My question is regarding the Laplace transform and it's inversion formula given by the "Mellin", "Bromwich" or "Fourier-Mellin" integrals (found on wikipedia). Consider for simplicity, the two sided Laplace transform
$$
{F}(s)= \int_{-\infty}^{+\infty} f(t)e^{-st}dt
$$
of course, accompanied by its region of convergence (ROC). The inversion formula in question is the following
$$
f(t) = \lim_{\Omega\to\infty}\frac{1}{2\pi i}\int_{\sigma-i\Omega }^{\sigma+i\Omega}{F}(s)e^{st}ds
$$
with $\sigma$ chosen to be greater than the real part of all singularities of $\hat{f}(s)$. The question is regarding the last part. Why do we require such condition on $\sigma$?
Let me give a little more details. Recall the Fourier transform pair (written in terms of angular frequency $\omega$):
$$
\hat{f}(\omega)= \mathcal{F}\{f\}=\int_{-\infty}^{+\infty}f(t)e^{-i\omega t}dt, \ \ \ f(t)=\mathcal{F}^{-1}\{\hat{f}\}= \frac{1}{2\pi}\int_{-\infty}^{+\infty}\hat{f}(\omega)e^{i\omega t}d\omega
$$
Now, let $s=\sigma+i\omega$. Then, the Laplace transform:
$$
F(s) = \int_{-\infty}^{+\infty}\left(f(t)e^{-\sigma t}\right)e^{-i\omega t}dt = \mathcal{F}\{f(t)e^{-\sigma t}\}
$$
Let $f_\sigma(t)=f(t)e^{-\sigma t}$ and $\hat{f}_\sigma(\omega) = \mathcal{F}\{f_{\sigma}(t)\}$. Then, $F(s)=F(\sigma+i\omega) = \hat{f}_\sigma(\omega)$. Therefore, using Fourier inversion formula
$$
f(t) = e^{\sigma t}{f}_\sigma(t)=e^{\sigma t}\mathcal{F}^{-1}\{\hat{f}_\sigma(\omega)\} = e^{\sigma t}\frac{1}{2\pi}\int_{-\infty}^{+\infty}\hat{f}_\sigma(\omega)e^{i\omega t}d\omega = \lim_{\Omega\to\infty}\frac{1}{2\pi}\int_{-\Omega}^{+\Omega} \hat{f}_\sigma
(\omega)e^{(\sigma + i\omega)t}d\omega$$
And we make the change of variables $s = \sigma+i\omega$ where $ds = id\omega$ since $\sigma$ was a constant chosen inside the region of convergence of $F(s)$. Then,
$$
f(t) = \lim_{\Omega\to\infty}\frac{1}{2\pi i}\int_{\sigma-i\Omega}^{\sigma+i\Omega}\hat{f}_\sigma(s)e^{st}ds=\lim_{\Omega\to\infty}\frac{1}{2\pi i}\int_{\sigma-i\Omega}^{\sigma+i\Omega}F(s)e^{st}ds
$$
Obtaining the Mellin inversion integral with the requirement of $\sigma$ in the ROC of $F(\sigma+i\omega)$. Then, a more detailed version of my question would be: since any $\sigma$ in the ROC would do the work, why require $\sigma$ greater than the real part of the singularities of $F(s)$?
I believe that the answer may lie in the fact that such integral is computed in practice using residue theory in complex analysis. Let the contour $\Gamma_\Omega = \Gamma^1_\Omega+\Gamma^2_\Omega$ where $\Gamma_\Omega^1$ the line from $s=\sigma-i\Omega$ to $s=\sigma+i\Omega$ and $\Gamma_\Omega^2$ is the half-circle to the left centered at $s=\sigma$ with radius $\Omega$. Clearly $\Gamma_\Omega$ is a closed contour and
$$
\frac{1}{2\pi i}\oint_{\Gamma_{\Omega}}F(s)e^{st}ds = \frac{1}{2\pi i}\int_{\Gamma_{\Omega}^1}F(s)e^{st}ds + \frac{1}{2\pi i}\int_{\Gamma_{\Omega}^2}F(s)e^{st}ds
$$
where the integral over $\Gamma_{\Omega}^1$ corresponds to the inversion formula we derived before as $\Omega\to\infty$. Moreover, assume that the integral over $\Gamma_{\Omega}^2$ vanishes. Then, the inversion formula becomes
$$
f(t) = \frac{1}{2\pi i}\oint_{\Gamma_{\Omega}}F(s)e^{st}ds
$$
and we can use residue theorem to compute the integral. This is,
$$
f(t) = \sum_{i=1}^n \text{Res}(F(s)e^{st}; s_i)
$$
where $s_1,\dots,s_n$ are the singularities of $F(s)$ contained inside the closed contour $\lim_{\Omega\to\infty}\Gamma_\Omega$, this is, over all singularities with real parts less than $\sigma$. Here, we get a condition which resembles the one of my original question. Now, the final version of my question is the following.
Is the condition that ALL real parts of the singularities of $F(s)$ be less than $\sigma$ posed such that the contour integral over $\lim_{\Omega\to\infty}\Gamma_\Omega^2$ vanishes?
Assuming the opposite, if such integral vanishes for arbitrary sigma in the ROC, we would obtain different answers for $f(t)$ since different amount of residues would be included in the formula. Then, it must be the case that such integral only vanishes for $\sigma$ great enough to use all residues to compute $f(t)$. However, besides this, I don't see understand the reason (perhaps intuition) behind this vanishing only for $\sigma$ satisfying this conditions (i.e. by analysing the integral over $\Gamma_{\Omega}^2$ by itself).
Moreover, this reasoning is only valid if the goal was to use residues. However, the integral over $\Gamma_{\Omega}^1$ already gave the correct answer (although hard to compute without residue theory). So, its the condition over $\sigma$ posed only to compute the inversion formula using residues? Or is there something wrong with my reasoning regarding just $\sigma$ in the ROC and computing the inversion with the integral over $\Gamma_{\Omega}^1$.?
Best Answer
Just as someone suggested in the comments: The condition on $\sigma$ greater than all the singularities of $F(s)$ and its relation with the ROC, is precisely that the ROC is always of the form Re$\{s\}>\sigma$ for some $\sigma$. see the wikipedia description of the ROC. Intuitively, this is so, since for the integral to converge we need $e^{-st}$ sufficiently small (i.e. $s=\sigma+i\omega$ and $\sigma$ sufficiently big) to overcome the growth of $f(t)$. Moreover, the singularities cannot be in the ROC, and they occur precisely because they are outside of it. (This explanation is due to @Yves Daoust in the comments)
Hence, answering to my questions:
Q: Why do we require such condition on $\sigma$? A: To ensure that the line $s=\sigma$ lies in the ROC.
Q: Since any $\sigma$ in the ROC would do the work, why require $\sigma$ greater than the real part of the singularities of $F(s)$? A: since the ROC and "$\sigma$ greater than the real part of the singularities of $F(s)$" are the same set.
Q: Is the condition that ALL real parts of the singularities of $F(s)$ be less than $\sigma$ posed such that the contour integral over $\lim_{\Omega\to\infty}\Gamma_\Omega^2$ vanishes? A: Not necessarily.
Q: So, its the condition over $\sigma$ posed only to compute the inversion formula using residues? Or is there something wrong with my reasoning regarding just $\sigma$ in the ROC and computing the inversion with the integral over $\Gamma_\Omega^2$.? A: As stated before, the condition is not due to the used of the residue theorem, but because the ROC is described with such condition. Hence, the reasoning on the ROC is fine.