What happens if we combine fractional part problems and randomness?
For a start here are two examples.
Let $r$ be a random variable with a given probability distribution function (PDF) $f(r)$ and define $x = \{1/r\}$ where $\{z\}$ is the fractional part of $z$.
For the two cases
(a) $f(r)=
\begin{cases}
1 & \quad \text{if}\hspace{2mm} 0\le r \le 1\\
0 & \quad \text{elsewhere}
\end{cases} $
(b) $f(r)=
\begin{cases}
\exp(-r) & \quad \text{if}\hspace{2mm} r\ge 0\\
0 & \quad \text{elsewhere}
\end{cases} $
calculate the PDF $h(x)$ of $x$ and the first two moments. Express the results in terms of known functions and constants as far as possible.
Best Answer
I start here with a self answer treating case (a). Maybe others join in. Case (b) is more challenging.
1. General
The PDF is given by the formula
$$h(x) = \int_{a}^b \delta (x-\{\frac{1}{r}\}) f(r) \,dr\tag{1}$$
Here $\delta(z)$ is the Dirac delta function and $a$ and $b$ are the limits of the integration region.
The procedure to deal with the fractional part was described in detail for instance in this example (Closed form of integral over fractional part $\int_0^1 \left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\}\,dx$)
Substituting $1/r\to y$ the integral becomes ($A = \frac{1}{a}, B=\frac{1}{b}$)
$$h(x) = \int_{B}^{A}\,dy\; \delta (x-\{y\}) f(\frac{1}{y})\frac{1}{y^2} \tag{2}$$
2. Case (a)
Here $f=1, A=\infty, B=1$, hence
$$h_{a}(x) = \int_{1}^{\infty} \,dy \;\delta (x-\{y\}) \frac{1}{y^2} =\sum_{k=1}^\infty a_k\tag{a1}$$
where, setting $y=k+\xi$ whence $\{y\}=\xi$, the summands become
$$a_k = \int_{0}^1 \delta (x-\xi) \frac{1}{(k+\xi)^2} \,d\xi = \frac{\theta (1-x) \theta (x)}{(k+x)^2}\tag{a2}$$
The heaviside functions make $a_k=0$ for $x<0$ or $x>1$. We remember this and drop $\theta$ henceforth.
Performing the sum gives
$$h_{a}(x) = \sum_{k=1}^\infty \frac{1}{(k+x)^2} = \psi ^{(1)}(x+1)\tag{a3}$$
Hence the PDF of $x=\{1/r\}$ for a flat Distribution of $r$ is given by a polygamma function. Vice versa we have obtained an interesting interpretation of $\psi ^{(1)}$.
The first few moments, defined as
$$\mu(q) = \int_0^1 x^q \psi ^{(1)}(x+1) \, dx$$
are given in the format $\{q,\mu(q)\}$ by
$$\left( \begin{array}{cc} 0 & 1 \\ 1 & 1-\gamma \\ 2 & \log (2 \pi)-\gamma -1 \\ 3 & \frac{1}{2} (\log (8)-12 \log (A)+3 \log (\pi )-2 \gamma -1) \\ 4 & \frac{1}{3} \left(\log (64)-36 \log (A)+6 \log (\pi )+\frac{9 \zeta (3)}{\pi ^2}-3 \gamma -1\right) \\ \end{array} \right)$$
Here $A$ is Glaisher's constant, which satisfies the relation $\log(A) = \frac{1}{12} - \zeta'(-1)$
Notice that the nomalization is ok, the average (q=1) is the well known result for $\int_{0}^1 \{\frac{1}{x}\} \,dx$