Laplace Transform of Elementary and Special Functions

1. Introduction to Laplace Transform

The Laplace transform is a powerful mathematical tool used extensively in engineering and physics, particularly in solving differential equations. It transforms a function of time, $f(t)$, into a function of a complex variable, $s$. This transformation often simplifies problems by converting differential equations into algebraic equations, which are easier to solve. The variable $s$ is a complex number, typically written as $s = \sigma + i\omega$, where $\sigma$ is the real part and $\omega$ is the imaginary part.

2. Definition of Laplace Transform

The Laplace transform of a function $f(t)$, denoted as $\mathcal{L}\{f(t)\}$ or $F(s)$, is defined by the improper integral:

$F(s) = \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) dt$

This integral is defined for all values of $s$ for which it converges. For the transform to exist, the function $f(t)$ must be of exponential order, meaning there exist constants $M$ and $a$ such that $|f(t)| \le Me^{at}$ for all $t \ge 0$. The integral converges if the real part of $s$, $\sigma$, is greater than $a$.

3. Laplace Transform of Elementary Functions

Let's derive the Laplace transforms of some basic and frequently used functions. These form the building blocks for transforming more complex functions.

3.1 Laplace Transform of a Constant Function ($f(t) = c$)

Let $f(t) = c$, where $c$ is a constant.

$\mathcal{L}\{c\} = \int_{0}^{\infty} e^{-st} (c) dt = c \int_{0}^{\infty} e^{-st} dt$

Evaluating the integral:

$c \left[ \frac{e^{-st}}{-s} \right]_{0}^{\infty} = c \left( 0 - \frac{e^{0}}{-s} \right) = c \left( 0 - \frac{1}{-s} \right) = \frac{c}{s}$

This is valid for $s > 0$.

$\mathcal{L}\{c\} = \frac{c}{s}, \quad s > 0$

3.2 Laplace Transform of $f(t) = t^n$ (where $n$ is a non-negative integer)

Let $f(t) = t^n$.

$\mathcal{L}\{t^n\} = \int_{0}^{\infty} e^{-st} t^n dt$

This integral is related to the Gamma function, $\Gamma(z)$, which is defined as $\Gamma(z) = \int_{0}^{\infty} x^{z-1} e^{-x} dx$. By substituting $x = st$, we get $dx = s dt$, so $dt = \frac{dx}{s}$. When $t=0$, $x=0$. When $t \to \infty$, $x \to \infty$.

$\mathcal{L}\{t^n\} = \int_{0}^{\infty} e^{-x} \left(\frac{x}{s}\right)^n \frac{dx}{s} = \frac{1}{s^{n+1}} \int_{0}^{\infty} x^n e^{-x} dx$

The integral $\int_{0}^{\infty} x^n e^{-x} dx$ is $\Gamma(n+1)$. For integer $n$, $\Gamma(n+1) = n!$.

$\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, \quad n = 0, 1, 2, \dots, \quad s > 0$

For $n=0$: $\mathcal{L}\{t^0\} = \mathcal{L}\{1\} = \frac{0!}{s^{0+1}} = \frac{1}{s}$, which matches our previous result for a constant $c=1$.

For $n=1$: $\mathcal{L}\{t\} = \frac{1!}{s^{1+1}} = \frac{1}{s^2}$.

For $n=2$: $\mathcal{L}\{t^2\} = \frac{2!}{s^{2+1}} = \frac{2}{s^3}$.

3.3 Laplace Transform of $f(t) = e^{at}$

Let $f(t) = e^{at}$, where $a$ is a constant.

$\mathcal{L}\{e^{at}\} = \int_{0}^{\infty} e^{-st} e^{at} dt = \int_{0}^{\infty} e^{-(s-a)t} dt$

Evaluating the integral:

$\left[ \frac{e^{-(s-a)t}}{-(s-a)} \right]_{0}^{\infty} = 0 - \frac{e^{0}}{-(s-a)} = \frac{1}{s-a}$

This is valid for $s-a > 0$, i.e., $s > a$.

$\mathcal{L}\{e^{at}\} = \frac{1}{s-a}, \quad s > a$

Example: $\mathcal{L}\{e^{2t}\} = \frac{1}{s-2}$ for $s > 2$.

If $a=0$, $\mathcal{L}\{e^{0t}\} = \mathcal{L}\{1\} = \frac{1}{s-0} = \frac{1}{s}$, which is consistent.

3.4 Laplace Transform of $f(t) = \sin(\omega t)$

Let $f(t) = \sin(\omega t)$. We use Euler's formula: $\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$.

$\mathcal{L}\{\sin(\omega t)\} = \mathcal{L}\left\{\frac{e^{i\omega t} - e^{-i\omega t}}{2i}\right\}$

Using the linearity property of the Laplace transform: $\mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\}$.

$\mathcal{L}\{\sin(\omega t)\} = \frac{1}{2i} \left( \mathcal{L}\{e^{i\omega t}\} - \mathcal{L}\{e^{-i\omega t}\} \right)$

Using the result for $\mathcal{L}\{e^{at}\}$ with $a = i\omega$ and $a = -i\omega$:

$\mathcal{L}\{e^{i\omega t}\} = \frac{1}{s - i\omega}$ for $s > \text{Re}(i\omega) = 0$.

$\mathcal{L}\{e^{-i\omega t}\} = \frac{1}{s - (-i\omega)} = \frac{1}{s + i\omega}$ for $s > \text{Re}(-i\omega) = 0$.

Substituting these back:

$\mathcal{L}\{\sin(\omega t)\} = \frac{1}{2i} \left( \frac{1}{s - i\omega} - \frac{1}{s + i\omega} \right)$

$= \frac{1}{2i} \left( \frac{(s + i\omega) - (s - i\omega)}{(s - i\omega)(s + i\omega)} \right) = \frac{1}{2i} \left( \frac{2i\omega}{s^2 - (i\omega)^2} \right)$

$= \frac{1}{2i} \left( \frac{2i\omega}{s^2 - (-\omega^2)} \right) = \frac{\omega}{s^2 + \omega^2}$

This is valid for $s > 0$.

$\mathcal{L}\{\sin(\omega t)\} = \frac{\omega}{s^2 + \omega^2}, \quad s > 0$

3.5 Laplace Transform of $f(t) = \cos(\omega t)$

Let $f(t) = \cos(\omega t)$. Using Euler's formula: $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$.

$\mathcal{L}\{\cos(\omega t)\} = \mathcal{L}\left\{\frac{e^{i\omega t} + e^{-i\omega t}}{2}\right\}$

$= \frac{1}{2} \left( \mathcal{L}\{e^{i\omega t}\} + \mathcal{L}\{e^{-i\omega t}\} \right)$

$= \frac{1}{2} \left( \frac{1}{s - i\omega} + \frac{1}{s + i\omega} \right)$

$= \frac{1}{2} \left( \frac{(s + i\omega) + (s - i\omega)}{(s - i\omega)(s + i\omega)} \right) = \frac{1}{2} \left( \frac{2s}{s^2 + \omega^2} \right)$

$= \frac{s}{s^2 + \omega^2}$

This is valid for $s > 0$.

$\mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2}, \quad s > 0$

3.6 Laplace Transform of $f(t) = \sinh(at)$

Let $f(t) = \sinh(at)$. Using the definition $\sinh(x) = \frac{e^x - e^{-x}}{2}$.

$\mathcal{L}\{\sinh(at)\} = \mathcal{L}\left\{\frac{e^{at} - e^{-at}}{2}\right\}$

$= \frac{1}{2} \left( \mathcal{L}\{e^{at}\} - \mathcal{L}\{e^{-at}\} \right)$

$= \frac{1}{2} \left( \frac{1}{s-a} - \frac{1}{s+a} \right)$

$= \frac{1}{2} \left( \frac{(s+a) - (s-a)}{(s-a)(s+a)} \right) = \frac{1}{2} \left( \frac{2a}{s^2 - a^2} \right)$

$= \frac{a}{s^2 - a^2}$

This is valid for $s > |a|$.

$\mathcal{L}\{\sinh(at)\} = \frac{a}{s^2 - a^2}, \quad s > |a|$

3.7 Laplace Transform of $f(t) = \cosh(at)$

Let $f(t) = \cosh(at)$. Using the definition $\cosh(x) = \frac{e^x + e^{-x}}{2}$.

$\mathcal{L}\{\cosh(at)\} = \mathcal{L}\left\{\frac{e^{at} + e^{-at}}{2}\right\}$

$= \frac{1}{2} \left( \mathcal{L}\{e^{at}\} + \mathcal{L}\{e^{-at}\} \right)$

$= \frac{1}{2} \left( \frac{1}{s-a} + \frac{1}{s+a} \right)$

$= \frac{1}{2} \left( \frac{(s+a) + (s-a)}{(s-a)(s+a)} \right) = \frac{1}{2} \left( \frac{2s}{s^2 - a^2} \right)$

$= \frac{s}{s^2 - a^2}$

This is valid for $s > |a|$.

$\mathcal{L}\{\cosh(at)\} = \frac{s}{s^2 - a^2}, \quad s > |a|$

4. Laplace Transform of Special Functions

Beyond the elementary functions, the Laplace transform is also applied to various special functions that arise in different fields of study.

4.1 Laplace Transform of the Unit Step Function (Heaviside Function)

The unit step function, denoted by $u(t)$ or $H(t)$, is defined as:

$u(t) = \begin{cases} 0 & \text{if } t < 0 \\ 1 & \text{if } t \ge 0 \end{cases}$

Note that the value at $t=0$ doesn't affect the integral.

$\mathcal{L}\{u(t)\} = \int_{0}^{\infty} e^{-st} (1) dt$

This is the same integral as for $\mathcal{L}\{1\}$, so:

$\mathcal{L}\{u(t)\} = \frac{1}{s}, \quad s > 0$

4.2 Laplace Transform of the Delayed Unit Step Function

A delayed unit step function is defined as $u(t-c)$ for $c > 0$:

$u(t-c) = \begin{cases} 0 & \text{if } t < c \\ 1 & \text{if } t \ge c \end{cases}$

$\mathcal{L}\{u(t-c)\} = \int_{0}^{\infty} e^{-st} u(t-c) dt = \int_{c}^{\infty} e^{-st} (1) dt$

Evaluating the integral:

$\left[ \frac{e^{-st}}{-s} \right]_{c}^{\infty} = 0 - \frac{e^{-sc}}{-s} = \frac{e^{-sc}}{s}$

This is valid for $s > 0$.

$\mathcal{L}\{u(t-c)\} = \frac{e^{-sc}}{s}, \quad c > 0, s > 0$

This result is crucial for understanding the Laplace transform of functions that are zero for an initial period.

4.3 Laplace Transform of the Dirac Delta Function

The Dirac delta function, $\delta(t)$, is a generalized function characterized by the property:

$\delta(t) = \begin{cases} \infty & \text{if } t = 0 \\ 0 & \text{if } t \ne 0 \end{cases}$ and $\int_{-\infty}^{\infty} \delta(t) dt = 1$.

For Laplace transform purposes, we consider $\delta(t)$ for $t \ge 0$. The defining property relevant here is $\int_{0}^{\infty} \delta(t) f(t) dt = f(0)$ for a continuous function $f(t)$.

$\mathcal{L}\{\delta(t)\} = \int_{0}^{\infty} e^{-st} \delta(t) dt$

Applying the property of the delta function with $f(t) = e^{-st}$:

$\mathcal{L}\{\delta(t)\} = e^{-s(0)} = e^0 = 1$

This is valid for all $s$.

$\mathcal{L}\{\delta(t)\} = 1$

4.4 Laplace Transform of the Ramp Function ($f(t) = t$)

The ramp function is $f(t) = t$ for $t \ge 0$. We've already derived this in section 3.2 for $n=1$.

$\mathcal{L}\{t\} = \frac{1!}{s^{1+1}} = \frac{1}{s^2}, \quad s > 0$

4.5 Laplace Transform of $f(t) = t^n e^{at}$

This can be derived using the first shifting theorem (also known as the frequency shifting theorem). The theorem states:

We know $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$.

Applying the first shifting theorem with $f(t) = t^n$:

$\mathcal{L}\{e^{at} t^n\} = F(s-a)$, where $F(s) = \frac{n!}{s^{n+1}}$.

So, replace $s$ with $(s-a)$ in $F(s)$:

$\mathcal{L}\{t^n e^{at}\} = \frac{n!}{(s-a)^{n+1}}, \quad s > a$

Example: $\mathcal{L}\{t e^{3t}\} = \frac{1!}{(s-3)^{1+1}} = \frac{1}{(s-3)^2}$ for $s > 3$.

4.6 Laplace Transform of $e^{at}\sin(\omega t)$ and $e^{at}\cos(\omega t)$

Using the first shifting theorem again:

For $\mathcal{L}\{e^{at}\sin(\omega t)\}$: We know $\mathcal{L}\{\sin(\omega t)\} = \frac{\omega}{s^2 + \omega^2}$. Let $F(s) = \frac{\omega}{s^2 + \omega^2}$. Then $\mathcal{L}\{e^{at}\sin(\omega t)\} = F(s-a) = \frac{\omega}{(s-a)^2 + \omega^2}$. This is valid for $s > a$.

For $\mathcal{L}\{e^{at}\cos(\omega t)\}$: We know $\mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2}$. Let $F(s) = \frac{s}{s^2 + \omega^2}$. Then $\mathcal{L}\{e^{at}\cos(\omega t)\} = F(s-a) = \frac{s-a}{(s-a)^2 + \omega^2}$. This is valid for $s > a$.

4.7 Laplace Transform of $t\sin(\omega t)$ and $t\cos(\omega t)$

These can be derived using the differentiation in the $s$-domain theorem. The theorem states:

For $\mathcal{L}\{t\sin(\omega t)\}$: Let $f(t) = \sin(\omega t)$, so $F(s) = \frac{\omega}{s^2 + \omega^2}$. $\mathcal{L}\{t\sin(\omega t)\} = -\frac{d}{ds} \left( \frac{\omega}{s^2 + \omega^2} \right)$ $= -\omega \frac{d}{ds} (s^2 + \omega^2)^{-1} = -\omega (-1)(s^2 + \omega^2)^{-2} (2s)$ $= \frac{2\omega s}{(s^2 + \omega^2)^2}$. This is valid for $s > 0$.

For $\mathcal{L}\{t\cos(\omega t)\}$: Let $f(t) = \cos(\omega t)$, so $F(s) = \frac{s}{s^2 + \omega^2}$. $\mathcal{L}\{t\cos(\omega t)\} = -\frac{d}{ds} \left( \frac{s}{s^2 + \omega^2} \right)$ $= -\frac{(s^2 + \omega^2)(1) - s(2s)}{(s^2 + \omega^2)^2} = -\frac{s^2 + \omega^2 - 2s^2}{(s^2 + \omega^2)^2}$ $= -\frac{\omega^2 - s^2}{(s^2 + \omega^2)^2} = \frac{s^2 - \omega^2}{(s^2 + \omega^2)^2}$. This is valid for $s > 0$.

5. Table of Common Laplace Transforms

It is highly beneficial to memorize the Laplace transforms of common functions. Here is a summary table:

$f(t)$	$\mathcal{L}\{f(t)\} = F(s)$	Region of Convergence (ROC)
$c$ (constant)	$\frac{c}{s}$	$s > 0$
$t^n$ ($n=0, 1, 2, \dots$)	$\frac{n!}{s^{n+1}}$	$s > 0$
$t^a$ ($a > -1$)	$\frac{\Gamma(a+1)}{s^{a+1}}$	$s > 0$
$e^{at}$	$\frac{1}{s-a}$	$s > a$
$\sin(\omega t)$	$\frac{\omega}{s^2 + \omega^2}$	$s > 0$
$\cos(\omega t)$	$\frac{s}{s^2 + \omega^2}$	$s > 0$
$\sinh(at)$	$\frac{a}{s^2 - a^2}$	$s > |a|$
$\cosh(at)$	$\frac{s}{s^2 - a^2}$	$s > |a|$
$u(t)$ (unit step)	$\frac{1}{s}$	$s > 0$
$u(t-c)$ ($c>0$)	$\frac{e^{-cs}}{s}$	$s > 0$
$\delta(t)$ (Dirac delta)	$1$	All $s$
$t^n e^{at}$	$\frac{n!}{(s-a)^{n+1}}$	$s > a$
$e^{at}\sin(\omega t)$	$\frac{\omega}{(s-a)^2 + \omega^2}$	$s > a$
$e^{at}\cos(\omega t)$	$\frac{s-a}{(s-a)^2 + \omega^2}$	$s > a$
$t\sin(\omega t)$	$\frac{2\omega s}{(s^2 + \omega^2)^2}$	$s > 0$
$t\cos(\omega t)$	$\frac{s^2 - \omega^2}{(s^2 + \omega^2)^2}$	$s > 0$

6. Properties of Laplace Transforms

These properties are essential for manipulating Laplace transforms and finding transforms of more complex functions.

6.1 Linearity Property

For constants $a$ and $b$, and functions $f(t)$ and $g(t)$:

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

This property was used in deriving transforms for $\sin(\omega t)$ and $\cos(\omega t)$.

6.2 First Shifting Theorem (Frequency Shifting)

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

This theorem is used to find the Laplace transform of functions multiplied by an exponential term.

6.3 Second Shifting Theorem (Time Shifting)

If $\mathcal{L}\{f(t)\} = F(s)$, then $\mathcal{L}\{f(t-c)u(t-c)\} = e^{-cs}F(s)$ for $c > 0$.

This theorem is useful for functions that are zero for an initial period or are defined piecewise.

Example: Find $\mathcal{L}\{f(t)\}$ where $f(t) = \begin{cases} 0 & \text{if } t < 2 \\ (t-2)^2 & \text{if } t \ge 2 \end{cases}$. Here, $f(t) = (t-2)^2 u(t-2)$. Let $g(t) = t^2$. Then $\mathcal{L}\{g(t)\} = \frac{2!}{s^3} = \frac{2}{s^3}$. Using the second shifting theorem with $c=2$, $\mathcal{L}\{(t-2)^2 u(t-2)\} = e^{-2s} \mathcal{L}\{t^2\} = e^{-2s} \frac{2}{s^3}$.

6.4 Differentiation in the $s$-domain

If $\mathcal{L}\{f(t)\} = F(s)$, then $\mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s)$.

For $n=1$: $\mathcal{L}\{t f(t)\} = -\frac{d}{ds} F(s)$. This was used for $t\sin(\omega t)$ and $t\cos(\omega t)$.

6.5 Integration in the $s$-domain

If $\mathcal{L}\{f(t)\} = F(s)$, then $\mathcal{L}\left\{\int_{0}^{t} f(\tau) d\tau\right\} = \frac{1}{s} F(s)$.

This property is the dual of differentiation in the $s$-domain and is useful for finding the Laplace transform of integrals.

6.6 Differentiation in the $t$-domain

If $\mathcal{L}\{f(t)\} = F(s)$, then $\mathcal{L}\{f'(t)\} = sF(s) - f(0)$.

For the second derivative: $\mathcal{L}\{f''(t)\} = s\mathcal{L}\{f'(t)\} - f'(0) = s(sF(s) - f(0)) - f'(0) = s^2F(s) - sf(0) - f'(0)$.

In general, for the $n$-th derivative: $\mathcal{L}\{f^{(n)}(t)\} = s^nF(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \dots - f^{(n-1)}(0)$.

This property is fundamental for solving linear ordinary differential equations with constant coefficients.

6.7 Convolution Theorem

The convolution of two functions $f(t)$ and $g(t)$, denoted by $(f*g)(t)$, is defined as:

$(f*g)(t) = \int_{0}^{t} f(\tau) g(t-\tau) d\tau$

The Convolution Theorem states:

This theorem is particularly useful for finding the inverse Laplace transform of a product of two functions $F(s)G(s)$.

7. Conclusion

The Laplace transform provides a systematic way to analyze linear time-invariant systems and solve differential equations. Understanding the transforms of elementary and special functions, along with their properties, is crucial for applying this powerful technique effectively. Mastery of these foundational elements enables the transformation of complex problems into simpler algebraic forms, facilitating their solution and analysis.