Suppose I have the following function: $$ M(s) = \int_{0}^{\infty}\frac{f(E)}{s+2\pi i /T(E)}\text{d}E \tag{1} $$ Where $f(E)$ is some potentially complex function, $T(E)$ is some real function and $s$ is some complex variable. I want to use the following procedure: $$ M(x) = \mathcal{L}^{-1}[M(s),x] = \int_{0}^{\infty}f(E)e^{-2\pi i x/T(E)}\text{d}E $$ Since the inverse Laplace transform of $(s - a)^{-1}$ is $e^{ax}$. But then, I could get back to $s$ space with a normal Laplace transform: $$ M(s) = \mathcal{L}[M(x),s] = \int_{0}^{\infty}\int_{0}^{\infty}f(E)e^{-2\pi i x/T(E)}\text{d}E\, e^{-x s} \text{d} x \tag{2} $$
My question is: can you just do that? The motivation for this procedure is to circumvent the difficulties that might arise at the poles of $(1)$, since $(2)$ has no poles. These functions have to be computed numerically so my main concern is whether or not $(1)$ and $(2)$ are equivalent in this case.
I think you are confusing yourself. So long as you can change the order of integration in (2) above, then the Laplace transform is pretty simple:
$$M(s) = \int_0^{\infty} dE \, f(E) \, \int_0^{\infty} dx \, e^{-[s+i 2 \pi/T(E)] x} = \int_0^{\infty} dE \, \frac{f(E)}{s+i 2 \pi/T(E)}$$