On Wikipedia's article on Laplace transform the following property is listed: $$\mathcal{L}\left\{f(t)/t\right\} = \int_s^\infty F(\sigma)d\sigma. $$ A question about this property has been asked before but does not address my concerns.

Since $s$ is complex, the integral and in particular its limits don't make too much sense. Is the idea to treat $s$ and $\sigma$ as real parameters, and then extend to $s$ complex by analytic continuation?

  • $\begingroup$ $F(s) = L[f(t)/t](s) = \int_0^\infty \frac{f(t)}{t}e^{-st}dt$. Assuming it converges without problems (for example $f(t) = 0$ for $t < 1$) then $F'(s) = -\int_0^\infty f(t) e^{-st}dt = L[f(t)](s)$. Now differentiate the integral you wrote, and show the integral is equal to the LHS for one value of $s$, for example at the limit $s = +\infty$ $\endgroup$
    – reuns
    Commented Sep 30, 2016 at 14:56
  • $\begingroup$ And assuming $F(s) \to 0$ fast enough as $Re(s) \to \infty$, you can replace $\int_s^\infty F(\sigma)d\sigma$ by $\int_0^\infty F(s+\sigma)d\sigma = \int_s^{s+\infty} F(\sigma) d\sigma$ (a contour integral in the complex plane) $\endgroup$
    – reuns
    Commented Sep 30, 2016 at 14:58
  • $\begingroup$ And you have an example here with a closed contour (instead of a path $s \to \infty$) $\endgroup$
    – reuns
    Commented Sep 30, 2016 at 15:07
  • $\begingroup$ @user1952009 Thank you for your comments! I'm afraid I don't see how they address my question, namely what we mean by the integral I wrote when $s$ and $\sigma$ are complex. Also, I don't quite understand how the linked example (integral of $z^{-1}$ around the unit circle) is relevant. $\endgroup$ Commented Sep 30, 2016 at 17:32
  • $\begingroup$ Everything is explained in the link and in complex analysis courses. You have to understand that $\int_\gamma h(z)dz = \int_a^b h(\gamma(t)) \gamma'(t)dt$ (change of variable $z= \gamma(t)$, $dz = \gamma'(t)dt$) so that $\int_s^{+\infty} h(z) dz = \int_\gamma h(z)dz = \int_a^b h(\gamma(t))\gamma'(t)dt$ where $\gamma$ is any $C^1$ curve such that $\gamma(a) = s, \gamma(b) = +\infty$, and the integral doesn't depend on the chosen curve $\gamma(a) \to \gamma(b)$ whenever $h(z)$ is holomorphic (analytic) $\endgroup$
    – reuns
    Commented Sep 30, 2016 at 19:00


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