Consider the Hilbert space $L^2(\mathbb R)$ and the multiplication operator $A \colon D(A) \to L^2(\mathbb R)$ defined by $$(Af)(x) := (1+x^2)f(x)$$ with domain $$D(A) := \left \{f \in L^2(\mathbb R): (1+x^2)f \in L^2(\mathbb R), \int_{\mathbb R} f(x) \, dx = 0 \right \}.$$ One can show that $A$ is a densely defined, closed operator that is not self-adjoint. In fact, there are functions such as $x\mapsto e^{-x^2} \in D(A^*)$, but not in $D(A)$. One can also show that the quadratic form domain of $A$ is given by $$Q(A) = \left \{ f \in L^2(\mathbb R): \sqrt{1+ x^2}f\in L^2(\mathbb R): \int_{\mathbb R} f(x) \, dx = 0\right \}.$$ From this, we see that one self-adjoint extension is given by the Friedrichs extension $A_F$ with domain $D(A_F) \subset Q(A)$. Furthermore, we have the obvious self-adjoint extension given just by the multiplication operator with the unconstrained maximal domain.
Question: The domain of the adjoint operator $D(A^*)$ seems to be very complicated and I couldn't really figure out how it should look like. Is there any possibility to give a more or less precise description of it?