I was trying to calculate the distributional derivative of $H(x)\log(x)$, where $H(x)$ is the Heaviside Function.
$$ \langle DH(x)\log(x), \phi(x) \rangle = -\langle H(x)\log(x), \phi^{\prime}(x) \rangle $$
$$ = -\int_{-\infty}^\infty H(x)\log(x)\phi^{\prime}(x) \, dx $$
$$ = -\int_0^\infty \log(x)\phi^{\prime}(x) \, dx $$
$$ =-\left[ \log(x)\phi(x)\Big|^\infty_0 - \int_0^\infty \frac{1}{x} \phi(x)\,dx \right] $$
$$ =-\Big[\log(x)\phi(x)\Big|^\infty_0\Big] + \int_{-\infty}^\infty H(x) \frac{1}{x} \phi(x) \, dx $$
Now I can write the second term as $\left\langle \frac{H(x)}{x}, \phi(x) \right\rangle$ but how to solve the first term. I am stuck at this step. Is there some other approach to solve this kind of integral? Or if my current working is correct up to this point, how do I proceed further?
You almost did it. You just have to be careful at $x=0.$ For example, you can use a limit: $$ \langle \frac{d}{dx}(H(x) \ln(x)), \phi(x) \rangle = - \langle H(x) \ln(x), \phi'(x) \rangle = - \int_0^\infty \ln(x) \, \phi'(x) \, dx \\ = - \lim_{\epsilon\to 0} \int_\epsilon^\infty \ln(x) \, \phi'(x) \, dx = - \lim_{\epsilon\to 0} \left( \left[ \ln(x) \, \phi(x) \right]_\epsilon^\infty - \int_\epsilon^\infty \frac{1}{x} \, \phi(x) \, dx \right) \\ = \lim_{\epsilon\to 0} \left( \ln(\epsilon) \, \phi(\epsilon) + \int_\epsilon^\infty \frac{1}{x} \, \phi(x) \, dx \right). $$