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 denoted ${\displaystyle {\hat {\mathbb {C} ))}$ 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

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

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:

${\displaystyle \operatorname {Res} (f(z),\infty )=-\sum _{k}\operatorname {Res} \left(f\left(z\right),a_{k}\right).}$

Motivation

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