My problem is to find a function $g(s)$, solution of the equation: $$ \frac{1}{t} \mathcal{L}^{-1} \left\{ f(s)\right\}(t)=\mathcal{L}^{-1} \{ g(s)\}(t)$$

I know the general property: $$ \mathcal{L}^{-1} \{ g(s) \}(t)=\frac{(-1)^n}{t^n} \mathcal{L}^{-1}\left\{\frac{d^ng(s)}{ds^n}\right\}(t)$$ from which, for $n=1$ we get: $$ \mathcal{L}^{-1} \{ g(s) \}(t)=-\frac{1}{t} \mathcal{L}^{-1}\{g'(s)\}(t)$$ So, if I take, for example: $$g(s)=\int f(s) ds$$ from the previous property I get: $$\mathcal{L}^{-1}\{ g(s)\}(t)=-\frac{1}{t} \mathcal{L}^{-1}\{ g'(s)\}(t)=-\frac{1}{t} \mathcal{L}^{-1} \{ f(s)\}(t)$$ so: $$\frac{1}{t} \mathcal{L}^{-1} \{ f(s)\}(t)=-\mathcal{L}^{-1} \{ g(s)\}(t)= \mathcal{L}^{-1} \left\{- \int f(s)ds\right\}(t)$$ My problem is that the function $g(s)$ is not uniquely determined because indefinite integral is defined up to a constant; for all type of functions in the form, for example: $$g_1(s)=g(s)+C, \; C \; \text{constant}$$ the result is the same.

The are known results or conditions on $f$ to uniquely determine $g(s)$?


  • 1
    $\begingroup$ Since $g$ is Laplace transform of some function $g(s) \to 0$ as $s \to \infty $. This determines the value of the constant. $\endgroup$ – Kavi Rama Murthy Jun 20 '18 at 11:51
  • $\begingroup$ A necessary condition for a function to admit Inverse Laplace Transform is to be of exponential type. Your limit is due to this condition? $\endgroup$ – Papemax89 Jun 20 '18 at 13:52

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