---
title: "Cosine transform of e^{-mx}; evaluate integral"
subject: "Applied Mathematics"
chapter: 4
key: "semester_4/applied_mathematics:fourier-cosine-transform-e-mx-beta"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/fourier-cosine-transform-e-mx-beta
site: BCT Engineering
---

# Cosine transform of e^{-mx}; evaluate integral

Applied Mathematics · Chapter 4: Fourier Transform

- Asked in: 2070 Magh, 5 marks; 2071 Bhadra; 2068 Magh, 5 marks (asked 3 times)
- Video: [Fourier Cosine and Sine Transforms of f(x)=e^{-ax}](https://www.youtube.com/watch?v=BUZ1Zvhhjys) — Math Alchemist
- 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-transform-e-mx-beta
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-4

## Question

Find the Fourier cosine transform of f(x) = e^{-mx} (m > 0) and hence show that ∫₀^∞ cos py/(y² + β²) dy = (π/(2β)) e^{-pβ} (β > 0, p > 0).

## Short answer

F_c(s) = √(2/π) m/(m²+s²). Inversion gives e^{-mx} = (2m/π) ∫₀^∞ cos(sx)/(s²+m²) ds. Renaming x→p, s→y, m→β yields ∫₀^∞ cos(py)/(y²+β²) dy = (π/(2β)) e^{-pβ}.

## Answer

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

#### Cosine transform of e^{−mx}

Using ∫₀^∞ e^{−mx} cos sx dx = m/(m² + s²):

**F_c(s) = √(2/π) · m/(m² + s²)**

(Derivation: integrating I = ∫₀^∞ e^{−mx} cos sx dx by parts twice gives I = m/(m² + s²).)

#### Deducing the integral

By the inversion formula, for x > 0:

```text
e^{−mx} = √(2/π) ∫₀^∞ [ √(2/π) m/(m² + s²) ] cos sx ds
        = (2m/π) ∫₀^∞ cos sx /(s² + m²) ds
```

So ∫₀^∞ cos sx/(s² + m²) ds = (π/2m) e^{−mx}.

Now rename: x → p, s → y and m → β (β > 0, p > 0):

**∫₀^∞ cos py /(y² + β²) dy = (π/(2β)) e^{−pβ}** ∎

## Check yourself

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

1. What is Fc of e^{-mx}?
   A. sqrt(2/pi) s/(m^2 + s^2)
   B. m/(m^2 + s^2)
   C. sqrt(2/pi) m/(m^2 + s^2)
   D. sqrt(2/pi) m/(m^2 - s^2)
   Answer: C. Because int e^{-mx} cos sx dx = m/(m^2+s^2). (Key point: Cosine transform)

2. Which formula links the transform back to the integral?
   A. Parseval's identity
   B. Convolution theorem
   C. Fourier sine inversion
   D. Inversion: f(x) = sqrt(2/pi) int Fc(s) cos sx ds
   Answer: D. Applying the cosine inversion to Fc gives e^{-mx} as an integral. (Key point: Inversion)

3. What does inversion give for the integral of cos sx/(s^2+m^2)?
   A. (pi/(2m)) e^{-mx}
   B. (pi/m) e^{-mx}
   C. (2m/pi) e^{-mx}
   D. (pi/(2m)) e^{mx}
   Answer: A. e^{-mx} = (2m/pi) int ..., so the integral = (pi/2m) e^{-mx}. (Key point: Integral result)

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