Sufficient conditions for the Fourier transform to be integrable

In this article, the condition for the Mellin inversion theorem (MIT) is that the inverse Mellin transform $$ f(x) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} F(s) x^{-s} \, ds $$ converges absolutely and uniformly for $c \in (a,b)$, with $F(s)$ being some function that is analytic in the strip $a < \Re(s) < b$. This is so that the Fourier inversion theorem (FIT) can be used to prove the former theorem. However, I am struggling to that the aforementioned condition is sufficient. In my attempted proof of the MIT, I arrive at $$ \mathcal{F}^{-1}\{F_\xi\}(t) = e^{-ct} \mathcal{M}^{-1}\{F\}\left(e^{-t}\right), $$ where $s = c + 2\pi i \xi$ and $F_\xi (\xi) = F(s)$. This is where I am to apply the FIT and then proceed to easily complete the proof. The version of the FIT that I am working with requires $F$ to be piecewise continuous, as well as $F_\xi$ and $\mathcal{F}\{F_\xi\}$ being $\mathcal{L}^1(\mathbb{R})$. Since $F_\xi$ is analytic, it is piecewise continuous. It is also easy to show that it is integrable from the convergence of the inverse Mellin transform. How do I show that $\mathcal{F}\{F_\xi\}$ is also integrable? I have seen examples on this site of integrable functions with fourier transfroms that are not integrable.