# Laplace Transforms and the Differentiation Property

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?

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

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).