Does this Laplace transform exist? I had a final in differential equations with the first question being:
"1. Does the Laplace transform of $\displaystyle \frac{1}{(1+t)}$ exist? Why or why not?"
and number 2 was
"2. If number one was true, then what is this transform?"
At first I thought it was true because the definition of the Laplace says that it must be of exponential order (it is I believe) and must be piece wise continuous from $[0,\infty)$. That equation satisfies both of those properties, but It's not defined or has complex components I'm reading now? Can somebody set me straight on this problem.
 A: It does and follows from the limit
$$
\lim_{t\to\infty} \frac{f(t)}{e^{\alpha t}} = \frac{1}{(1+t)e^{\alpha t}} = 0 \qquad \forall \alpha\geq 0
$$ Roughly speaking, this is only to check that it is slower than some exponential function such that the Laplace transform integral does not blow up. 
Computing it only requires a variable change but some terminology about Exponential Integral, use $1+t=x$
$$
F(s) = \int_0^\infty{\frac{e^{-st}}{1+t}}dt = \int_1^\infty{\frac{e^{-sx}e^s}{x}}dx = e^s\int_1^\infty{\frac{e^{-sx}}{x}}dx = -e^s\operatorname{Ei}(-s) = e^s\operatorname{E_1}(s)
$$
The last two equalities are something of a naming convention and you can find more details about it online, e.g. Mathworld page
A: $$ f(t)=\frac{1}{1+t} $$
$$ f(t)(t+1)=1 $$
$$ \downarrow \mathcal{L} $$
$$ F(s)-\frac{d}{ds}F(s)=\frac{1}{s} $$
$$ \frac{d}{ds}F(s)-F(s)=\frac{-1}{s}  $$
we have first order differential equation:
$$ F(s)=e^{-\int(-1)ds}\left[ \int \frac{-1}{s}e^{\int (-1)ds} ds+C\right] $$
$$ F(s)=e^s \left[ \int \frac{-1}{s} e^{-s} ds +C \right]  $$
