Module I: Laplace Transform

1. Definition of Laplace Transform

Definition

Let \( f(t) \) be a function defined for \( t \geq 0 \). The Laplace Transform of \( f(t) \) is:

\[ \mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} e^{-st} f(t) \, dt \]

Where \( s \) is a complex number \( s = \sigma + i\omega \)

Find \( \mathcal{L}\{1\} \)

\( \mathcal{L}\{1\} = \int_0^{\infty} e^{-st} \cdot 1 \, dt \)
\( = \left[ \frac{e^{-st}}{-s} \right]_0^{\infty} = 0 - \left( \frac{-1}{s} \right) \)
\( = \frac{1}{s} \), where \( s > 0 \)

\( \mathcal{L}\{1\} = \frac{1}{s} \)

2. Conditions of Existence

Sufficient Conditions for Laplace Transform to Exist

\( f(t) \) is piecewise continuous on every finite interval \([0, T]\)
\( f(t) \) is of exponential order: \( |f(t)| \leq Me^{at} \) for constants \( M, a \) and large \( t \)

If these conditions are satisfied, \( \mathcal{L}\{f(t)\} \) exists for \( s > a \)

3. Laplace Transform of Standard Functions

\( f(t) \)	\( \mathcal{L}\{f(t)\} = F(s) \)	Condition
\( 1 \)	\( \frac{1}{s} \)	\( s > 0 \)
\( e^{at} \)	\( \frac{1}{s-a} \)	\( s > a \)
\( t^n \) (n ≥ 0, integer)	\( \frac{n!}{s^{n+1}} \)	\( s > 0 \)
\( \sin(at) \)	\( \frac{a}{s^2 + a^2} \)	\( s > 0 \)
\( \cos(at) \)	\( \frac{s}{s^2 + a^2} \)	\( s > 0 \)
\( \sinh(at) \)	\( \frac{a}{s^2 - a^2} \)	\( s > \|a\| \)
\( \cosh(at) \)	\( \frac{s}{s^2 - a^2} \)	\( s > \|a\| \)

Derivation of \( \mathcal{L}\{e^{at}\} \)

Find \( \mathcal{L}\{e^{at}\} \)

\( \mathcal{L}\{e^{at}\} = \int_0^{\infty} e^{-st} \cdot e^{at} \, dt = \int_0^{\infty} e^{-(s-a)t} \, dt \)
\( = \left[ \frac{e^{-(s-a)t}}{-(s-a)} \right]_0^{\infty} \)
For \( s > a \): Upper limit → 0, Lower limit → \( \frac{-1}{-(s-a)} = \frac{1}{s-a} \)

\( \mathcal{L}\{e^{at}\} = \frac{1}{s-a} \)

Derivation of \( \mathcal{L}\{\sin(at)\} \)

Find \( \mathcal{L}\{\sin(at)\} \)

Use \( \sin(at) = \frac{e^{iat} - e^{-iat}}{2i} \)
\( \mathcal{L}\{\sin(at)\} = \frac{1}{2i}\left[ \mathcal{L}\{e^{iat}\} - \mathcal{L}\{e^{-iat}\} \right] \)
\( = \frac{1}{2i}\left[ \frac{1}{s-ia} - \frac{1}{s+ia} \right] = \frac{1}{2i} \cdot \frac{2ia}{s^2+a^2} \)

\( \mathcal{L}\{\sin(at)\} = \frac{a}{s^2 + a^2} \)

Derivation of \( \mathcal{L}\{t^n\} \)

Find \( \mathcal{L}\{t^n\} \) for n = positive integer

\( \mathcal{L}\{t^n\} = \int_0^{\infty} e^{-st} t^n \, dt \)
Substitute \( st = u \), so \( t = \frac{u}{s} \), \( dt = \frac{du}{s} \)
\( = \int_0^{\infty} e^{-u} \left(\frac{u}{s}\right)^n \frac{du}{s} = \frac{1}{s^{n+1}} \int_0^{\infty} e^{-u} u^n \, du \)
Using Gamma function: \( \Gamma(n+1) = n! \)

\( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \)

4. Properties of Laplace Transform

4.1 Linearity Property

\[ \mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\} \]

Find \( \mathcal{L}\{3e^{2t} + 5\sin(4t)\} \)

Apply linearity: \( = 3\mathcal{L}\{e^{2t}\} + 5\mathcal{L}\{\sin(4t)\} \)
\( = 3 \cdot \frac{1}{s-2} + 5 \cdot \frac{4}{s^2+16} \)

\( = \frac{3}{s-2} + \frac{20}{s^2+16} \)

4.2 First Shifting Theorem

If \( \mathcal{L}\{f(t)\} = F(s) \), then \( \mathcal{L}\{e^{at}f(t)\} = F(s-a) \)

Replace every \( s \) with \( (s-a) \) in \( F(s) \)

Find \( \mathcal{L}\{e^{3t}\sin(2t)\} \)

We know \( \mathcal{L}\{\sin(2t)\} = \frac{2}{s^2+4} \)
By First Shifting Theorem, replace \( s \) with \( (s-3) \)

\( \mathcal{L}\{e^{3t}\sin(2t)\} = \frac{2}{(s-3)^2+4} \)

Find \( \mathcal{L}\{e^{-2t}t^3\} \)

We know \( \mathcal{L}\{t^3\} = \frac{3!}{s^4} = \frac{6}{s^4} \)
By First Shifting Theorem, replace \( s \) with \( (s+2) \)

\( \mathcal{L}\{e^{-2t}t^3\} = \frac{6}{(s+2)^4} \)

4.3 Change of Scale Property

If \( \mathcal{L}\{f(t)\} = F(s) \), then \( \mathcal{L}\{f(at)\} = \frac{1}{a}F\left(\frac{s}{a}\right) \)

Find \( \mathcal{L}\{\sin(3t)\} \) using change of scale, given \( \mathcal{L}\{\sin(t)\} = \frac{1}{s^2+1} \)

Here \( a = 3 \), so \( \mathcal{L}\{\sin(3t)\} = \frac{1}{3} \cdot \frac{1}{(s/3)^2+1} \)
\( = \frac{1}{3} \cdot \frac{1}{\frac{s^2}{9}+1} = \frac{1}{3} \cdot \frac{9}{s^2+9} \)

\( \mathcal{L}\{\sin(3t)\} = \frac{3}{s^2+9} \)

4.4 Multiplication by t

\[ \mathcal{L}\{t \cdot f(t)\} = -\frac{d}{ds}F(s) \]

In general: \( \mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n}F(s) \)

Find \( \mathcal{L}\{t \cdot \sin(2t)\} \)

\( \mathcal{L}\{\sin(2t)\} = \frac{2}{s^2+4} \)
\( \mathcal{L}\{t\sin(2t)\} = -\frac{d}{ds}\left(\frac{2}{s^2+4}\right) \)
\( = -2 \cdot \frac{-2s}{(s^2+4)^2} \)

\( \mathcal{L}\{t\sin(2t)\} = \frac{4s}{(s^2+4)^2} \)

Find \( \mathcal{L}\{t^2 e^{3t}\} \)

\( \mathcal{L}\{e^{3t}\} = \frac{1}{s-3} \)
\( \mathcal{L}\{t^2 e^{3t}\} = (-1)^2 \frac{d^2}{ds^2}\left(\frac{1}{s-3}\right) \)
First derivative: \( \frac{d}{ds}\left(\frac{1}{s-3}\right) = \frac{-1}{(s-3)^2} \)
Second derivative: \( \frac{d}{ds}\left(\frac{-1}{(s-3)^2}\right) = \frac{2}{(s-3)^3} \)

\( \mathcal{L}\{t^2 e^{3t}\} = \frac{2}{(s-3)^3} \)

4.5 Division by t

\[ \mathcal{L}\left\{\frac{f(t)}{t}\right\} = \int_s^{\infty} F(u) \, du \]

Condition

\( \lim_{t \to 0} \frac{f(t)}{t} \) must exist

Find \( \mathcal{L}\left\{\frac{\sin(t)}{t}\right\} \)

\( \mathcal{L}\{\sin(t)\} = \frac{1}{s^2+1} \)
\( \mathcal{L}\left\{\frac{\sin(t)}{t}\right\} = \int_s^{\infty} \frac{1}{u^2+1} \, du \)
\( = \left[\tan^{-1}(u)\right]_s^{\infty} = \frac{\pi}{2} - \tan^{-1}(s) \)

\( \mathcal{L}\left\{\frac{\sin(t)}{t}\right\} = \cot^{-1}(s) = \tan^{-1}\left(\frac{1}{s}\right) \)

Find \( \mathcal{L}\left\{\frac{e^{at} - e^{bt}}{t}\right\} \)

\( \mathcal{L}\{e^{at} - e^{bt}\} = \frac{1}{s-a} - \frac{1}{s-b} \)
\( \mathcal{L}\left\{\frac{e^{at} - e^{bt}}{t}\right\} = \int_s^{\infty} \left(\frac{1}{u-a} - \frac{1}{u-b}\right) du \)
\( = \left[\ln(u-a) - \ln(u-b)\right]_s^{\infty} = \left[\ln\frac{u-a}{u-b}\right]_s^{\infty} \)
At \( u = \infty \): \( \ln(1) = 0 \). At \( u = s \): \( \ln\frac{s-a}{s-b} \)

\( \mathcal{L}\left\{\frac{e^{at} - e^{bt}}{t}\right\} = \ln\frac{s-b}{s-a} \)

5. Laplace Transform of Derivatives

\[ \mathcal{L}\{f'(t)\} = sF(s) - f(0) \] \[ \mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0) \]

Find \( \mathcal{L}\{\cos(at)\} \) using derivative property

Let \( f(t) = \sin(at) \), so \( f'(t) = a\cos(at) \)
\( f(0) = \sin(0) = 0 \), and \( \mathcal{L}\{\sin(at)\} = \frac{a}{s^2+a^2} \)
\( \mathcal{L}\{a\cos(at)\} = s \cdot \frac{a}{s^2+a^2} - 0 \)
\( a \cdot \mathcal{L}\{\cos(at)\} = \frac{as}{s^2+a^2} \)

\( \mathcal{L}\{\cos(at)\} = \frac{s}{s^2+a^2} \)

6. Laplace Transform of Integrals

\[ \mathcal{L}\left\{\int_0^t f(u) \, du\right\} = \frac{F(s)}{s} \]

Find \( \mathcal{L}\left\{\int_0^t e^{2u} \, du\right\} \)

\( \mathcal{L}\{e^{2t}\} = \frac{1}{s-2} \)
Apply integral property: \( \mathcal{L}\left\{\int_0^t e^{2u} \, du\right\} = \frac{1}{s} \cdot \frac{1}{s-2} \)

\( \mathcal{L}\left\{\int_0^t e^{2u} \, du\right\} = \frac{1}{s(s-2)} \)

7. Evaluation of Real Integrals using Laplace Transform

Key Technique

Use \( \int_0^{\infty} e^{-st}f(t)\,dt = F(s) \), then substitute specific value of \( s \)

Evaluate \( \int_0^{\infty} te^{-3t}\sin(2t) \, dt \)

We know \( \mathcal{L}\{t\sin(2t)\} = \frac{4s}{(s^2+4)^2} \)
So \( \int_0^{\infty} e^{-st} \cdot t\sin(2t) \, dt = \frac{4s}{(s^2+4)^2} \)
Put \( s = 3 \): \( \int_0^{\infty} e^{-3t} \cdot t\sin(2t) \, dt = \frac{4(3)}{(9+4)^2} \)

\( \int_0^{\infty} te^{-3t}\sin(2t) \, dt = \frac{12}{169} \)

Evaluate \( \int_0^{\infty} \frac{\sin(t)}{t} \, dt \)

We know \( \mathcal{L}\left\{\frac{\sin(t)}{t}\right\} = \tan^{-1}\left(\frac{1}{s}\right) \)
This means \( \int_0^{\infty} e^{-st} \cdot \frac{\sin(t)}{t} \, dt = \tan^{-1}\left(\frac{1}{s}\right) \)
Put \( s = 0 \): \( \int_0^{\infty} \frac{\sin(t)}{t} \, dt = \tan^{-1}(\infty) \)

\( \int_0^{\infty} \frac{\sin(t)}{t} \, dt = \frac{\pi}{2} \)

Evaluate \( \int_0^{\infty} e^{-t}\frac{1 - \cos(t)}{t} \, dt \)

First find \( \mathcal{L}\{1 - \cos(t)\} = \frac{1}{s} - \frac{s}{s^2+1} = \frac{s^2+1-s^2}{s(s^2+1)} = \frac{1}{s(s^2+1)} \)
\( \mathcal{L}\left\{\frac{1-\cos(t)}{t}\right\} = \int_s^{\infty} \frac{1}{u(u^2+1)} \, du \)
Using partial fractions: \( \frac{1}{u(u^2+1)} = \frac{1}{u} - \frac{u}{u^2+1} \)
\( = \left[\ln|u| - \frac{1}{2}\ln(u^2+1)\right]_s^{\infty} = \left[\ln\frac{u}{\sqrt{u^2+1}}\right]_s^{\infty} \)
\( = 0 - \ln\frac{s}{\sqrt{s^2+1}} = \frac{1}{2}\ln\frac{s^2+1}{s^2} \)
Put \( s = 1 \): \( \int_0^{\infty} e^{-t}\frac{1-\cos(t)}{t} dt = \frac{1}{2}\ln\frac{2}{1} \)

\( \int_0^{\infty} e^{-t}\frac{1 - \cos(t)}{t} \, dt = \frac{1}{2}\ln 2 \)

Quick Reference: All Formulas

Property/Transform	Formula
Definition	\( \mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st}f(t)\,dt \)
Linearity	\( \mathcal{L}\{af + bg\} = a\mathcal{L}\{f\} + b\mathcal{L}\{g\} \)
First Shifting	\( \mathcal{L}\{e^{at}f(t)\} = F(s-a) \)
Change of Scale	\( \mathcal{L}\{f(at)\} = \frac{1}{a}F(s/a) \)
Multiplication by t	\( \mathcal{L}\{tf(t)\} = -\frac{d}{ds}F(s) \)
Division by t	\( \mathcal{L}\{f(t)/t\} = \int_s^{\infty}F(u)\,du \)
Derivative	\( \mathcal{L}\{f'(t)\} = sF(s) - f(0) \)
Second Derivative	\( \mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0) \)
Integral	\( \mathcal{L}\{\int_0^t f(u)du\} = F(s)/s \)