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Contour Integration by Residues

Learning goal

See an "impossible" real integral fall to complex analysis — and see why the trick works, by watching the arc contribution actually vanish instead of being told that it does.

Everything in the table is computed. Each residue is \(\frac{1}{2\pi i}\oint f\,dz\) around a small circle enclosing the pole, integrated numerically; the arc term is integrated along the semicircle; the real-axis term is direct quadrature. Nothing is looked up.

Things to try

  1. Start with \(\int dx/(1+x^2)\). One simple pole at \(z = i\) inside the contour, residue \(1/2i\), and \(2\pi i \times \frac{1}{2i} = \pi\). You know this one from elementary calculus (\(\arctan\)), which makes it the right place to check that the machinery agrees with something you already trust.

  2. Slide R and watch the arc. This is the step every textbook asserts and no textbook shows. For \(1/(1+x^2)\) the arc falls only as \(1/R\) — slowly. For \(1/(1+x^4)\) it falls as \(1/R^3\), and for \(1/(1+x^6)\) as \(1/R^5\). The rule is that the arc dies whenever \(f\) decays faster than \(1/|z|\), and you can watch the exponent in action.

  3. Try \(\int dx/(1+x^6) = 2\pi/3\). Three poles inside. Then consider doing it by real methods: factor \(1+x^6\) into three quadratics, partial-fraction it, integrate each. It is a genuinely unpleasant afternoon, and the contour method takes three lines. That contrast is the argument for complex analysis.

  4. Look at the double pole, \(\int dx/(x^2+a^2)^2\). The usual formula for a higher-order residue requires differentiating \((z-ia)^2f(z)\) — fiddly and error-prone. The contour definition \(\frac{1}{2\pi i}\oint f\,dz\) handles any order with no change at all, which is a good reason to regard the contour integral, not the limit formula, as the definition.

  5. Try \(\int \cos x/(1+x^2)\,dx = \pi/e\). Note what the widget integrates: \(e^{iz}/(1+z^2)\), not \(\cos z/(1+z^2)\). This matters enormously. On the upper semicircle \(e^{iz} = e^{ix}e^{-y}\) decays, while \(\cos z = \tfrac12(e^{iz} + e^{-iz})\) contains \(e^{-iz}\), which grows like \(e^{+y}\) and destroys the arc estimate. Integrate the complex exponential and take the real part at the end — this is Jordan's lemma, and it is the most common place to go wrong.

  6. Notice the poles below the axis. They are drawn faintly and contribute nothing: the contour does not enclose them. Close the contour in the lower half-plane instead (which you must do for \(e^{-iz}\)) and the other set contributes, with an extra minus sign from the reversed orientation.

The theorem being used

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

for poles \(z_k\) enclosed by \(C\). The strategy is always the same three steps:

  1. Close the real integral into a contour by adding an arc.
  2. Show the arc contributes nothing in the limit (decay estimate, or Jordan's lemma).
  3. The contour integral is then equal to the real integral, and the residue theorem evaluates it.

The reason this works at all is that analyticity is such a strong constraint — a function's behaviour on a contour is determined by a handful of numbers inside it. That is a genuinely strange fact, and it has no analogue for real functions.

Where this returns in physics

  • Landau damping. The velocity integral in the plasma dispersion relation has a pole on the real axis at \(v = \omega/k\). How you go around it is not a matter of taste — causality (the Laplace-transform derivation) fixes the contour, and the resulting imaginary part is the damping rate. Getting this contour wrong gives the wrong physics, and it took the field some years to settle.
  • Green's functions. Retarded, advanced and Feynman propagators solve the same equation and differ only in how the inversion contour passes the poles.
  • Kramers–Kronig relations. Causality means analyticity in the upper half-plane, which by Cauchy's theorem relates the real and imaginary parts of any response function. Refractive index and absorption are not independent.
  • The Casimir effect, where a divergent mode sum is regularised by contour rotation.

Residue theorem · Contour integration · Cauchy's theorem · Laurent series · Classifying singularities · Landau damping (PC368)