Laplace Transform Definition - laplace-transform.com

1. Intuitive introduction

In elementary calculus, we learned that a function $f(x)$ can be transformed into another one if we use differentiation or integration operations. For example, the function $f(x) = x$ becomes:

$g(x)= \displaystyle \frac{df}{dx}=1$ or $h(x)=\displaystyle\int f(x)dx=\frac{x^{2}}{2}+c$

Let $f(t)$ be a function in the variable $t$ . When we evaluate the integral with respect to $t$ :

\int_{a}^{b}N(s,t)\cdot f(t)dt

it transforms the function $f$ into a function $F(s)$ in the variable $s$ . We call $N(s,t)$ the kernel of the integral transform.

For example, let $N(s,t)=3s^{2}t^{3}$ and $f(t)=t^2$ . Then, $\int_{1}^{3}3s^{2}t^{3}\cdot t^{2}dt=\int_{1}^{3}3s^{2}t^{5}dt$

And since $3s^{2}$ is a constant, we have:

$\begin{aligned} \int_{1}^{3}3s^{2}t^{5} & = 3s^{2}\int_{1}^{3}t^{5}dt \\ & = 3s^{2}\cdot\left.\frac{t^{6}}{6}\right|_{t=1}^{t=3} \\& = 3s^{2}\cdot\left(\frac{3^{6}}{6}-\frac{1^{6}}{6}\right) \\ & = 3s^{2}\cdot\left(\frac{729}{6}-\frac{1}{6}\right) \\ & = 3s^{2}\cdot\frac{728}{6}\\ \end{aligned}$

Therefore, the integral transformed the function $f(t)=t^{2}$ into the function:

F(s)=364s^{2}

However, we are more interested in the case when the integral is improper:

\int_{0}^{\infty}N(s,t)f(t)dt=\lim_{b\to\infty}\int_{0}^{b}N(s,t)f(t)

And if:

N(s,t)=e^{-st}

we have the Laplace transform definition.

2. Definition of Laplace transform

We also note the Laplace transform $\mathscr{L}\{f(t)\}$ by $F(s)$ .

Laplace transform of f(t)=1

Let’s use the definition of the Laplace transform to find $F(s)$ of the function $f(t)=1$ . We have:

\mathscr{L}\{1\}=\int_{0}^{\infty}e^{-st}\cdot1dt=\lim_{b\to\infty}\int_{0}^{b}e^{-st}dt

In the integral above, if $u=-st$ then:

du=-sdt\Rightarrow dt=-\frac{du}{s}

So, changing the variables:

$\begin{aligned} \mathscr{L}\{1\} & = \lim_{b\to\infty}\int e^{u}\cdot\left(-\frac{du}{s}\right) \\ & = \lim_{b\to\infty}\int -\frac{e^{u}}{s}du \end{aligned}$

As $\displaystyle-\frac{1}{s}$ is constant due to the fact we integrate in respect to $u$ , then

\begin{align*} \mathscr{L}\{1\} & = \lim_{b\to\infty}-\frac{1}{s}\int e^{u}du \\ & = \lim_{b\to\infty}-\frac{1}{s}e^{u} \\ & = \lim_{b\to\infty}\left.-\frac{1}{s}e^{-st}\right|_{t=0}^{t=b} \end{align*}

because $u=-st$ . Now, evaluating in the boundaries:

\begin{align*} \mathscr{L}\{1\} & = \lim_{b\to\infty}-\frac{1}{s}(e^{-s\cdot b}-\overbrace{e^{{-s\cdot0}}}^{=1}) \\ & = \lim_{b\to\infty}-\frac{1}{s}(e^{-sb}-1) \\ & = \frac{1}{s}\lim_{b\to\infty}(-e^{-sb}+1) \\ & = \frac{1}{s}(\lim_{b\to\infty}-e^{-sb}+\lim_{b\to\infty}1) \end{align*}

Since the limit:

\lim_{b\to\infty}-e^{-sb}=0

for $s>0$ and

\lim_{b\to\infty}1=1

then, the Laplace Transform of $f(t)=1$ is the function:

F(s)=\frac{1}{s}

which is defined for every $s$ in $(0,+\infty)$ .

For a more general case, check the laplace transform of a constant a.

Laplace transform of f(t)=e^{at}

Using the definition, we have

\mathscr{L}\{e^{at}\}=\int_{0}^{\infty}e^{-st}\cdot e^{at}dt=\lim_{b\to\infty}\int_{0}^{b}e^{(-s+a)t}dt

And just like we did in the previous example, calling: $u=(-s+a)t$ , so

du=(-s+a)dt\Rightarrow dt=\frac{du}{-s+a}

Changing the variables, we get

$\begin{aligned} \mathscr{L}\{e^{at}\} & = \lim_{b\to\infty}\int e^{u}\frac{du}{-s+a} \\ & = \frac{1}{-s+a}\lim_{b\to\infty}\int e^{u}du \\ & = \frac{1}{-s+a}\lim_{b\to\infty}e^{u} \\ & = \frac{1}{-s+a}\left.\lim_{b\to\infty}e^{(-s+a)t}\right|_{t=0}^{t=b} \end{aligned}$

When $t=0$ , the exponential is:

e^{(-s+a)\cdot0}=e^{0}=1

And then

$\begin{aligned} \mathscr{L}\{e^{at}\} & = \frac{1}{-s+a}\lim_{b\to\infty}(e^{(-s+a)b}-1) \\ & = \frac{1}{-s+a}(\lim_{b\to\infty}e^{(-s+a)b}-\lim_{b\to\infty}1) \end{aligned}$

For $-s+a<0\Rightarrow s>a$ , we have:

\lim_{b\to\infty}e^{(-s+a)b}=0

and as $\displaystyle\lim_{b\to\infty}1=1$ , we conclude that

F(s)=\frac{1}{s-a}

is the Laplace Transform of $e^{at}$ . By the steps showed before, $F$ is defined for every $s$ such that $s>a$ .

Laplace transform of f(t)=sin(at)

To find the Laplace transform of $f(t)=\sin(at)$ , where $a\in\mathbb{R}$ , we will proceed in the same way.

\mathscr{L}\{\sin(at)\}=\int_{0}^{\infty}e^{-st}\sin(at)dt=\lim_{b\to\infty}\int_{0}^{b}e^{-st}\sin(at)dt

Since exponential and sinus function are from different families, the integration by parts is the most indicated in this case. Taking:

u=\sin(at)\Rightarrow du=a\cos(at)dt

and

dv=e^{-st}dt\Rightarrow v=\frac{-e^{-st}}{s}

and using the formula

\int udv=uv-\int v du

we get

\begin{aligned} \int_{0}^{b}e^{-st}\sin(at)dt & = \sin(at)\cdot\left.\left(\frac{-e^{-st}}{s}\right)\right|_{t=0}^{t=b}-\int_{0}^{b}\frac{-e^{-st}}{s}\cdot a\cos(at)dt \\ & =-\sin(ab)\cdot\frac{e^{-sb}}{s}+\sin(0)\cdot\frac{e^{-0}}{s}+\frac{a}{s}\int_{0}^{b}e^{-st}\cos(at)dt \\ & =-\sin(ab)\cdot\frac{e^{-sb}}{s}+\frac{a}{s}\int_{0}^{b}e^{-st}\cos(at)dt\end{aligned}

because $\sin(0)=0$ . Now, on the integral in the last line, if

u=\cos(at)\Rightarrow du=-a\sin(at)dt

and

dv=e^{-st}dt\Rightarrow v=\frac{-e^{-st}}{s}

we integrate by parts one more time:

$\begin{aligned} \int_{0}^{b}e^{-st}\cos(at)dt & = \cos(at)\left.\cdot\left(\frac{-e^{-st}}{s}\right)\right|_{t=0}^{t=b}-\int_{0}^{b}\frac{-e^{-st}}{s}\cdot(-a\sin(at))dt \\ & = -\cos(ab)\cdot\frac{e^{-sb}}{s}+\cos(0)\cdot\frac{e^{-0}}{s}-\frac{a}{s}\int_{0}^{b}e^{-st}\sin(at)dt \\ & = -\cos(ab)\cdot\frac{e^{-sb}}{s}+\frac{1}{s}-\frac{a}{s}\int_{0}^{b}e^{-st}\sin(at)dt \end{aligned}$

since $\cos(0)=e^{-0}=1$ . If we call

I=\int_{0}^{b}e^{-st}\sin(at)dt

we have

\begin{aligned} &I=-\sin(ab)\cdot\frac{e^{-sb}}{s}+\frac{a}{s}\cdot\left(-\cos(ab)\cdot\frac{e^{-sb}}{s}+\frac{1}{s}-\frac{a}{s}\cdot I\right) \\ &I=-\sin(ab)\cdot\frac{e^{-sb}}{s}-\frac{a}{s}\cos(ab)\cdot\frac{e^{-sb}}{s}+\frac{a}{s^{2}}-\frac{a^{2}}{s^{2}}\cdot I \\ &I+\frac{a^{2}}{s^{2}}I=-\sin(ab)\cdot\frac{e^{-sb}}{s}-\frac{a}{s}\cos(ab)\cdot\frac{e^{-sb}}{s}+\frac{a}{s^{2}} \end{aligned}

Now, once

|\sin(x)|\leq1,\qquad\text{and}\qquad|\cos(x)|\leq1

|\sin(ab)\cdot e^{-sb}|\leq e^{-sb},\qquad\text{and}\qquad|\cos(ab)\cdot e^{-sb}|\leq e^{-sb}

and since

\lim_{b\to\infty}e^{-sb}=0

for $s>0$ , the squeeze theorem says

\lim_{b\to\infty}-\sin(ab)\cdot\frac{e^{-sb}}{s}=-\frac{1}{s}\lim_{b\to\infty}\sin(ab)e^{-sb}=0

and also

\lim_{b\to\infty}-\frac{a}{s}\cos(ab)\cdot\frac{e^{-sb}}{s}=-\frac{a}{s^{2}}\lim_{b\to\infty}\cos(ab)e^{-sb}=0

And as

I+\frac{a^{2}}{s^{2}}I=\left(\frac{s^{2}+a^{2}}{s^{2}}\right)I

we have

\lim_{b\to\infty}\left(\frac{s^{2}+a^{2}}{s^{2}}\right)I=\lim_{b\to\infty}\left(-\sin(ab)\cdot\frac{e^{-sb}}{s}-\frac{a}{s}\cos(ab)\cdot\frac{e^{-sb}}{s}+\frac{a}{s^{2}}\right)

which give us

\lim_{b\to\infty}\left(\frac{s^{2}+a^{2}}{s^{2}}\right)I=\frac{a}{s^{2}}

i.e.,

\lim_{b\to\infty}I=\frac{a}{s^{2}}\cdot\frac{s^{2}}{s^{2}+a^{2}}

\Rightarrow\lim_{b\to\infty}I=\frac{a}{s^{2}+a^{2}}

So,

F(s)=\mathscr{L}\{\sin(at)\}=\lim_{b\to\infty}\int_{0}^{b}e^{-st}\sin(at)dt=\frac{a}{s^{2}+a^{2}}

is the Laplace Transform of $\sin(at)$ ; the function $F(s)$ is defined for every $s>0$ .

A simple method of calculating the Laplace transform of sine using Euler’s formula can be found here!

Conclusion

In this lesson, we presented the Laplace transform definition, known also as the integral definition of Laplace transform. We applied the Laplace transformation definition to the examples: $f(t)=1$ , $f(t)=e^{at}$ and $f(t)=\sin(at)$ .