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I am having trouble computing the following Laplace transform: $\frac {f(x)}{x}$.
From Wikipedia it should be equivalent to this: $\int_s^\infty F(\sigma) \,d\sigma$ .
What I've done so far is substitute in $-\int_s^\infty e^{-sx} ds$ for $\frac {e^{-sx}}{x}$.
After simplification I end up with this: $-\int_s^\infty F(s)\, ds$ where $F(s)$ is the Laplace transform of $f(x)$
I am unsure of how to simplify this further to get the result of Wikipedia.
Let $g(t) = \frac{f(t)}t$.
We compute:
$$\begin{align*} \int_s^\infty F(\sigma) \,d\sigma &= \int_s^\infty \left( \int_0^\infty e^{-\sigma t}f(t) \,dt\right) \,d\sigma \\ &= \int_0^\infty \left( \int_s^\infty e^{-\sigma t}f(t) \,d\sigma\right) \,dt \\ &= \int_0^\infty f(t) \left(\int_s^\infty e^{-\sigma t}\, d\sigma\right) \,dt \\ &= \int_0^\infty f(t) \left[-\frac{1}t e^{-\sigma t}\right]_{\sigma=s}^{\sigma=\infty} \,dt \\ &= \int_0^\infty f(t) \left[0 - \left(-\frac{e^{-st}}{t}\right)\right] \,dt \\ &= \int_0^\infty f(t) \frac{e^{-st}}{t} \,dt \\ &= (\mathcal Lg)(s). \end{align*}$$
Of course there are various questions about changing the order of integration (q.v. Fubini's theorem) and convergence of the integrals, but this is essentially the computation.
EDIT: Expanded calculation of the integral.
