Integral of delta function and derivative of delta function Can anyone rigorously prove this?
$$\int dx \, \delta(x-\alpha)\delta^{\prime} (x-\beta) = \delta^{\prime} (\alpha-\beta).$$
 A: 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 $\beta$.
