---
title: "Second shifting theorem"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:second-shifting-theorem"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/second-shifting-theorem
site: BCT Engineering
---

# Second shifting theorem

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2069 Chaitra, 5 marks
- Video: [Laplace Transform: Second Shifting Theorem](https://www.youtube.com/watch?v=ns2F_WPZxvQ) — Dr Chris Tisdell
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/second-shifting-theorem
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

State and prove the second shifting theorem of the Laplace transform.

## Short answer

If L[f(t)]=F(s) then L[f(t-a)u(t-a)]=e^(-as)F(s). Proof: L[f(t-a)u(t-a)] = ∫_a^∞ e^(-st)f(t-a)dt, substitute τ=t-a → e^(-as)∫_0^∞ e^(-sτ)f(τ)dτ = e^(-as)F(s).

## Answer

#### Statement
If L[f(t)] = F(s) and

```text
g(t) = f(t − a)   for t > a
g(t) = 0          for t < a        (a > 0)
```

i.e. g(t) = f(t − a) u(t − a) with u the unit step function, then

```text
L[g(t)] = e^(−as) F(s)
```

Equivalently, L⁻¹[e^(−as) F(s)] = f(t − a) u(t − a).

#### Proof
```text
L[g(t)] = ∫₀^∞ e^(−st) g(t) dt
        = ∫₀^a e^(−st)·0 dt + ∫ₐ^∞ e^(−st) f(t − a) dt
```

In the second integral put τ = t − a. Then t = τ + a, dt = dτ, and the limits become τ = 0 to ∞:

```text
L[g(t)] = ∫₀^∞ e^(−s(τ + a)) f(τ) dτ
        = e^(−as) ∫₀^∞ e^(−sτ) f(τ) dτ
        = e^(−as) F(s)
```

∎

#### Example
L[u(t − 2)(t − 2)] = e^(−2s)/s². Also, L[u(t − a)] = e^(−as)/s (take f(t) = 1).

## Check yourself

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

1. L[f(t-a)u(t-a)] = ?
   A. e^(as) F(s)
   B. F(s-a)
   C. F(s)/a
   D. e^(-as) F(s)
   Answer: D. Second shifting. (Key point: Statement)

2. g(t) for t < a is?
   A. 0
   B. f(t)
   C. f(-t)
   D. f(a)
   Answer: A. Unit step. (Key point: Meaning of g)

3. After tau = t - a the limits become?
   A. a to infinity
   B. 0 to infinity
   C. -a to 0
   D. 0 to a
   Answer: B. t = a -> 0. (Key point: Substitution)

4. What is L[u(t-2)(t-2)]?
   A. e^(-2s)/s
   B. e^(2s)/s^2
   C. e^(-2s)/s^2
   D. 1/s^2
   Answer: C. f(t) = t. (Key point: Example)

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