I am working on calculating Coulomb interaction for excitonic absorption spectra that involves integral of two functions $$ \int_{V_b} ( f_1(|\vec{ a} - \vec b|) \cdot f_2(|\vec b|) d^3b $$ The first function depends on difference between two vectors $\vec{a}$ and $\vec{b}$ , while the second function depends on only vector $b$. The integral is over the volume $V_b$ of vector $\vec{b}$.
Any clue or suggestions or references ? That would be greatly appreciated.
Heuristically, if the volume $V_b$ has spherical symmetry, we can proceed as follows: we are free to orient our original cartesian coordinates in any way we want, since the integral does not depend on that due to spherical symmetry. In particular, orient the system such that $\vec{a}$ lies on the $z$-axis. Now plug in the spherical coordinate system. Note that the distance from $\vec{b}$ to $\vec{a}$ does not depend on the angle $\phi$, but it does depend on the azimuthal angle $\theta$. However, we can find an expression for $AB=|\vec{b}-\vec{a}|$ because this is the only unknown side of the triangle OAB with sides $OA=|\vec{b}|=r$, $OB=|a|$, where the angle between the sides $OA$ and $OB$ is $\theta$, the azimuthal angle. If you do the algebra, you'll find that we have $$ |\vec{b}-\vec{a}|^2 = (r-|\vec{a}|\cos \theta)^2+|\vec{a}|^2 \sin^2 \theta = r^2 +|\vec{a}|^2-2r|\vec{a}|\cos\theta. $$ So we can rewrite the integral at least as (assuming your volume $V_b$ is infinite) $$ 2\pi \int_0^{\infty}dr \int_0^\pi d\theta \sin \theta f_1\left( \sqrt{r^2 +|\vec{a}|^2-2r|\vec{a}|\cos\theta}\right)f_2(r) = \\ 2\pi \int_0^{\infty}dr \int_{-1}^1 d(\cos\theta) f_1\left( \sqrt{r^2 +|\vec{a}|^2-2r|\vec{a}|\cos\theta}\right)f_2(r) = \\ 2\pi \int_0^{\infty}dr \int_{-1}^1 dy f_1\left( \sqrt{r^2 +|\vec{a}|^2-2r|\vec{a}|y}\right)f_2(r), $$ where the last two forms of the integral show clearly that the dependence on $\theta$ is of a very simple form and is treatable as long as $f_1$ is not overly complicated.
Without more details, I think you cannot do much more.