Sum of a Poisson point process Let $Z_1, Z_2, \dots$ be a Poisson point process on $[0, 1]$ with intensity function $1/z$. What is the distribution of the sum $Z = \sum_{i=1}^\infty Z_i$?
One can construct $Z_1, Z_2, \dots$ by taking a standard Poisson point process $X_1, X_2, \dots$ on $[0, \infty)$ with intensity $1$ (so the number of points in any interval $I$ is Poisson-distributed with mean $|I|$) and setting $Z_i = \exp(-X_i)$.
The sum $Z$ seems to have mean $1$ and variance $1/2$. Empirical CDF and distribution function below. It looks like $Z$ is uniformly distributed on $[0,1]$ and exponentially distributed above $1$. Is something like that true?


 A: There is a formula for the Laplace transform of any additive functional of a Poisson process with intensity measure $\lambda$. Specifically, for any non-negative measurable function $f$,
$$E\left[e^{-\sum_{i=1}^\infty f(Z_i)}\right]=e^{-\int_0^\infty(1-e^{-f(u)})\lambda(du)}.$$
In your case, fixing $\theta>0$ and choosing $f(u)=\theta u$ yields
$$E\left[e^{-\theta Z}\right]=e^{-\int_0^\theta\left(\frac{1-e^{- u}}{u}\right) du}.$$
You can then easily compare this with the Laplace transform of any guess you might have for the distribution of $Z$. For example, if  $Z$ had a density of the form
$$f_Z(z) \propto 1_{(0,1)}(z)+e^{-\beta(z-1)}1_{(1,\infty)}(z),$$
for some constant $\beta>0$ as you suggest, then one would have
$$E\left[e^{-\theta Z}\right]\propto \frac{1-e^{-\theta}}{\theta}+\frac{e^{-\theta}}{\theta+\beta}.$$
Those two expressions do not seem to match, so the answer to your question should be no...
A: The density function of $Z$ is the Dickman $\rho$ function, normalized (that is, divided by its mass, $e^{\gamma}$).
This function, $\rho\colon [0,\infty)\to (0,\infty)$, is defined via $\rho(t)=1$ for $t \in [0,1]$ and $\rho'(t)=-\rho(t-1)/t$ for $t \ge 1$. Equivalently, $u\rho(u) = \int_{u-1}^{u} \rho(t)dt$. This already shows $\rho(t) \le 1/\Gamma(t+1)$, so it decays superexponentially.
Its Laplace transform is given by
$$\hat{\rho}(s):=\int_{0}^{\infty} e^{-st}\rho(t)dt = e^{\gamma+I(-s)}$$
where $I(s) = \int_{0}^{s} \frac{e^t-1}{t}dt$ (see Theorem III.5.10 in Tenenbaum's book, "Introduction to Analytic and Probabilistic Number Theory"; the same chapter contains a wealth of information on $\rho$). Plugging $s=0$ we get that $\int_{0}^{\infty} \rho(t)dt = e^{\gamma}$, so $\rho(t) e^{-\gamma}$ is indeed a density function.
From DRJ's answer, we know that if $f$ is your density function then
$$\int_{0}^{\infty} e^{-st} f(t)dt = e^{-\int_{0}^{s} \frac{1-e^{-u}}{u}du}=e^{I(-s)}$$
by change of variables $u=-v$. Standard uniqueness properties imply $f \equiv \rho e^{-\gamma}$.

This function features prominently in number theory and probability. In number theory $\rho(u)$ arises as the probability that a number $x$ is $x^{1/u}$-smooth (or friable). Equivalently, the CDF of the random variable $\log P(n)/\log n$ ($n$ random from $\mathbb{Z}\cap[1,x]$, $P(n)$ the largest prime factor of $n$) tends to $\rho(1/\cdot)$ as $x \to \infty$. Relatedly, in probability, $\rho(1/\cdot)$ arises as the CDF of the first coordinate of a Poisson-Dirichlet process.
In the last examples it arises as a CDF but in your situation it is a density function, so let me give an example where it arises as the latter. It is the density function for the limit of $\sum_{k=1}^{n} k Z_k/n$, where $Z_k$ are independent Poisson with parameters $1/k$. This is intuitive if we work with Laplace transforms:
$$\mathbb{E} e^{-s \sum_{k=1}^{n} k Z_k/n} = \prod_{i=1}^{n} \mathbb{E} e^{-s kZ_k/n} = \prod_{k=1}^{n} e^{\frac{1}{k}\left( e^{-sk/n}-1\right)} = e^{\sum_{k=1}^{n} \frac{1}{k}(e^{-sk/n}-1)}$$
and the exponent tends to $I(-s)$ with $n$. This is mentioned in "On strong and almost sure local limit theorems for a probabilistic model of the Dickman distribution" by La Bretèche and Tenenbaum, along with references.
A related nice fact: if you condition on $\sum_{k=1}^{n} k Z_k=n$, then the vector $(Z_1,\ldots,Z_k)$ becomes distributed like $(C_1(\pi_n),\ldots,C_n(\pi_n))$ where $\pi_n$ is a permutation chosen uniformly at random from $S_n$ and $C_i$ is its number of cycles of size $i$. This appears in the book "Logarithmic Combinatorial Structures" by Arartia, Barbour and Tavaré, specifically equation (1.15) and the discussion on page 26. (It could be that the book includes a discussion of the previous fact as well but I couldn't find it.)
