Consider a real Sturm-Liouville operator $L$ on $[0,+\infty)$ and use the following notations : https://www.encyclopediaofmath.org/index.php/Titchmarsh-Weyl_m-function
Assume $a = 0$, $\alpha \in [0,\pi)$ is fixed, $w \equiv 1$ and let's say that we are in the limit-point case so that $m(\lambda)$ is unique and for $\beta \in [0,\pi)$ fixed, $$m(\lambda) = \ell_{\infty}(\lambda) = \lim \limits_{b \to +\infty}\ell_b(\lambda)=\lim \limits_{b \to +\infty}-\frac{\cot\beta\,\phi(b,\lambda)+p(b)\phi'(b,\lambda)}{\cot\beta\,\psi(b,\lambda)+p(b)\psi'(b,\lambda)}$$ For $0 < b \leq +\infty$, consider the Green function on $[0,b]$ $$G_b(x,y,\lambda) := \begin{cases} \phi(x,\lambda)\chi_b(y,\lambda) \quad 0 \leq x \leq y \leq b\\ \chi_b(x,\lambda)\phi(y,\lambda) \quad b \geq x > y \geq 0 \\ 0 \quad x,y \gt b \end{cases}$$ In Sturm-Liouville and Dirac Operators from Levitan & Sargsjan, at page 57, it is said that the Fourier transform of $$\dfrac{dG}{dx}(x,y,z)$$ (where $G = G_{\infty}$) is exactly $$\int_0^{+\infty}\dfrac{dG}{dx}(x,y,z)\phi(y,\lambda)dy = \dfrac{\phi'(x,\lambda)}{(z-\lambda)}$$ but at this point in the book (at this point, we have proved the limit-point and limit-circle theorem, an integral expansion theorem for arbitrary functions and an integral representation for the resolvant. For example, the surjectivity of the Fourier transform is not proved), it is unclear why this is true. Indeed, in the regular case, we have $$\int_0^{b}\dfrac{dG_b}{dx}(x,y,z)\phi(y,\lambda_{n,b})dy = \dfrac{\phi'(x,\lambda_{n,b})}{(z-\lambda_{n,b})}$$ where the $\lambda_{n,b}$'s are the eigenvalues of the Sturm-Liouville operator $L$ restricted to $[0,b]$. In the book, it is said that it suffices "to take the limit $b \to +\infty$", but I do not understand why this should work.
