On the resolvent of the Gauss operator (Q2436393)

Let \[ \alpha=a_0+\dfrac{1}{a_1+\dfrac{1}{a_2+\ddots}}=:[a_0;a_1,a_2,\dots] \quad \text{where } a_n, n \in \mathbb{N}, \; \alpha \in \mathbb{R}_+, \tag{1} \] be the representation of numbers \(\alpha\) by a regular continued fraction. Let \(\eta_n:=[0;a_{n+1},a_{n+2},\dots] \in [0,1)\). Denote \[ F_n(x):=\mathrm{mes} \{\alpha: \eta_n(\alpha)<x\}=\mathrm{mes} \left\{\alpha : a_{n+1}(\alpha) \geq \big[\tfrac{1}{x} \big] \right\}. \] The Gauss operator \(G\) is defined as the distribution function \(F\) of measures \[ G[F(x)]:=\sum_{j=1}^{\infty} F\Big(\tfrac{1}{j}\Big)- F\Big(\tfrac{1}{j+x}\Big). \] The representation of the kernel of the Gauss-Babenko integral operator through a numerator \(p_n\) and a denominator \(q_n\) of the convergence \(p_n/q_n:=[0;a_1,a_2,\dots,a_n]\) of the continued fraction (1), the Bessel function and modified Bessel function are established.