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

# Fourier cosine integral of e^{-kx}

Applied Mathematics · Chapter 4: Fourier Transform

- Asked in: 2073 Bhadra, 5 marks; 2068 Bhadra (asked 2 times)
- Video: [Lecture 3 ~ Example Problem of Fourier Integral (FOURIER COSINE INTEGRAL)](https://www.youtube.com/watch?v=OsE-o_4sj-A) — SukantaNayak edu
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/fourier-cosine-integral-e-kx
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-4

## Question

Find the Fourier cosine integral representation of the function f(x) = e^{-kx} (x > 0, k > 0) and hence show that ∫₀^∞ cos ωx dω/(k² + ω²) = (π/(2k)) e^{-kx}, x > 0, k > 0.

## Short answer

A(ω)=∫₀^∞ e^{-kv} cos ωv dv = k/(k²+ω²). The cosine integral representation is e^{-kx} = (2k/π)∫₀^∞ cos ωx/(k²+ω²) dω. So ∫₀^∞ cos ωx/(k²+ω²) dω = (π/(2k)) e^{-kx}.

## Answer

**Fourier cosine integral.** For an even extension of f defined on x > 0:

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

**Step 1: find A(ω)** for f(v) = e^{−kv}, k > 0.

```text
A(ω) = ∫₀^∞ e^{−kv} cos ωv dv
```

Integrating by parts twice (or using the standard result):

A(ω) = k/(k² + ω²)

**Step 2: representation.**

**e^{−kx} = (2k/π) ∫₀^∞ cos ωx /(k² + ω²) dω**, x > 0.

**Step 3: deduction.** Multiply both sides by π/(2k):

**∫₀^∞ cos ωx dω/(k² + ω²) = (π/(2k)) e^{−kx}**, x > 0, k > 0 ∎

At x = 0 this gives ∫₀^∞ dω/(k² + ω²) = π/(2k), which is correct.

## Check yourself

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

1. What is the Fourier cosine integral formula?
   A. f(x) = (2/pi) int A(w) cos wx dw, A(w) = int from 0 to inf f(v) cos wv dv
   B. f(x) = (1/pi) int A(w) sin wx dw
   C. f(x) = (2/pi) int A(w) sin wx dw
   D. f(x) = int A(w) cos wx dw
   Answer: A. It corresponds to the even extension of f. (Key point: Cosine integral)

2. What is A(w) for e^{-kv}?
   A. w/(k^2 + w^2)
   B. k/(k^2 + w^2)
   C. 1/(k^2 + w^2)
   D. k/(k^2 - w^2)
   Answer: B. Standard integral of e^{-kv} cos wv. (Key point: A(w))

3. What multiplication gives the required identity?
   A. Multiply by 2k/pi
   B. Multiply by pi/2
   C. Multiply by pi/(2k)
   D. Multiply by k
   Answer: C. e^{-kx} = (2k/pi) int ..., so multiplying by pi/(2k) gives the result. (Key point: Deduction)

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