I stumbled across a property of the Laplace Transform that can transform an improper integral of a product of 2 function.
$$\int_0^{\infty} fg = \int_0^{\infty} \mathcal{L}(f) \mathcal{L}^{-1}(g)$$
This is very interesting and you can even use it to compute the dirichlet integral extremely easily:
$$2\int_0^{\infty} \frac{\sin(x)}{x}dx = 2\int_0^{\infty} \mathcal{L}(\sin(x))\mathcal{L}^{-1}(x^{-1}) = 2\arctan(x) \bigg\rvert_{0}^{\infty}=\pi$$
A pretty neat tool! But how would I go about proving that this is true? I've searched around but haven't found anything about this and can't even find the original place I first learned about this property.
A rigorous formulation of the identity that is easy to prove states that $$ \int_0^{\infty} ({\cal L} f)(s) g(s) \ ds= \int _0^{\infty} f(s) {\cal L}(g)(s) \ ds$$
(Your version reduces to this one when you symbolically replace $g\to {\cal L}(g$).
Sketch of proof.
The proof begins by re-writing the LHS as
$$ \int_0^{\infty} ({\cal L} f)(s) g(s) \ ds = \int_{s=0}^{s=\infty}\left (\int _{x=0}^{\infty} f(x) e^{-xs} g(s) dx\right) \ ds$$
and then swapping the roles of the two dummy variables of integration to convert it into the RHS.