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

# Fourier sine integral of e^{-kx}

Applied Mathematics · Chapter 4: Fourier Transform

- Asked in: 2071 Magh, 5 marks
- Video: [How to solve Fourier cosine integral? || How to solve Fourier sine integral?|| Fourier integral ||](https://www.youtube.com/watch?v=5i6Ptd4t_jg) — Mathematics Solutions Dr Mohan
- 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-kx
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-4

## Question

Find the Fourier sine integral of the function f(x) = e^{-Kx} and hence show that ∫₀^∞ λ sin λx/(λ² + K²) dλ = (π/2) e^{-Kx}, x > 0, K > 0.

## Short answer

For f(x)=e^{-Kx} (x>0, K>0), the sine integral coefficient is B(λ)=λ/(K²+λ²). Thus e^{-Kx} = (2/π)∫₀^∞ λ sin λx/(λ²+K²) dλ. Hence ∫₀^∞ λ sin λx/(λ²+K²) dλ = (π/2) e^{-Kx} for x>0.

## Answer

**Fourier sine integral.** For f(x) defined for x > 0 (extended as an odd function):

f(x) = (2/π) ∫₀^∞ B(λ) sin λx dλ , where B(λ) = ∫₀^∞ f(v) sin λv dv.

**Step 1: find B(λ)** for f(v) = e^{−Kv}, K > 0.

```text
B(λ) = ∫₀^∞ e^{−Kv} sin λv dv = λ/(K² + λ²)
```

(Integration by parts twice gives I = λ/K² − (λ²/K²) I, hence I = λ/(K² + λ²).)

**Step 2: representation.**

**e^{−Kx} = (2/π) ∫₀^∞ λ sin λx /(λ² + K²) dλ**, x > 0

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

**∫₀^∞ λ sin λx /(λ² + K²) dλ = (π/2) e^{−Kx}**, x > 0, K > 0 ∎

At x = 0 the integral is 0, while e⁰ = 1: the odd extension of f has a jump at x = 0 (value 0, the average of −1 and 1), so the formula holds for x > 0 only.

## Check yourself

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

1. What is the sine integral formula?
   A. f(x) = (2/pi) int B(l) cos lx dl
   B. f(x) = (1/pi) int B(l) sin lx dl
   C. f(x) = (2/pi) int B(l) sin lx dl, B(l) = int from 0 to inf f(v) sin lv dv
   D. f(x) = int B(l) sin lx dl
   Answer: C. Odd extension of f. (Key point: Sine integral)

2. What is B(l) for e^{-Kv}?
   A. K/(K^2 + l^2)
   B. 1/(K^2 + l^2)
   C. l/(K^2 - l^2)
   D. l/(K^2 + l^2)
   Answer: D. Integration by parts twice. (Key point: B)

3. What do you multiply by?
   A. pi/2
   B. 2/pi
   C. pi/(2K)
   D. K
   Answer: A. e^{-Kx} = (2/pi) int ... (Key point: Deduction)

4. Why does the formula hold only for x>0?
   A. The integral diverges
   B. The odd extension jumps at 0, so the integral gives 0 not 1
   C. K = 0 at x = 0
   D. It holds for all x
   Answer: B. At the jump the value is the average of -1 and 1. (Key point: x = 0)

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