---
title: "Integral theorem of Laplace transform"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:integral-theorem-of-laplace-transform"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/integral-theorem-of-laplace-transform
site: BCT Engineering
---

# Integral theorem of Laplace transform

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2067 Asar
- Video: [Laplace Transforms Of Integrals | Problems | Laplace Transform | Engineering Mathematics](https://www.youtube.com/watch?v=4Jmh_FmVN6U) — Gautam Varde
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/integral-theorem-of-laplace-transform
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

State and prove the integral theorem of the Laplace transform.

## Short answer

If L[f(t)] = F(s), then L[∫₀ᵗ f(u) du] = F(s)/s. Proof: let g(t) = ∫₀ᵗ f(u) du, then g′(t)=f(t), g(0)=0, so sG(s) = F(s) giving G(s)=F(s)/s.

## Answer

#### Statement
If L[f(t)] = F(s), then the Laplace transform of the integral of f is

```text
L[ ∫₀ᵗ f(u) du ] = F(s)/s
```

(Integration in the t-domain corresponds to division by s.)

#### Proof
Let g(t) = ∫₀ᵗ f(u) du. Then g(0) = 0 and g′(t) = f(t). Take the Laplace transform of g′(t) = f(t) using L[g′] = s G(s) − g(0):

```text
s G(s) − g(0) = F(s)
s G(s) = F(s)          (since g(0) = 0)
G(s) = F(s)/s
```

Hence L[∫₀ᵗ f(u) du] = F(s)/s. ∎

(Direct proof: G(s) = ∫₀^∞ e^(−st) [∫₀ᵗ f(u) du] dt. Integrate by parts with dv = e^(−st) dt: the boundary term vanishes, leaving (1/s) ∫₀^∞ e^(−st) f(t) dt = F(s)/s.)

#### Related form (division by t)
If L[f(t)] = F(s) and lim f(t)/t exists as t → 0, then L[f(t)/t] = ∫ₛ^∞ F(u) du.

#### Example
L[∫₀ᵗ sin u du] = L[1 − cos t] = 1/s − s/(s²+1) = 1/(s(s²+1)) = (1/s)·1/(s²+1) ✓.

## Check yourself

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

1. The integral theorem states L[integral from 0 to t of f(u) du] =
   A. F(s)/s
   B. s F(s)
   C. F(s) - f(0)
   D. F'(s)
   Answer: A. Integration in t corresponds to division by s. (Key point: Statement)

2. In the proof, g(t) = integral of f has which properties?
   A. g(0) = f(0) and g' = F
   B. g(0) = 0 and g'(t) = f(t)
   C. g(0) = 1 and g' = f
   D. g(0) = 0 and g'' = f
   Answer: B. Then L[g'] = sG - g(0) = F gives G = F/s. (Key point: Proof idea)

3. Why does s G(s) = F(s) lead to G = F/s?
   A. Because f(0) = 0
   B. Because F(s) = 1
   C. Because g(0) = 0 so no extra term remains
   D. Because s = 1
   Answer: C. The g(0) term vanishes. (Key point: Why division)

4. The related result for f(t)/t is
   A. L[f(t)/t] = F(s)/s
   B. L[f(t)/t] = -F'(s)
   C. L[f(t)/t] = F(1/s)
   D. L[f(t)/t] = integral of F(u) from s to infinity
   Answer: D. It needs lim f(t)/t as t to 0 to exist. (Key point: Division by t form)

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