There are a number of questions about this topic, but none are very clear and none has a good answer. For functions $f,g:\mathbb{R}\to\mathbb{R}$, let their Laplace transforms be $$ \mathcal{L}[f] = F(s) = \int_0^\infty e^{-sx}f(x)\,\mathrm{d}x,\qquad \mathcal{L}[g] = G(s) = \int_0^\infty e^{-sx}g(x)\,\mathrm{d}x. $$ Suppose we are interested in their product, $(fg)(x) = f(x)g(x)$. Wikipedia states that the Laplace transform of their product is given by $$ \mathcal{L}[fg] = \frac{1}{2\pi i}\lim_{T\to\infty}\int_{c-iT}^{c+iT}F(\sigma)G(s-\sigma)\,\mathrm{d}\sigma. $$ Unfortunately the only reference is to a book that I don't currently have access to, and, surprisingly, I can barely find any information about this formula elsewhere on the web. Consequently, I have a number of closely related questions:
How can this formula be derived?
Is the formula the same (or does it have an equivalent) for the two-sided Laplace transform, $\mathcal{L}_2[f] = \int_{-\infty}^\infty e^{sx}f(x)\,\mathrm{d}x$? My main interest is really in the two-sided case, so I'm keen to know this.
How does this formula relate to the inverse Laplace transform? It seems almost like the inverse Laplace transform formula, except that it's missing an $e^{\sigma x}$ inside the integral. Can we express this formula in terms of the inverse Laplace transform somehow, and therefore perhaps make some more intuitive sense of it?
(This one might be wishful thinking) does the formula simplify in the case where $f$ and $g$ are nonnegative? In my case they are probability measures.
(Note: I'm actually trying to solve this other question of mine and figured this might be one approach.)