Change of the order of integration including Dirac delta

[samalkhaiat]

Oh... my... Gawd!

I'd love to see you try and make that rigorous.

Schwartz said it is impossible. In fact he showed that it is impossible to define (associative and commutative) multiplication in the class of generalized functions (distributions). Indeed, it is easy to show that such multiplication leads to contradictions: Consider the two most common distributions, the Dirac delta function $\delta (x)$ and the Principal value function $\mathscr{P}(1/x)$. We can rigorously prove the following relations $$x \ \delta (x) = \delta (x) \ x = 0, \ \ \ \ x \ \mathscr{P}(\frac{1}{x}) = \mathscr{P}(\frac{1}{x}) \ x = 1 .$$ If a product existed, then, using these relations, we would have the following contradictory chain of equalities $$0 = 0 \ \mathscr{P}(\frac{1}{x}) = \left( \delta (x) \ x \right) \mathscr{P}(\frac{1}{x}) = \delta (x) \left( x \mathscr{P}(\frac{1}{x}) \right) = \delta (x) .$$
So, in order to define a product of two distributions $f$ and $g$, it is necessary that they have the following properties: $f$ must be just as irregular in the neighbourhood of an arbitrary point as $g$ is regular in that neighbourhood, and vice versa.

 

[samalkhaiat]

Thank you, @samalkhaiat! I think this would be the best correct answer, even though I didn't understand it well. I will try to understand it along this direction. Thank you again.

If I know how much you know about the subject, I might be able to explain it better for you. Feel free to ask.