Let \(f(t)\) be defined for \(t\geq 0\) and \(s\) represent a real number. Then the Laplace transform of the function \(f(t)\) is the function \(\displaystyle F(s)=\int _{0}^{\infty }e^{-st}f(t)\,dt\) on its domain of convergence (meaning where the improper integral converges). The functions \(f\) and \(F\) are called a Laplace transform pair.

Notation: \(\mathcal {L}(f)=F\) or we use the identification \(f\leftrightarrow F\).

Note: Technically the improper integral must be turned in a limit, but we will be lazy and never formally do this. We will allow \(``\infty \)" to appear as though it were a real number (but with the understanding that we are taking a limit).

Let’s calculate a few basic Laplace transforms:

Consider the constant function \(f(t)=k\) for \(k\neq 0\). Then \(F(0)\) diverges (to \(\pm \infty \)) and for \(s\neq 0\), \(\displaystyle F(s)=\int _{0}^{\infty }ke^{-st}\,dt=\left .-\frac {k}{s}e^{-st}\right |_{0}^{\infty }=\) \(\displaystyle \left \{ \begin{array}{lr} \dfrac {k}{s} & \mbox { if }s>0\\ \\ \pm \infty & \mbox { if }s<0 \end{array}\right .\). So \(\mathcal {L}(k)=\dfrac {k}{s}\) (\(s>0\)).

Consider \(f(t)=t\). Then \(F(0)\) diverges (to \(\infty \)) and for \(s\neq 0\),
\(\displaystyle F(s)=\int _{0}^{\infty }te^{-st}\,dt=\left .-\frac {t}{s}e^{-st}\right |_{0}^{\infty }+\dfrac {1}{s}\int _{0}^{\infty }e^{-st}\,dt=\) \(\displaystyle \left (\left .-\frac {t}{s}e^{-st}-\frac {1}{s^2}e^{-st}\right |_{0}^{\infty }\right )\) \(\displaystyle =\left \{ \begin{array}{lr} \dfrac {1}{s^{2}} & \mbox { if }s>0\\ \\ \infty & \mbox { if }s<0 \end{array}\right .\).
So \(\mathcal {L}(t)=\dfrac {1}{s^2}\) (\(s>0\)).

Consider \(f(t)=e^{at}\).

\(\displaystyle F(s)=\int _{0}^{\infty }e^{-st}e^{at}\,dt=\int _{0}^{\infty }e^{-(s-a)t}\,dt=\) \(\displaystyle \left .-\dfrac {1}{s-a}e^{-(s-a)t}\right |_{0}^{\infty }=\) \(\displaystyle \left \{ \begin{array}{lr} \dfrac {1}{s-a} & \mbox { if }s>a\\ \\ \infty & \mbox { if }s< a \end{array}\right .\).

Note: F(a) diverges.

So \(\mathcal {L}(e^{at})=\dfrac {1}{s-a}\) (\(s>a\)).

Linearity of Laplace transforms: If \(\mathcal {L}(f)\) and \(\mathcal {L}(g)\) both exist for \(s>s_{0}\), where \(f,g\) are functions and \(s_0\) is a real number, then for any constants \(c_{1},c_{2}\) we have

(The proof is left as an exercise for the reader.)

Let’s find the Laplace transform of \(\sinh {(at)}=\dfrac {e^{at}-e^{-at}}{2}\) where \(a>0\). We know that \(\mathcal {L}(e^{at})=\dfrac {1}{s-a}\) for \(s>a\) and \(\mathcal {L}(e^{-at})=\dfrac {1}{s+a}\) for \(s>-a\). So by linearity,
\(\mathcal {L}(\sinh {(at)})=\dfrac {1}{2}\left ( \dfrac {1}{s-a}-\dfrac {1}{s+a}\right )=\dfrac {a}{s^{2}-a^{2}}\) for \(s>a\).

1st Shifting Property: Let \(f(t)\) be a function and \(\mathcal {L}(f)=F\), which means
\(\displaystyle F(s)=\int _{0}^{\infty }e^{-st}f(t)\,dt\). Let’s determine \(\mathcal {L}(e^{at}f(t))\) where \(a\) is a constant. Thus we get
\(\displaystyle \mathcal {L}(e^{at}f(t))=\int _{0}^{\infty }e^{-st}(e^{at}f(t))\,dt=\) \(\displaystyle \int _{0}^{\infty }e^{-(s-a)t}f(t)\,dt=F(s-a).\)

The identity \(\mathcal {L}(e^{at}f(t))=F(s-a)\) is the first shifting property of the Laplace transform.

One can show, using integration by parts twice, and then solving for the integral as an algebraic variable, that \(\displaystyle \mathcal {L}(\sin {(\omega t)})=\frac {\omega }{s^2+\omega ^2}\). Therefore by the first shifting property, \(\displaystyle \mathcal {L}(e^{at}\sin {(\omega t)})=\frac {\omega }{(s-a)^2+\omega ^2}\)

A function \(f\) is of exponential order \(s_{0}\) if there are constants \(M\) and \(t_{0}\) such that \(\displaystyle |f(t)|\leq Me^{s_{0}t}\mbox { for all }t\geq t_{0}\). Often we say a function is of \(``\)exponential order" without a reference to a particular \(s_{0}\).

Theorem: Suppose \(f\) is of exponential order \(s_{0}\). Then \(\mathcal {L}(f)\) is defined for \(s>s_{0}\).

(The proof is left as an exercise left for the reader.)