I have a question with the composition of two operators.
To contextualize. Let $p_1 = p_1 (x,D_x):= 1-\partial_ {x}^2 $ be the operator $ p_1: D(p_1)\subset L^2 \to L^2 $ where $$ D(p_1) = \left\{u \in L^2: p_1u\in L^2\right\}$$
It can be verified that $p_1u(x) = \mathcal {F}^{-1}((1+ \xi^2)\widehat {u}(\xi))(x)$ and if $p_2 = p_2 (x, D_x)u(x): = p_1(x, D_x)^{-1}u(x)$ ($p_2$ the resolvent) , it can be proved that $p_2u(x)=\mathcal {F}^{-1} ((1+\xi^2)^{-1}\widehat{u}(\xi))(x)$.
Since $p_1\circ p_2 u(x)=u(x)$, that is, $p_1\circ p_2 = Id$, the identity operator symbol is $1$ (I suppose this is true since $Id(u(x)) = u(x)$ does not appear any differentiation operator).
My doubt leaves now. In general, if $p_1$ and $p_2$ are two operators, the symbols of the composition $p_1\circ p_2$ is given by $$(p_1\#p_2) (x, \xi): = \int \int e^{-ix'\xi'} p_1 (x, \xi + \xi ') p_2 (x + x' , \xi) dx'd \xi'$$ (as you can see in, for example, Pseudo differential and singular integral operators by the author Abels).
Given the above, it should happen that $$\int \int e^{-ix '\xi'} (1 + (\xi + \xi ')^2) (1+ \xi)^{-1} dx'd \xi' = 1$$
This is true?