Integrating product of Dirac delta function and it's derivative

Question

Is the following true: $$ \int \delta(x-a) \frac{\partial^l \delta(x)}{\partial x^l} dx = \frac{\partial^l \delta(x)}{\partial x^l} \Bigg |_{x=a}$$

If not, is there a correct way to evaluate the left hand side?

Thanks!

Edit/Addition:

Related: Why/How does WolframAlpha evaluate the following, and in general for odd derivatives? $$ \int \delta(x) \frac{\partial \delta(x)}{\partial x} dx = \frac{1}{2} (\delta(x))^2 + C $$ WolframAlpha

Answer 1

Multiplying distributions is a highly dubious activity. There is no reasonable way to extend multiplication to the space of distributions. $\delta(x)^2$ is not a meaningful quantity. So your identities don't make sense. I suspect

If Wolfram alpha is giving an answer for the delta function resulting in a square this is rather worryingly wrong.

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In general, you can convolve a Delta function with anything $$\int \delta(x-y) f(y) dy = f(x)$$

so you could interpret you original identity in this sense.

Answer 2

The convention of calling $\delta$ as 'delta function' need not justify its nature. It is what physicists call a generalized function and what mathematicians call a distribution, which is defined as continuous linear functional on $C_c^{\infty}(\Bbb{R}^d)$.

One way of giving a meaning to the left-hand side is to interpret them as the following convolution

$$ (\delta * \delta^{(n)} )(a) \text{ $``=$''} \int_{\Bbb{R}} \delta(a-x)\delta^{(n)}(x) \, \mathrm{d}x. $$

Since both are compactly supported distributions, the convolution is well-defined. Then we can apply test functions to determine $\delta * \delta^{(n)}$. To this end, let $\varphi \in C_c^{\infty}(\Bbb{R})$. Then

$$ \langle \delta * \delta^{(n)}, \varphi \rangle = \langle \delta^{(n)}, \delta * \varphi \rangle = \langle \delta^{(n)}, \varphi \rangle $$

and hence $\delta * \delta^{(n)} = \delta^{(n)}$. This is not surprising, since we expect that $\delta$ behaves like an identity element for the convolution operation.