Convergence of the Laplace Transform The bilateral Laplace transform is defined as
$$X(s) = \int_{-\infty}^{\infty}x(t)e^{-st}dt$$where $s = \sigma +j\omega$. I'm trying to prove rigorously

The ROC(region of convergence) of $X(s)$ consists of strips parallel to the $j\omega$-axis in the $s$-plane.

In Oppenheim's book it's proved by noting that the ROC of $X(s)$ consists of
those values of $s = \sigma +j\omega$ for which the Fourier transform of $x(t)e^{-\sigma t}$ converges and this means that  the ROC of the Laplace transform of $x(t)$ consists of those values of $s$ for which $x(t)e^{-\sigma t}$ is absolutely integrable:
$$\int_{-\infty}^{\infty}|x(t)|e^{-\sigma t}dt\lt\infty$$Because this condition depends only on $\sigma$, the real part of $s$, we have proved the mentioned statement.
My problem with this proof is that $$\text{convergence of the Fourier transform} \implies \text{absolute integrability }$$In fact the converse of this implication is true and absolute integrability is a sufficient condition for the convergence of the Fourier transform. So I think this proof is wrong. Is my understanding correct? Also is the mentioned statement correct? How can we prove that rigorously?
 A: The first thing to get straight is that convergence of the Fourier transform of $f$ does not imply absolute integrability, i.e., that $f \in L^1(\mathbb{R})$.
For a counterexample, take $f(x) = \sin (x^2)$ where
$$\int_{-\infty}^\infty\sin(x^2)e^{ikx} \, dx = 2\int_0^\infty\sin(x^2) \cos(kx) \, dx = \int_0^\infty(\sin(x^2+kx)+ \sin(x^2-kx)) \, dx$$
Making the change of variables $x = u \mp \frac{k}{2}$ we get
$$\begin{align} \int_0^\infty \sin(x^2\pm kx)\, dx &= \int_0^\infty \sin\left(\left(u\mp \frac{k}{2}\right)^2 \pm k\left(u \mp \frac{k}{2}\right)\right)\, du \\ &= \int_0^\infty\sin\left(u^2 - \frac{k^2}{2}\right)\, du \\ &= \cos \left(\frac{k^2}{4}\right)\int_0^\infty \sin (u^2) \, du - \sin \left(\frac{k^2}{4}\right)\int_0^\infty \cos (u^2) \, du \end{align}$$
The well-known Fresnel integrals on the RHS are $\displaystyle \int_0^\infty \cos (u^2) \, du= \int_0^\infty \sin (u^2) \, du = \sqrt{\frac{\pi}{8}}$.
Substituting above, we get
$$\hat{f}(k) = \int_{-\infty}^\infty\sin(x^2)e^{ikx} \, dx = \sqrt{\frac{\pi}{2}}\left( \cos \left(\frac{k^2}{4}\right)-\sin \left(\frac{k^2}{4}\right)\right), $$
and the Fourier transform converges.  However, $f \not\in L^1(\mathbb{R})$.  This can either be shown directly or inferred from the fact that
$$\lim_{k \to \infty} \left( \cos \left(\frac{k^2}{4}\right)-\sin \left(\frac{k^2}{4}\right)\right)\,\, \text{DNE},$$
If $f \in L^1(\mathbb{R})$ then by the Riemann-Lebesgue lemma we would have $\lim_{k \to \infty}\hat{f}(k) = 0$

ROC Proof
Suppose $x$ is locally integrable and for some $s_0 \in \mathbb{C}$ we have convergence of the unilateral Laplace transform
$$X_+(s_0) = \int_0^\infty x(t) e^{-s_0t} \, dt$$
Let $\alpha(t) = \int_0^t x(u)e^{-s_0 u} \, du$. For any $T>0$ it follows from a basic property of the Riemann-Stieltjes integral (proved here) that
$$\int_0^T x(t) e^{-st}\, dt = \int_0^T e^{-(s-s_0)t} x(t) e^{-s_0t}\, dt = \int_0^T e^{-(s-s_0)t} \, d\alpha(t)$$
Integrating by parts on the RHS, we get
$$\tag{*}\int_0^T x(t) e^{-st}\, dt = \alpha(T)e^{-(s-s_0)T} + (s-s_0)\int_0^T e^{-(s-s_0)t} \alpha(t) \, dt$$
Since $X_+(s_0)$ converges, it follows that $|\alpha(t)|$ is bounded for all $t \geqslant 0$. This implies that for all $s$ such that $\Re(s) > \Re(s_0)$, we have
$$\lim_{T \to \infty}\alpha(T)e^{-(s-s_0)T} = 0,$$
and the integral on the RHS of (*) is absolutely convergent.
Thus,
$$X_+(s) = \lim_{T \to \infty}\int_0^T x(t) e^{-st}\, dt= \lim_{T\to \infty}\int_0^T e^{-(s-s_0)t} \, d\alpha(t) = \int_0^\infty e^{-(s-s_0)t} \, d\alpha(t),$$
where $X_+(s)$ converges for all $s$ such that $\Re(s) > \Re(s_0)$.
In a similar way we can show that if we have convergence of
$$X_-(s_1) = \int_{-\infty}^0 x(t) e^{-s_1t} \, dt,$$
then $X_-(s)$ converges for all $s$ such that $\Re(s) < \Re(s_1)$.
