Solution. (a) \(f(t)=1\):
\[
F(s)=\int_0^\infty\ee^{-st}(1)\,\dd t=\Bigl[\frac{\ee^{-st}}{-s}\Bigr]_0^\infty=\frac{-1}{s}\bigl[\ee^{-\infty}-\ee^0\bigr]=\frac1s,
\qquad L\{1\}=\frac1s.
\]
(b) \(f(t)=\ee^{at}\):
\[
F(s)=\int_0^\infty\ee^{at}\cdot\ee^{-st}\,\dd t=\Bigl[\frac{\ee^{(a-s)t}}{a-s}\Bigr]_0^\infty=\frac1{a-s}\bigl[\ee^{(a-s)\infty}-\ee^0\bigr].
\]
If \(s>a\),
\[
F(s)=\frac1{a-s}[0-1]=\frac1{s-a}.
\]
The Laplace transform is linear:
\[
L\bigl[af(t)+bg(t)\bigr]=aLf(t)+bLg(t).
\]
(c) \(f(t)=\sinh at\):
\[
L(\sinh at)=L\Bigl\{\frac{\ee^{at}-\ee^{-at}}2\Bigr\}=\frac12\Bigl[\frac1{s-a}-\frac1{s+a}\Bigr]
=\frac12\Bigl[\frac{s+a-s+a}{s^2-a^2}\Bigr]=\frac{a}{s^2-a^2}.
\]
(d) \(f(t)=\cosh at\):
\[
L\{\cosh at\}=L\Bigl\{\frac{\ee^{at}+\ee^{-at}}2\Bigr\}=\frac12\Bigl[\frac1{s-a}+\frac1{s+a}\Bigr]
=\frac12\Bigl[\frac{s+a+s-a}{s^2-a^2}\Bigr]=\frac{s}{s^2-a^2}.
\]
(e) \(f(t)=\sin\omega t\):
\begin{align*}
L\{\sin\omega t\}&=L\Bigl\{\frac{\ee^{\ii\omega t}-\ee^{-\ii\omega t}}{2\ii}\Bigr\}
=\frac1{2\ii}\bigl[L(\ee^{\ii\omega t})-L(\ee^{-\ii\omega t})\bigr]=\frac1{2\ii}\Bigl(\frac1{s-\ii\omega}-\frac1{s+\ii\omega}\Bigr)\\
&=\frac1{2\ii}\Bigl(\frac{s+\ii\omega-s+\ii\omega}{s^2+\omega^2}\Bigr)=\frac{2\ii\omega}{2\ii(s^2+\omega^2)},
\qquad L\{\sin\omega t\}=\frac{\omega}{s^2+\omega^2}.
\end{align*}
