Laplace transform of $f(t)=\frac{\sin(t)}t$ for $t>0$, $f(t)=0$ for $t \le 0$ Determine Laplace transform of $$f(t)=\begin{cases} \frac{\sin(t)}{t} \qquad t>0 \\ \\ 0 \qquad \ \ \ \ \ \ \ t  \le 0 \end{cases}$$


I know Laplace transform of $\frac{\sin(t)}{t}$ is $F(s)=\arctan(\frac{1}{s})$ (from another topic on this forum)


The Laplace transform of $$g(t)=\begin{cases} f(t-a) \qquad t \ge a \\ 0 \qquad \ \ \ \ \ \ \ \ \ \ \ \  t<a  \end{cases}$$ is $$G(s)=F(s) \ e^{-sa}$$
How can I determine the Laplace transform considering the different conditions on $t$ ?

Thanks!
 A: From Wikipedia we have:
The Laplace transform of a function $f(t)$, defined for all real numbers $t \ge 0$,
is the function $F(s)$, which is a unilateral transform [ emphasis by us ]
defined by
$$
F(s) = \int_0^\infty f(t)e^{-st}\,dt
$$
The unilateral transform ($t \ge 0$) of the sine is (table @ Wikipedia):
$$
\sin(\omega t)\cdot u(t) \; \to \; \frac{\omega}{s^2+\omega^2}
\quad \Longrightarrow \quad
\sin(t)\cdot u(t) \; \to \; \frac{1}{1+s^2}
$$
Where the Heaviside step function $u(t)$ is defined by:
$$
u(t) =
\begin{cases} 0 & \mbox{for} & t < 0\\ 1 & \mbox{for} & t \ge 0\end{cases}
$$
The unilateral transform ($t \ge 0$) of a function $f(t)/t$ is (table @ Wikipedia):
$$
\frac{1}{t}f(t) \; \to \; \int_s^\infty F(\sigma)\,d\sigma
$$
Combining these:
$$
f(t) = \frac{\sin(t)}{t}\cdot u(t) \; \to \; \int_s^\infty \frac{d\sigma}{1+\sigma^2} =
\left[\arctan(\sigma)\right]_s^\infty = \frac{\pi}{2}-\arctan(s) = F(s)
$$
Forget about the rule $\;G(s)=F(s) \ e^{-sa}\;$, because for $\,a=0\,$ it adds nothing new here : $G(s)=F(s) \ e^{-s0} = F(s)$ .
