---
title: "Contour integral ∫dx/(1+x²)²"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:real-integral-1-over-1-x2-sq"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/real-integral-1-over-1-x2-sq
site: BCT Engineering
---

# Contour integral ∫dx/(1+x²)²

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2074 Bhadra, 5 marks
- Video: [Evaluation of Improper Integrals by Contour Integration](https://www.youtube.com/watch?v=Vb_g0lsPDq8) — Maths by Nidhi
- 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-1-over-1-x2-sq
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Using contour integration evaluate ∫_{-∞}^{∞} dx/(1 + x²)².

## Short answer

Consider f(z)=1/(z²+1)². Pole of order 2 at z=i, residue 1/(4i). Integral over the closed contour gives 2πi·1/(4i) = π/2, arc vanishes. The real integral is π/2.

## Answer

**Method.** Let I = ∫_{−∞}^{∞} dx/(1 + x²)². Take f(z) = 1/(z² + 1)² = 1/((z − i)²(z + i)²) and integrate around the closed contour C formed by the segment [−R, R] and the upper semicircle Γ: |z| = R, R > 1.

Inside C there is a single pole, **z = i, of order 2**.

**Residue at z = i:**

Res = d/dz [ (z − i)² f(z) ] at z = i = d/dz [ (z + i)⁻² ] = −2 (z + i)⁻³ at z = i

= −2/(2i)³ = −2/(−8i) = 1/(4i)

**Residue theorem:**

∮_C f dz = 2πi · 1/(4i) = π/2.

**Arc integral:** on Γ, |f(z)| ≤ 1/(R² − 1)², so |∫_Γ f dz| ≤ πR/(R² − 1)² → 0 as R → ∞.

Therefore ∫_{−∞}^{∞} dx/(1 + x²)² = π/2.

**Answer: I = π/2**

(Check: ∫ dx/(1 + x²)² = ½[x/(1 + x²) + tan⁻¹x], which over (−∞, ∞) gives ½ · π = π/2.)

## Check yourself

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

1. What is the pole inside the upper semicircle?
   A. z = i simple
   B. z = -i of order 2
   C. z = i of order 2
   D. z = i and -i simple
   Answer: C. (z²+1)² = (z-i)²(z+i)². (Key point: Pole order)

2. What is the residue at z = i?
   A. 1/(2i)
   B. -1/(4i)
   C. 1/4
   D. 1/(4i)
   Answer: D. d/dz (z+i)^-2 = -2(z+i)^-3 = 1/(4i) at i. (Key point: Residue)

3. What is the integral?
   A. π/2
   B. π
   C. π/4
   D. 2π
   Answer: A. 2πi/(4i) = π/2. (Key point: Result)

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