3.37. Find the image of the following (functions: a) $f(t)=1$; b) $f(t)=t$; c) $f(t)=e^{-k t}$.
Solution.
a) $\bar{f}(p)=\int_{0}^{-1} e^{p t} d t=\lim _{\alpha \rightarrow+\infty} \int_{0}^{\alpha} e^{-p t} d t=$ $=-\left.\frac{1}{p} \lim _{\alpha \rightarrow+\infty} e^{-p t}\right|_{0} ^{\alpha}=-\frac{1}{p} \lim _{\alpha \rightarrow+\infty}\left(e^{-p \alpha}-1\right)=\frac{1}{p}$

(since $\lim _{\alpha \rightarrow+\infty} e^{-p \alpha}=0, \operatorname{Re} p>0$ ).

b) We find the image of $f(t)$ by integration by parts:

$$
\begin{gathered}
u=t \Rightarrow d u=d t, \quad d v=e^{-p t} d t \Rightarrow v=-\frac{1}{p} e^{-p t} \\
\begin{aligned}
& \bar{f}(p)=\int_{0}^{+\infty} e^{-p t} t d t=-\left.\frac{1}{p} \lim _{\alpha \rightarrow+\infty}\left(t e^{-p t}\right)\right|_{0} ^{\alpha}+\frac{1}{p} \lim _{\alpha \rightarrow+\infty} \int_{0}^{\alpha} e^{-p t} d t= \\
&=0+\frac{1}{p} \cdot \frac{1}{p}=\frac{1}{p^{2}}
\end{aligned} \\
\text { c) } \bar{f}(p)=\int_{0}^{+\infty} e^{-p t} e^{-k t} d t=\lim _{\alpha \rightarrow+\infty} \int_{0}^{\alpha} e^{-(p+k) t} d t=\frac{1}{p+k}, \quad(k \neq-p)
\end{gathered}
$$