RoC and Stability of a Rectangular Signal If we have a system with an impulse defined as: $$h(t)=u(t)-u(t-2)$$
Then the Laplace Transform of h(t) would be the transfer function: $$H(s)=\frac{1}{s}-\frac{e^{-2s}}{s}, \quad Re(s)&gt;0$$
We also know that a stable system must have a RoC region that passes through the imaginary axis. In our case for $h(t)$, $\textrm{Re}(s)&gt;0$ so it doesn't pass the imaginary axis, indicating that the system is not stable.
However, the system is stable because $$\int_{-\infty}^{\infty}|h(t)| \leq M$$
So my question is how is this possible? Am I missing something in understanding the stability condition in Laplace Transform?
Note that for $s\to 0$, the transfer function $H(s)$ doesn't have a pole because also the numerator approaches zero. In fact, you have
$$\lim_{s\to 0}H(s)=2\tag{1}$$
The ROC is the whole $s$-plane, except for $\textrm{Re}(s)\to -\infty$.