I would like to deal with the following integral:
$$ f(\vec y) = \int\limits_{\mathbb R^3}\text{d}^3\vec x\, H(a-|\vec x|)\,H(b-|\vec x+\vec y|) \quad a,b>0 $$
where $H(x)$ is the Heaviside step function.
I think changing to spherical coordinates ($x,\theta,\varphi$) might be helpful, then assuming that $\vec x \cdot \vec y = |\vec x||\vec y|\, \cos(\theta)$ might help to simplify the $|\vec x+\vec y|$. However, this did not lead me to a result.
edit: could partial integration be useful? i.e. integrating the first $H$, turning the other into a delta function..?
edit 2: as for convergence, I am pretty sure it converges. I have a result for when $a=b$ (sadly no proof), and now I would like to generalize it:
$$ f(\vec y) = \int\limits_{\mathbb R^3}\text{d}^3\vec x\, H(a-|\vec x|)\,H(a-|\vec x+\vec y|) = \frac{2\pi }{3} a^3 \left(1-\frac{|\vec y|}{2a}\right)^2 \left(\frac{|\vec y|}{2a}+2\right) H\left(1-\frac{|\vec y|}{2a}\right) $$
Term $H(a-|x|)$ defines a sphere at origin with radius $a$. Term $H(b-|x+y|)$ defines a sphere at $y$ with radius $b$. You are basically asking for a volume of intersection. Which is:
$$ \frac{\pi}{12|y|}\Big(a+b-|y|\Big)^2\Big(y^2+2|y|(a+b)-3(a-b)^2\Big)H(a+b-|y|) $$