I have a function $\theta (x-x') $ which is zero when $x-x'$ is negative and one when $x-x'$ is positive. I am to prove, $$\delta (x-x') = d/dx [\theta(x-x')] $$
The following is my reasoning:
I consider the action of $\theta$ on an arbitrary function $f$ whose anti-derivative $F$ is finite at infinity.
$$\int_{-\infty}^{\infty} \theta(x-x') f(x') dx'= \int_{-\infty}^{x} f(x') dx' = F(x) - F(-\infty) \tag{1}$$
Expanding the above integral by parts I get, $$\theta(x-x') F(x') \Biggr|_{-\infty}^{\infty} + \int_{-\infty}^{\infty} d/dx[\theta(x-x')] F(x') dx'\tag{2}$$ Since, $$d/dx[\theta(x-x')] = - d/dx'[\theta(x-x')]$$
The first term of (2) is just $-F(-\infty)$. On combining (1) and (2) we get, $$\int_{-\infty}^{\infty} d/dx[\theta(x-x')] F(x') dx'\tag{3} = F(x)$$ Eqn.(3) is identical to the known, $$\int_{-\infty}^{\infty}\delta(x-x')F(x') dx'\tag{3} = F(x)$$ We therefore conclude, $$\delta (x-x') = d/dx [\theta(x-x')] $$
I would like to know if there is any flaw in the reasoning.