The Laplace Transform: Introduction

The Laplace transform is an integral transform that allows us to solve a linear differential equation by converting it to algebraic expressions such as rational functions.

The Laplace transform $\mathcal{L}\{F(t)\}$ of a function $F(t)$ is defined by

\label{eq:laplace}
\mathcal{L}\{F(t)\}=\int_0^\infty e^{-st}F(t)dt

Since the improper integral in \eqref{eq:laplace} is a function of $s$, we also write $\mathcal{L}\{F(t)\}=f(s)$.

Example. Let $F(t)=1$, $t>0$.
$$\mathcal{L}\{F(t)\}=\int_0^\infty e^{-st}dt=\frac{1}{2}$$
provided $s>0$.

Example. Let $F(t)=e^{kt}$, $t>0$ and $k$ a constant.
\begin{align*}
\mathcal{L}\{F(t)\}&=\int_0^\infty e^{-st}e^{kt}dt\\
&=\int_0^\infty e^{-(s-k)}tdt\\
&=\frac{1}{s-k}
\end{align*}
provided $s>k$.

Example.
\begin{align*}
\mathcal{L}\{t\}&=\int_0^\infty te^{-st}dt\\
&=\frac{1}{s^2}
\end{align*}
for $s>0$.

Example. Again by the definition of the Laplace transform one obtains
\begin{align*}
\mathcal{L}\{t^2\}&=\frac{2!}{s^3}\\
\mathcal{L}\{t^3\}&=\frac{3!}{s^4}
\end{align*}
by simply calculating the required improper integrals. There appears to be a pattern and one may expect the formula
$$\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$$
This is indeed true and we will prove this later.

Example. By the definition of the Laplace transform one obtains
\begin{align}
\label{eq:laplace2}
\mathcal{L}\{\sin kt\}&=\frac{k}{s^2+k^2}\\
\label{eq:laplace3}
\mathcal{L}\{\cos kt\}&=\frac{s}{s^2+k^2}
\end{align}

The Laplace transform is linear, namely
\begin{align*}
\mathcal{L}\{F(t)+G(t)\}&=\mathcal{L}\{F(t)\}+\mathcal{L}\{G(t)\}\\
\mathcal{L}\{cF(t)\}&=c\mathcal{L}\{F(t)\}
\end{align*}

Example. $\sinh kt=\frac{e^{kt}-e^{-kt}}{2}$ so by the linearity of the Laplace transform
\begin{align*}
\mathcal{L}\{\sinh kt\}&=\frac{1}{2}(\mathcal{L}\{e^{kt}\}-\mathcal{L}\{e^{-kt}\}\\
&=\frac{1}{2}\left(\frac{1}{s-k}-\frac{1}{s+k}\right)\\
&=\frac{k}{s^2-k^2}
\end{align*}
Hence we have

\label{eq:laplace4}
\mathcal{L}\{\sinh kt\}=\frac{k}{s^2-k^2}

Similarly we also obtain

\label{eq:laplace5}
\mathcal{L}\{\cosh kt\}=\frac{s}{s^2+k^2}