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The Residue Theorem

Source lecture(s): PHY622 Ch. 5

Intuition

Cauchy's theorem says a contour integral of an analytic function vanishes. The residue theorem is what happens when it does not: the integral is determined entirely by the singularities inside, and each contributes one number.

\[\oint_C f(z)\,dz = 2\pi i\sum_k\operatorname{Res}(f, z_k)\]

An integral around a closed loop — an infinite amount of information about \(f\) on the boundary — collapses to a finite sum of local data. That is a very strong statement, and it is only possible because analyticity is so restrictive.

What a residue is

The coefficient \(a_{-1}\) in the Laurent expansion about \(z_0\). Equivalently, and more robustly,

\[\operatorname{Res}(f,z_0) = \frac{1}{2\pi i}\oint_{|z-z_0|=\epsilon}f(z)\,dz\]

Practical formulas:

Case Formula
Simple pole \(\displaystyle\lim_{z\to z_0}(z-z_0)f(z)\)
Simple pole of \(p/q\) with \(q(z_0)=0\), \(q'(z_0)\neq0\) \(p(z_0)/q'(z_0)\)
Pole of order \(m\) \(\displaystyle\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}\left[(z-z_0)^mf(z)\right]\)
Essential singularity expand and read \(a_{-1}\) — no shortcut

The \(p/q'\) form is the one to remember; it saves considerable algebra for rational functions.

The standard manoeuvre

To evaluate a real integral \(\int_{-\infty}^{\infty}f(x)\,dx\):

  1. Close the contour with a large semicircle in the upper (or lower) half-plane.
  2. Kill the arc. It vanishes as \(R\to\infty\) if \(|f|\) decays faster than \(1/|z|\). For oscillatory integrands use Jordan's lemma: \(\oint e^{iaz}f(z)dz\) over the upper arc vanishes for \(a>0\) provided \(f\to0\), because \(|e^{iaz}| = e^{-ay}\) decays.
  3. Sum the residues inside and multiply by \(2\pi i\).

Watch this happen — including the arc genuinely dying — in the contour integrator.

The single most common error is integrating \(\cos z\) instead of \(e^{iz}\). Since \(\cos z = \tfrac12(e^{iz}+e^{-iz})\) and \(e^{-iz}\) grows like \(e^{+y}\) in the upper half-plane, the arc estimate fails completely. Always use the exponential, then take the real part at the end.

Poles on the contour: the principal value

If a pole sits on the real axis the integral diverges, and you must say what you mean. Indent the contour with a small semicircle around it: a half-circle detour contributes half the residue,

\[\lim_{\epsilon\to0}\int_{\text{semicircle}}f\,dz = \pm i\pi\operatorname{Res}\]

with the sign set by which way you go around. The result is the Cauchy principal value plus (or minus) \(i\pi\times\) residue — the Sokhotski–Plemelj formula:

\[\frac{1}{x - x_0 \mp i\epsilon} = \mathrm{P}\frac{1}{x-x_0} \pm i\pi\delta(x-x_0)\]

This is not a technicality. In Landau damping the pole at \(v = \omega/k\) sits exactly on the integration path, and the choice of detour is fixed by causality, not convenience. The \(i\pi\delta\) term is the damping. Landau's insight was precisely that the Laplace-transform derivation dictates the contour — and getting it wrong gives a wave that does not damp.

Where it shows up

  • Real integrals that resist every real method (see the widget's \(\int dx/(1+x^6)\)).
  • Green's functions. Retarded vs advanced is a choice of contour around the poles.
  • Kramers–Kronig relations. Causality ⇒ analyticity in the upper half-plane ⇒ real and imaginary parts of any response function determine each other.
  • The Casimir effect, where a divergent mode sum is tamed by contour rotation.
  • Summing series. \(\oint \pi\cot(\pi z)f(z)dz\) has residues at every integer, converting \(\sum_n f(n)\) into a contour integral — which is how \(\sum 1/n^2 = \pi^2/6\) falls out.

Common mistakes

  • Using \(\cos z\) rather than \(e^{iz}\). See above.
  • Forgetting to check the arc. If \(f\) decays only as \(1/z\) the arc does not vanish and the method needs modification.
  • Missing poles. Especially those from a \(\sin\) or \(\cos\) in a denominator, which give infinitely many.
  • Taking a residue at a branch point. There is no Laurent series there — see singularities and branch points.
  • Ignoring orientation. Counter-clockwise is positive; closing in the lower half-plane introduces a minus sign.

Knowledge graph position

Prerequisites: Cauchy's theorem, Laurent series, contour integration. Leads to: dispersion relations, Green's functions, Landau damping, scattering theory.

Quiz

Q1 (computational). Evaluate \(\displaystyle\int_{-\infty}^{\infty}\frac{dx}{x^2+4}\).

Answer

Poles at \(z = \pm2i\); only \(z = 2i\) is in the upper half-plane. Using \(p/q'\): \(\operatorname{Res} = 1/(2z)|_{z=2i} = 1/4i\). The arc vanishes (\(f \sim 1/z^2\)). So the integral is \(2\pi i \times \frac{1}{4i} = \pi/2\).

Q2 (conceptual). Why must you integrate \(e^{iz}/(1+z^2)\) rather than \(\cos z/(1+z^2)\)?

Answer

On the upper arc \(|e^{iz}| = e^{-y} \to 0\), so Jordan's lemma applies. But \(\cos z = \tfrac12(e^{iz}+e^{-iz})\) contains \(e^{-iz}\), whose modulus is \(e^{+y}\) — it grows exponentially on that arc and the estimate collapses. Integrate the exponential and take the real part at the end.

Q3 (MCQ). A pole lying exactly on the integration contour is handled by:

  • (a) ignoring it
  • (b) indenting the contour, which contributes \(\pm i\pi\) times the residue — the Sokhotski–Plemelj formula
  • (c) moving the contour to the other side arbitrarily
  • (d) declaring the integral divergent
Answer

(b). A half-circle detour contributes half of \(2\pi i\operatorname{Res}\). Which side you pass is not arbitrary — in physics it is fixed by causality, and in Landau damping that choice is the damping rate.