Residue theorem and standard contour integrals

In one sentence

The residue is the single Laurent coefficient $a_{-1}$, the residue theorem says a contour integral is $2\pi i$ times the residues enclosed, and the exam's real content is the handful of standard contours that turn a real integral into that sum.

Why the exam asks it

Because it converts hard real integrals into arithmetic, and because each standard contour has a convergence condition that decides whether the method works at all. A candidate who applies the semicircle without checking the degree condition will get a confident wrong answer, which is exactly what the options are for.

The idea

For an isolated singularity at $a$, $\operatorname{Res}(f, a)$ is the coefficient $a_{-1}$ in the Laurent expansion on a punctured disc around $a$. It is the only coefficient that survives integration, because $\oint (z-a)^n dz = 0$ for every $n \ne -1$.

Residue theorem. If $f$ is holomorphic inside and on a positively oriented simple closed contour $\gamma$ except for finitely many isolated singularities $a_k$ inside, then

$$\oint_\gamma f(z)\,dz = 2\pi i \sum_k \operatorname{Res}(f, a_k).$$

Computing residues

• Simple pole: $\operatorname{Res}(f,a) = \lim_{z\to a}(z-a)f(z)$.
• Simple pole of a quotient: if $f = p/q$ with $p(a)\ne0$ and $q$ having a simple zero at $a$, then $\operatorname{Res} = \dfrac{p(a)}{q'(a)}$. This is the one worth memorising; it avoids factorising the denominator.
• Pole of order $m$: $$\operatorname{Res}(f,a) = \frac{1}{(m-1)!}\lim_{z\to a}\frac{d^{m-1}}{dz^{m-1}}\Big[(z-a)^m f(z)\Big].$$
• Essential singularity: no formula. Expand and read $a_{-1}$. For $e^{1/z}$ the coefficient of $1/z$ is $1$, so the residue is $1$.

Getting $m$ wrong is the commonest error — the formula silently returns a number for the wrong $m$ rather than failing.

Trigonometric integrals

For $\int_0^{2\pi} R(\cos\theta,\sin\theta)\,d\theta$, set $z = e^{i\theta}$, so

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

and the integral becomes a contour integral round $|z| = 1$. For example $\int_0^{2\pi}\frac{d\theta}{2+\cos\theta} = \frac{2\pi}{\sqrt3}$, from the single pole inside the unit circle.

Rational integrals over the line

Close with a semicircle in the upper half plane and let $R \to \infty$. The arc has length $\pi R$, so by ML it vanishes only if the integrand decays faster than $1/R$:

Condition: $\deg(\text{denominator}) \ge \deg(\text{numerator}) + 2$.

Then $\int_{-\infty}^{\infty} f = 2\pi i \sum \operatorname{Res}$ over poles in the upper half plane only. For $\frac{1}{1+x^2}$: the pole at $i$ has residue $\frac{1}{2i}$, so the integral is $2\pi i \cdot \frac{1}{2i} = \pi$, which is $\arctan$ agreeing.

Closing in the lower half plane is equally valid and picks up the other poles with a sign change; the answers agree, and that is a useful check.

Oscillatory integrals

For $\int_{-\infty}^{\infty} f(x)e^{iax}dx$ with $a > 0$, Jordan's lemma allows the weaker condition that $f \to 0$ on the arc — degree of denominator exceeding numerator by $1$ suffices, because $|e^{iaz}| = e^{-a\operatorname{Im}z}$ decays in the upper half plane.

The sign of $a$ decides which half plane to close in. Getting it wrong closes the contour where the exponential grows, and the arc no longer vanishes.

Poles on the contour

$\int_0^\infty \frac{\sin x}{x}dx$ has an integrand whose complex counterpart $e^{iz}/z$ has a pole at the origin, sitting on the path. Indent the contour with a small semicircle around it. A small semicircle around a simple pole contributes $i\pi \operatorname{Res}$ — half of $2\pi i$, because it turns through half a revolution. The result is $\pi/2$.

Note this computes a principal value. $\int_{-\infty}^{\infty}\frac{\sin x}{x}dx$ converges as an improper Riemann integral, but many principal values exist for integrands that do not converge at all, and the option will not always distinguish them.

Where intuition breaks

"Apply the semicircle to any rational function." Only when the denominator's degree exceeds the numerator's by at least $2$. For $\int \frac{x}{1+x^2}$ the arc does not vanish and the integral does not converge — though its principal value is $0$ by symmetry, which is a different claim.

"The residue is the value of the function at the pole." It is the coefficient of $(z-a)^{-1}$. The function has no value at the pole.

"Use the order-$m$ formula; if $m$ is too big it will still work." Too large an $m$ gives a wrong number, not an error — the derivative of a function with a removable singularity is perfectly finite. Establish the order first.

"Count all the poles." Only those inside the contour. Half the options in this topic move a pole across the axis.

"A pole on the contour just contributes its residue." An indentation round a simple pole contributes $i\pi\operatorname{Res}$, not $2\pi i \operatorname{Res}$ — half the turn, half the contribution. And for a pole of order $\ge 2$ on the contour the indented integral diverges.

"$e^{iaz}$ decays in the upper half plane whatever the sign of $a$." $|e^{iaz}| = e^{-a\operatorname{Im}z}$ decays above the axis only for $a > 0$. For $a<0$, close below.

"The principal value is the integral." It agrees when the integral converges, and exists in many cases where it does not. $\int_{-\infty}^{\infty} x\,dx$ has principal value $0$ and no value.

The exam's angle

1. Identify the poles and which are enclosed. That is most of the work; the residues are then mechanical.
2. Use $p(a)/q'(a)$ for simple poles. It is faster than factorising and it is right whenever $q$ has a simple zero.
3. Check the decay condition before closing a contour. Degree gap of $2$ for rational, $1$ with an oscillatory factor via Jordan.
4. For $\int_0^{2\pi}$, substitute $z = e^{i\theta}$ immediately. The unit circle is the contour and $d\theta = dz/(iz)$.
5. Watch for a pole on the real axis — that means an indentation and a factor of $i\pi$, and it means the answer is a principal value.

The night before

• $\operatorname{Res}(f,a) = a_{-1}$. $\oint = 2\pi i\sum\operatorname{Res}$ over enclosed singularities.
• Simple pole: $\lim (z-a)f(z)$, or $p(a)/q'(a)$ for a quotient.
• Order $m$: $\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-a)^mf(z)]$. Establish $m$ first — a wrong $m$ returns a number, not an error.
• Essential: expand and read $a_{-1}$. $\operatorname{Res}(e^{1/z},0) = 1$.
• $\int_0^{2\pi}$: put $z = e^{i\theta}$, $d\theta = dz/(iz)$, integrate round $|z|=1$.
• Rational over $\mathbb{R}$: semicircle above, denominator degree $\ge$ numerator $+\,2$, count upper-half poles only. $\int\frac{dx}{1+x^2} = \pi$.
• With $e^{iax}$: Jordan, degree gap of $1$ suffices, close above for $a>0$ and below for $a<0$.
• Pole on the contour: indent, and a simple pole contributes $i\pi\operatorname{Res}$. $\int_0^\infty \frac{\sin x}{x} = \frac{\pi}{2}$.
• Principal value exists more often than the integral does.