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 ;-)
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 $x$.
The Fourier and Laplace transforms are defined by testing the given function f by special functions (characters in the case of Fourier, exponentials in the case of Laplace).
These special functions happen to be eigenfunctions of translation: if one translates a character or an exponential, one gets a scalar multiple of that character or an exponential.
As a consequence, the Fourier or Laplace transforms diagonalise the translation operation (formally, at least).
Whenever two linear operations commute, they are simultaneously diagonalisable (in principle, at least). As such, one expects the Fourier or Laplace transforms to also diagonalise other linear, translation-invariant operations.
Differentiation and integration are linear, translation-invariant operations. This is why they are diagonalised by the Fourier and Laplace transforms.
Diagonalisation is an extremely useful tool; it reduces the non-abelian world of operators and matrices to the abelian world of scalars.
You can think of integral transforms as a change of coordinates. One of the key tricks in physics is to pick a coordinate system that makes your problem simpler. For example, you may set your coordinates so that the action you're interested in happens along an axis.
You could think of a Fourier transform as a rotation in a function space. Differentiation is particularly simple in the rotated coordinate system, just as forces are simpler when the coordinate system lines up with the force.
Fourier transform really is a rotation of sorts (an "orthogonal transformation"). If you apply Fourier transform four times, you get back your original function, just as you get sine back when you differentiate four times.
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.
One unifying way to look at many of the transforms is through the eyes of quantum theory. For example the Fourier transform is a change of basis of the quantum Hilbert space between the coordinate and momentum representations. The unitarity of the transform is an expression of the fact that they preserve quantum probabilities and there is no difference in the physics of the problem if you use either representation.
The theory of geometric quantization is actually the rigorous way to express this unified point of view. There are many transforms for example the Fourier-Wiener and the Berezin transform which share this property (conservation of quantum probability).
After having read all the answers (and not having understood most of them completely) I finally came to the following conclusion:
The mentioned integral transforms involve the exponential function, e.g.: $e^{nx}$
Differentiating this means $n e^{n x}$ - which is simply multiplication by $n$
Integrating this means $\frac{1}{n} e^{nx}$ - which is simply division by $n$
This holds true for power series which are the discrete form of integral transforms (sort of) and originally stems from the power rule (for differentiation) - although you would have a pesky division by the base using "ordinary" power terms. This is prevented by using the exponential function.
So, if this makes any sense please vote - otherwise please comment..
Thank you all for your help!
