Let $H^i$ denote the $i$-th Sobolev space on $\mathbb{R}^n$. Let $T$ be a first-order differential operator on $\mathbb{R}^n$, so that we may view $T$ as a bounded operator $H^i\rightarrow H^{i-1}$ for any $i>0$. Suppose that $T$ has an inverse, which can be viewed as a bounded operator $H^{i-1}\rightarrow H^i$ for each $i>0$.
Let $S$ be a second-order differential operator with positive spectrum, so that a bounded inverse $S^{-1}:H^{i-2}\rightarrow H^i$ exists for each $i>1$. Suppose that, for any $\psi\in H^0$, the integral
$$A\psi:=\int_0^\infty T(S+\lambda^2)^{-1}\psi\,\,d\lambda$$
converges in in the $H^0$-norm and defines a bounded operator $A: H^0\rightarrow H^0$.
Question: Is it possible to conclude that $A$ has a bounded inverse $A^{-1}:H^0\rightarrow H^0$, perhaps by defining it explicitly in terms of $T^{-1}, S$ and $\lambda$?