I have been trying to prove that $\partial _x\theta(-x) = -\delta(x)$ from the fact that $\theta(x) = \int_{-\infty}^xdy$ $\delta(y)$, where $\theta(x)$ is the Heaviside step function and $\delta (x)$ is the Dirac delta. So far, I have done the following change of variables:
$$\theta(-x)=\int_{-\infty}^{-x}dy\delta(y)=-\int_{\infty}^{x}dt\delta(-t)=\int_{\infty}^{x}dt(-\delta(t))$$
where $t=-y$. Though I think I can conclude my goal from that last equality, I am not comfortable with the lower bound of the integral. I have already checked this post Relation between Heaviside step function to Dirac Delta function, though it hasn't been truly useful. Any piece of advice?
Best Answer
$$ \theta(-x) = \int_{-\infty}^{-x} \delta(y) \, dy = \{ y=-z \} = \int_{\infty}^{x} \delta(-z) \, (-dz) = -\int_{\infty}^{x} \delta(-z) \, dz \\= \{ \text{$\delta$ is even} \} = -\int_{\infty}^{x} \delta(z) \, dz $$ $$ \frac{d}{dx}\theta(-x) = \frac{d}{dx} \left( -\int_{\infty}^{x} \delta(z) \, dz \right) = -\delta(x) . $$