---
title: "Fourier integral of e^{-x}; ∫cos sx/(s²+1)"
subject: "Applied Mathematics"
chapter: 4
key: "semester_4/applied_mathematics:fourier-integral-e-x-cos-sx"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/fourier-integral-e-x-cos-sx
site: BCT Engineering
---

# Fourier integral of e^{-x}; ∫cos sx/(s²+1)

Applied Mathematics · Chapter 4: Fourier Transform

- Asked in: 2072 Asoj, 5 marks; 2069 Bhadra; 2071 Bhadra (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/fourier-integral-e-x-cos-sx
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-4

## Question

Find the Fourier integral representation of f(x) = e^{-x}, x ≥ 0 with f(-x) = f(x) and hence evaluate ∫₀^∞ cos(sx)/(s² + 1) ds.

## Short answer

Since f is even, f(x)=e^{-|x|}. The inner integral A(s)=1/(1+s²). The cosine integral is e^{-|x|}=(2/π)∫₀^∞ cos(sx)/(1+s²) ds. Hence ∫₀^∞ cos(sx)/(s²+1) ds = (π/2)e^{-x} for x≥0.

## Answer

**Fourier integral.** For an even function f(x) the Fourier integral reduces to the cosine form:

f(x) = (2/π) ∫₀^∞ cos sx [ ∫₀^∞ f(v) cos sv dv ] ds

**Here** f(x) = e⁻ˣ for x ≥ 0 and f(−x) = f(x), so f(x) = e^{−|x|}, an even function.

**Step 1: inner integral.**

A(s) = ∫₀^∞ e^{−v} cos sv dv = 1/(1 + s²)

**Step 2: representation.**

**e^{−|x|} = (2/π) ∫₀^∞ cos sx /(1 + s²) ds**

**Step 3: the integral.** Multiplying by π/2 and taking x ≥ 0:

**∫₀^∞ cos sx /(s² + 1) ds = (π/2) e^{−x}**, x ≥ 0.

Since f is continuous everywhere (including x = 0), the representation holds for all x. At x = 0 it gives ∫₀^∞ ds/(1 + s²) = π/2 ✓.

(Numerical check at x = 1.5: the integral is 0.35049 = (π/2)e^{−1.5}.)

## Check yourself

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

1. Why is the cosine form used?
   A. f is odd
   B. f(-x) = f(x) makes f even, so the sine part vanishes
   C. f is periodic
   D. f is zero for x<0
   Answer: B. For even f the Fourier integral reduces to the cosine integral. (Key point: Even function)

2. What is A(s) = int from 0 to infinity of e^{-v} cos sv dv?
   A. s/(1 + s^2)
   B. 1/(1 - s^2)
   C. 1/(1 + s^2)
   D. pi/(1 + s^2)
   Answer: C. Standard result a/(a^2+s^2) with a = 1. (Key point: Inner integral)

3. What representation results?
   A. e^{-|x|} = (1/pi) int cos sx/(1+s^2) ds
   B. e^{-|x|} = (2/pi) int sin sx/(1+s^2) ds
   C. e^{-|x|} = (2/pi) int s cos sx/(1+s^2) ds
   D. e^{-|x|} = (2/pi) int cos sx/(1+s^2) ds
   Answer: D. Fourier cosine integral with A = 1/(1+s^2). (Key point: Representation)

4. What is int cos sx/(s^2+1) ds for x>=0?
   A. (pi/2) e^{-x}
   B. (2/pi) e^{-x}
   C. pi e^{-x}
   D. (pi/2) e^{x}
   Answer: A. Multiply the representation by pi/2. (Key point: Evaluation)

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