Residue theorem

A closed-contour integral of a function with finitely many isolated singularities reduces to a finite sum of residues. That’s the whole content of the residue theorem, and it’s the main computational tool in complex analysis.

Statement

Let $f$ be analytic on a simply connected domain $\Omega$ except at finitely many isolated singularities $z_1,\dots ,z_k$. Let $C$ be a positively oriented simple closed contour in $\Omega$ that passes through none of the singularities. Let $z_1,\dots ,z_m$ be the singularities lying inside $C$. Then

${\oint }_{C}f(z)dz=2\pi i{\sum }_{j=1}^m\mathrm{Res}(f,z_j).$

Singularities outside $C$ contribute nothing.

Sketch of proof

Connect $C$ to small circles ${\gamma }_j$ around each interior singularity $z_j$ via thin “keyhole” corridors traversed in both directions. The composite contour ($C$ counterclockwise, each ${\gamma }_j$ clockwise, plus the keyhole bridges) bounds a region on which $f$ is analytic, so its integral is zero by Cauchy’s theorem. The keyhole bridges cancel in pairs.

What remains: ${\oint }_{C}fdz={\sum }_j{\oint }_{{\gamma }_j}fdz$ with each ${\gamma }_j$ now counterclockwise around $z_j$.

On each ${\gamma }_j$, expand $f$ in its Laurent series around $z_j$ and integrate term by term. By the fundamental integral $\oint (z-z_j{)}^{n}dz=2\pi i$ if $n=-1$ and $0$ otherwise, only the $a_{-1}$ term contributes. Sum gives $2\pi i\sum \mathrm{Res}(f,z_j)$. ∎

What the theorem unifies

Every earlier closed-contour result is a special case:

• Cauchy’s theorem: no singularities inside $C$ $\Rightarrow$ sum is empty $\Rightarrow$ integral $=0$.
• Cauchy integral formula: $f(z)=g(z)/(z-z_0)$ with $g$ analytic, $z_0$ inside $C$. Simple pole, $\mathrm{Res}=g(z_0)$. Integral $=2\pi ig(z_0)$.
• Generalized CIF: $f(z)=g(z)/(z-z_0{)}^{n+1}$. Pole of order $n+1$, residue computed via the formula gives $g^{(n)}(z_0)/n!$. Integral $=2\pi i{g}^{(n)}(z_0)/n!$.

All of them collapse to “compute residues, sum, multiply by $2\pi i$.”

The closed-contour decision tree

For ${\oint }_{C}f(z)dz$ with $C$ positively oriented:

1. Find all singularities of $f$ inside $C$. If none, integral is $0$.
2. Classify each one: removable, pole of order $m$, or essential.
3. Compute residues by the matching formula:
   • Removable: $0$.
   • Simple pole ($m=1$): ${\lim }_{z\to z_0}(z-z_0)f(z)$, or $p(z_0)/q^{\prime }(z_0)$ if applicable.
   • Pole of order $m$: $\frac{1}{(m-1)!}\lim \frac{d^{m-1}}{d{z}^{m-1}}[(z-z_0{)}^{m}f(z)]$.
   • Essential: read $a_{-1}$ from Laurent expansion.
4. Sum residues and multiply by $2\pi i$.

Every closed-contour integral collapses to this procedure.

Worked example 1

${\oint }_{|z|=2}\frac{e^z}{z^2-1}dz$.

Singularities: $z=\pm 1$, both inside $|z|=2$.

Residue at $z=1$ (simple pole, $p/q^{\prime }$): $p=e^z$, $q=z^2-1$, $q^{\prime }=2z$. $\mathrm{Res}=e^1/(2\cdot 1)=e/2$.

Residue at $z=-1$: $\mathrm{Res}=e^{-1}/(2\cdot (-1))=-1/(2e)$.

Sum: $e/2-1/(2e)$. Result: $2\pi i\cdot (e/2-1/(2e))=\pi i(e-1/e)=2\pi i\sinh 1$.

Worked example 2

${\oint }_{|z|=3}\frac{z}{(z-1{)}^2(z-2)}dz$.

Both singularities inside.

Residue at $z=1$ (pole of order 2): $\mathrm{Res}={\lim }_{z\to 1}\frac{d}{dz}[z/(z-2)]={\lim }_{z\to 1}\frac{(z-2)-z}{(z-2{)}^2}={\lim }_{z\to 1}\frac{-2}{(z-2{)}^2}=-2$.

Residue at $z=2$ (simple pole): $\mathrm{Res}=z/(z-1{)}^2{|}_{z=2}=2/1=2$.

Sum: $0$. Result: $0$.

Real trig integrals via the unit circle

For ${\int }_0^{2\pi }R(\cos \theta ,\sin \theta )d\theta$ with $R$ rational in $\cos ,\sin$, substitute $z=e^{i\theta }$:

$\cos \theta =\frac{z+1/z}{2},\sin \theta =\frac{z-1/z}{2i},d\theta =\frac{dz}{iz}.$

As $\theta$ runs $0$ to $2\pi$, $z$ traverses $|z|=1$ counterclockwise, so the real integral becomes a closed contour integral on $|z|=1$.

Example: ${\int }_0^{2\pi }d\theta /(2+\cos \theta )$.

Substitute: $\cos \theta =(z+1/z)/2=(z^2+1)/(2z)$. $d\theta =dz/(iz)$.

Denominator: $2+\cos \theta =(4z+z^2+1)/(2z)=(z^2+4z+1)/(2z)$.

Integral becomes $\oint \frac{2z}{z^2+4z+1}\cdot \frac{dz}{iz}=\oint \frac{2dz}{i(z^2+4z+1)}=\frac{2}{i}\oint \frac{dz}{z^2+4z+1}$.

Roots of $z^2+4z+1=0$: $z=-2\pm \sqrt{3}$. Only $z=-2+\sqrt{3}\approx -0.27$ lies inside $|z|=1$ (the other is $\approx -3.73$).

Residue at $z=-2+\sqrt{3}$ (simple pole, $p/q^{\prime }$): $\mathrm{Res}=1/(2z+4){|}_{z=-2+\sqrt{3}}=1/(2\sqrt{3})$.

Multiply: $(2/i)(2\pi i)(1/(2\sqrt{3}))=2\pi /\sqrt{3}$. Result: $2\pi /\sqrt{3}$.

Real improper integrals: rational functions

For ${\int }_{-\infty }^{\infty }R(x)dx$ with $R$ rational, $\mathrm{deg}\text{denom}\ge \mathrm{deg}\text{numer}+2$, no real poles:

• Close the contour with a large upper semicircle of radius $R$.
• As $R\to \infty$, the semicircle contribution vanishes by ML estimate (since $|R(z)|$ decays as $1/R^2$ or faster on $|z|=R$).
• The real-axis integral plus the semicircle integral equals $2\pi i\sum \mathrm{Res}$ (upper half-plane).

Example: ${\int }_{-\infty }^{\infty }dx/(x^2+1)$.

Poles at $z=\pm i$. Upper half-plane: $z=i$. Residue ($p/q^{\prime }$ with $p=1,q=z^2+1,q^{\prime }=2z$): $1/(2i)=-i/2$.

Real integral $=2\pi i(-i/2)=\pi$. ✓ (Direct check: $\arctan (\infty )-\arctan (-\infty )=\pi$.)

Real improper integrals with cos x or sin x

For ${\int }_{-\infty }^{\infty }R(x)\cos (ax)dx$ with $R$ rational and decaying: replace $\cos (ax)$ with $e^{iaz}$ (take real part at end), close contour in the upper half-plane (so $|e^{iaz}|=e^{-ay}$ decays for $y>0$). The semicircle contribution vanishes by Jordan’s lemma.

Example: ${\int }_{-\infty }^{\infty }\cos x/(x^2+1)dx$.

Consider $\int e^{iz}/(z^2+1)$. Pole in UHP: $z=i$. Residue: $e^{i\cdot i}/(2i)=e^{-1}/(2i)=-i{e}^{-1}/2$.

Contour integral: $2\pi i\cdot (-i{e}^{-1}/2)=\pi /e$.

Real part: $\pi /e$. So $\int \cos x/(x^2+1)dx=\pi /e$. (Imaginary part vanishes by parity: $\sin x/(x^2+1)$ is odd.)

Where it applies

• Closed-contour integrals of meromorphic functions.
• Real trig integrals ${\int }_0^{2\pi }R(\cos ,\sin )d\theta$ via unit-circle contour.
• Real improper integrals ${\int }_{-\infty }^{\infty }R(x)dx$ via large semicircle.
• Real improper integrals with $\cos$ or $\sin$ multipliers via Jordan’s lemma.

Computing the residues is mechanical. The hard part is choosing the right contour and verifying the negligibility of any auxiliary pieces (semicircles, indentations).