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?
Best Answer
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 $\Re(s) < \Re(s_1)$.