Suppose I have a series expansion
$$S_1(\omega,k) = z+\frac{1}{4} z^3 \left(k ^2-\omega ^2\right)+\frac{1}{64} z^5 \left(k ^2-\omega ^2\right)^2+\frac{z^7 \left(k ^2-\omega ^2\right)^3}{2304}+O\left(z^{9}\right)$$
I also have another series
$$S_2 (\omega_r,k_r) = z+\left(k_r^2-\omega _r^2\right) z^3+\frac{z^5}{4} \left(k_r^4-2 \left(\omega _r^2+1\right) k_r^2+\left(\omega _r^2-1\right){}^2\right)+\frac{z^7}{36} \left(k_r^6-\left(3 \omega _r^2+10\right) k_r^4+\left(3 \omega _r^4+25\right) k_r^2-\omega _r^2 \left(\omega _r^2-5\right){}^2\right)+O\left(z^{9}\right)$$
The two series have the property that at every order in $z$, if we set $$k_r = \frac{k}{2}, \omega_r = \frac{\omega}{2}$$
in $S_2$, then at a general order in $z$ the coefficient is a general function of $\omega$ and $k$, but the terms with the largest exponents are the same as the coefficient of $S_1$. In both cases, assume $0<z<1$ and $\omega, k \in \mathbb{R}$.
Question: I was wondering if there is a simple integral transform that performs the following order by order $$S_1(\omega,k) = \int_{-\infty}^{\infty}d\omega_r \int_{-\infty}^{\infty}dk_r F(\omega,k; \omega_r,k_r) S_2 (\omega_r,k_r)$$ using the property above, that holds true $\forall \omega, k \in \mathbb{R}$? I know about the "large" $\omega, k \gg 1$ solution, but am missing the general solution. Stated differently, is there a transform which "simply throws" away the subdominant powers of $\omega$ and $k$ at arbitrary order in $z$?
Note: The series expansions are that of a Bessel function and a $_2F_1$ hypergeometric function respectively but I am hoping we do not need the explicit form, and whether the above property could determine the same.