Residue at infinity

In complex analysis, a branch of mathematics, the residue at infinity is a residue of a holomorphic function on an annulus having an infinite external radius. The infinity $$\infty$$ is a point added to the local space $$\mathbb C $$ in order to render it compact (in this case it is a one-point compactification). This space denoted $$ \hat{\mathbb C} $$ is isomorphic to the Riemann sphere. One can use the residue at infinity to calculate some integrals.

Definition
Given a holomorphic function f on an annulus $$ A(0, R, \infty) $$ (centered at 0, with inner radius $$R$$ and infinite outer radius), the residue at infinity of the function f can be defined in terms of the usual residue as follows:


 * $$ \operatorname{Res}(f,\infty) = -\operatorname{Res}\left( {1\over z^2}f\left({1\over z}\right), 0 \right)$$

Thus, one can transfer the study of $$ f(z) $$ at infinity to the study of $$ f(1/z) $$ at the origin.

Note that $$\forall r > R$$, we have


 * $$ \operatorname{Res}(f, \infty) = {-1\over 2\pi i}\int_{C(0, r)} f(z) \, dz$$

Since, for holomorphic functions the sum of the residues at the isolated singularities plus the residue at infinity is zero, it can be expressed as:

$$ \operatorname{Res}(f(z), \infty) = -\sum_k \operatorname{Res}\left(f\left(z\right), a_k\right).$$

Motivation
One might first guess that the definition of the residue of $$f(z)$$ at infinity should just be the residue of $$f(1/z)$$ at $$z=0$$. However, the reason that we consider instead $$-\frac{1}{z^2}f\left(\frac{1}{z}\right)$$ is that one does not take residues of functions, but of differential forms, i.e. the residue of $$f(z)dz$$ at infinity is the residue of $$f\left(\frac{1}{z}\right)d\left(\frac{1}{z}\right)=-\frac{1}{z^2}f\left(\frac{1}{z}\right)dz$$ at $$z=0$$.