Derivatives of the Dirac delta function

Question

From what I understand the Dirac's Delta derivatives have the meaning $$\int_{-\infty}^{\infty}\delta^{(k)}(x)\phi(x)dx=(-1)^k\int_{-\infty}^{\infty}\delta(x)\phi^{(k)}(x)dx$$ Assuming, of course that the function $\phi$ is differentiable up to order $k$. If that's true, you can then say $$\phi^{(k)}(x_0)=(-1)^k\int_{-\infty}^{\infty}\delta^{(k)}(x-x_0)\phi(x)dx$$ Is this correct? And also, could it be useful in any circumstance?

Answer 1

It is correct provided that one understand the notation $\int_{-\infty}^\infty\delta(x)\phi(x)\ dx$ correctly. This is by no means an integral and $\delta$ is not a function with real variables.

Also, one should specify in what function space is the function $\phi$ so that your identities would make sense.

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Any distribution $T$ is infinitely differentiable in the sense of distribution by the definition $$ \langle \partial^\alpha T,\varphi\rangle:=(-1)^{|\alpha|}\langle \partial^\alpha\varphi,T\rangle. $$

However, it is not true that $T$ has arbitrary order weak derivative. For instance, the Heaviside step function is not weakly differentiable but its distributional derivative is the delta function.

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I have never seen the word "derivable" used in that way before. (Due to change of OP)

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In the theory of electromagnetism, the first derivative of the delta function represents a point magnetic dipole situated at the origin. As for the higher order distributional derivatives, can I say that it is useful for demonstrating an example of distributions of which the $k$-th order derivative can be explicitly calculated?