Derivative of the Dirac delta function. The Fourier transform of this function is the a-th derivative of the Dirac delta function. How to interprete?

Question

Evaluating this integral $$\int_{-\infty}^{\infty} x^{a} e^{-i(k-k')x}dx=\sqrt{2\pi}(-i)^a\delta^{(a)}(k-k')$$ I got this result that lead to the definition of $$\delta^{(a)}(k-k')=\sqrt{2\pi}(i)^a\int_{-\infty}^{\infty} x^{a} e^{-i(k-k')x}dx$$ Obtained applying this property and iterating $$\int f(k)\delta^{(a)}(k-k')dk=-\frac{\partial f}{\partial k}\delta^{(a-1)}(k-k') $$ What's the difference between the cases $a\in \mathbb{N}$ and $a\in \mathbb{R}/\mathbb{N}$) How could i interprete this result? How is defined the derivative of the Dirac delta function?

Answer 1

The $\delta$ distribution is defined by $$ \langle \delta, \phi \rangle = \phi(0) $$ or using integral notation, $$ \int_{-\infty}^{\infty} \delta(x) \, \phi(x) \, dx = \phi(0) $$ for every $\phi\in C^\infty_c(\mathbb R).$

The derivative $\delta'$ is defined by $$ \langle \delta', \phi \rangle = -\langle\delta, \phi'\rangle = -\phi'(0). $$

For higher order derivatives we have $$ \langle \delta^{(k)}, \phi \rangle = (-1)^k \langle \delta, \phi^{(k)}\rangle = (-1)^k \phi^{(k)}(0). $$