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It's trivial that the Laplace Transform of a positive function is a positive function on $s$ domain. What about the inverse thought? What can we say about the positiveness of the inverse Laplace Transform of a positive function of form $s^{-\alpha}$?

The answer is trivial when $\alpha>0$, because $\mathcal{L}\{t^{\alpha-1}\}=\Gamma(\alpha) s^{-\alpha}$ when $\alpha>0$.

But and if $\alpha<0$? The exact function is not known by any people, but could we say something about positiveness?

Thank you so much.

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    $\begingroup$ I must take issue with the claim that the exact function is not known by any people. This is simply not true. The standard formula holds for all values of the parameter. The reason for the restriction in the values of the latter in many tables is that they only consider the transforms of functions. But the expression $s^\alpha$ defines a distribution on the posiitive half-line for any real $\alpha$ and the Laplace tranformation of distributions has been studied in detail, starting with Schwartz in the 50's. $\endgroup$
    – bathalf15320
    Commented Feb 18, 2021 at 14:35
  • $\begingroup$ @bathalf15320 I thank you so much, I did not know. Thank you! $\endgroup$
    – Quiet_waters
    Commented Feb 18, 2021 at 15:01

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The inverse Laplace transform of $s$ is the derivative $\delta'(x)$ of the Dirac delta function (a distribution). This is not a positive "function", as you can check by evaluating the integral with a test function $f(x)$, $$\int_{-\infty}^\infty f(x)\delta'(x)\,dx=-\int_{-\infty}^\infty f'(x)\delta(x)\,dx=-\lim_{x\rightarrow 0}f'(x).$$ So even if $f(x)\geq 0$ for all $x$, this integral will be negative if $f'(0)>0$.

The same applies to higher integer powers of $s$, the inverse Laplace transform of $s^p$ is the $p$-th derivative of a delta function, which is not a positive function, in the sense that the integral with a positive test function can be negative, $$\int_{-\infty}^\infty f(x)\delta^{(p)}(x)\,dx=(-1)^p\int_{-\infty}^\infty f^{(p)}(x)\delta(x)\,dx=-\lim_{x\rightarrow 0}f^{(p)}(x).$$

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  • $\begingroup$ Thank you so much! I've noticed that you've used the bilateral Laplace Transform. Does it matter? Thank you $\endgroup$
    – Quiet_waters
    Commented Feb 15, 2021 at 13:32
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    $\begingroup$ I defined the Laplace transform as $\int_{0^-}^\infty f(t)e^{-st}\,dt$, as in these notes $\endgroup$
    – Carlo Beenakker
    Commented Feb 15, 2021 at 14:01
  • $\begingroup$ Thank you so much, I got confused with using the test function, now I understand, thank you $\endgroup$
    – Quiet_waters
    Commented Feb 15, 2021 at 14:28

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