I am reading an engineering book (equations 2.21, 2.22) where the author claims the following :-
If $y,u,g :[0,\infty] \to \Re$, $y(t) = \int_0^\infty g(r) u(t-r)dr$ and $Y(s)=G(s)U(s)$ (the laplace transforms), then $y(t) = G(p)u(t)$ where $p$ is the differentiation operator $\frac{d}{dt}$.
As an example, if I choose $G(s)= \frac{1}{1+s}$, then I get $y(t)=\frac{1}{1+p}u(t) = (1 + p - p^{2} + \cdots) u(t)$.
The following concocted example shows that the above might not be true :- $u(t) = 1, t \geq 1$ otherwise $u(t)=0$. Then $U(s) = \frac{e^{-s}}{s}$. Let $G(s) = \frac{1}{1+s}$. Then $Y(s) = \frac{e^{-s}}{(1+s)s}$. By inverting we get $y(t) = (1-e^{-(t-1)})*H(t-1)$ where $H$ is the Heaviside function.
Now, $y(2) = 1 - e^{-1}$ from this method whereas by the differentiaion operator way above $y(2) = 1$.
I wanted to know if the infinite series expansion $1+p^{1}-p^{2}+\cdots$ is mathematically sound. If it is so, what is the reason for discrepancy in the two values for $y(2)$ ? (assuming my calculations are correct!)
Thanks, Phanindra