Laplace transform of a function divided by t 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?

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 $N \to \infty$.
