---
title: "Contour integral ∫x²/((x²+a²)(x²+b²))"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:real-integral-x2-a-b"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/real-integral-x2-a-b
site: BCT Engineering
---

# Contour integral ∫x²/((x²+a²)(x²+b²))

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2071 Bhadra
- Video: [By Using Contour Integration Type 02 Problems 01 -Complex Integration](https://www.youtube.com/watch?v=_CxdH_-Yfks) — SK Maths Tutorial
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/real-integral-x2-a-b
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Using contour integration, evaluate real integral: ∫_{-∞}^{+∞} x² dx / ((x² + a²)(x² + b²)).

## Short answer

Integrate f(z)=z²/((z²+a²)(z²+b²)). Simple poles at ia and ib inside. Residues sum to 1/(2i(a+b)), giving π/(a+b) by residue theorem.

## Answer

**Method.** Let I = ∫_{−∞}^{∞} x²/((x² + a²)(x² + b²)) dx, with a > 0, b > 0, a ≠ b. Take f(z) = z²/((z² + a²)(z² + b²)) around the closed contour C made of [−R, R] and the upper semicircle Γ: |z| = R (R > max(a, b)).

Simple poles inside C: **z = ia and z = ib**.

**Residues:**

- At z = ia: Res = (ia)² / ( 2ia (b² − a²) ) = −a² / ( 2ia (b² − a²) ) = −a / ( 2i (b² − a²) )
- At z = ib: Res = (ib)² / ( (a² − b²) · 2ib ) = −b / ( 2i (a² − b²) ) = b / ( 2i (b² − a²) )

Sum = (b − a) / ( 2i (b² − a²) ) = 1 / ( 2i (a + b) ).

**Residue theorem:**

∮_C f dz = 2πi · 1 / (2i (a + b)) = π/(a + b).

**Arc:** |f(z)| ≤ R²/((R² − a²)(R² − b²)) ~ 1/R², so ∫_Γ f dz ≤ πR/R² → 0 as R → ∞.

**Answer: I = π/(a + b)**

Special case a = b gives π/(2a). (Check: a = 1, b = 2 gives π/3, which agrees with a direct partial-fraction evaluation.)

## Check yourself

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

1. Which poles lie inside the contour?
   A. z = ±ia
   B. z = ia and z = ib
   C. z = a and b
   D. z = -ia and -ib
   Answer: B. Upper half-plane poles. (Key point: Poles inside)

2. What is the sum of residues?
   A. 1/(2i(a-b))
   B. 1/(2i(b²-a²))
   C. 1/(2i(a+b))
   D. (a+b)/(2i)
   Answer: C. (b-a)/(2i(b²-a²)) simplifies to 1/(2i(a+b)). (Key point: Residue sum)

3. What is the integral?
   A. π/(a-b)
   B. π/(2ab)
   C. 2π/(a+b)
   D. π/(a+b)
   Answer: D. 2πi × 1/(2i(a+b)). (Key point: Result)

4. What does it give for a = b?
   A. π/(2a)
   B. π/a
   C. π/(4a)
   D. 2π/a
   Answer: A. π/(a+a) = π/(2a). (Key point: Special case)

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