# 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.

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I don't know if I quite understand the last part of your question, but I think the reason the second question is worded the way it is is to avoid simply asking what the transform is (which would give part of the answer to the first question, although it's still pretty obvious that it should be "yes"). – Calvin McPhail-Snyder Dec 18 '11 at 5:56

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