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?
 A: 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 $\alpha \approx 0.68$.
A: Set $x = e^{-y}$ for $y \in \Bbb R$. Thus, we look for probability distribution functions $p(x) = f (\ln x)$ for positive $x$ such that
$f = p \circ \exp$
is an even function.
Now, let us check that
$$
1 = \int_0^\infty p(x)\, dx = \int_0^1 p(x)\, dx + \int_1^\infty p(x)\, dx = \int_0^1 p(x) \left(1+ x^{-2}\right) dx \, ,
$$
where we have used the change of variable $z=1/ x$. This condition can be rewritten as
$$
1 = \int_{0}^{\infty} f(y)\left(e^{-y}+e^{y}\right) dy \, .
$$
For instance, we note that the function defined by $f(y) = e^{-(1+\sqrt{2}) |y|}$ is a suitable choice, and we have $$
p(x) = e^{-(1+\sqrt{2}) |\!\ln x|} \Bbb{I}_{x>0} = \left\lbrace
\begin{aligned}
&x^{(1+\sqrt{2})} && 0<x<1\\
&x^{-(1+\sqrt{2})} && x \geq 1
\end{aligned}
\right.
$$
in this case (see figure below). Thus there are obviously various possibilities, even if we restrict the study to continuous $p$.

NB. The function $p(x) = \exp(-\alpha e^{|\!\ln x|}) \Bbb{I}_{x>0}$ proposed by @molarmass corresponds to $f(y) = \exp(-\alpha e^{|y|})$.
