Residue at infinity is the ordinary residue after inversion, with the sign supplied by $dz=-w^{-2}dw$.Suggest a slogan
Let \(R>0\) and let \(f\) be holomorphic on \(\{z : \abs{z}>R\}\). Define \(g(z)\coloneqq-\dfrac{1}{z^2}f\Bigl(\dfrac1z\Bigr)\) for \(0<\abs{z}<1/R\). Then the residue of \(f\) at \(\infty\) is \[ \Res_{z=\infty}f=\Res_{z=0}g=-\frac{1}{2\pi i}\int_{\abs{z}=\rho}f(z)\dz \] for every \(\rho>R\), the circle \(\abs{z}=\rho\) being oriented counterclockwise.
Residues belong to the differential form \(f(z)\dz\) rather than to the function \(f\). Under \(w=1/z\) the form \(f(z)\dz\) becomes \(f(1/w)\,d(1/w)=-\frac{1}{w^2}f(1/w)\dw=g(w)\dw\), and the counterclockwise circle \(\abs{z}=\rho\) becomes the clockwise circle \(\abs{w}=1/\rho\), which accounts for the sign.