In the theory of distributions, if $H(x)$ is the Heaviside function, we have seen that the distributional derivative of $H$ can be found as follows, for a test-function $\varphi$: \begin{align*} \left\langle H'(x),\varphi \right \rangle&=\left\langle H,-\varphi' \right\rangle\\ &=-\int_0^{\infty}\varphi'(x)dx\\ &=\varphi(0) \\ &=\left\langle \delta,\varphi\right\rangle \,. \end{align*} Now, my question is, when using a multiplication operator in $\mathcal{D}'(\mathbb{\Omega})$ for some function $f \in \mathcal{C}^\infty(\Omega)$, how do I calculate $\left\langle fH',\varphi \right\rangle = \left\langle H',f\varphi\right\rangle =\left\langle \delta,f\varphi\right\rangle $, which is also equal to $-\left\langle H,(f\varphi)' \right\rangle$? If I just write out this latter integral I end up circling around in partial integration, and I am unable to find a meaningful expression.
Am I missing something? Do I just evaluate $f(0)\varphi(0)$?
You don't need to integrate here, just use the definition of $\delta$:
$$\langle fH', \varphi \rangle = \langle H', f\varphi \rangle = \langle \delta, f\varphi \rangle = (f\varphi)(0) = f(0) \varphi(0).$$