p(x) = \begin{cases} \exp(-\alpha/x)&\text{for $0<x<1$},\\
\exp(-\alpha x) & \text{for $x \geq 1$},\end{cases}$$
where $\alpha$ is a parameter such that
$$\left[\int^1_0\exp(-\alpha/x)\, dx\right] + \left[\int^\infty_1 \exp(-\alpha x) \, dx\right] = 1.$$
A quick numerical check yields $\alpha \approx 0.68$.
