If a function \(f(t)\) is defined for \(t \ge 0\text{,}\) then its Laplace transform is denoted \(\La{f}\) or \(F(s)\) and defined by the integral: \begin{equation} F(s) = \La{f(t)} = \int_0^\infty e^{-st}f(t)\; dt \label{eqn-laplace-transform}\tag{7.1.1} \end{equation} for all values of \(s\) for which the improper integral converges.
Notice that the last sentence of the above definition reminds us that improper integrals do not necessarily converge, i.e. equal a finite number. Thus when computing Laplace transforms of functions, we must be careful to state any assumptions on \(s\) which we make to ensure convergence. This just means that the domain of a transformed function, \(F(s)\text{,}\) is almost never the whole real number line, but in practice it doesn't matter.
The inverse Laplace transform is also defined via an integral, but a special kind of integral called a contour integral. In a contour integral, the integration follows a path or contour in the complex plane. Since most students have not learned how to compute such integrals at this point in their careers, we will not define the inverse Laplace transform. Instead we will build a dictionary or table of Laplace transforms and compute inverse Laplace transforms via table lookup or pattern matching. Since the Laplace transform method of solving differential equations requires both the forward transform, \(\Lapl\text{,}\) and its inverse, \(\Lapl^{-1}\text{,}\) we must first create a table of Laplace transforms of common functions.
\(\La{k} = \dfrac{k}{s} \qquad s>0\)
Recall that both the integral and the limit operator are linear, so we can pull constants outside of these operations. \begin{align*} \La{k} \amp = \int_0^{\infty} e^{-st} k \; dt\\ \amp = k \lim_{b \rightarrow \infty} \int_0^b e^{-st} \; dt\\ \amp = k \lim_{b \rightarrow \infty} \left[-\frac{1}{s} e^{-st}\right]_{t=0}^{t=b}\\ \amp = k \lim_{b \rightarrow \infty} \left[\cancelto{0}{-\frac{1}{s} e^{-sb}} + \frac{1}{s} \right] \quad s>0\\ \amp = \frac{k}{s} \quad \text{ for } s>0. \end{align*}
\(\La{e^{kt}} = \dfrac{1}{s-k} \qquad s>k\)
\begin{align*} \La{e^{kt}} \amp = \int_0^{\infty} e^{-st} e^{kt} \; dt\\ \amp = \int_0^{\infty} e^{(-s+k)t} \; dt \qquad \text{let } u = (-s+k)t \quad du = -(s-k)dt\\ \amp = \int_0^{\infty} \dfrac{-e^{u}}{s-k} \; du\\ \amp = \lim_{b \rightarrow \infty} \left[ \dfrac{-e^{-(s-k)t}}{s-k} \right]_{t=0}^{t=b}\\ \amp = \lim_{b \rightarrow \infty} \left[ \cancelto{0}{\dfrac{-e^{-(s-k)b}}{s-k}} + \dfrac{1}{s-k} \right] \qquad s-k>0 \quad \Leftrightarrow \quad s>k\\ \amp = \dfrac{1}{s-k} \quad \text{for } s>k. \end{align*}
\(\La{t} = \dfrac{1}{s^2} \qquad s>0\)
\begin{align*} \La{t} \amp = \int_0^{\infty} e^{-st}\, t \; dt\\ \amp = \left[ \dfrac{-t\,e^{-st}}{s} \right]_{t=0}^{t=\infty} - \int_0^{\infty} \dfrac{-e^{-st}}{s} \; dt\\ \amp = \lim_{b \rightarrow \infty} \cancelto{0}{\left[ \dfrac{-b\,e^{-sb}}{s} \right]} - \int_0^{\infty} \dfrac{-e^{-st}}{s} \; dt \quad s>0\\ \amp = -\lim_{b \rightarrow \infty} \left[ \dfrac{e^{-st}}{s^2} \right]_{t=0}^{t=b} \quad s>0\\ \amp = -\lim_{b \rightarrow \infty} \left[ \cancelto{0}{\dfrac{e^{-sb}}{s^2}} -\dfrac{1}{s^2} \right] \quad s>0\\ \amp = \dfrac{1}{s^2} \quad \text{for } s>0. \end{align*}
Use the definition of the Laplace transform to compute the transforms in exercises 1 and 2.