I would like to prove that $$ \hat{f}_I(s) = \exp \left\{- \lambda_1 2 \pi \int_0^{\infty} t(1-\hat{f}_F (s/l(t))) dt \right\} $$ is square integrable provided that $\hat{f}_F$ is square integrable and $l(t) = (At)^{\beta}$, $A>0$, $\beta >2$. Variable $F \in \mathbb{R}_+ $ corresponds to some probability density here.
This is a result from this monograph: https://hal.inria.fr/inria-00403040v4/document, "Stochastic Geometry and Wireless Networks", page 9. It is proclaimed as a "easy to see" result, but I am nevertheless confused.
Let me answer my own question. However, HAL monograph implies that the result would apply for general probability variable $F$ (i.e. does not have to admit a density). I would appreciate further insights.
Let $I$ be the interference and $F$ the virtual power. Virtual power $F$ admits a density $f, $ and $\mathbb{P}\{F >0 \} >0.$ Consider a path-loss model $l(r) = (A r)^{\beta},$ where $A>0, \beta>1.5.$ We have that \begin{align*} | \mathcal{L}_I(i s) | &= \exp \left\{\Re \left(- \lambda_1 2 \pi \int_0^{\infty} t\left(1- \mathcal{L}_F(is/(A t)^{\beta}) \right) dt \right) \right\} \\ &= \exp \left\{- \lambda_1 2 \pi \int_0^{\infty} t\left(1- \Re (\mathcal{L}_F(is (A t)^{-\beta})) \right) dt \right\} \\ &= \exp \left\{ -\lambda_1 2 \pi \int_0^{\infty} t\left(1- \int_0^{\infty}\Re \left( e^{-\zeta i s (At)^{-\beta}}\right) f(\zeta) d\zeta \right) dt \right\} \\ &= \exp \left\{- \lambda_1 2 \pi \int_0^{\infty} t\left(1- \int_0^{\infty} \cos(\zeta s (At)^{-{\beta}}) f(\zeta) d\zeta \right) dt \right\}. \\ \end{align*}
For some $\tau_s,$ which depend of given s and $f$, we have for $t \geq \tau_s$ $$ 1 \geq \int_{0}^{\infty} \cos(\zeta s (At)^{-\beta}) f(\zeta) d\zeta \geq 1-1/t^3, (*) $$ which implies that the tail is integrable: $$ \int_{\tau_s}^{\infty} t \left(1- \int_0^{\infty} \cos ( \zeta s (At)^{-\beta}) f(\zeta) d\zeta \right) dt \leq \int_{\tau_s}^{\infty} 1/t^2 = 1/\tau_s, $$ hence $$ \int_0^{\infty} t\left(1- \int_0^{\infty} \cos ( \zeta s (At)^{-\beta}) f(\zeta) d\zeta \right) dt$$ is (absolutely) integrable for every $s \in \mathbb{R}.$
Notice that by Riemann–Lebesgue lemma $$\int_{0}^{\infty}\cos(\psi) f(\zeta) d\zeta \rightarrow 0 \text{ as } |\psi| \rightarrow \infty,$$ hence we can pick large enough $\psi> 0$ so that $$ -1/2 <\int_0^{\infty} \cos(\zeta s (At)^{-\beta}) f(\zeta) d\zeta < 1/2 $$ for all $t $ and $s$ approving $|s| t^{-\beta} \geq \psi.$ In this case \begin{align*} &\int_{0}^{\infty} t\left(1- \int_0^{\infty} \cos(\zeta s (At)^{-{\beta}}) f(\zeta) d\zeta \right) dt \\ & \geq \int_{0}^{(|s|\psi^{-1})^{1/\beta}} t\left(1- \int_0^{\infty} \cos(\zeta s (At)^{-{\beta}}) f(\zeta) d\zeta \right) dt \\ & \geq \int_{0}^{(|s|\psi^{-1})^{1/\beta}} t\left(1- 1/2 \right) dt = 1/4 (|s| \psi^{-1})^{1/\beta} \end{align*} for any given s.
Then \begin{align*} &\int_{-\infty}^{\infty} \exp \left\{- \lambda_1 2 \pi \int_0^{\infty} t\left(1- \int_0^{\infty} \cos(\zeta s (At)^{-{\beta}}) f(\zeta) d\zeta \right) dt \right\} ds \\ &\leq \int_{-\infty}^{\infty} \exp\left\{-\frac{2\pi\lambda_1}{4} (|s| \psi^{-1})^{1/\beta}\right\} ds \\ &= \frac{2^{1+\beta/2} (2 \pi \lambda_1)^{-\beta/2} \Gamma(1+\beta/2)}{\psi}. \end{align*}
We conclude that $|\mathcal{L}_I(is)|$ is integrable. Clearly it is also square integrable.
Proof of (*): Assume that second moment $\int_0^{\infty} \zeta^2 f(\zeta) d\zeta $ exists. After some sufficiently large $\tau_s \in \mathbb{R}_+$ \begin{align*} &1- \int_{0}^{\infty}\cos(\zeta s (At)^{-\beta}) f(\zeta) d\zeta \\ &=\ 1- \int_{0}^{\infty} \left(1- \frac{(\zeta s (At)^{-\beta})^2}{2}+ \frac{(\zeta s (At)^{-\beta})^3}{3!}- \dots \right) f(\zeta) d\zeta \\ &\leq \int_{0}^{\infty} \frac{(\zeta s (At)^{-\beta})^2}{2} f(\zeta) d\zeta < 1/t^3, \end{align*} for all $t > \tau_s,$ because $\beta > 1.5 $ by assumption.