---
title: "Evaluate ∫ dx/x³ from ∞ to α"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:integral-dx-over-x3-infinity"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-dx-over-x3-infinity
site: BCT Engineering
---

# Evaluate ∫ dx/x³ from ∞ to α

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2069 Asar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-dx-over-x3-infinity
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Evaluate ∫_∞^α dx/x³.

## Short answer

∫ dx/x³ = −1/(2x²). Evaluate from ∞ to α: ∫_∞^α dx/x³ = [−1/(2x²)]_∞^α = −1/(2α²) − 0 = −1/(2α²).

## Answer

**Note:** the limits are printed as ∞ (lower) and α (upper). We evaluate exactly as written, and also give the usual version ∫_α^∞.

∫ dx/x³ = ∫ x⁻³ dx = −1/(2x²).

**As written** (taking α > 0):

```text
∫_∞^α dx/x³ = [−1/(2x²)]_∞^α
            = −1/(2α²) − lim_(x→∞)(−1/(2x²))
            = −1/(2α²) − 0
            = −1/(2α²)
```
Reversing the limits changes the sign, so this equals −∫_α^∞ dx/x³.

**The usual version** ∫_α^∞ dx/x³ = [−1/(2x²)]_α^∞ = 0 + 1/(2α²) = **1/(2α²)**, which converges (for α > 0).

**Answer: ∫_∞^α dx/x³ = −1/(2α²)** (and ∫_α^∞ dx/x³ = 1/(2α²)).

## Check yourself

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

1. The integral of x^-3 is
   A. -1/(3x^2)
   B. 1/(2x^2)
   C. -2/x^2
   D. -1/(2x^2)
   Answer: D. Power rule: x^-2/(-2). (Key point: Antiderivative)

2. With limits ∞ (lower) and α (upper), the value is
   A. -1/(2α^2)
   B. 1/(2α^2)
   C. 0
   D. ∞
   Answer: A. Reversed limits change sign. (Key point: As written)

3. The integral from α to ∞ equals
   A. -1/(2α^2)
   B. 1/(2α^2)
   C. 1/α^2
   D. diverges
   Answer: B. The term at ∞ vanishes. (Key point: Usual version)

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