I have a little question that has been bugging ever since I took my Laplace Transform course a couple of months ago. If you see heaviside's function $H(t)$ pop up a lot it's because I'm using the Bilateral transform.
In this course, when we got to the Inverse Laplace Transform, we showed that any rational $R(z)$ function in the frequency domain corresponded to a time domain function of the form
$$ r(t) = \sum _{k} c_k \delta ^{(k)} (t) + H(t) \sum _{\alpha \in \mathbb{C}} \textrm{Res}({e^{tz} R(z)};{\alpha}). $$
And that's all and good, but then when we jump to the Bormwich integral, there's something I can't quite get. I remember reading that under the assumption
$$ |R(z)| < \frac{M}{|z^2|}, \qquad z \rightarrow \infty $$
which is simply a condition so that we can assert the convergence of a contour integral, then the Bromwich integral states
$$ r(t) = \frac{1}{2 \pi i} \lim _{\omega \rightarrow \infty} \int _{\sigma - i\omega}^{\sigma - i\omega} e^{tz} R(z) \, dz. $$
But this formula rests upon the assumption on $|R(z)|$. From the calculations I saw in the lecture notes, this integral came straight from finding an integral form for
$$ H(t) \sum _{\alpha \in \mathbb{C}} \textrm{Res}({e^{tz} R(z)};{\alpha}). $$
My problem is that I keep reading everywhere else that this Bromwich integral is applicable to an apparently arbitrarily large class of functions, without it even mattering that $|R(z)| < M/|z^2|$. Am I missing something here? Is the requirement that
$$ |R(z)| < \frac{M}{|z^2|}, \qquad z \rightarrow \infty $$
actually necessary or are authors just lazy enough to omit this?
I'm really confused, because I see it in pretty much every place that says anything about the ILT, and it's nothing like I remember from my lecture notes. Any help will be kindly appreciated!