Does this Laplace transform exist?

Question

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.

Answer 1

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

Answer 2

$$ f(t)=\frac{1}{1+t} $$ $$ f(t)(t+1)=1 $$ $$ \downarrow \mathcal{L} $$ $$ F(s)-\frac{d}{ds}F(s)=\frac{1}{s} $$ $$ \frac{d}{ds}F(s)-F(s)=\frac{-1}{s} $$ we have first order differential equation: $$ F(s)=e^{-\int(-1)ds}\left[ \int \frac{-1}{s}e^{\int (-1)ds} ds+C\right] $$ $$ F(s)=e^s \left[ \int \frac{-1}{s} e^{-s} ds +C \right] $$