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.


1 Answer 1


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.

  • $\begingroup$ I see it now, I got confused by subbing in the integral with 'x', I should have chosen some other arbitrary variable instead. But I am still slightly confused as to why a negative integral is not subbed in since $\int_s^\infty e^{-\bar st}f(t) \,d\bar s$ is $\frac{-e^{-\bar st} f(t)}{x} $ $\endgroup$ Mar 29, 2015 at 22:54
  • $\begingroup$ I'm not sure exactly what you're asking, but you're definitely doing the integral wrong, because you can get negative results from putting non-negative functions into what you've computed. $\endgroup$
    – kahen
    Mar 30, 2015 at 1:01

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