https://i.stack.imgur.com/5ca3G.jpg (I just made an account so I can't post images directly.)
First some background: This image is from a reader on statistical physics, specifically it is about the calculation of the second virial coefficient of a real gas. But my question is purely related to the mathematics (i.e. integration) involved.
In Eq. 9.24 line 2 (line 1 is irrelevant here) there is a multiple integral over a convex region $\mathbb{V} \subset \mathbb{R^3}$ with measure (total volume) $V$.
$\mathbf{r_1},\mathbf{r_2} \in \mathbb{V}$ are vectors and the integrand depends only on the distance $r_{12}=|\mathbf{r_1}-\mathbf{r_2}|\equiv r$ between the vectors. $\beta$ is a positive real constant and $\phi(r_{12})$ a real scalar function. $f(r)$ is a new (not important) definition of the integrand in line 2.
I don't understand why it is justified to go from line 2 to line 3. The substitution $\mathbf{r}=\mathbf{r_1}-\mathbf{r_2}$ is made and the integration limits should change accordingly, which is not explicitly noted as the integration domain is yet unspecified. This is all fine by me. What bothers me though is that the volume $V$ is pulled out of the double integral (it falls away against the $V$ in the denominator of line 2). Because the limits of integration have changed after the transformation of variables, I don't see how this can happen.
For example, take the simple one-dimensional case of integration on a subspace $[0,a]$ of the real line:
$$-\frac{1}{2a}\int_{0}^{a}\bigg{(} \int_{0}^{a} f(|r_1-r_2|) \ dr_1 \bigg{)} \ dr_2$$
Here $V=a$, the vectors have only one component and thus are scalars, and they both can vary from $0$ to $a$. Then $r_1-r_2$ can vary from $-a$ to $a$. To make a transformation of variables, because this is a double integral, we need an extra integration variable, choose $r'= r_2$, which can vary from $0$ to $a$. The integration domain is a parallelogram, if we integrate over $r'$ first, the integration limits of $r$ are $-r'$ and $a-r'$, respectively. The Jacobian determinant is $1$ for this transformation of variables. The new integral becomes:
$$-\frac{1}{2a}\int_{-r'}^{a-r'} \ \bigg{(} \int_{0}^{a} f(|r|) \ dr'\bigg{)} \ dr$$
Is this equal to:
$$-\frac{1}{2a}\bigg{(}\int_{-r'}^{a-r'} \ dr' \bigg{)} \bigg{(} \int_{0}^{a} f(|r|) \ dr \bigg{)}=-\frac{1}{2}\int_{0}^{a}f(|r|) \ dr$$?
If that is the case, the generalization to three dimensions is straightforward and my problem is solved.