Dear @Abdelmalek, no I have not tried it. I can give it a try and see if some kind of change of order of integration really helps.

Dear @Asaf, yes of course you are right. It is just that the summation I obtained of the K-Bessel functions is kind of messy and I would be happy to have a simpler formula to work with. In fact what one naturally finds is a linear combination of higher derivatives of K-Bessel functions but then using the usual cylinder identities (for cylinder functions) we can rewrite the higher derivatives in terms of $K_{\nu}$ where the order $\nu$ has been shifted.

Ok got it since after all you want your numerator to be a again a polynomial in $x$ after having divided it by $x$.

@Iosif, should you not multiply $r_0(y)$ by $x^{m-k}$ if you want the leading term in $x$ to cancel ?

Thanks @Wojowu for this great example!

Thanks @Iosif for the nice inductive argument. I also just realized that my original statement is a bit sloppy since it may well happen that for a fixed $z_2$, $[z_1\mapsto f(z_1,z_2)]=\infty$ which is not rational per se (e.g. $1/z_2$). The same issue arises in more variables....it is not completely clear to me how to correct that. May be $(z_2,\ldots,z_{n} )$ should be allowed to avoid a finite union of analytic subvarieties in $\mathbf{C}^{n-1}$....