Integral of delta function and derivative of delta function

Question

Can anyone rigorously prove this? $$\int dx \, \delta(x-\alpha)\delta^{\prime} (x-\beta) = \delta^{\prime} (\alpha-\beta).$$

Answer 1

In THIS ANSWER, I provided a primer on the Dirac Delta and the Unit Doublet distributions.

Here, we have for any test function $f$, the distribution $\int_{-\infty}^\infty\delta(x-\alpha)\delta'(x-\beta)\,dx$ satisfies the following:

$$\begin{align} \int_{-\infty}^\infty f(\alpha)\left(\int_{-\infty}^\infty\delta(x-\alpha)\delta'(x-\beta)\,dx\right)\,d\alpha&=\int_{-\infty}^\infty \delta'(x-\beta)\left(\int_{-\infty}^\infty f(\alpha)\,\delta(x-\alpha)\,d\alpha\right)\,dx\\\\ &=\int_{-\infty}^\infty \delta'(x-\beta)\,f(x)\,dx\\\\ &=-f'(\beta)\\\\ &=\int_{-\infty}^\infty \delta'(\alpha-\beta)\,f(\alpha)\,d\alpha \end{align}$$

We also have

$$\begin{align} \int_{-\infty}^\infty f(\beta)\left(\int_{-\infty}^\infty\delta(x-\alpha)\delta'(x-\beta)\,dx\right)\,d\beta&=\int_{-\infty}^\infty \delta(x-\alpha)\left(\int_{-\infty}^\infty f(\beta)\,\delta'(x-\beta)\,d\beta\right)\,dx\\\\ &=\int_{-\infty}^\infty \delta(x-\alpha)\,f'(x)\,dx\\\\ &=f'(\alpha)\\\\ &=\int_{-\infty}^{\infty}\delta'(\alpha -\beta)f(\beta)\,d\beta \end{align}$$

Therefore, in terms of Generalized Functions, we have the equivalence

$$\delta'(x-\beta)=\int_{-\infty}^\infty\delta(x-\alpha)\delta'(x-\beta)\,dx$$

Using a more compact notation, we have

$$\begin{align} \langle f,\langle \delta'_{\beta},\delta_{\alpha}\rangle \rangle_{\alpha}&=\langle \delta'_{\beta},\langle f,\delta_{\alpha}\rangle_{\alpha} \rangle\\\\ &=\langle \delta'_{\beta},f\rangle\\\\ &=-f'(\beta) \end{align}$$

A similar notation is applicable where we take the distribution over $\beta$.