I don't completely understand how to find the original time signal when I'm given a Laplace transform and its region of convergence. For instance, if I'm given the Laplace transform: $$ X(s) = \frac{s}{s^2+9} $$ R.o.C.: $\Re(s)<0$
There is no table which provides me with the inverse Laplace transform of this (due to the R.o.C.: being $\Re(s) < 0$). How would I know what the bounds of integration are to compute the inverse Laplace transform?
The inverse Laplace transform is $$ \mathcal{L}^{-1}\{X(s)\} = \frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}X(s)e^{st}ds = \sum\operatorname{Res}\{X(s);s_j\} $$ Let $X(s) = \frac{s}{s^2+9}$. Then there are two simple poles are $s=\pm 3i$. Let's consider the radius of convergence given. The integral will have a problem with convergence if it blows up. The exponential term is what we must worry about. In order to keep the exponential from running off to infinity, $st < 0\iff s < 0$. Now $s\in\mathbb{C}$ so $s=\sigma + i\omega$. Suppose $\lvert X(s)\rvert < M$ for some $M > 0$. Then $$ \Biggl\lvert\frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}X(s)e^{st}ds\Biggr\rvert\leq \frac{1}{2\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}\lvert X(s)\rvert\lvert e^{st}\rvert ds=\frac{M}{2\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}\lvert e^{\sigma t}\rvert\lvert e^{it\omega}\rvert ds $$ What is $\lvert e^{\sigma t}\rvert\lvert e^{it\omega}\rvert$? Well $\lvert\cdot\rvert = \sqrt{\Re\{z\}^2+\Im\{z\}^2}$. In the case of $e^{\sigma t}$, we simple get $e^{\sigma t}$. For the second exponential, recall that $e^{ix}=\cos(x)+i\sin(x)$. Therefore, we have $\sqrt{\cos^2(t\omega)+\sin^2(t\omega)} = \sqrt{1} =1$. Thus, our integral converges when $e^{t\sigma}\not\to\infty$ or $\sigma < 0\iff\Re\{s\} < 0$. Now we can go about solving the ILT as usually. \begin{align} \frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{se^{st}}{s^2+9}ds &= \lim_{s\to -3i}(s+3i)\frac{se^{st}}{s^2+9}+\lim_{s\to 3i}(s-3i)\frac{se^{st}}{s^2+9}\\ &= \frac{e^{-3it}}{2}+\frac{e^{3it}}{2}\\ &= \cos(3t) \end{align}