Show for $\mathcal{L}$, the Laplace transform, that $$\mathcal{L}\left\{\frac{1}{t}f(t)\right\} = \int_{s}^{\infty}F(u)du.$$
I know that $\mathcal{L}\left\{ t^n f(t) \right \} = (-1)^n \frac{d^n}{ds^n} F(s)$ and this is sort of suggestive for $n=-1$, but I can't seem to show the result. Some help would be appreciated.
\begin{align}\int^{\infty}_{s}F(u)\,du&=\int^{\infty}_{s}\int^{\infty}_{0}f(\tau)e^{-u\tau}\,d\tau\,du\\&=\int^{\infty}_{0}f(\tau)\int^{\infty}_{s}e^{-u\tau}\,du\,d\tau\\&= \int^{\infty}_{0}f(\tau)\frac{1}{\tau}\int^{\infty}_{s\tau}e^{-v}\,dv\,d\tau\\&=\int^{\infty}_{0}f(\tau)\frac{1}{\tau}e^{-s\tau}\,d\tau\\&=\mathcal{L}\{\frac{f(\tau)}{\tau}\}\end{align}