I Change of Variables in Double Volume Integral

[pherytic]

In Greiner's Classical Electromagnetism book (page 126) he has a derivation equivalent to the following.
$$\int_V d^3r^{'} \nabla \int_V d^3r^{''}\frac {f(\bf r^{''})}{|\bf r + \bf r^{'}- \bf r^{''}|}$$

$$ \bf z = \bf r^{''} - \bf r^{'} $$

$$\int_V d^3r^{'} \nabla \int_V d^3z \frac {f(\bf z + \bf r^{'})}{|\bf r - \bf z|}$$$$\nabla\int_V d^3z \frac{1}{|\bf r - \bf z|} \int_V d^3r^{'} {f(\bf z + \bf r^{'})}$$

The book says that after the change of variables to z, it is okay to just move the 1/|r-z| term out of the r' integral. But this doesn't make sense to me, given z is still a function of r'. Am I missing something or is this mathematically incorrect?

 

[pasmith]

The integration with respect to \mathbf{r}' is with \mathbf{z} held constant. You have replaced the independent variables \mathbf{r}' and \mathbf{r}'' with the independent variables \mathbf{r}' and \mathbf{z} by replacing \mathbf{r}'' with \mathbf{r}' + \mathbf{z}. It then makes sense to do the integral over \mathbf{r}' first, precisely because you can regard \|\mathbf{r} - \mathbf{z}\| as constant while doing so. The domain of integration will have changed because we require both \mathbf{r}' \in V and \mathbf{r}' + \mathbf{z} \in V.

 

[pherytic]

Thank you, that really helped. If the two original variables are independent wrt each other, that has to still be true after substitution.