Laplace transform of a function divided by t

Question

Using the formula $$\mathcal{L}\left\{\frac{f(t)}{t}\right\}=\int_s^\infty F(u)~du$$ I'm trying to determine the transform with $f(t)=1-e^{-t}$.

The formula gives me $$\mathcal{L}\left\{\frac{1-e^{-t}}{t}\right\}=\int_s^\infty \left(\frac{1}{u}-\frac{1}{u+1}\right)~du$$ which doesn't seem to converge, yet Mathematica insists the answer is $$\ln\left(1+\frac{1}{s}\right)$$ Am I using the formula incorrectly, or did I some silly mistake in my computation?

Answer 1

This certainly converges. Write the integral as

$$\begin{align}\lim_{N \to \infty} \int_s^N du \left( \frac1{u}-\frac1{u+1} \right ) &= \lim_{N \to \infty} \left [\log{N}-\log{s} - \log{(N+1)}+\log{(s+1)}\right ]\\ &= \log{\left (1+\frac1{s}\right)} - \lim_{N\to\infty} \log{\left (1+\frac1{N}\right)}\end{align}$$

This approaches the correct limit as $N \to \infty$.

Answer 2

Why do you say "doesn't seem to converge"? Note that $$\frac{1}{u} - \frac{1}{u+1} = \frac{1}{u^2+u} \leq \frac{1}{u^2}$$ so the integral must converge.

Moreover, $$\frac{d}{du} \log \left( 1 + \frac{1}{u}\right) = - \frac{1}{u^2+u} $$ so the software is right.