In some of the videos I've watched on the Laplace transform, the authors say that if the exponential is decreasing, the area calculated by the transform is finite, and in control theory we can say that the system is stable at the complex variable s.
If we take the function 1/x, we know that the area between 1 and +infinity is not finite, it evolves very slowly (ln(x)) but it is infinite.
The area under the curve of the function e^-x is 1 between 0 and +infinity (Why does $e^{-x}$ integrate to $1$ in interval $(0, \infty)$?)
The curves of the two functions have asymptotes y = 0, I'm trying to understand intuitively why the area of one curve converges and not the other. Is it possible to understand this intuitively without calculations ?
Over 300 years ago, Leibniz explained why the area under the hyperbola is infinite roughly as follows (I will use some modern terminology that Leibniz didn't use). The key is that the hyperbola is invariant under the area-preserving linear transformation $L_t(x,y)=(\frac{x}t, ty)$. We start with the unit square with vertices at $(0,0), (1,0), (0,1), (1,1)$, of area $1$. Notice that the last point is on the hyperbola. Now we apply the transformation $L_t$ with small $t$. We get a thin rectangle of the same area $1$, but now its base is very long and its height is small. Notice that the intersection between the square and the rectangle is negligible (only $t$). Therefore we have found a region under the hyperbola given by the union of the square and the rectangle, of total area close to $2$. We continue this process by applying $L_t$ with an even smaller value of the parameter which is negligible compared to $t$, say $t^2$, and get an even thinner rectangle of area $1$ whose intersection with the previous rectangle is negligible. This gives us a combined area close to $3$, etc. Doing this sufficiently many times we can identify a region under the hyperbola of arbitrarily large area. This implies that the area under the hyperbola is infinite.
The argument does not work for the exponential curve because it is not preserved by the transformation $L_t$.