Relation between laplace transform and differentiation operator

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

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The equation $\dfrac{1}{1+p} = 1 - p + p^2 - \ldots$ is not true. For numbers, $\dfrac{1}{1+x} = \sum_{j=0}^\infty (-x)^j$ is true only for $|x| < 1$: otherwise the series on the right diverges. Similarly when applied to the function $f(x) = e^{\alpha x}$, $(1 - p + p^2 - \ldots) f(x) = (1 - \alpha + \alpha^2 - \ldots) f(x)$, which converges only if $|\alpha| < 1$.