Contradiction between Fourier and Laplace transforms? Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function that has both Fourier and Laplace transforms. Also let $f(t)=0$ for all $t<0$. The Fourier transform of $f$ is
$$\mathcal{F}(\omega)=\mathcal{F}[f](\omega)=\int_0^\infty e^{-2\pi i\omega t}f(t)dt,\;for \;\omega\in\mathbb{R}$$
The Laplace transform of $f$ on the other hand is
$$\mathcal{L}(s)=\mathcal{L}[f](s)=\int_0^{\infty}e^{-st}f(t)dt,\;for \;s\in\mathbb{C}$$
If we set $s=2\pi i \omega$. It seems like $\mathcal{F}(\omega)=\mathcal{L}(2\pi i\omega)$ must hold, right? Well, these transforms also have the properties
$$\mathcal{F}\left[f'\right](\omega)=2\pi i\omega\mathcal{F}(\omega)$$
$$\mathcal{L}\left[f'\right](s)=s\mathcal{L}(s)-f(0)$$ 
And once again setting $s=2\pi i\omega$, we have
$$\mathcal{L}\left[f'\right](2\pi i \omega)=2\pi i\omega \mathcal{F}(\omega)-f(0)\neq 2\pi i\omega \mathcal{F}(\omega)$$
Unless $f(0)=0$. Furthermore, 
$$\mathcal{F}\left[f^{(n)}\right](\omega)=(2\pi i\omega)^n\mathcal{F}(\omega)$$
$$\mathcal{L}\left[f^{(n)}\right](s)=s^n\mathcal{L}(s)-\sum_{k=1}^{n} s^{n-k}f^{(k-1)}(0)$$
Which in general (unless $f^{(k)}(0)=0$ for all $k<n$) implies
$$\mathcal{L}\left[f^{(n)}\right](2\pi i \omega)\neq \mathcal{F}\left[f^{(n)}\right](\omega)$$ 
And $f^{(n)}(t)$, if $f$ is derivable to the $n^{th}$ order, is nothing but a function, so what the hell?
This really puzzles me, so any answer will be a great contribution to my mental peace, thanks.
 A: Mr. Moonshine, there is no contradiction at all, you just messed it up.
Let me explain. As Tryss and John Martin pointed out, the unilateral Fourier transform doesn't indeed has the property
$$\mathcal{F}\left[f^{(n)}\right](\omega)=(2\pi i \omega)^n\mathcal{F}(\omega)$$ but rather
$$\mathcal{F}\left[f^{(n)}\right](\omega)=(2\pi i \omega)^n\mathcal{F}(\omega)-\sum_{k=1}^{n}(2\pi i\omega)^{n-k}f^{(k-1)}(0)$$
Something that can be checked integrating by parts and using some induction. This matches the property of the unilateral Laplace transform when $s=2\pi i\omega$. But, as JeanMarie commented, this transform hardly has any relevance because it is never used. So, to describe the situation on the question, the best is to follow the advice of user1952009 and do the Fourier and Laplace bilateral transforms of $g(t)=f(t)\theta(t)$, where $\theta(t)$ is the Heaviside unitary step. Let's remember that 
$$g'(t)=f'(t)\theta(t)+f(t)\delta(t)$$
Where $\delta(t)$ is the Dirac delta. If we take the Laplace transform of $g$,
$$
\mathcal{L}\left[g'\right](s)=\mathcal{L}\left[f'\theta\right](s)+f(0)
$$
$$
=\int_{-\infty}^{\infty}f'(t)\theta(t)e^{-st}dt+f(0)=I(s)+f(0)
$$
And then do the integral by parts, choosing $du=f'(t)dt$, $u=f(t)$, $v=\theta(t)e^{-st}$ and $dv=e^{-st}\left(\delta(t)-s\theta(t)\right)dt$, we have
$$I(s)=\left[f(t)\theta(t)e^{-st}\right]_{-\infty}^{\infty}-\int_{-\infty}^{\infty}f(t)\delta(t)e^{-st}dt+s\int_{-\infty}^{\infty}f(t)\theta(t)e^{-st}dt$$
Substituting back $g$ and the trivial integral with the $\delta$,
$$I(s)=\left[g(t)e^{-s t}\right]_{-\infty}^{\infty}+s\int_{-\infty}^{\infty}g(t)e^{-st}dt-f(0)$$
For the transform to exist, we must demand $g(t)e^{-st}$ goes to zero at $\pm\infty$, and hence
$$\mathcal{L}\left[g'\right](s)=s\mathcal{L}\left[g\right](s)$$
The trick here is that in the unilateral Laplace transform property
$$\mathcal{L}\left[f^{(n)}\right](s)=s^n\mathcal{L}(s)-\sum_{k=1}^{n}s^{n-k}f^{(k-1)}(0)$$
Is not applicable to a function like $g(t)=f(t)\theta(t)$ because it requires that the derivatives are defined up to the $n^{th}$ order at $0$, which doesn't hold for $g$.
Summarizing then: either you compare the unilateral Fourier transform with the unilateral Laplace transform, in which case the property of the FT for the derivatives in the question doesn't hold, or you compare bilateral with bilateral, in which case the derivatives property of the Laplace in the question doesn't hold.
I sincerely wish you the return of your mental peace.
