## 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

so

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).