I wonder under what condition the Laplacian transform is interchangeable with integrals.
Specifically, let $f(t, w)$ be continuous with repsect to both $w$ and $t$ and $g(t) = \int_0^1 f(t, w) \mathrm{d}w$. Then, when do we have \begin{align*} g(s) = \mathcal{L}(g(t)) &= \mathcal{L} \left(\int_0^1 f(t, w) \mathrm{d}w \right) \\ &= \int_0^1 \mathcal{L} (f(t, w)) \mathrm{d} w = \int_0^1 f(s, w) \mathrm{d}w, \end{align*} where $g(s)$ and $f(s, w)$ are Laplacian transforms of $g(t)$ and $f(t, w)$, respectively?
You can always do this (given the conditions satisfied by your function), since this is nothing more than a change in order of integration. That is, you want $$\int_0^{+\infty} e^{-st}\int_0^1 f(t,w)dw dt=\int_0^1 \int_0^{+\infty} e^{-st}f(t,w)dt dw,$$ or in other words $$\int_0^{+\infty} \int_0^1 e^{-st} f(t,w)dw dt=\int_0^1 \int_0^{+\infty} e^{-st}f(t,w)dt dw.$$