I have shown that if $f$ is a continuous function with scaling symmetry given by $f(mt)=m^kf(t)$ for any $m>0$, than $(\mathcal{Lf})(s) = m^{-k-1}(\mathcal{Lf})(\frac{s}{m})$.
For this $f$, what can be said about $\int_{0}^{\infty}f(t)\,dt$ and what is the behaviour of $\mathcal{L}f$ as $s \to \infty$
My initial thought is that the integral $\int_{0}^{\infty}f(t)\,dt$ diverges, because $f$ is defined as a function with scaling symmetry - it increases exponentially and therefore since you integrate on $\mathbb{R_+}$ it will diverge - I'm not sure if that argument is actually correct though.
As for the other part of the question - I think it converges to $0$, because of the definition: $\mathcal{L}(f) = \int_{0}^{\infty}e^{-st}f(t)\,dt $