Consider the convolution product:
$$H(x)\ast\operatorname{Pf}\dfrac{H(x)}{x},$$
where $\operatorname{Pf}$ denotes pseudo function. This means, that $\operatorname{Pf}\dfrac{H(x)}{x}$ is, as defined in the notes I'm reading, the following distribution:
$$\left(\operatorname{Pf}\frac{H(x)}{x},\phi\right)=\int_0^1 x^{-1}(\phi(x)-\phi(0))dx+\int_1^\infty x^{-1}\phi(x)dx.$$
In that case, I've tried to compute the convolution product directly by the definition:
$$\left(H(x)\ast \operatorname{Pf}\dfrac{H(x)}{x},\phi\right)=\left(H(x),\left(\operatorname{Pf}\dfrac{H(y)}{y},\phi(x+y)\right)\right).$$
For $x$ fixed, we have
$$\left(\operatorname{Pf}\dfrac{H(y)}{y},\phi(x+y)\right)=\int_0^1y^{-1}(\phi(x+y)-\phi(x))dy+\int_1^\infty y^{-1}\phi(x+y)dy.$$
Thus when we apply $H(x)$ we get
$$\left(H(x)\ast \operatorname{Pf}\dfrac{H(x)}{x},\phi\right)=\int_0^\infty \int_0^1y^{-1}(\phi(x+y)-\phi(x))dy dx+\int_0^\infty \int_1^\infty y^{-1}\phi(x+y)dy dx.$$
Now I don't know how to solve this. I've tried substitution it didn't get very far.
I know that the answer is $\ln x H(x)$, but I don't know how to get this.
How can I proceed my computation and show that this product is just $\ln x H(x)$?