Unilateral Laplace Transform's Differentiation Property I've read in numerous places that the unilateral laplace transform is extermely useful in solving differential equations with initial conditions based on the differentiation property of the unilateral transform:
$\mathscr{L}{f′(t)}=sF(s)−f(0_−)$
What i don't understand is why this is possible to do in Laplace but not in the Fourier Transform?
Is this related to the dacaying exponential added by the Laplace transform?
I would be grateful for an in-depth answer involving intuition.
You need to look at the derivation of that property. Integration by parts gives
$$\begin{align}\mathcal{L}\{f'(t)\}&amp;=\int_{0^-}^{\infty}f'(t)e^{-st}dt\\&amp;=f(t)e^{-st}\Big|_{0^-}^{\infty}+s\underbrace{\int_{0^-}^{\infty}f(t)e^{-st}dt}_{F(s)}\\&amp;=\lim_{t\to\infty}f(t)e^{-st}-f(0^-)+sF(s)\tag{1}\end{align}$$
The first term in $(1)$ is only guaranteed to vanish for $t\to\infty$ if $\text{Re}\{s\}&gt;\alpha$ for some value $\alpha$. So even if you define a unilateral Fourier transform, that term may not vanish in general.
However, if $\lim_{t\to\infty}f(t)=0$ holds, we could use the same property with a unilateral Fourier transform. It's just much more common to use the well-established (unilateral) Laplace transform in cases where non-zero initial conditions need to be taken into account.