Answer
The Laplace transforms of the functions f are:
(a)
$$
F(s)=\frac{1}{s}
$$
and the domain of $F$ is $ {s: s\gt 0} $.
(b)
$$
F(s)=\frac{1}{s-1}
$$
and the domain of $F$ is $ {s: s\gt 1} $.
(c)
$$
F(s)=\frac{1}{s^{2}}
$$
and the domain of $F$ is $ {s: s\gt 0} $
Work Step by Step
To find the Laplace transforms of the following functions:
(a) $f(t)=1$
the Laplace transform of $f$ is the function $F $ defend by
$$
F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t
$$
so
$$
\begin{split}
F(s)&=\int_{0}^{\infty} f(t) e^{-s t} d t \\
&=\int_{0}^{\infty} e^{-s t} d t \\
&=\lim _{n \rightarrow \infty}\left[-\frac{e^{-s t}}{s}\right]_{0}^{n} \\
&=\lim _{n \rightarrow \infty}\left(\frac{e^{-s n}}{-s}+\frac{1}{s}\right) \quad\left[\begin{array}{c}{ \text {only if } s \gt 0 }\end{array}\right] \\
&=0+\frac{1}{s}\\
&=\frac{1}{s}
\end{split}
$$
Therefore $F(s)=\frac{1}{s} $ with domain $ {s: s\gt 0} $
(b) $f(t)=e^{t}$
the Laplace transform of $f$ is the function $F $ defend by
$$
F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t
$$
so
$$
\begin{split}
F(s)&=\int_{0}^{\infty} f(t) e^{-s t} d t \\
&=\int_{0}^{\infty} e^{ t} e^{-s t} d t \\
&=\int_{0}^{\infty} e^{-(s-1) t} d t \\
&=\lim _{n \rightarrow \infty}\left[\frac{e^{-(s-1)t}}{1-s}\right]_{0}^{n} \\
&=\lim _{n \rightarrow \infty}\left(\frac{e^{-(s-1)n}}{1-s}-\frac{1}{1-s}\right) \quad\left[\begin{array}{c}{ \text {only if } (s-1) \gt 0 }\end{array}\right] \\
&=0+\frac{1}{s-1}\\
&=\frac{1}{s-1}
\end{split}
$$
Therefore $F(s)=\frac{1}{s-1} $ with domain $ {s: s\gt 1} $
(c) $f(t)=t$
the Laplace transform of $f$ is the function $F $ defend by
$$
F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t
$$
so
$$
\begin{split}
F(s)&=\int_{0}^{\infty} f(t) e^{-s t} d t \\
&=\int_{0}^{\infty} t e^{-s t} d t \\
&=\lim _{n \rightarrow \infty} \int_{0}^{n} t e^{-s t} d t \\
& \quad\quad\quad\left[\text { use integration by parts with }
\right] \\
&\quad\quad\quad \left[\begin{array}{c}{u=t, \quad\quad dv= e^{-st}dt } \\ {d u= dt, \quad\quad v= \frac{1}{-s}e^{-st} }\end{array}\right] , \text { then }\\
&=\lim _{n \rightarrow \infty}\left(\frac{-te^{-st}}{s}]_0^{n}+\frac{1}{s}\int_{0}^{n} e^{-s t} d t \right) \\
&=\lim _{n \rightarrow \infty}\left[\frac{-te^{-st}}{s}-\frac{1}{s^{2}} e^{-s t} \right] _{0}^{n} \\
&=\lim _{n \rightarrow \infty}\left[\frac{-ne^{-sn}}{s}+\frac{1}{s^{2}} e^{-s n} +0+\frac{1}{s^{2}}\right] \quad\left[\begin{array}{c}{ \text {only if } s \gt 0 }\end{array}\right] \\ \\
&=\frac{1}{s^{2}}
\end{split}
$$
Therefore $F(s)=\frac{1}{s^{2}} $ with domain $ {s: s\gt 0} $
