TITLE: Intuition for Integral Transforms
QUESTION [45 upvotes]: 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 ;-)

REPLY [4 votes]: Simply because the exponential function $\exp(xy)$ as a function of $x$ is an eigenfunction of the derivative operator and also the integration operator so we have:
$$ \frac{\mathrm{d}}{\mathrm{d}x} \exp(xy)=y \exp(xy)$$
If we think of integration as the inverse of the derivative operation we have:
$$\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^{-1} \exp(xy)=\frac{1}{y} \exp(xy)$$
The situation is likely as we are working in a "continuous" basis $\exp(xy)$ indexed by the continuous parameter $y$, the derivation an integration being the diagonal matrices $diag(y)$ and $diag(1/y)$ respectively. So  because the family  $\exp(xy)$ with $y=-i\omega$ in the case of the fourier transform, is a "basis" for functions, the operations of differentiation and integration are reduced to multiplication and division for functions admitting such decompositions.