---
title: "Residues of a function"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:residues-of-a-function-three"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/residues-of-a-function-three
site: BCT Engineering
---

# Residues of a function

Mathematics III · Chapter 2: Line, Surface and Volume Integrals

- Asked in: 2082 Kartik (new course), 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/residues-of-a-function-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate the real integral ∫_{−∞}^{∞} x²/((x² + 1)(x² + 4)) dx using Cauchy's residue theorem.

## Short answer

Use f(z)=z^2/(( z^2+1)(z^2+4)). Upper-plane poles at i and 2i have residues -1/(6i) and 1/(3i); sum 1/(6i). Contour integration gives the real integral = 2πi·1/(6i) = π/3.

## Answer

Let I = ∫_{−∞}^{∞} x²/((x² + 1)(x² + 4)) dx. Consider f(z) = z²/((z² + 1)(z² + 4)) over a closed contour C: the real axis from −R to R and the upper semicircle Γ of radius R.

Poles of f: z = ±i, ±2i. In the upper half plane: z = i and z = 2i (simple poles).

```text
Res at z = i:   z²/[(z + i)(z² + 4)] at z = i = (−1)/(2i · 3) = −1/(6i)
Res at z = 2i:  z²/[(z² + 1)(z + 2i)] at z = 2i = (−4)/((−3)(4i)) = 1/(3i)
Sum = −1/(6i) + 2/(6i) = 1/(6i)
```

On Γ, |f(z)| ≈ R²/R⁴ = 1/R², so |∫_Γ f dz| ≤ (1/R²)(πR) = π/R → 0 as R → ∞.

By the residue theorem:

```text
I = 2πi × 1/(6i) = π/3
```

**Answer: I = π/3.** (This agrees with the formula π/(a + b) for ∫ x²/((x² + a²)(x² + b²)) dx with a = 1, b = 2.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The contour used is:
   A. Unit circle
   B. Lower semicircle
   C. Real axis plus upper semicircle
   D. Square
   Answer: C. Standard for integrals over the real line. (Key point: Contour)

2. Poles in the upper half plane:
   A. z = -i and -2i
   B. z = i only
   C. z = 2 and -2
   D. z = i and z = 2i
   Answer: D. Simple poles. (Key point: Poles)

3. Why does the semicircle integral vanish?
   A. |f| ~ 1/R^2 so the integral is bounded by pi/R -> 0
   B. f is zero on it
   C. it has no poles
   D. the length is 0
   Answer: A. ML inequality. (Key point: Semicircle)

4. The value is:
   A. 2pi/3
   B. pi/3
   C. pi/6
   D. pi
   Answer: B. 2 pi i * 1/(6i) = pi/3. (Key point: Result)

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