For any given (suitable) function \(\f\), the Laplace transform of \(\f\) is the function \(\F\) defined by
\[
\F(s) = \int_0^\infty \e^{-st}\f(t)\d t
\quad \quad \, (s>0)
\,.
\]
Show that the Laplace transform of \(\e^{-bt}\f(t)\), where \(b>0\),
is \(\F(s+b)\).
Show that
the Laplace transform of \(\f(at)\), where \(a>0\), is \(a^{-1}\F(\frac s a)\,\).
Show that the Laplace
transform of \(\f'(t)\) is \(s\F(s) -\f(0)\,\).
In the case \(\f(t)=\sin t\), show that \(\F(s)= \dfrac 1 {s^2+1}\,\).
Using only these four results, find the Laplace transform of
\(\e^{-pt}\cos{qt}\,\), where \(p>0\) and \(q>0\).
Solution:
\begin{align*}
\mathcal{L}\{e^{-bt}f(t)\}(s) &= \int_0^{\infty}e^{-st}\{ e^{-bt}f(t) \} \d t \\
&= \int_0^{\infty} e^{-(s+b)t}f(t) \d t \\
&= F(s+b)
\end{align*}
\begin{align*}
\mathcal{L}\{f(at)\}(s) &= \int_0^{\infty} e^{-st}f(at) \d t \\
&= \int_{u=0}^{\infty}e^{-s \frac{u}{a}} f\left(a \tfrac{u}{a}\right)\frac{1}{a} \d u \\
&= \int_0^{\infty}e^{-su/a}f(u) a^{-1} \d u \\
&= a^{-1} \int_0^{\infty} e^{-(s/a)u}f(u) \d u \\
&= a^{-1} F\left (\frac{s}{a} \right)
\end{align*}