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What needs to be assumed for a function $f(t)$ to be used with a Laplace transform? The reason I'm asking is because of this property. If, $$ \mathscr{L}(f(t)) = F(s) = \int_{0} ^{\infty} f(t)e^{-st}dx $$ then, $$ \mathscr{L}(f'(t)) =sF(s)-f(0^{+}). $$ When I was trying to prove the property using integration by parts I got this, $$ \mathscr{L}(f'(t)) = \int_{0} ^{\infty} f'(t)e^{-st}dx = sF(s) + \lim_{t \to \infty} {e^{-st}f(t) } - f(0^{+}). $$ If $f(t) = e^{t^{2}}$ the limit doesn't approach zero as t approaches infinity. If $f(t)=1/t$ then $f(0^{+})$ will approach infinity. What do I need to assume about $f(t)$ for this property to work?

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    $\begingroup$ You need to have convergence of the integral, that is, that $F(s)$ exists. If it exists, then we have that $\lim_{t \to \infty} {e^{-st}f(t) } =0$. $\endgroup$
    – KBS
    Commented Apr 24, 2022 at 22:58

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We only formally define the Laplace transform for exponential and subexponential functions such that the integral will converge (as user KBS notes): $$\mathcal{L}[f(t)](s)=\int\limits^{\infty}_0{e^{-st}f(t)dt}$$

Typically, for superexponential functions (e.g. $t!$ or $e^{t^2}$) we just do not define the Laplace transform, since the integral will diverge. The convergence of the integral implies that the limit in question is zero (divergence test for improper integrals).

Some generalizations of the Laplace transform have been studied which extend to superexponential functions (here). These hold similar properties to the Laplace transform.

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