If you "fill in" the indefinite integral with a (definite integral with) bounds -infinity to x, then I think the first step works. I'm not 100% how to deal with the 1/(1-integral) step, but my guess would be to transform to the Laplace domain because Laplace transform analysis is "aware of" convergence issues, i.e. the region of convergence "pops out" when calculating the transform, or it's present in the lookup table entry.