Proving an integral relation (isotropic function) In Hansen-McDonald's book Theory of Simple Liquids the following relation is often used:
We want to evaluate the integral
$$\int_V f(\vec r_1, \vec r_2) d \vec r_1$$
We observe that if the function $f$ is isotropic, i.e.
$$f(\vec r_1, \vec r_2) = f(\mid \vec r_1- \vec r_2 \mid) $$
then
$$\int_V f(\vec r_1, \vec r_2) d \vec r_1 = \int_V f(\mid \vec r_1- \vec r_2 \mid) d \vec r_1 =  f(r) \ V $$
where
$$V= \int_V d \vec r$$
This makes sense to me, because if the function only depends on the distance I can choose to fix $\vec r_1$ everywhere in the volume (obtaining a volume term) and I will be left with $f(r)=f(\mid \vec r_1- \vec r_2 \mid)$.
But how can I prove the last relation in a rigorous way? I guess it must be some kind of change of variables, but I cannot figure out which one.
Update
This relation is used for example in equation (2.5.15):



(notice that $\rho = N/V$. Also, $g(r) \to 1$ for $r\to \infty$)
I also just found out that a similar relation, which we can write as
$$\int_V \int_V f(\mid \vec r_1-\vec r_2\mid) d\vec r_1 d\vec r_2 = V \int_V f(r) d\vec r$$
is used in equation (2.5.18-19), with the following explanation:



(notice that in this case $v(r) \to 0$ for $r \to \infty$) 
It would be also nice to start by proving this in two dimensions, i.e. for
$$\int_I \int_I f(\mid x-y\mid) dx dy \ \ \text{and} \ \ \int_I f(\mid x-y\mid) dy$$
where $I$ is a closed interval, even if I'm starting to suspect (see John Barber's answer) that this result is not rigorous.
 A: I think something must be wrong here. Counterexample:
Suppose
$$
f\left(\vec{r}_1, \vec{r}_2\right) = 
f\left(\left| \vec{r}_1 - \vec{r}_2\right|\right) = 
\exp\left(-\frac{1}{2}\frac{{\left|\vec{r}_1 - \vec{r}_2\right|}^2}{\epsilon^2}\right)
$$
for some small $\epsilon$. Suppose the length scale of $V$ is much larger than $\epsilon$, and consider the quantity
$$
g\left(\vec{r}_2\right) = \int_V d\vec{r}_1\, f\left(\vec{r}_1, \vec{r}_2\right)\, .
$$
The integrand is a narrow Gaussian centered at $\vec{r}_2$. If $\vec{r}_2$ is more than several multiples of $\epsilon$ away from the boundary of $V$, then essentially all of the Gaussian is integrated over, and 
$$
g\left(\vec{r}_2\right) \approx {\left(2\pi \epsilon^2\right)}^{d/2}\, ,
$$
where $d$ is the dimensionality of the space.
If $\vec{r}_2$ is on the boundary of $V$, then about half the Gaussian is integrated over, and we have
$$
g\left(\vec{r}_2\right) \approx \frac{1}{2}{\left(2\pi \epsilon^2\right)}^{d/2}\, .
$$
These bear no resemblance to (what appears to be) the asserted relationship
$$
g\left(\vec{r}_2\right) = V\, f\left( \left| \vec{r}_2 \right| \right)\, .
$$
What section of H&M does this appear in? I've got a copy of it around here somewhere. I'm wondering if perhaps the actual relationship might be
$$
\int_V d\vec{r}_1 \int_V d\vec{r}_2\, f\left(\vec{r}_1-\vec{r}_2\right)
 = V \int_V d\vec{r}_1\, f\left(\vec{r}_1\right)\, ,
$$
which should be perfectly valid so long as $f$ is sufficiently strongly peaked or localized. Since it's H&M, I would imagine that $f$ is something like $f(r) = \exp\left[-U(r)/k_B T\right]$ or $f(r) = 1 - \exp\left[-U(r)/k_B T\right]$, where $U$ is an intermolecular potential?
Response to Update:
The thing to keep in mind is that there are some assumptions implicit in H&M that are relevant here. These integral identities don't apply to just any radial function, but only to the sort relevant in molecular thermodynamics. This assumes that the function $f(\vec{r})$ varies on a molecular length scale $\epsilon$, and that the domain of integration $V$ is more like a macroscopic length scale. So we have a radical difference of scales here. Outside a length scale $\epsilon$, the function $f$ will typically be either zero or constant. Under these assumptions, some of these integral "identities" are approximately true.
I'm not going to get into H&M Equations (2.5.15) and (2.5.18) copied above. They depend on the precise definition of the two-particle distribution function, which is beyond the scope of a math.stackexchange post.
I do want to address the last part of your update. Define:
$$
g(y) = \int_I dx\,  f(x - y)\, .
$$
Here $I$ is an interval of length $L$, and we assume that $f$ is compactly supported (more or less) over an interval of length on the order of $\epsilon$, a small number. An example would be:
$$
f(x) = \exp\left(-\frac{1}{2}\frac{x^2}{\epsilon^2}\right)\, .
$$
By the same argument as above,
$$
g(y) = \mathrm{constant} \approx \int_{-\infty}^{+\infty} dx\,  f(x)
$$
unless $y$ is within a distance of about $\epsilon$ of one of the boundaries of $I$. Thus the assertion in the original post is actually wrong, even for H&M-type functions, and isn't what is being used in the equations from H&M.
However:
\begin{align}
\int_I dy\, g(y) &\approx \int_I dy \int_{-\infty}^{+\infty} dx\,  f(x) \\
&\approx L \int_I dx\, f(x) \;+\; \text{An error of order $\epsilon$}\, .
\end{align}
Here I've implicitly assumed that the domain $I$ is sufficiently "centered" that 
$$
\int_{-\infty}^{+\infty} dx\,  f(x) \approx \int_I dx\,  f(x)\, .
$$
As long as $f$ falls off sufficiently rapidly (i.e. don't try this with particles mediated by a Coulomb potential!), we can ignore the error term, and the "double integral" identity should approximately hold.
Update to address Equation 2.5.15 above:
Theory of Simple Liquids can be found here on Google Books. The relevant page is 31.
I've gone over the steps leading to equation 2.5.15, and I don't think it makes sense as written. I'm convinced it's a typo. 2.5.15 doesn't seem to follow from 2.5.14. I think the argument of the integral in 2.5.15 is supposed to be:
$$
g^{(2)}_N(r' + r, r') = g^{(2)}_N\left(\left|r' + r - r'\right|\right) = g^{(2)}_N\left(\left|r\right|\right) = g(r)
$$
That makes much more sense to me, and leads to the desired result.
