It is well known that the operations of differentiation and integration are reduced to multiplication and division after being transformed by an integral transform (like e.g. Fourier or Laplace Transforms).


My question: Is there any intuition why this is so? It can be proved, ok - but can somebody please explain the big picture (please not too technical - I might need another intuition to understand that one then, too ;-)



Intuition for Integral Transforms
It might help you to think about a discrete model: consider complex valued functions on $Z/n$. The discrete Fourier transform takes $f(k)$ to 
$g(j) :=\sum\_{k=1}^n f(k) \zeta^{jk}$ where $\zeta=e^{2 \pi i/n}$. It is pretty easy to see that, if we change $f(k)$ to $f(k+1)$, we change $g(j)$ to $g(j)\*\zeta^j$.


Similarly, changing $f(k)$ to $f(k+1)-f(k)$ changes $g(j)$ to $g(j)\*(\zeta^j-1)$. So, in this discrete model, taking a difference becomes multiplication by $(\zeta^j-1)$. In a similar way, in the continuous setting, taking a derivative becomes multiplication by $x$.


