Proof. We can extract the residues by Laurent coefficient \(a_{-1}\). Rewrite the inner form:

$$ \frac{c}{t} \left( 1+zt+\frac{\left(zt\right)^{2}}{2!}+\frac{\left(zt\right)^{3}}{3!} \dots \right) + \frac{b}{t^2} \left( 1+zt+\frac{\left(zt\right)^{2}}{2!}+\frac{\left(zt\right)^{3}}{3!}\dots\right) + \frac{2a}{t^3} \left( 1+zt+\frac{\left(zt\right)^{2}}{2!}+\frac{\left(zt\right)^{3}}{3!}\dots\right) $$

Distribute each quotient:

$$ \left( \frac{c}{t}+z+\frac{z^{2}tc}{2!}+\frac{z^{3}t^{2}c}{3!} \dots \right) + \left( \frac{b}{t^{2}}+\frac{zb}{t}+\frac{z^{2}b}{2!}+\frac{z^{3}tb}{3!}\dots\right) + \left( \frac{2a}{t^{3}}+\frac{2za}{t^{2}}+\frac{2z^{2}a}{2!t}+\frac{2z^{3}a}{3!} \dots\right) $$

The \(\frac{1}{t}\) term emerges as \(az^2+bz+c\); the residue at \(t=0\) is \(az^2+bz+c\). Considering the other Hankel rays vanish from the meromorphic structure, We can now invoke the Residue Theorem on the original expression:

$$ \begin{aligned} \frac{1}{2\pi i } \left(2\pi i \sum \text{Res}(f,t_k) \right)\\ = az^2+bz+c \end{aligned} $$

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