The Residue Theorem & the Evaluation of Integrals

the residue

Picture a function that misbehaves at a single point — it blows up there, or oscillates wildly — but is perfectly well-behaved on a tiny punctured ring around it. The residue is one number that captures the single most important thing about that bad point as far as integration is concerned. Of all the infinitely many terms in the function's local expansion, exactly one survives when you integrate around the point, and the residue is the size of that survivor.

Precisely: near an isolated singularity z_0, a holomorphic function has a Laurent series f(z) = sum a_n (z - z_0)^n where n runs over all integers, positive and negative. The residue of f at z_0, written Res(f, z_0), is the coefficient a_{-1} — the coefficient of the single term 1 / (z - z_0). Why that one term is special: if you integrate (z - z_0)^n around a small circle centered at z_0, you get 0 for every integer n except n = -1, for which the integral is 2 pi i. So integrating the whole Laurent series term by term kills everything but the a_{-1} term, leaving the integral over the circle of f(z) dz = 2 pi i times a_{-1}. The residue is exactly what an integral around the singularity reports back.

This is the keystone of the most powerful computational machinery in the subject. It lets you replace a global integral around a curve by a sum of local numbers, one per enclosed singularity (that is the residue theorem). For practical work you rarely compute a_{-1} by writing out the whole Laurent series — there are quick formulas for poles. But the definition as a_{-1} is the honest one, and for essential singularities, where no pole formula applies, expanding the series is exactly how you find the residue.

For f(z) = e^z / z^2 near z_0 = 0, expand e^z = 1 + z + z^2/2 + ..., so f(z) = 1/z^2 + 1/z + 1/2 + .... The coefficient of 1/z is 1, so Res(f, 0) = 1.

Reading the residue straight off the Laurent series: it is the coefficient of 1 / (z - z_0).

Only the a_{-1} coefficient is the residue; the more singular terms like a_{-2} / (z - z_0)^2 contribute nothing to a closed-loop integral. A function can blow up violently yet have residue zero (for example 1 / z^2 at 0).

Also called
殘數residue at a singularity