Consider the following integral
$\int_{-\infty}^{+\infty} dx \frac{\partial}{\partial x} \left[ f(x) \delta(x-y) \right]$
where f is a continuous function and x and y are real variables. Intuitively, I would say that the integral is zero, because $\left[ f(x) \delta(x-y) \right] = 0$ when $x=\pm \infty$ (unless also y goes to infinity).
Am I correct? Is there a more formal way to justify it?
Notice that $f(x)\delta(x-y)$ defines a compactly supported distribution. (This roughly means that $f(x)\delta(x-y)$ "vanishes" for large $|x|$.)
Here, a compactly supported distribution is a continuous linear functional on the sapce $\mathcal{E}(\mathbb{R})$, the set $C^{\infty}(\mathbb{R})$ endowed with a certain topology. Now for any compactly supported distribution $\eta$, we can regard the integral of $\eta$ as the pairing
$$ \int_{-\infty}^{\infty} \eta(x) \, dx \quad \text{$``$}=\text{''} \quad \langle \eta, 1 \rangle, $$
where $1 \in \mathcal{E}(\mathbb{R})$ is the constant function with value 1. Of course, this coincides with the usual integral when $\eta$ is a compactly supported integrable function, justifying the notation.
Then it follows that
$$ \int_{-\infty}^{\infty} \eta'(x) \, dx = \left\langle \frac{d}{dx}\eta', 1 \right\rangle = -\left\langle \eta, \frac{d}{dx}1 \right\rangle = -\left\langle \eta, 0 \right\rangle = 0. $$