The Laplace Transform: Further Properties

Let $\mathcal{L}\{F(t)\}=f(s)$ i.e. $f(s)=\int_0^\infty e^{-st}F(t)dt$. Then
\begin{equation}
\label{eq:laplace11}
e^{-bs}f(s)=\int_0^\infty e^{-s(t+b)}F(t)dt
\end{equation}
where $b$ is a positive constant. Using the substitution $\tau=t+b$ \eqref{eq:laplace11} can be written as
$$e^{-bs}f(s)=\int_b^\infty e^{-s\tau}F(\tau-b)d\tau$$
Define a function $F_b(\tau)$ by
\begin{equation}
F_b(\tau)=\left\{\begin{aligned}
0,\ &0<\tau<b\\
F(\tau-b),\ &\tau>b
\end{aligned}\right.\label{eq:laplace12}
\end{equation}
Then
$$e^{-bs}f(s)=\int_0^\infty e^{-s\tau}F_b(tau)d\tau$$

Theorem. If $\mathcal{L}\{F(t)\}=f(s)$ then for any positive constant $b$,
$$e^{-bs}f(s)=\mathcal{L}\{F_b(t)\}$$
where $F_b(t)$ is defined by \eqref{eq:laplace12}.

Let
$$S_b(t)=\left\{\begin{aligned}
0,\ &0<t<b\\
1,\ &t>b
\end{aligned}\right.$$
for any positive constant $b$. Define $S_0(t)=1$ for $t>0$. Then $S_b(t)$ is called a \emph{unit step function}. Using this unit step function, $F_b(t)$ can be written as
$$F_b(t)=S_b(t)F(t-b)$$

Example.
\begin{align*}
F_b(t)&=S_b(t)\sin k(t-b)\\
&=\mathcal{L}^{-1}\left\{\frac{ke^{-bs}}{s^2+k^2}\right\}
\end{align*}

\begin{align*}
\mathcal{L}\{F(at)\}&=\int_0^\infty e^{-st}F(at)dt\ (a>0)\\
&=\frac{1}{a}\int_0^\infty e^{-\frac{s}{a}\tau}F(\tau)d\tau\ (\tau=at)\\
&=\frac{1}{a}f\left(\frac{s}{a}\right)
\end{align*}

Theorem. If $\mathcal{L}\{F(t)\}=f(s)$ when $s>\alpha$, then
$$\mathcal{L}\{F(at)\}=\frac{1}{a}f\left(\frac{s}{a}\right)\ (s>a\alpha,\ a>0)$$

Example. Given that
$$\frac{s}{s^2+1}=\mathcal{L}\{\cos t\}$$
\begin{align*}
\frac{s}{s^2+k^2}&=\frac{1}{k^2}\frac{s}{\left(\frac{s}{k}\right)^2+1}\\
&=\frac{1}{k}\frac{\frac{s}{k}}{\left(\frac{s}{k}\right)^2+1}\\
&=\mathcal{L}\{\cos kt\}
\end{align*}

With the translation of transform (equation (1) in here) and the above theorem the effect of a general linear substitution for $s$ can be shown in the following formula.
\begin{align*}
f(as-b)&=f\left[a\left(s-\frac{b}{a}\right)\right]\\
&=\mathcal{L}\left\{\frac{1}{a}e^{\frac{b}{a}t}F\left(\frac{t}{a}\right)\right\}\ (a>0)
\end{align*}

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