# 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 ${\displaystyle \infty }$ is a point added to the local space ${\displaystyle \mathbb {C} }$ in order to render it compact (in this case it is a one-point compactification). This space noted ${\displaystyle {\hat {\mathbb {C} }}}$ is isomorphic to the Riemann sphere.[1] One can use the residue at infinity to calculate some integrals.

## Definition

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

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

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

Note that ${\displaystyle \forall r>R}$ , we have

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

## 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 -f(1/z)/z2 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(1/z)d(1/z)=-f(1/z)dz/z2 at z=0.