Residue theorem — computing real integrals: Worked examples

The scenario matrix

Every real integral this chapter attacks falls into one of these cells. Each worked example below is labelled with the cell it fills.

Cell	What makes it special	The danger it hides	Example
A Basic rational	gap $= de g Q - de g P \geq 2$ , simple poles	picking the wrong half-plane	Ex 1
B Higher-order pole	denominator has a repeated factor	must differentiate residue	Ex 2
C Oscillatory (cos/sin)	integrand has $cos a x$ or $sin a x$	use $e^{ia z}$ , not $cos z$	Ex 3
D Odd integrand → sine	the "imaginary part" branch	real part is $0$ , sine is the answer	Ex 4
E Trig over $[0, 2 π]$	already a loop → unit circle	the mandatory $\frac{1}{i z}$ factor	Ex 5
F Degenerate: pole on the real axis	contour passes through a spike	indent — take a principal value	Ex 6
G Gap $= 1$ with no oscillation	arc does not die	recipe fails — must recognise it	Ex 7
H Word problem	physics/engineering dressing	translate to a known cell	Ex 8

We now visit all eight. Watch the labels.

Ex 1 — Cell A: the honest rational integral

Worked example

$I = \int_{- \infty}^{\infty} \frac{x ^{2}}{x ^{4} + 1} d x$ Forecast: The integrand is even and positive, so the answer must be a positive real number, and (from the algebra to come) it will involve $2$ . Guess before reading on.

Step 1 — check the arc dies. Why this step? The boxed formula is only legal if the big semicircle contributes nothing. Here $P = x^{2}$ so $de g P = 2$ , and $Q = x^{4} + 1$ so $de g Q = 4$ ; gap $= 4 - 2 = 2 \geq 2$ . ✓ Arc dies, machine is legal.

Step 2 — find the poles. Why? We need every spike the contour can trap. Solve $z^{4} + 1 = 0 \Rightarrow z^{4} = - 1 = e^{iπ}$ . The four fourth-roots are $z_{k} = e^{i (π + 2 π k) /4}, k = 0, 1, 2, 3,$ i.e. angles $4 5^{\circ}, 13 5^{\circ}, 22 5^{\circ}, 31 5^{\circ}$ .

What to see in the figure: the four poles sit evenly spaced on a circle. The yellow path is our contour — the real axis plus the upper arc $Γ_{ρ}$ . The two pink dots ( $4 5^{\circ}$ and $13 5^{\circ}$ ) are inside the loop (positive imaginary part), so they count; the two blue crosses ( $22 5^{\circ}, 31 5^{\circ}$ ) are below the axis, outside the loop, so we ignore them. The two survivors are $z_{0} = e^{iπ /4} = \frac{1 + i}{2}, z_{1} = e^{i 3 π /4} = \frac{- 1 + i}{2} .$

Step 3 — residues (simple poles, use $g / h^{'}$ ). Why $g / h^{'}$ ? Factoring a quartic is ugly; the formula $Res = g (z_{0}) / h^{'} (z_{0})$ sidesteps it. Here $g = z^{2}$ , $h = z^{4} + 1$ , $h^{'} = 4 z^{3}$ , so $Res_{z_{k}} = \frac{z _{k}^{2}}{4 z _{k}^{3}} = \frac{1}{4 z _{k}} .$ Step 4 — evaluate at each UHP pole. Since $z_{k}$ lies on the unit circle, $\frac{1}{z _{k}} = \overline{z_{k}} = e^{- i a r g z_{k}}$ . So $Res_{z_{0}} = \frac{1}{4} e^{- iπ /4}, Res_{z_{1}} = \frac{1}{4} e^{- i 3 π /4} .$ Step 5 — sum and multiply by $2 π i$ . Why? That's the residue theorem's verdict. Write each exponential in rectangular form so we can add them: $e^{- iπ /4} = \frac{1 - i}{2}, e^{- i 3 π /4} = \frac{- 1 - i}{2} .$ Why this helps: the real parts $+ \frac{1}{2}$ and $- \frac{1}{2}$ cancel, and the imaginary parts $- \frac{i}{2}$ each add, leaving a purely imaginary sum: $e^{- iπ /4} + e^{- i 3 π /4} = \frac{- 2 i}{2} = - i 2 .$ Hence the residue sum is $\frac{1}{4} (- i 2) = - \frac{i 2}{4}$ , and $I = 2 π i \cdot (- \frac{i 2}{4}) = \frac{2 π 2}{4} = \frac{π}{2} = \frac{π 2}{2} .$ (Note $2 π i \cdot (- i) = 2 π$ , since $- i^{2} = 1$ — that's what turns the imaginary sum into a real answer.)

Verify: Answer is positive and carries $2$ — matches the forecast. Numerically $π / 2 \approx 2.221$ , and $\int$ of a bump of height $\leq \frac{1}{2}$ over a width $\sim$ few units is plausibly $\approx 2.2$ . ✓

Ex 2 — Cell B: a repeated factor forces a derivative

Worked example

$I = \int_{- \infty}^{\infty} \frac{d x}{( x ^{2} + 4 ) ^{2}}$ Forecast: Even, positive, and the poles sit at $\pm 2 i$ instead of $\pm i$ — the answer should be smaller than the $\int \frac{d x}{( x ^{2} + 1 ) ^{2}} = π /2$ case because the bump is flatter. Guess a fraction of $π$ .

Step 1 — arc check. $P = 1$ so $de g P = 0$ ; $Q = (x^{2} + 4)^{2}$ so $de g Q = 4$ ; gap $= 4 - 0 = 4 \geq 2$ . Arc dies. ✓

Step 2 — poles. $(z^{2} + 4)^{2} = (z - 2 i)^{2} (z + 2 i)^{2}$ . UHP pole is $z = 2 i$ , order $m = 2$ . Why order 2? The factor $(z - 2 i)$ appears squared.

Step 3 — order-2 residue formula. Why differentiate once? For an order- $m$ pole we take $m - 1 = 1$ derivative of $(z - 2 i)^{2} f$ : $Res_{2 i} = \frac{1}{1 !} lim_{z \to 2 i} \frac{d}{d z} [(z - 2 i)^{2} \cdot \frac{1}{( z - 2 i ) ^{2} ( z + 2 i ) ^{2}}] = \frac{d}{d z} (z + 2 i)^{- 2}_{2 i} .$ Step 4 — do the derivative. $\frac{d}{d z} (z + 2 i)^{- 2} = \frac{- 2}{( z + 2 i ) ^{3}} .$ At $z = 2 i$ : $(2 i + 2 i)^{3} = (4 i)^{3} = 64 i^{3} = - 64 i$ , so $Res_{2 i} = \frac{- 2}{- 64 i} = \frac{1}{32 i}$ .

Step 5 — assemble. $I = 2 π i \cdot \frac{1}{32 i} = \frac{2 π}{32} = \frac{π}{16} .$

Verify: $π /16 \approx 0.196$ , which is much smaller than $π /2 \approx 1.571$ — matches "flatter bump" forecast. Sanity scaling: substituting $x = 2 u$ in $\int \frac{d x}{( x ^{2} + 4 ) ^{2}}$ gives $\frac{1}{2 ^{3}} \int \frac{d u}{( u ^{2} + 1 ) ^{2}} = \frac{1}{8} \cdot \frac{π}{2} = \frac{π}{16}$ . ✓

Ex 3 — Cell C: cosine → use $e^{ia z}$ , Jordan's Lemma earns its keep

Worked example

$I = \int_{- \infty}^{\infty} \frac{cos ( 3 x )}{x ^{2} + 4} d x$ Forecast: Positive (the tail of $cos$ is damped by $1/ x^{2}$ and dominated by the peak near $0$ where $cos \approx 1$ ). The decay factor $e^{- a \cdot 2} = e^{- 6}$ makes it tiny. Guess something like $\frac{π}{2} e^{- 6}$ .

Step 1 — replace $cos 3 x$ by $e^{i 3 z}$ . Why not $cos z$ ? Because $cos (3 z) = \frac{1}{2} (e^{i 3 z} + e^{- i 3 z})$ and $e^{- i 3 z}$ explodes in the UHP (its modulus is $e^{+ 3 y}$ ). Only $e^{i 3 z}$ decays there (Jordan's Lemma). So set $f (z) = \frac{e ^{i 3 z}}{z ^{2} + 4}$ , $a = 3$ .

Step 2 — Jordan's Lemma legality. Here the rational leftover after peeling off the exponential is $R (z) = \frac{1}{z ^{2} + 4}$ (recall $R$ from the notation box: the non-oscillating part). It $\to 0$ as $∣ z ∣ \to \infty$ , and $a = 3 > 0$ . Jordan says the arc integral $\to 0$ — even though the gap is only $2$ , we get the bonus that we'd have it even at gap 1. ✓

Step 3 — UHP pole & residue. $z = 2 i$ (simple). $g = e^{i 3 z}$ , $h^{'} = 2 z$ : $Res_{2 i} = \frac{e ^{i 3 (2 i)}}{2 ( 2 i )} = \frac{e ^{- 6}}{4 i} .$ Why $e^{- 6}$ ? $i \cdot 3 \cdot 2 i = 6 i^{2} = - 6$ .

Step 4 — full complex integral, then take real part. $\int_{- \infty}^{\infty} \frac{e ^{i 3 x}}{x ^{2} + 4} d x = 2 π i \cdot \frac{e ^{- 6}}{4 i} = \frac{π}{2} e^{- 6} .$ This is already real, so Real part $= \frac{π}{2} e^{- 6}$ and (imag part $0$ ) $\int \frac{sin 3 x}{x ^{2} + 4} d x = 0$ — consistent with the integrand being odd.

Verify: $\frac{π}{2} e^{- 6} \approx 1.5708 \times 0.00248 \approx 0.00389$ — positive and tiny, exactly as forecast. ✓

Ex 4 — Cell D: odd integrand, the answer lives in the sine

Worked example

$I = \int_{- \infty}^{\infty} \frac{x sin x}{x ^{2} + 1} d x$ Forecast: Note $x sin x$ is even (odd $\times$ odd), so the integral is nonzero and positive. This is the case where the sine (imaginary part) carries the answer. Guess $π / e$ -ish.

Step 1 — build the complex integrand. Why? $sin x = Im (e^{i x})$ , so use $f (z) = \frac{z e ^{i z}}{z ^{2} + 1}$ , $a = 1$ .

Step 2 — arc / Jordan check. The rational leftover is $R (z) = \frac{z}{z ^{2} + 1} \to 0$ as $∣ z ∣ \to \infty$ (gap $= 2 - 1 = 1$ ), and $a = 1 > 0$ . Jordan's Lemma applies — the arc dies precisely because the exponential kills it even at gap $1$ . This is why we needed Jordan and not the naive gap- $2$ rule.

Step 3 — UHP pole & residue. $z^{2} + 1 = 0 \Rightarrow z = i$ (UHP). Simple pole, $g = z e^{i z}$ , $h^{'} = 2 z$ : $Res_{i} = \frac{i e ^{i \cdot i}}{2 i} = \frac{e ^{- 1}}{2} .$ Why: $e^{i \cdot i} = e^{i^{2}} = e^{- 1}$ , and the $i$ in $g$ cancels the $2 i$ down to $\frac{1}{2}$ .

Step 4 — assemble the complex integral. $\int_{- \infty}^{\infty} \frac{x e ^{i x}}{x ^{2} + 1} d x = 2 π i \cdot \frac{e ^{- 1}}{2} = \frac{π i}{e} .$ Step 5 — take the imaginary part. Why imaginary? Because we want $sin$ , and $sin x = Im (e^{i x})$ , so $I = Im (\frac{π i}{e}) = \frac{π}{e} .$ The real part is $0$ — matching that $\frac{x cos x}{x ^{2} + 1}$ is odd, so its integral vanishes.

Verify: $π / e \approx 1.156$ , positive as forecast. Cross-check: the pairing "real part $\to$ cos $\to$ 0 (odd), imag part $\to$ sin $\to$ answer" is exactly the parity bookkeeping. ✓

Ex 5 — Cell E: a trig integral over $[0, 2 π]$ , the $\frac{1}{i z}$ is mandatory

Worked example

$I = \int_{0}^{2 π} \frac{d θ}{5 + 4 cos θ}$ Forecast: The denominator ranges over $5 \pm 4 = [1, 9]$ , always positive, so $I > 0$ . Since the average of $\frac{1}{5 + 4 c o s θ}$ is less than $1$ over length $2 π$ , expect $I < 2 π \approx 6.28$ . Guess around $2 π /3$ .

Step 1 — substitute $z = e^{i θ}$ . Why? The integral is already a loop; on the unit circle it becomes a contour integral we can attack with residues. Use (parent): $cos θ = \frac{z + z ^{- 1}}{2}, d θ = \frac{d z}{i z} .$ Why the $\frac{1}{i z}$ ? Because $d z = i e^{i θ} d θ = i z d θ$ . Forgetting it is Cell-E's classic trap.

Step 2 — rewrite. $I = \oint_{∣ z ∣ = 1} \frac{1}{5 + 2 ( z + z ^{- 1} )} \cdot \frac{d z}{i z} = \oint \frac{d z}{i z ( 5 + 2 z + 2 z ^{- 1} )} = \oint \frac{d z}{i ( 2 z ^{2} + 5 z + 2 )} .$ Why did $z^{- 1}$ vanish? Multiplying numerator and denominator by $z$ cleared it; note this can add a pole at $z = 0$ — check it! Here $z = 0$ makes $2 z^{2} + 5 z + 2 = 2 \neq = 0$ , so no extra pole at the origin.

Step 3 — poles inside $∣ z ∣ = 1$ . Solve $2 z^{2} + 5 z + 2 = 0$ : $z = \frac{- 5 \pm 25 - 16}{4} = \frac{- 5 \pm 3}{4}$ , giving $z = - \frac{1}{2}$ and $z = - 2$ .

What to see in the figure: the yellow unit circle is our new contour. The pink dot at $z = - \frac{1}{2}$ sits inside it (distance $0.5 < 1$ ), so it counts; the blue cross at $z = - 2$ is outside (distance $2 > 1$ ), so it's discarded. Only one residue matters.

Step 4 — residue at $z_{0} = - \frac{1}{2}$ (simple). With $g = \frac{1}{i}$ , $h = 2 z^{2} + 5 z + 2$ , $h^{'} = 4 z + 5$ : $Res_{- 1/2} = \frac{1/ i}{4 ( - \frac{1}{2} ) + 5} = \frac{1/ i}{3} = \frac{1}{3 i} .$ Step 5 — assemble. $I = 2 π i \cdot \frac{1}{3 i} = \frac{2 π}{3} .$

Verify: $\frac{2 π}{3} \approx 2.094$ , positive and below $2 π$ — matches forecast. General formula check: $\int_{0}^{2 π} \frac{d θ}{a + b c o s θ} = \frac{2 π}{a ^{2} - b ^{2}}$ gives $\frac{2 π}{25 - 16} = \frac{2 π}{3}$ . ✓

Ex 6 — Cell F: a pole sitting on the real axis (degenerate → principal value)

Worked example

$I = P.V. \int_{- \infty}^{\infty} \frac{sin x}{x} d x$ Forecast: This is the famous "sinc" integral; the integrand is even, positive-on-average, and the answer is a clean multiple of $π$ . Guess $π$ .

Step 1 — the degeneracy. Why is this special? $f (z) = \frac{e ^{i z}}{z}$ has a pole at $z = 0$ , which is on our contour, not inside it. The straight recipe is illegal; we indent the contour with a tiny semicircle $γ_{ε}$ of radius $ε$ that detours over the origin (see Principal Value Integrals).

What to see in the figure: the yellow path runs along the real axis but makes a small pink half-loop $γ_{ε}$ around the blue pole at the origin, then closes with the big upper arc $Γ_{ρ}$ . Because the detour steps around $z = 0$ , no pole is enclosed — the whole loop integrates to zero. The answer will come from the tiny arc, not from an enclosed residue.

Step 2 — the closed contour encloses no pole. Why? The only pole $z = 0$ is excluded by the tiny detour, and none lies strictly inside. So by Cauchy's Integral Theorem & Formula: $\int_{real, indented} \frac{e ^{i z}}{z} d z + \int_{γ_{ε}} \frac{e ^{i z}}{z} d z + \int_{Γ_{ρ}} \frac{e ^{i z}}{z} d z = 0.$ Step 3 — big arc dies (Jordan). $R (z) = 1/ z \to 0$ , $a = 1 > 0$ ⟹ $\int_{Γ_{ρ}} \to 0$ .

Step 4 — small arc contributes a half residue. Why half? The tiny semicircle sweeps an angle of $π$ (half a full loop) clockwise around the simple pole $z = 0$ . For a simple pole, a small arc of angle $ϕ$ contributes $i ϕ \cdot Res$ ; here $ϕ = - π$ (clockwise). $Res_{0} \frac{e ^{i z}}{z} = e^{0} = 1$ , so $\int_{γ_{ε}} \frac{e ^{i z}}{z} d z \to - iπ \cdot 1 = - iπ .$ Step 5 — collect. The indented real integral tends to the principal value $P.V. \int \frac{e ^{i x}}{x} d x$ : $P.V. \int_{- \infty}^{\infty} \frac{e ^{i x}}{x} d x - iπ + 0 = 0 \Rightarrow P.V. \int \frac{e ^{i x}}{x} d x = iπ .$ Step 6 — take the imaginary part. $sin x = Im (e^{i x})$ : $\int_{- \infty}^{\infty} \frac{s i n x}{x} d x = Im (iπ) = π .$ (The real part gives $P.V. \int \frac{c o s x}{x} d x = 0$ — correct, that integrand is odd.)

Verify: $π \approx 3.1416$ , matches forecast. Standard result: $\int_{0}^{\infty} \frac{s i n x}{x} d x = \frac{π}{2}$ , doubled by evenness $= π$ . ✓

Ex 7 — Cell G: the arc does NOT die — recognise the trap

Worked example

$I = \int_{- \infty}^{\infty} \frac{x}{x ^{2} + 1} d x$ — is the machine legal? Forecast: The integrand is odd, so the principal value is $0$ , but the ordinary integral diverges (like $\int \frac{d x}{x}$ at large $∣ x ∣$ ). The naive residue recipe would hand you a wrong finite number — spot the fraud.

Step 1 — degree/decay check FIRST. Why first? Because this is the guardrail. $P = x$ so $de g P = 1$ ; $Q = x^{2} + 1$ so $de g Q = 2$ ; gap $= 2 - 1 = 1$ . With no oscillating $e^{ia z}$ , Jordan does not apply. So the arc integral does not vanish — the boxed formula is illegal here.

Step 2 — show the arc really survives. On $Γ_{ρ}$ , $z = ρ e^{i θ}$ : $\frac{z}{z ^{2} + 1} \sim \frac{1}{ρ}, arc length = π ρ \Rightarrow bound \sim \frac{1}{ρ} \cdot π ρ = π \neq \to 0.$ The estimate stalls at a constant — the arc contributes and cannot be dropped.

Step 3 — what IS true. The principal value exists by oddness: $P.V. \int_{- \infty}^{\infty} \frac{x}{x ^{2} + 1} d x = lim_{R \to \infty} \int_{- R}^{R} \frac{x d x}{x ^{2} + 1} = lim_{R \to \infty} [\frac{1}{2} ln (x^{2} + 1)]_{- R}^{R} = 0,$ because the integrand is odd and the limits are symmetric. But the ordinary two-sided limit diverges.

Verify: Antiderivative $\frac{1}{2} ln (x^{2} + 1)$ is even, so evaluated symmetrically it gives $0$ ; let the upper and lower limits go to infinity independently and it blows up. Lesson: always run the arc/decay check before summing residues. ✓ (P.V. $= 0$ .)

Ex 8 — Cell H: a word problem in disguise

Worked example Signal energy of a damped resonance

An electrical filter's response is $H (ω) = \frac{1}{ω ^{2} + ω _{0}^{2}}$ with resonance parameter $ω_{0} > 0$ . The total spectral weight is $I = \int_{- \infty}^{\infty} H (ω) d ω = \int_{- \infty}^{\infty} \frac{d ω}{ω ^{2} + ω _{0}^{2}}$ . Find it, and evaluate for $ω_{0} = 2$ . Forecast: More damping (larger $ω_{0}$ ) should spread and lower the curve, so $I$ should decrease as $ω_{0}$ grows — expect $I \propto 1/ ω_{0}$ .

Step 1 — recognise the cell. Why? Rename $ω \to x$ : this is Cell A, the basic rational integral. Here $P = 1$ ( $de g P = 0$ ) and $Q = ω^{2} + ω_{0}^{2}$ ( $de g Q = 2$ ), so gap $= 2 \geq 2$ — the machine is legal. Poles at $\pm i ω_{0}$ . Step 2 — arc check. Gap $= 2$ . ✓ Step 3 — UHP pole & residue. $z = i ω_{0}$ (simple), $h^{'} = 2 z$ : $Res_{i ω_{0}} = \frac{1}{2 i ω _{0}} .$ Step 4 — assemble. $I = 2 π i \cdot \frac{1}{2 i ω _{0}} = \frac{π}{ω _{0}} .$ Step 5 — plug $ω_{0} = 2$ . $I = \frac{π}{2}$ .

Verify: $I = π / ω_{0}$ indeed shrinks as $ω_{0}$ grows — matches the "more damping → lower weight" forecast. Units: $π / ω_{0}$ carries the $1/ ω_{0}$ scaling, consistent with a broader, lower Lorentzian. Numerically $π /2 \approx 1.571$ , and the elementary check $\int \frac{d ω}{ω ^{2} + 4} = \frac{1}{2} arctan \frac{ω}{2}_{- \infty}^{\infty} = \frac{1}{2} \cdot π = \frac{π}{2}$ . ✓

Which cell did we skip, and why not?

Recall We covered every cell in the matrix

A basic rational ::: Ex 1 ( $π / 2$ ) B higher-order pole ::: Ex 2 ( $π /16$ ) C cosine via $e^{ia z}$ ::: Ex 3 ( $\frac{π}{2} e^{- 6}$ ) D odd $\to$ sine, imaginary part ::: Ex 4 ( $π / e$ ) E trig over $[0, 2 π]$ ::: Ex 5 ( $2 π /3$ ) F pole on the axis, principal value ::: Ex 6 ( $π$ ) G arc survives — recipe illegal ::: Ex 7 (P.V. $= 0$ ) H word problem ::: Ex 8 ( $π / ω_{0}$ )

Connections

Residue theorem — computing real integrals (parent — the machine these examples drive)
Jordan's Lemma (Cells C, D, F: why oscillatory arcs die)
Principal Value Integrals (Cell F: indenting around a real-axis pole)
Laurent Series & Classification of Singularities (why $c_{- 1}$ is the residue in Ex 2)
Cauchy's Integral Theorem & Formula (Ex 6: no enclosed pole ⟹ loop is $0$ )
Contour Integration & Branch Cuts (next terrain: fractional powers, keyholes)
Argument Principle & Rouché's Theorem (locating which poles fall inside a contour)