---
title: "Laplace transforms of functions (1/t)(e^(at)−cos6t)"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:laplace-transforms-of-functions-1-t-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transforms-of-functions-1-t-two
site: BCT Engineering
---

# Laplace transforms of functions (1/t)(e^(at)−cos6t)

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2080 Bhadra, 1+2+2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transforms-of-functions-1-t-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

State the condition for a function to exist its Laplace transform. Find the Laplace transform of: a) (1/t)(e^(at) − cos 6t)  b) f(t) = sin t for 0 < t < π; t for t > π.

## Short answer

Existence: f piecewise continuous and of exponential order. (a) ln( sqrt(s^2+36) / (s-a) ); (b) (1+e^(-π s))/(s^2+1) + e^(-π s)(π/s + 1/s^2).

## Answer

**Existence condition.** The Laplace transform L{f(t)} = ∫₀^∞ e^(−st) f(t) dt exists for s > k if f(t) is piecewise continuous on every finite interval [0, T] and of exponential order, that is |f(t)| ≤ M e^(kt) for all t ≥ T. (These conditions are sufficient, not necessary.)

#### a) L{(1/t)(e^(at) − cos 6t)}
Use L{f(t)/t} = ∫ₛ^∞ F(u) du with F(s) = 1/(s − a) − s/(s² + 36).

```text
∫ₛ^∞ [1/(u − a) − u/(u² + 36)] du = [ln((u − a)/√(u² + 36))]ₛ^∞
 = 0 − ln((s − a)/√(s² + 36))
 = ln(√(s² + 36)/(s − a))        (s > a)
```

#### b) f(t) = sin t (0 < t < π); t (t > π)
Split the integral:

```text
L{f} = ∫₀^π e^(−st) sin t dt + ∫_π^∞ t e^(−st) dt

First:  [e^(−st)(−s sin t − cos t)/(s² + 1)]₀^π = (1 + e^(−πs))/(s² + 1)
Second: [−t e^(−st)/s − e^(−st)/s²]_π^∞ = e^(−πs) (π/s + 1/s²)
```

**Answers:** a) ln(√(s² + 36)/(s − a)); b) (1 + e^(−πs))/(s² + 1) + e^(−πs)(π/s + 1/s²).

## Check yourself

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

1. Which conditions guarantee that L{f(t)} exists for s > k?
   A. f is differentiable everywhere and bounded
   B. f is piecewise continuous on every finite interval and of exponential order |f(t)| <= M e^(kt)
   C. f is periodic and continuous at t = 0
   D. f is continuous with f(0) = 0 and f(t) -> 0 as t -> infinity
   Answer: B. The answer states these two conditions: piecewise continuity on each [0, T] and exponential order. They are sufficient, and the transform then exists for s > k. (Key point: Conditions for the Laplace transform to exist)

2. How do you find L{f} for f = sin t on (0, pi) and t for t > pi?
   A. Use only the sin t part
   B. Use only the t part
   C. Split the integral at t = pi
   D. Differentiate each piece first
   Answer: C. Each piece is integrated over its own interval: 0 to pi and pi to infinity. (Key point: Handling a piecewise function)

3. What is L{(1/t)(e^(at) - cos 6t)}?
   A. ln( (s-a)/sqrt(s^2+36) )
   B. ln( sqrt(s^2+36)(s-a) )
   C. (1/2) ln(s^2+36) + ln(s-a)
   D. ln( sqrt(s^2+36)/(s-a) )
   Answer: D. Integral from s to infinity of F gives -ln((s-a)/sqrt(s^2+36)). (Key point: Part (a))

4. What is the integral from pi to infinity of t e^(-st) dt?
   A. e^(-pi s)(pi/s + 1/s^2)
   B. e^(-pi s)(pi/s - 1/s^2)
   C. pi/s + 1/s^2
   D. e^(pi s)(pi/s + 1/s^2)
   Answer: A. The antiderivative -t e^(-st)/s - e^(-st)/s^2 evaluated from pi to infinity gives the stated value. (Key point: Second integral in part (b))

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