Let $f$ and $P$ be continuous, integrable functions $\mathbb{R} \to \mathbb{C}$ vanishing at $\pm \infty%$. Concisely, $f,P \in C_0(\mathbb{R}) \cap L^1(\mathbb{R})$. Also, assume that $P$ is real-valued.
Define an integral kernel $K: \mathbb{R}^2 \to \mathbb{C}$ by $$ K(x,y) = f(x-y) \left( \exp\left(i \int_x^y P(z) \ dz \right) - 1 \right).$$ It's not hard to see that $K$ defines bounded operator $\xi \mapsto K \cdot \xi$ on $L^2(\mathbb{R})$ by \begin{align*} (K \cdot \xi)(x) = \int K(x,y) \xi(y) \ dy && \xi \in L^2(\mathbb{R}). \end{align*} For example, one can apply Theorem 1.6 in Conway's A Course in Functional Analysis. I have reason to believe this operator is trace class. Any thoughts about how this could be proved would be appreciated. I'm not sure how necessary all these hypotheses are.