Contour integration of real integrals using residues

At a Glance

• Frequency: 9 sub-parts across 9 of 13 years (2013, 2014, 2017, 2018, 2020, 2021, 2022, 2023, 2025)
• Priority tier: T1
• Marks (count): 15 (3), 20 (6)
• Average solve time: ~15 min
• Difficulty mix: medium 6, hard 3
• Section: A | Dominant type: computation

Why This Chapter Matters

Contour integration is the single highest-mark topic in Paper 2 Complex Analysis — every question so far has been worth 15 or 20 marks, and the pattern holds year after year. There are exactly two question types and each has a four-step algorithm. The hard part is not the method (it is mechanical) but the residue calculation, especially for double poles or when two poles are enclosed simultaneously. A student who can reliably compute residues at simple and order-2 poles and correctly apply Jordan’s lemma will bank 20 marks on a predictable question.

Minimum Theory

Residue theorem. If $f$ is analytic inside and on a simple closed contour $C$ except at finitely many isolated singularities, then $\oint_C f(z)\,dz = 2\pi i\sum_k \operatorname{Res}_{z=z_k} f$, where the sum is over all poles inside $C$, and $C$ is traversed counterclockwise.

Computing residues. At a simple pole $z=a$: $\operatorname{Res}_a f = \lim_{z\to a}(z-a)f(z)$; if $f=p/q$ with $q(a)=0$, $p(a)\neq0$, $q'(a)\neq0$ then $\operatorname{Res}_a f = p(a)/q'(a)$. At an order-$m$ pole: $\operatorname{Res}_a f = \dfrac{1}{(m-1)!}\lim_{z\to a}\dfrac{d^{m-1}}{dz^{m-1}}[(z-a)^m f(z)]$.

Jordan’s lemma and arc estimates. For the upper semicircle $\Gamma_R$: if $g(z)\to0$ uniformly on $\Gamma_R$ as $R\to\infty$ and $m>0$, then $\int_{\Gamma_R}g(z)e^{imz}\,dz\to0$ (Jordan’s lemma — valid for $e^{imz}$ even when $|g|\sim C/R$ rather than $C/R^2$). For integrands without an oscillatory factor, the cruder ML estimate suffices: if $|f|\le M/R^{1+\varepsilon}$ on $\Gamma_R$ then $|\int_{\Gamma_R}f|\le \pi R\cdot M/R^{1+\varepsilon}\to0$. Indent around a simple pole on the real axis: a small upper semicircle of radius $\varepsilon$ traversed clockwise (from $-\varepsilon$ to $+\varepsilon$ on the real axis) contributes $-\pi i\cdot\operatorname{Res}_0 f$ in the limit $\varepsilon\to0$.

Question Archetypes

Two patterns cover every question in the corpus. Classify in seconds by looking at the integration limits and the integrand type.

real-line-integral-contour (5 question(s); 2017, 2018, 2021, 2022, 2025)

Recognition Cues

• Limits $\int_0^\infty$ or $\int_{-\infty}^\infty$; the integrand is rational, or rational times $\sin mx$ / $\cos mx$.
• The answer is a closed form involving $\pi$, $e^{-ma}$, or both.
• If $\sin$ or $\cos$ is present, replace with $\operatorname{Im}(e^{imz})$ or $\operatorname{Re}(e^{iz})$ before setting up the contour.

Solution Template

1. Symmetrise. If the integrand is even, write $I = \tfrac{1}{2}\int_{-\infty}^{\infty}$.
2. Replace trig. Replace $\sin mx$ by $\operatorname{Im}(e^{imz})$, or $\cos x$ by $\operatorname{Re}(e^{iz})$. This gives a complex function $f(z)$ that decays in the upper half-plane.
3. Choose contour. $C_R = [-R,R]\cup\Gamma_R$ (upper semicircle). Close in the upper half-plane (use lower if $m<0$).
4. Identify poles of $f$ in the upper half-plane (those with $\operatorname{Im}z>0$). Check if any lie on the real axis (requiring an indent).
5. Vanishing of $\Gamma_R$. For $e^{imz}$ factor with $m>0$: apply Jordan’s lemma. Without: use ML estimate (denominator degree $\geq$ numerator degree $+2$).
6. Compute residues at each enclosed pole (simple or order-2 formula).
7. Apply residue theorem: $\int_{-\infty}^{\infty}f(x)\,dx = 2\pi i\sum\operatorname{Res}$ (plus $-\pi i\cdot\operatorname{Res}_0$ if $z=0$ is indented).
8. Extract the real part (for $\cos$) or imaginary part (for $\sin$). Halve if symmetrised in step 1.

Worked Example(s)

2017 Paper 2, 2017-P2-Q2b (15 marks)

Prove $\displaystyle\int_0^\infty\frac{x\sin mx}{a^2+x^2}\,dx=\frac{\pi}{2}e^{-ma}$ ($m,a>0$).

1 — Symmetrise. Integrand is even, so $I=\tfrac12\int_{-\infty}^\infty\dfrac{x\sin mx}{a^2+x^2}\,dx$.

2 — Replace. Let $J=\int_{-\infty}^\infty\dfrac{xe^{imx}}{a^2+x^2}\,dx$; then $I=\tfrac12\operatorname{Im}J$.

3–4 — Contour and poles. $f(z)=\dfrac{ze^{imz}}{(z-ia)(z+ia)}$. Upper-half-plane pole: $z=ia$ only.

5 — Arc vanishes. $|z/(a^2+z^2)|\to0$ on $\Gamma_R$; Jordan’s lemma (with $m>0$) kills the arc.

6 — Residue. Simple pole at $z=ia$: $\operatorname{Res}_{z=ia}f=\frac{ia\,e^{im(ia)}}{2ia}=\frac{e^{-ma}}{2}.$

7–8 — Conclude. $J=2\pi i\cdot\frac{e^{-ma}}{2}=\pi i\,e^{-ma}.\qquad\operatorname{Im}J=\pi e^{-ma}.\qquad I=\frac{\pi}{2}e^{-ma}.\quad\blacksquare$

2018 Paper 2, 2018-P2-Q3b (15 marks)

Show $\displaystyle\int_0^\infty\frac{dx}{(x^2+a^2)^2}=\frac{\pi}{4a^3}$ ($a>0$).

1. Even integrand: $I=\tfrac12\int_{-\infty}^\infty f(x)\,dx$ with $f(z)=\dfrac{1}{(z^2+a^2)^2}=\dfrac{1}{(z-ia)^2(z+ia)^2}$.

5. $|f|\le C/R^4$ on $\Gamma_R$, so $|\int_{\Gamma_R}|\le \pi R\cdot C/R^4\to0$ (ML estimate).

6 — Order-2 residue at $z=ia$: $\operatorname{Res}_{z=ia}f=\lim_{z\to ia}\frac{d}{dz}\frac{1}{(z+ia)^2}=\frac{-2}{(2ia)^3}=\frac{-2}{-8ia^3}=\frac{1}{4ia^3}.$

7–8. $\int_{-\infty}^\infty f\,dx=2\pi i\cdot\frac{1}{4ia^3}=\frac{\pi}{2a^3}.\qquad I=\frac12\cdot\frac{\pi}{2a^3}=\boxed{\;\frac{\pi}{4a^3}.\;}\quad\blacksquare$

2021 Paper 2, 2021-P2-Q4b (20 marks)

Evaluate $\displaystyle\int_{-\infty}^\infty\frac{\sin x}{x(x^2+a^2)}\,dx$ ($a>0$).

Setup. $f(z)=\dfrac{e^{iz}}{z(z-ia)(z+ia)}$. The real-axis pole $z=0$ forces an indented contour: standard $C_R$ but with a small upper semicircle of radius $\varepsilon$ bypassing $z=0$.

Poles in upper half-plane: $z=ia$ (simple). Pole at $z=0$ is on the real axis — indented around.

Residues. $\operatorname{Res}_{z=ia}f=\frac{e^{-a}}{ia\cdot 2ia}=\frac{e^{-a}}{-2a^2},\qquad\operatorname{Res}_{z=0}f=\frac{e^0}{(-ia)(ia)}=\frac{1}{a^2}.$

Apply contour theorem. Jordan kills $\Gamma_R$; small CW semicircle at $0$ contributes $-\pi i\cdot\operatorname{Res}_0 f=-\pi i/a^2$: $\int_{-\infty}^\infty f(x)\,dx+\left(-\frac{\pi i}{a^2}\right)=2\pi i\cdot\frac{e^{-a}}{-2a^2}=-\frac{\pi i e^{-a}}{a^2}.$ $\int_{-\infty}^\infty f(x)\,dx=\frac{\pi i}{a^2}-\frac{\pi i e^{-a}}{a^2}=\frac{\pi i(1-e^{-a})}{a^2}.$

Take Im: $\boxed{\;\int_{-\infty}^\infty\frac{\sin x}{x(x^2+a^2)}\,dx=\frac{\pi(1-e^{-a})}{a^2}.\;}$

2022 Paper 2, 2022-P2-Q2c (20 marks)

Evaluate $\displaystyle\int_{-\infty}^\infty\frac{\cos x}{(x^2+a^2)(x^2+b^2)}\,dx$ ($a>b>0$).

Setup. Replace $\cos x$ with $\operatorname{Re}(e^{iz})$. $f(z)=\dfrac{e^{iz}}{(z^2+a^2)(z^2+b^2)}$. Upper-half-plane poles: $z=ia$ and $z=ib$ (both simple).

Residues ($a>b>0$ so $a^2>b^2$): $\operatorname{Res}_{z=ia}f=\frac{e^{-a}}{2ia(b^2-a^2)}=\frac{-e^{-a}}{2ia(a^2-b^2)},\qquad\operatorname{Res}_{z=ib}f=\frac{e^{-b}}{2ib(a^2-b^2)}.$

Sum and multiply by $2\pi i$ (Jordan kills arc): $\int_{-\infty}^\infty f\,dx=2\pi i\cdot\frac{1}{2i(a^2-b^2)}\!\left(\frac{e^{-b}}{b}-\frac{e^{-a}}{a}\right)=\frac{\pi}{a^2-b^2}\!\left(\frac{e^{-b}}{b}-\frac{e^{-a}}{a}\right).$

This is already real, so: $\boxed{\;\int_{-\infty}^\infty\frac{\cos x}{(x^2+a^2)(x^2+b^2)}\,dx=\frac{\pi}{a^2-b^2}\!\left(\frac{e^{-b}}{b}-\frac{e^{-a}}{a}\right).\;}$

2025 Paper 2, 2025-P2-Q2c (20 marks)

Prove $\displaystyle\int_{-\infty}^\infty\frac{x^2-x+2}{x^4+10x^2+9}\,dx=\frac{5\pi}{12}$.

Factor denominator. $x^4+10x^2+9=(x^2+1)(x^2+9)=(z-i)(z+i)(z-3i)(z+3i)$.

Upper-half-plane poles: $z=i$ and $z=3i$ (both simple). Use $q'(z)=4z^3+20z$.

Residues via $p(z_0)/q'(z_0)$:

At $z=i$: $p(i)=i^2-i+2=1-i$; $q'(i)=4i^3+20i=-4i+20i=16i$. $\operatorname{Res}_{z=i}f=\frac{1-i}{16i}=\frac{(1-i)(-i)}{16}=\frac{-i-1}{16}\cdot\frac{-1}{-1}=-\frac{1}{16}-\frac{i}{16}.$

At $z=3i$: $p(3i)=(3i)^2-3i+2=-7-3i$; $q'(3i)=4(-27i)+60i=-48i$. $\operatorname{Res}_{z=3i}f=\frac{-7-3i}{-48i}=\frac{(7+3i)(-i)}{48}=\frac{3-7i}{48}=\frac{1}{16}-\frac{7i}{48}.$

Arc estimate. $|f|\le CR^2/R^4=C/R^2$, so $|\int_{\Gamma_R}f|\le\pi C/R\to0$.

Sum residues and multiply by $2\pi i$: $\text{Sum}=\left(-\frac{1}{16}+\frac{1}{16}\right)+i\left(-\frac{1}{16}-\frac{7}{48}\right)=i\left(-\frac{3+7}{48}\right)=-\frac{10i}{48}=-\frac{5i}{24}.$ $\int_{-\infty}^\infty f\,dx=2\pi i\cdot\left(-\frac{5i}{24}\right)=\frac{10\pi}{24}=\boxed{\;\frac{5\pi}{12}.\;}\quad\blacksquare$

Common Traps

• Use $e^{imz}$, not $\sin mz$ or $\cos mz$ directly: $\sin mz$ blows up in the upper half-plane (imaginary part of $mz$ is positive there), while $e^{imz}=e^{im(x+iy)}=e^{-my}e^{imx}$ decays. Take $\operatorname{Im}$ or $\operatorname{Re}$ only after evaluating the contour integral.
• Double poles need one derivative: $(z-ia)^2f$ must be differentiated once (not the simple-pole formula). Track $i^3=-i$ and $(2ia)^3=-8ia^3$ carefully — sign errors here are common and costly.
• Halve for $0$ to $\infty$: the contour gives $-\infty$ to $\infty$; use evenness to write $I=\tfrac12\int_{-\infty}^\infty$. Only the 2021 question already runs to $\pm\infty$ and needs no halving.
• Indented contour sign: the small CW semicircle around a simple real-axis pole contributes $-\pi i\cdot\operatorname{Res}_0$ (minus, not plus) because it is traversed clockwise while the full-circle contribution would be $+2\pi i\cdot\operatorname{Res}$.
• Include only upper-half-plane poles: $z=-ia,-ib,-i,-3i$ etc. are in the lower half-plane; they lie outside the contour and contribute zero to the sum.

trig-integral-contour (4 question(s); 2013, 2014, 2020, 2023)

Recognition Cues

• Integration limits $[0,\pi]$ or $[0,2\pi]$; integrand is a rational function of $\cos\theta$ and/or $\sin\theta$.
• For $[0,\pi]$: use the symmetry $\cos(2\pi-\theta)=\cos\theta$ or $\sin^n(\theta+\pi)=\sin^n\theta$ to double to $[0,2\pi]$ first.
• If the integrand involves $\cos n\theta$ or $\sin n\theta$ (as in 2023), express those via $z^n+z^{-n}$ or $z^n-z^{-n}$ — not via $(z\pm z^{-1})^n$.

Solution Template

1. Extend to $[0,2\pi]$ if necessary, using a symmetry of the integrand. Record the factor (often $1/2$).
2. Substitute $z=e^{i\theta}$: $\cos\theta=\tfrac{z+z^{-1}}{2}$, $\sin\theta=\tfrac{z-z^{-1}}{2i}$, $d\theta=\tfrac{dz}{iz}$. Clear all $z^{-n}$ by multiplying numerator and denominator by $z^n$.
3. Factor the denominator of the resulting rational integrand in $z$. Find the poles.
4. Check which poles lie inside $|z|=1$ (if $|z_0|<1$ it is inside; $|z_0|>1$ outside).
5. Compute residues at interior poles (simple or order-2).
6. Apply residue theorem: $\oint_{|z|=1}g(z)\,dz=2\pi i\sum\operatorname{Res}$.
7. Undo the extension (divide by 2 if step 1 doubled the range). Box the answer.

Worked Example(s)

2013 Paper 2, 2013-P2-Q4b (15 marks)

Evaluate $\displaystyle\int_0^\pi\sin^4\theta\,d\theta$ using Cauchy’s residue theorem.

1 — Extend. $\sin^4(\theta+\pi)=\sin^4\theta$, so $\int_0^{2\pi}\sin^4\theta\,d\theta=2I$.

2 — Substitute. $\sin\theta=\dfrac{z-z^{-1}}{2i}$, $d\theta=\dfrac{dz}{iz}$. Then $\sin^4\theta=\dfrac{(z-z^{-1})^4}{16}$.

Expand by binomial: $(z-z^{-1})^4=z^4-4z^2+6-4z^{-2}+z^{-4}$. After multiplying by $dz/(iz)$: $2I=\frac{1}{16i}\oint_{|z|=1}\!\left(z^3-4z+\frac{6}{z}-\frac{4}{z^3}+\frac{1}{z^5}\right)dz.$

3–4 — Pole and residue. Only $z=0$ is inside $|z|=1$. The only $z^{-1}$ term has coefficient $6$, so $\operatorname{Res}_{z=0}(\text{integrand})=6$.

5–7. $2I=\dfrac{1}{16i}\cdot2\pi i\cdot6=\dfrac{12\pi}{16}=\dfrac{3\pi}{4}$.

$\boxed{\;I=\int_0^\pi\sin^4\theta\,d\theta=\frac{3\pi}{8}.\;}$

2014 Paper 2, 2014-P2-Q3c (20 marks)

Evaluate $\displaystyle\int_0^\pi\frac{d\theta}{(1+\tfrac{1}{2}\cos\theta)^2}$ using residues.

1. $\cos\theta$ is even, so $\int_0^{2\pi}=2\int_0^\pi$.

2. $1+\tfrac12\cos\theta=\dfrac{z^2+4z+1}{4z}$, giving the contour integral $\int_0^{2\pi}(\cdots)\,d\theta=\frac{16}{i}\oint_{|z|=1}\frac{z\,dz}{(z^2+4z+1)^2}.$

3–4. Roots of $z^2+4z+1$: $z_{1,2}=-2\pm\sqrt3$. $|z_1|=2-\sqrt3\approx0.27<1$ (inside); $|z_2|=2+\sqrt3>1$ (outside). Factor: $(z^2+4z+1)^2=(z-z_1)^2(z-z_2)^2$. Order-2 pole at $z_1$.

5. Order-2 residue of $z/(z-z_1)^2(z-z_2)^2$ at $z_1$: $\operatorname{Res}_{z_1}=\frac{d}{dz}\!\left[\frac{z}{(z-z_2)^2}\right]\bigg|_{z=z_1}=\frac{-z_1-z_2}{(z_1-z_2)^3}=\frac{4}{(2\sqrt3)^3}=\frac{4}{24\sqrt3}=\frac{\sqrt3}{18}.$ (Used $z_1+z_2=-4$ and $z_1-z_2=2\sqrt3$.)

6–7. $\frac{16}{i}\cdot2\pi i\cdot\frac{\sqrt3}{18}=\frac{16\pi\sqrt3}{9}\qquad\Longrightarrow\qquad\int_0^\pi=\frac{8\pi\sqrt3}{9}.$

$\boxed{\;\int_0^\pi\frac{d\theta}{(1+\tfrac12\cos\theta)^2}=\frac{8\pi\sqrt3}{9}.\;}$

Standard-formula check: $\int_0^{2\pi}\frac{d\theta}{(a+b\cos\theta)^2}=\frac{2\pi a}{(a^2-b^2)^{3/2}}$ gives $\frac{16\pi\sqrt3}{9}$, halved is $\frac{8\pi\sqrt3}{9}$ ✓.

2020 Paper 2, 2020-P2-Q2c (20 marks)

Evaluate $\displaystyle\int_0^{2\pi}\frac{d\theta}{3+2\sin\theta}$ by contour integration.

1. Limits already $[0,2\pi]$. 2. $\sin\theta=(z-z^{-1})/(2i)$; after simplification: $3+2\sin\theta=3-i(z-z^{-1})\;\Longrightarrow\; I=\oint_{|z|=1}\frac{dz}{iz^2-3z-i}\cdot(-1).$

3–4. Quadratic $iz^2-3z-i=0$ gives $z_{1,2}=-\tfrac{i(3\mp\sqrt5)}{2}$ (purely imaginary). $|z_1|=(3-\sqrt5)/2\approx0.38<1$ (inside); $|z_2|\approx2.62>1$.

5. Simple-pole residue at $z_1$ (leading coeff $i$): $\operatorname{Res}_{z_1}=\frac{1}{i(z_1-z_2)}=\frac{1}{i\cdot i\sqrt5}=-\frac{1}{\sqrt5}.$ Accounting for the overall factor, the residue of the integrand is $-i/\sqrt5$.

6–7. $I=2\pi i\cdot\left(-\frac{i}{\sqrt5}\right)=\frac{2\pi}{\sqrt5}=\boxed{\;\frac{2\pi\sqrt5}{5}.\;}$

Check: standard formula $2\pi/\sqrt{a^2-b^2}$ with $a=3,b=2$ gives $2\pi/\sqrt5$ ✓.

2023 Paper 2, 2023-P2-Q2c (20 marks)

Evaluate $\displaystyle\int_0^{2\pi}\frac{\cos2\theta}{5+4\cos\theta}\,d\theta$ by contour integration.

2. Use $\cos\theta=\tfrac{z+z^{-1}}{2}$ and $\cos2\theta=\tfrac{z^2+z^{-2}}{2}$. After clearing denominators: $I=\oint_{|z|=1}\frac{z^4+1}{2iz^2(2z+1)(z+2)}\,dz.$

3–4. Poles: $z=0$ (order 2, inside), $z=-\tfrac12$ (simple, inside), $z=-2$ (simple, outside).

5 — Residue at $z=-\tfrac12$ (simple): numerator $(\tfrac{1}{2})^4+1=\tfrac{17}{16}$; denominator factor $2iz^2\cdot2\cdot(z+2)$ at $z=-\tfrac12$ gives $\tfrac{3i}{2}$. $\operatorname{Res}_{z=-1/2}=\frac{17/16}{3i/2}=\frac{17}{24i}=-\frac{17i}{24}.$

Residue at $z=0$ (order 2): let $h(z)=\dfrac{z^4+1}{2i(2z+1)(z+2)}$. Then $h(0)=\tfrac{1}{4i}$, $h'(0)=\dfrac{-5/4}{2i}=\dfrac{-5}{8i}=\dfrac{5i}{8}$. $\operatorname{Res}_{z=0}=h'(0)=\frac{5i}{8}.$

6–7. $I=2\pi i\!\left(-\frac{17i}{24}+\frac{5i}{8}\right)=2\pi i\cdot\frac{-17i+15i}{24}=2\pi i\cdot\frac{-2i}{24}=\frac{4\pi}{24}=\boxed{\;\frac{\pi}{6}.\;}$

Common Traps

• Extend to $[0,2\pi]$ before applying the method: the $z=e^{i\theta}$ substitution requires a full period. Use an appropriate symmetry (e.g., $\cos(2\pi-\theta)=\cos\theta$, or $\sin^4(\theta+\pi)=\sin^4\theta$) and record the halving factor.
• Only the $z^{-1}$ coefficient is the residue: when the integrand at $z=0$ is a Laurent series $\sum a_n z^n$, only $a_{-1}$ contributes. The terms $z^{-2},z^{-3},\ldots$ have well-defined primitives and integrate to zero round a closed curve.
• $(2z+1)$ has leading coefficient 2: for the simple-pole residue at $z=-\tfrac12$, the denominator has an implicit factor of 2 from $d(2z+1)/dz=2$ — equivalently, use $p(z_0)/q'(z_0)$ where $q$ is the full denominator polynomial.
• The extension factor must be applied at the end: if you wrote $\int_0^{2\pi}=2\int_0^\pi$ in step 1, remember to divide the contour-integral result by 2 when reporting the original integral.

Marks-Aware Writing

15-mark questions (2017, 2018, 2013): Five numbered steps are sufficient — symmetrize, complexify, state the pole(s) and their location, compute the residue explicitly, apply the theorem and halve. The arc-vanishing argument must be stated in one sentence (Jordan’s lemma / ML estimate) — omitting it loses method marks.

20-mark questions (2021, 2022, 2023, 2025, 2014, 2020): Identical structure but expect to compute two or more residues (2022, 2025), handle a double pole (2018 if it were 20 marks, 2014, 2023), or manage an indented contour (2021). Each residue computation should be a labelled sub-step. Show the $i$ cancellation explicitly: writing $2\pi i\cdot(k/i)=2\pi k$ is worth a mark.

For both mark levels: box the numerical answer; verify by the standard formula if one exists ($2\pi/\sqrt{a^2-b^2}$ for $\int_0^{2\pi}d\theta/(a+b\cos\theta)$, etc.).

Practice Set