As an example, consider the Klein-Gordon equation: $$(\partial_t^2 - \Delta + m^2)\psi = 0. \tag{1}$$ In physics one usually starts with an ansantz of plane wave solutions so begin by writing $\psi$ as an inverse Fourier transform: $$\psi(x,t) = \int \frac{1}{(2\pi)^3} e^{ip\cdot x} \hat{\psi}(p,t) dp. \tag{2}$$ After substituting (2) into (1) one gets $$(\partial_t^2 + \omega_p^2) \hat{\psi} = 0$$ where $\omega_p^2 = p^2 + m^2$. Solving the above with the assumption that $\psi$ must real one gets $$\psi = \frac{1}{(2\pi)^{3/2}} \int A(p)e^{-i(\omega_pt + p\cdot x)} + A^*(-p)e^{i(\omega_p t + p \cdot x)} dp \tag{3}$$ where $A(p)$ is a function and $A^*$ its complex conjugate.
Now in quantum field theory one does what's known as second quantization where the function $A$ above is replaced by an operator, $A \rightarrow \hat{A}$. Thus (3) is now an operator valued: $$\psi = \frac{1}{(2\pi)^{3/2}} \int \hat{A}(p)e^{-i(\omega_pt + p\cdot x)} + \hat{A}^*(-p)e^{i(\omega_p t + p \cdot x)} dp \tag{4}$$ where $\hat{A}^*$ is now the adjoint of $\hat{A}$.
Expressions like (4) are very common in physics and I am trying to make mathematical sense of them. I am aware that for this all to technically work out $\psi$ actually has to be an operator-valued distribution. But putting that issue aside for now, is there any proper mathematical meaning to (4)? I am only aware of operators being involved with integrals when we use the functional calculus where we have something of the form $$f(A) = \int_{\sigma(A)} f dP$$ where $dP$ is some projection-valued measure and $f$ some nice function. But in (4) we are integrating over all of $\mathbb{R}^n$ with what seems to be the ordinary Lebesgue measure so how can one make sense of it?