---
title: "Sine transform of e^{-x}; ∫x sin mx/(x²+1)"
subject: "Applied Mathematics"
chapter: 4
key: "semester_4/applied_mathematics:fourier-sine-transform-x-sin-mx"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/fourier-sine-transform-x-sin-mx
site: BCT Engineering
---

# Sine transform of e^{-x}; ∫x sin mx/(x²+1)

Applied Mathematics · Chapter 4: Fourier Transform

- Asked in: 2071 Magh, 5 marks; 2070 Bhadra, 5 marks (asked 2 times)
- Video: [#14 || Problem#13 ||  Fourier sine transform of ||𝒇(𝒙)=𝒆^−𝒙 || By Shafiqahmed.](https://www.youtube.com/watch?v=N3pJO1KLzyU) — Math E Connect
- 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-transform-x-sin-mx
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-4

## Question

Find the Fourier sine transform of e^{-x}, x ≥ 0 and hence show that ∫₀^∞ x sin mx/(x² + 1) dx = (π/2) e^{-m}, m > 0. [Evaluate for m = 1: ∫₀^∞ x sin x/(1 + x²) dx.]

## Short answer

F_s(s)=√(2/π) s/(1+s²). Inversion: e^{-x}=(2/π)∫₀^∞ s sin(sx)/(1+s²) ds, so ∫₀^∞ s sin(sx)/(1+s²) ds=(π/2)e^{-x}. Rename s→x, x→m: ∫₀^∞ x sin(mx)/(x²+1) dx = (π/2)e^{-m}. For m=1: π/(2e).

## Answer

**Definition.** F_s(s) = √(2/π) ∫₀^∞ f(x) sin sx dx, and the inversion is f(x) = √(2/π) ∫₀^∞ F_s(s) sin sx ds.

#### Sine transform of e⁻ˣ

∫₀^∞ e^{−x} sin sx dx = s/(1 + s²), so

**F_s(s) = √(2/π) · s/(1 + s²)**

#### Deducing the integral

By the inversion formula, for x > 0:

```text
e^{−x} = √(2/π) ∫₀^∞ [ √(2/π) s/(1 + s²) ] sin sx ds
       = (2/π) ∫₀^∞ s sin sx /(1 + s²) ds
```

So ∫₀^∞ s sin sx /(1 + s²) ds = (π/2) e^{−x}. Rename the variables: s → x and x → m (m > 0):

**∫₀^∞ x sin mx /(x² + 1) dx = (π/2) e^{−m}** ∎

#### The case m = 1

**∫₀^∞ x sin x /(1 + x²) dx = π/(2e) ≈ 0.5779**

## Check yourself

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

1. What is Fs of e^{-x}?
   A. sqrt(2/pi) 1/(1+s^2)
   B. s/(1+s^2)
   C. sqrt(2/pi) s/(1+s^2)
   D. sqrt(2/pi) s/(1-s^2)
   Answer: C. int e^{-x} sin sx dx = s/(1+s^2). (Key point: Sine transform)

2. What does the inversion formula give?
   A. e^{-x} = (2/pi) int cos sx/(1+s^2) ds
   B. e^{-x} = (1/pi) int s sin sx/(1+s^2) ds
   C. e^{-x} = int s sin sx/(1+s^2) ds
   D. e^{-x} = (2/pi) int s sin sx/(1+s^2) ds
   Answer: D. sqrt(2/pi) times sqrt(2/pi) = 2/pi. (Key point: Inversion)

3. How is the required form reached?
   A. Multiply by pi/2 and rename s to x and x to m
   B. Differentiate with respect to x
   C. Use m = 0
   D. Put x = infinity
   Answer: A. int s sin sx/(1+s^2) ds = (pi/2) e^{-x}; then rename. (Key point: Renaming)

4. What is int x sin x/(1+x^2) dx for m=1?
   A. pi/2
   B. pi/(2e)
   C. pi e/2
   D. 2/(pi e)
   Answer: B. (pi/2) e^{-1}. (Key point: m = 1)

## Close variants set in other exams

- [Sine transform of e^{-x}; ∫x²/(1+x²)²](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Afourier-sine-transform-x2-over-1-x2-sq) — Applied Mathematics, chapter 4, asked 4×

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