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How can we find the inverse Laplace transform of:

$[x]$ (floor function) ? My question isn't LLaplace transform of floor function i asked the "inverse" laplace transform of floor function $$\mathcal{L^{-1}}\left(\text{floor}({s})\right)$$

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  • $\begingroup$ What is $\lfloor 1+i\rfloor$? $\endgroup$ – Rodrigo de Azevedo May 31 '16 at 13:33
  • $\begingroup$ Is the Laplace transform of any $f$ ever constant on an interval? Seems not at a first intuitive glance, for me. $\endgroup$ – jdods May 31 '16 at 13:43
  • $\begingroup$ If $F(s)$ the Laplace transform of $f(x)$ exists on some line $Re(s) = \sigma$, then it can be anything, since it is the Fourier transform of $e^{-\sigma x} f(x)$. but if it exists on some strip, then it has to be analytic on the interior of that strip. Of course $\lfloor x \rfloor$ isn't analytic (on the real line), while it isn't even defined for $x$ complex. hence, did you mean the Laplace transform ? or the Mellin transform ? or the (inverse) Fourier transform ?.. ( @jdods ) $\endgroup$ – reuns May 31 '16 at 13:49
  • $\begingroup$ @user1952009, I don't know what op means, but I'm wondering about a real function $f$. Is there a real $f$ that has the Heaviside function or a square bump as its Laplace transform, e.g. like $(L(f))(s)=c$ on an interval and zero elsewhere? If so one could use that to construct the inverse Laplace transform of the floor function possibly. $\endgroup$ – jdods May 31 '16 at 14:23
  • $\begingroup$ @jdods ? I told you, the Laplace transform for a fixed $Re(s)$ is the Fourier transform : in a first time $f(x)$ has to be integrable, but then the Fourier transform is extended to so many things and nearly every function that you know which is increasing at most polynomially has a Fourier transform and hence can be in some sense the (bilateral) Laplace transform of some function on a vertical line of the complex plane. $\endgroup$ – reuns May 31 '16 at 14:36
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$\because\mathcal{L}^{-1}\{c\}=c\delta(t)$

$\therefore\mathcal{L}^{-1}\{\lfloor s\rfloor\}=\lfloor t\rfloor\delta(t)$

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