---
title: "Fourier sine integral of e^{-x}cos x"
subject: "Applied Mathematics"
chapter: 4
key: "semester_4/applied_mathematics:fourier-sine-integral-e-x-cosx"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/fourier-sine-integral-e-x-cosx
site: BCT Engineering
---

# Fourier sine integral of e^{-x}cos x

Applied Mathematics · Chapter 4: Fourier Transform

- Asked in: 2075 Baisakh, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/fourier-sine-integral-e-x-cosx
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-4

## Question

Obtain the Fourier sine integral representation of e^{-x} cos x and hence show that ∫₀^∞ ω³ sin ωx/(ω⁴ + 4) dω = (π/2) e^{-x} cos x, x > 0.

## Short answer

B(ω)=∫₀^∞ e^{-v} cos v sin ωv dv = ω³/(ω⁴+4). The sine integral is e^{-x} cos x = (2/π)∫₀^∞ ω³ sin ωx/(ω⁴+4) dω. Multiply by π/2 to obtain ∫₀^∞ ω³ sin ωx/(ω⁴+4) dω = (π/2)e^{-x} cos x.

## Answer

**Fourier sine integral.** For f defined on x > 0:

f(x) = (2/π) ∫₀^∞ B(ω) sin ωx dω , where B(ω) = ∫₀^∞ f(v) sin ωv dv.

**Step 1: find B(ω)** for f(v) = e^{−v} cos v.

Use sin ωv cos v = ½[sin (ω + 1)v + sin (ω − 1)v] and ∫₀^∞ e^{−v} sin bv dv = b/(1 + b²):

```text
B(ω) = ½ [ (ω + 1)/(1 + (ω + 1)²) + (ω − 1)/(1 + (ω − 1)²) ]
     = ½ [ (ω + 1)/(ω² + 2ω + 2) + (ω − 1)/(ω² − 2ω + 2) ]
```

Combine over the common denominator (ω² + 2ω + 2)(ω² − 2ω + 2) = ω⁴ + 4:

numerator = (ω + 1)(ω² − 2ω + 2) + (ω − 1)(ω² + 2ω + 2) = (ω³ − ω² + 2) + (ω³ + ω² − 2) = 2ω³

**B(ω) = ω³/(ω⁴ + 4)**

**Step 2: sine integral representation.**

**e^{−x} cos x = (2/π) ∫₀^∞ ω³ sin ωx /(ω⁴ + 4) dω**, x > 0

**Step 3: deduction.** Multiplying by π/2:

**∫₀^∞ ω³ sin ωx /(ω⁴ + 4) dω = (π/2) e^{−x} cos x**, x > 0 ∎

(Numerical check of B(ω) at ω = 2.1: the integral gives 0.39496 = ω³/(ω⁴ + 4).)

## Check yourself

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

1. Which formula is used?
   A. f(x) = (2/pi) int B(w) cos wx dw
   B. f(x) = (1/pi) int B(w) sin wx dw
   C. f(x) = (2/pi) int B(w) sin wx dw, B(w) = int from 0 to inf f(v) sin wv dv
   D. f(x) = int B(w) sin wx dw
   Answer: C. This is the Fourier sine integral. (Key point: Sine integral)

2. Which identity simplifies sin(wv) cos v?
   A. (1/2)[cos (w+1)v + cos (w-1)v]
   B. sin wv + cos v
   C. (1/2)[sin (w+1)v - sin (w-1)v]
   D. (1/2)[sin (w+1)v + sin (w-1)v]
   Answer: D. Then each term integrates with e^{-v}. (Key point: Product to sum)

3. What is B(w)?
   A. w^3/(w^4 + 4)
   B. w/(w^4 + 4)
   C. w^3/(w^4 - 4)
   D. 2 w^3/(w^4 + 4)
   Answer: A. Numerator sums to 2w^3 and the 1/2 cancels it; denominator is w^4+4. (Key point: B(w))

4. How is the required identity obtained?
   A. Multiply by 2/pi
   B. Multiply by pi/2
   C. Differentiate with respect to x
   D. Set x = 0
   Answer: B. e^{-x} cos x = (2/pi) int ..., so multiplying by pi/2 gives it. (Key point: Deduction)

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