---
title: "Fourier transform of 1-x²; evaluate integral"
subject: "Applied Mathematics"
chapter: 4
key: "semester_4/applied_mathematics:fourier-transform-1-x2-evaluate"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/fourier-transform-1-x2-evaluate
site: BCT Engineering
---

# Fourier transform of 1-x²; evaluate integral

Applied Mathematics · Chapter 4: Fourier Transform

- Asked in: 2070 Magh, 5 marks; 2069 Bhadra (asked 2 times)
- Video: [Fourier Transform of 1-x^2,](https://www.youtube.com/watch?v=7xBemsql-W4) — SNP MATH 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/fourier-transform-1-x2-evaluate
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-4

## Question

Find the Fourier transform of g(x) = 1 - x² if |x| < 1 (i.e. -1 < x < 1); 0 otherwise, and hence evaluate ∫₀^∞ ((x cos x - sin x)/x³) cos(x/2) dx.

## Short answer

F(s)=4(sin s - s cos s)/s³. Inversion: g(x)=(4/π)∫₀^∞ (sin s - s cos s)/s³ cos(sx) ds. At x=1/2, g=3/4 gives ∫₀^∞ (sin s - s cos s)/s³ cos(s/2) ds = 3π/16. With opposite sign, the required integral equals -3π/16.

## Answer

**Definition used.** F(s) = ∫_{−∞}^{∞} g(x) e^{−isx} dx, with inverse g(x) = (1/2π) ∫_{−∞}^{∞} F(s) e^{isx} ds.

#### Fourier transform of g(x) = 1 − x² on |x| < 1

g is even, so the sine part vanishes:

```text
F(s) = ∫₋₁¹ (1 − x²) cos sx dx = 2 ∫₀¹ (1 − x²) cos sx dx
```

Integrate by parts:

∫₀¹ (1 − x²) cos sx dx = [ (1 − x²) sin sx / s ]₀¹ + (2/s) ∫₀¹ x sin sx dx

The first term is 0. For the second:

∫₀¹ x sin sx dx = [ −x cos sx / s ]₀¹ + (1/s) ∫₀¹ cos sx dx = −cos s / s + sin s / s²

So ∫₀¹ (1 − x²) cos sx dx = (2/s)(sin s − s cos s)/s² = 2(sin s − s cos s)/s³, and

**F(s) = 4 (sin s − s cos s) / s³**

#### Evaluating the integral

Use inversion with g even:

```text
g(x) = (1/2π) ∫₋∞^∞ F(s) cos sx ds = (1/π) ∫₀^∞ F(s) cos sx ds
     = (4/π) ∫₀^∞ (sin s − s cos s)/s³ · cos sx ds
```

Put x = ½ (inside |x| < 1), where g(½) = 1 − ¼ = ¾:

```text
(4/π) ∫₀^∞ (sin s − s cos s)/s³ · cos(s/2) ds = 3/4
∫₀^∞ (sin s − s cos s)/s³ · cos(s/2) ds = 3π/16
```

The required integrand is (x cos x − sin x)/x³ cos(x/2), which has the opposite sign:

**∫₀^∞ [(x cos x − sin x)/x³] cos(x/2) dx = −3π/16 ≈ −0.589**

(Numerical check: the integral of (sin s − s cos s) cos(s/2)/s³ computes to 0.58905 = 3π/16 ✓.)

## Check yourself

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

1. Why is F(s) real and a cosine integral?
   A. g is odd
   B. g has period 2
   C. g is even, so the sine part vanishes
   D. g is zero outside (-1, 1)
   Answer: C. For even g the integral of g sin sx over symmetric limits is 0. (Key point: Even function)

2. What is F(s)?
   A. 4 (s cos s - sin s)/s^3
   B. 2 (sin s - s cos s)/s^2
   C. 4 (sin s - s cos s)/s^2
   D. 4 (sin s - s cos s)/s^3
   Answer: D. From integration by parts twice. (Key point: Result)

3. Which x is used in the inversion?
   A. x = 1/2, where g = 3/4
   B. x = 1, where g = 0
   C. x = 0, where g = 1
   D. x = 2, where g = -3
   Answer: A. cos(x/2) in the required integral matches s x with x = 1/2 inside |x|<1. (Key point: Inversion point)

4. What is the value of the required integral?
   A. 3 pi/16
   B. -3 pi/16
   C. 3 pi/4
   D. -3 pi/4
   Answer: B. The integrand is the negative of (sin s - s cos s)/s^3 cos(s/2), whose integral is 3 pi/16. (Key point: Sign)

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