In statistical mechanics, we see a lot of high-dimensional integrals. When using only a pairwise potential, we tend to "integrate away" one of the dimension by shifting it to the origin.
For example, in the energy equation, $$U^{ex} = \int \int u(|\mathbf{r}_1-\mathbf{r}_2|)g(|\mathbf{r}_1-\mathbf{r}_2|)d\mathbf{r}_1 d\mathbf{r}_2$$ Now they say, shifting $\mathbf{r}_1$ to be the origin, they say they can integrate out $\mathbf{r}_1$ and say that $$U^{ex} = V\int u(r)g(r)d\mathbf{r}$$
I want to know the step-by-step procedure of how something like this is allowed. I know the standard basics of integration by parts and substitution, but I have not been able to come up with ways to drag this high 3N (N=2) dimensional integral down to a one dimensional, isotropic integral.
Any advice you have would be appreciated.
Define $\vec{r}=\vec{r}_1-\vec{r}_2$, treating $\vec{r}_2$ as a constant in the inner integral. Then, $$ d^3r_1 = d^3u, $$ because the Jacobian of the transformation clearly has determinant one. Assuming we are integrating over "all space", the limits are the same, and under the transformation, $$ |\vec{r}_1-\vec{r}_2|=|\vec{r}|=r. $$ Then, the integrand no longer depends on $\vec{r}_2$, and so we just integrate a constant over "all space", which brings out a factor of the volume $V$ of "all space". (Note of course that we are doing some standard physics "tricks", which is partly to assume that we are integrating over a region large enough so that the shift in the coordinates doesn't make a difference. So, asking for "rigorous" in physics isn't quite right, but perhaps the transformation above is what you were looking for.)