When does the Laplace Transform converge?

Question

Consider the following function in the $s$-domain:

$$F(s) = \frac{1}{(s^2 + 1)(2s-1)}$$

My book concludes that the ROC (Region of convergence) must be $\Re(s) > 1/2$ because

a) The ROC can't contain poles

b)The ROC must be a right half plane.

The first point makes sense, but I don't get the second point. Could someone explain why it's necessary? Why isn't the ROC all $s$ such that $s \neq 1/2, j, -j$?

Answer 1

The Laplace Transform of a function $f$ is $$ F(s)=\int_0^\infty\,f(t)e^{-st}\,dt. $$ The imaginary part of $s$ bears no influence in whether the integral converges. And one can show that if the integral does not converge for a certain $s$, then it doesn't converge for all $s$ with smaller real part. In other words, the ROC is always of the form $\text{Re}(s)\geq c$ or $\text{Re}(s)>c$ for a given real number $c$.

In some detail, suppose that the integral converges for $s_0=c+ib$. Then, if $s=a+id$ with $a>c$, we have, integrating by parts, \begin{align} \int_0^\infty f(t)e^{-st}\,dt&=\int_0^\infty f(t)e^{-s_0t}\,e^{-(s-s_0)t}\,dt\\ \ \\ &=e^{-(s-s_0)t}\,\left.\left(\int_0^tf(v)e^{-s_0v}\,dv\right)\right|_0^\infty +(s-s_0)\int_0^\infty e^{-(s-s_0)t}\left(\int_0^tf(v)e^{-s_0v}\,dv\right)\,dt\\ \ \\ &=(s-s_0)\int_0^\infty e^{-(s-s_0)t}\left(\int_0^tf(v)e^{-s_0v}\,dv\right)\,dt\\ \ \\ &=(s-s_0)\int_0^\infty e^{-(a-c)t}e^{-(d-b)t\,i}\left(\int_0^tf(v)e^{-s_0v}\,dv\right)\,dt. \end{align} This last integral converges absolutely because the integral between brackets is bounded by the existence of $F(s_0)$ (dependent on $f$ being nice enough) and then the exponential with negative exponent coefficient $-(a-c)$ forces convergence. The imaginary part $e^{-(d-b)t\,i}$ does not influence the absolute convergence because $|e^{-(d-b)t\,i}|=1$ for all $d,b,t$.

Answer 2

The Laplace transform of a function has the following bound: take $s=\sigma+it$, $\sigma, t$ real $$ \lvert \mathcal{L}(f)(\sigma+it) \rvert = \left\lvert \int_0^{\infty} e^{-\sigma x-itx} f(x) \, dx \right\rvert \\ \leqslant \int_0^{\infty} \lvert e^{-\sigma x}e^{-itx} \rvert \lvert f(x) \rvert \, dx \\ = \int_0^{\infty} e^{-\sigma x} \lvert f(x) \rvert \, dx = \mathcal{L}(\lvert f \lvert )(\sigma) $$ Therefore if the Laplace transform of $\lvert f \rvert$ converges at $s=\sigma_0$, the Laplace transform of $f$ converges for all $s$ with real part $\sigma_0$. (And hence the region of convergence contains the half-plane $\sigma \geqslant \sigma_0$)

As for why the Laplace transform doesn't converge everywhere: if $f$ grows as fast as $e^{kx}$, then the integrand will not decay as $x \to \infty$ for $\sigma <k$, and so the integral cannot possibly converge. (On the other hand, if $f$ decays sufficiently fast, the Laplace transform is analytic: $e^{-x^2}$, for example, or a function with nonzero values on only a finite interval.)