I am working with an integral within the context of a Carleman estimate, and am trying to manifest its reality (with the later goal of finding a lower bound for $-S$ in the $L^2$ sense) but am having trouble. Although I believe the operator is symmetric from my calculations, there might be small errors, so I wanted to first ask the question of symmetricity of the operator $S$ that I will implicitly define below. If the answer to the question is yes, I am interested in how to manifest the reality of its corresponding $L^2$ weighted norm. For context, let $ f\in C_0^\infty(\mathbb{R}\times[0,1])$ take values in $\mathbb{C}$, $\alpha\in\mathbb{R}$, $\phi\in C^\infty([0,1])$, take values in $\mathbb{R}$ and define the function $\psi(x,t)=x+\phi(t)$. I am interested in the following integral $$ \begin{split} \int f^\dagger Sf &:= \alpha \int 192\alpha^4 \psi^3 f^\dagger \partial_x f + \alpha^248\psi^4 f^\dagger \partial_ {xx}f+6i\alpha^2 \psi^2 \phi'(t)f^\dagger \partial_x f - 48\alpha^2 \ \psi f^\dagger \partial_{xxx}f\\ &\quad-12 \alpha^2 \psi^2 f^\dagger \partial_{xxxx} f -12 \alpha^2 f^\dagger \partial_{xx}f+(1/2)i\phi'(t)f^\dagger \partial_{xxx}f + f^\dagger \partial^6_{x} f \\ &\quad+ \frac{1}{4}f^\dagger f\left(48\alpha^2 \psi^2 - 256 \alpha^6 \psi^6 +24i \alpha^2 \psi \phi'(t) -\frac{1}{4}\psi\phi'' - \frac{1}{4} \phi \right), \end{split}$$ where the integrals are computed over $[0,1]\times \mathbb{R}$, and the implicitly defined operator $S$ should be symmetric according to my calculations. My questions are:

Is the operator $S$ defined above indeed symmetric?

If yes, how to manifest the reality of the integral? In other words, how to show that the terms containing the imaginary unit $i$ either,

*e.g*., disappear or turn into the real/imaginary part of some expression.

Edit 1: I added a pre-factor of $\alpha$ inside the integral

Edit 2: I added a forgotten factor of $\alpha^2$ to the term $48\psi^4 f^\dagger \partial_ {xx}f$