At times, I am required to take the Fourier Transform of some function that does not decay quickly enough for the Fourier Transform to converge in the usual sense. For example, $$ \int_{-\infty}^{\infty} \sin(x) e^{-i k x} dx $$ where we must multiply by some "convergence factor" that decays as $\left| x \right| \to \infty$ and then the limit taken so that the "convergence factor" approaches 1. $$\lim_{\eta \to 0} \int_{-\infty}^{\infty} f(x,\eta) \sin(x) e^{-i k x} dx$$ $$\lim_{\eta \to 0} f(x, \eta) = 1$$ The 'magical' part of this procedure is that it seems to not depend on which function $f(x, \eta)$ we choose, so long as it fulfills the above properties.
Question: What assumptions do we need to make about $f(x, \eta)$ to be sure that: (1) The Fourier Transform of some smooth function $g(x)$ that does not grow too quickly as $\left|x\right| \to \infty$ (my guess is that $g(x)$ must grower slower than exponentially) exists and (2) that this transform is independent of our choice of $f(x, \eta)$?
I am trained in physics rather than mathematics, so my understanding of what is rigorously going on here is limited. My basic belief is that this procedure lives in a space larger than the usual elementary functions and includes some distributions. From browsing Wolfram MathWorld it seems like $f(x, \eta)$ being a Schwartz Function with the appropriate limit for $\eta$ is a sufficient condition, but is it necessary?