Double integral with $\vec{r}$ and $\vec{r}-\vec{r}'$ as independent variables I've seen double integrals of the type
$$\int d^3r\,d^3r'\, f(\vec{r})g(\vec{r}-\vec{r}')$$
being solved by making the substitution $\vec{u}=\vec{r}-\vec{r}'$:
$$\int d^3r\,d^3u \,\, f(\vec{r})g(\vec{u})=-\left (\int d^3r \,\, f(\vec{r}) \,\right)\,\,\left( \int d^3u \,\,g(\vec{u})\,\right)$$ 
I don't understand how can one possibly regard $\vec{r}$ and $\vec{u}$ as independent variables (which is done in this last equal sign).
Note: I post this here because I've seen this done by physicists.
 A: This is the integral of the convolution $f*g$ of functions $f$ and $g$. You have:
$$
\begin{align}
    \int d^3r'\,d^3r\, f(\vec{r})g(\vec{r}-\vec{r}') 
&= \int d^3r'\,\left(\int d^3r\, f(\vec{r})g(\vec{r}-\vec{r}') \right)\\
&\overset{\text{Fubini}}= \int d^3r\,\left(\int d^3r'\, f(\vec{r})g(\vec{r}-\vec{r}') \right)\\
&\overset{f(r) \text{ does not depend on }r'}= \int d^3r\,f(\vec{r})\left(\int d^3r'\, g(\vec{r}-\vec{r}') \right)\\
&\overset{\vec{u}=\vec{r}-\vec{r}'}= \int d^3r\,f(\vec{r})\left(\int d^3u\, g(\vec{u}) \right)\\
&= \left(\int d^3r\,f(\vec{r}) \right) \left(\int d^3u\, g(\vec{u}) \right).\\
\end{align}
$$
Notes:


*

*the inner integral of lines 2, 3, 4 is computed for fixed $r$,

*as mentioned by kryomaxim, the change of variable with fixed $r$ in the fourth line provides an inner integral independent on $r$ because the integration w.r.t. $r'$ (and $u$) is done on the whole space,

*there is no minus sign in the change of variable $\vec{u}=\vec{r}-\vec{r}'$ because $d^3u = |\text{Jacobian}|d^3r'$ (for $x'=-x$, think also about $\int_{\mathbb{R}} dxf(x)=\int_{\mathbb{R}} dx'f(-x')$ which is a property of the Lebesgue integral, to compare to the Riemann integral where $\int_{-\infty}^{+\infty} dxf(x)=-\int_{+\infty}^{-\infty} dx'f(-x')=\int_{-\infty}^{+\infty} dx'f(-x')$).
A: Initially, $r'$ and $r$ are Independent variables. After Substitution, $u$ is Independent on the other variable $r$, because $u$ is the variable $r'$, but shifted by a value $r$.
In General, this formula is true if the Integration bounds are set from $- \infty$ to $\infty$ in each direction. Integrals over an infinite Region are Translation invariant, therefore the shift by $r$ makes $u$ still an Independent variables.
If you want to know more of it mathematically you can look for "Theorem of Fubini".
