Contour integration Study Guide

📖 Core Concepts Contour – A finite concatenation of directed smooth curves whose end‑points match, giving a continuous oriented path in the complex plane. Directed smooth curve – A smooth (continuously differentiable, non‑zero derivative) map \(\gamma:[a,b]\to\mathbb C\) together with an orientation (the natural ordering of the parameter). Contour integral – For a continuous \(f\) on a directed smooth curve \(\gamma\) $$\int{\gamma} f(z)\,dz = \int{a}^{b} f\bigl(\gamma(t)\bigr)\,\gamma'(t)\,dt,$$ independent of the chosen parametrization. Residue theorem – If \(\Gamma\) is a closed contour enclosing isolated singularities \(\{zk\}\), $$\int{\Gamma} f(z)\,dz = 2\pi i \sumk \operatorname{Res}\bigl(f,zk\bigr).$$ A clockwise (negative) orientation flips the sign. Cauchy integral formula – For holomorphic \(f\) on and inside a simple closed contour \(C\) and \(z0\) inside \(C\), $$f(z0)=\frac{1}{2\pi i}\oint{C}\frac{f(z)}{z-z0}\,dz.$$ Branch cut & keyhole contour – To make a multi‑valued function (e.g., \(z^{1/2}\), \(\log z\)) single‑valued, cut the plane along a curve (often the positive real axis) and integrate around a “keyhole’’ that avoids the cut. --- 📌 Must Remember Integral of \(1/z\) around the unit circle (counter‑clockwise): \(\displaystyle\int{|z|=1}\frac{1}{z}\,dz = 2\pi i.\) Parametrization of the unit circle: \(z(t)=e^{it},\;t\in[0,2\pi],\;dz=i e^{it}dt.\) Zero integral rule: \(\displaystyle\int{|z|=1} z^{n}\,dz = 0\) for any integer \(n\neq -1.\) Orientation sign: Positive (CCW) → \(+2\pi i\); Negative (CW) → \(-2\pi i.\) Residue at a simple pole \(z0\): \(\operatorname{Res}(f,z0)=\lim{z\to z0}(z-z0)f(z).\) Branch‑cut factor for \(\sqrt{z}\): Going around the cut multiplies \(z^{1/2}\) by \(-1\) (since \(e^{2\pi i/2}=-1\)). --- 🔄 Key Processes Evaluating a real integral with residues Choose a contour that contains the real‑axis segment of interest and encloses the integrand’s poles. Identify all poles inside the contour. Compute each residue (simple pole limit or higher‑order formula). Use the residue theorem: \(\displaystyle\int{\text{contour}} f = 2\pi i\sum\text{Res}\). Show the contribution of the “extra’’ arcs (usually large semicircles) → 0 via the estimation/Jordan lemma. Conclude the original real integral equals the sum of residues (possibly multiplied by a factor like \(2\pi i\)). Parametrizing a circle – Set \(z(t)=re^{it}\) (\(r\) radius, \(t\) from \(0\) to \(2\pi\)); then \(dz = i r e^{it} dt\). Keyhole contour for branch‑cut integrals Place the cut (e.g., positive real axis). Build a contour that goes above the cut, circles a large radius \(R\), returns below the cut, and finally circles a small radius \(\varepsilon\) around the origin. Let \(R\to\infty,\;\varepsilon\to0\); the outer/inner arcs vanish. Relate the two integrals along the cut (they differ by the branch factor) to the desired real integral. --- 🔍 Key Comparisons Cauchy’s theorem vs. Residue theorem Cauchy: If \(f\) is holomorphic everywhere inside a closed contour, \(\displaystyle\oint f(z)\,dz = 0.\) Residue: If \(f\) has isolated singularities, the integral equals \(2\pi i\) times the sum of residues. Positive vs. Negative orientation CCW: Integral = \(+2\pi i\sum\text{Res}\). CW: Integral = \(-2\pi i\sum\text{Res}\). Simple pole vs. Higher‑order pole Simple: \(\operatorname{Res} = \lim{z\to z0}(z-z0)f(z).\) Order \(m>1\): Use \(\displaystyle\operatorname{Res} = \frac{1}{(m-1)!}\lim{z\to z0}\frac{d^{\,m-1}}{dz^{\,m-1}}\bigl[(z-z0)^m f(z)\bigr].\) Integral of \(z^{n}\) around unit circle \(n=-1\) → \(2\pi i\). \(n\neq -1\) → \(0\). --- ⚠️ Common Misunderstandings “All closed contours give zero.” Only true when the integrand is holomorphic inside the contour (Cauchy’s theorem). Orientation ignored. A clockwise contour flips the sign of the result. Exponent rule mis‑applied. The zero‑integral statement holds for powers \(z^{n}\) with \(n\neq -1\); it does not apply to \(\frac{1}{z^{k}}\) where \(k\neq 1\) (still zero). Branch cut contribution forgotten. The two sides of the cut generally differ by a multiplicative factor; neglecting it yields a wrong real‑integral value. --- 🧠 Mental Models / Intuition Residues as “electric charge’’ placed at poles; the contour integral measures the total “flux’’ (charge) times \(2\pi i\). Cauchy formula: The value of a holomorphic function inside a circle is the average of its values on the circle weighted by \(\frac{1}{z-z0}\). Think of the circle as a “sampling net’’ that captures the interior information. Keyhole contour: Imagine cutting a sheet of paper (the cut) and looping a string around the cut – the string picks up a twist (the factor \(-1\) for square‑root) each time it goes around. --- 🚩 Exceptions & Edge Cases Clockwise orientation → multiply the usual \(2\pi i\) result by \(-1\). Residue at infinity – When evaluating integrals over the whole complex plane, include \(\operatorname{Res}(f,\infty) = -\sum{k}\operatorname{Res}(f,zk).\) Branch cut not on the positive real axis – Choose the cut where the function is discontinuous; the factor around the cut depends on the chosen branch (e.g., \(e^{2\pi i\alpha}\) for \(z^\alpha\)). Poles on the contour – If a pole lies exactly on the chosen path, the standard residue theorem does not apply; deform the contour or take a Cauchy principal value. --- 📍 When to Use Which Direct parametrization (e.g., unit circle) → when the contour is a simple geometric shape and the integrand is easy to substitute. Residue theorem → whenever the integrand has isolated poles and the contour can be closed so that the unwanted arcs vanish. Cauchy integral formula → to evaluate an integral of the form \(\displaystyle\oint \frac{f(z)}{z-z0}\,dz\) or to obtain derivatives \(f^{(n)}(z0)\). Keyhole contour → for integrals containing branch‑point functions like \(z^{\alpha}\) or \(\log z\). Estimation/Jordan lemma → to justify that large semicircular arcs contribute zero (used in steps 5 of the residue process). --- 👀 Patterns to Recognize Rational function × \(e^{iz}\) → poles come only from the rational part; the exponential is entire. Trigonometric integral \(\int0^{2\pi}g(\sin t,\cos t)\,dt\) → substitute \(z=e^{it}\) → integral becomes a contour integral over \(|z|=1\). Square‑root or log in integrand → expect a branch cut; look for a keyhole or a “dog‑bone’’ contour. Integral of \(\frac{1}{z^k}\) around a closed loop → non‑zero only when \(k=1\). --- 🗂️ Exam Traps Missing the minus sign for a clockwise contour – answer choices may give \(+2\pi i\) when the contour is actually clockwise. Assuming zero integral just because the path is closed – forget to check for interior singularities. Choosing the wrong set of poles – e.g., for \(\frac{1}{(z^2+1)^2}\) only the simple poles inside the chosen contour count; a pole on the real axis may be excluded. Neglecting the branch‑cut factor – for \(\sqrt{z}\) the two sides of the cut differ by \(-1\); failing to include it yields a result off by a sign. Confusing exponent \(-1\) with power \(-1\) – the integral \(\int{|z|=1} z^{-2}dz\) is zero, not \(2\pi i\). Forgetting the residue at infinity when the contour is taken to enclose the whole plane – the sum of all finite residues plus the residue at infinity must be zero. ---