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How can one compute the following integral? $ \frac{\partial}{\partial r}\int_{\partial B(0,r)}f(r,x)dx. $ I know a similar integral $ \frac{\partial}{\partial r}\int_{B(0,r)}f(r)dx =\int_{\partial B(0,r)}f(r)dx+\int_{B(0,r)}\frac{\partial}{\partial r}f(r)dx. $ Maybe I don't fully understand the latter integral either. I would appreciate it if someone could kindly explain what is going on.

More specifically I want to compute something like

$ \frac{\partial}{\partial r}\int_{\partial B(0,r)}\frac{f(x)}{r}dx. $

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    Sorry, it should depend on other space variable $x$, of course.2012-10-09

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In many cases like this, it is best to do a change of variables so that the domain of integration remains invariant. Let me do the specific case, which should illustrate well the basic concepts.

  1. Given the integral $\int_{\partial B(0,r)} f(x)/r \mathrm{d} x$, the domain of integration is a function of $r$. This makes taking the derivative difficult. So we perform first a change of variables to uniformise the domain of integration. Notice that $\partial B(0,r)$ is $r$ times $\partial B(0,1)$ as a set in $\mathbb{R}^n$. So the change of variable $x = ry$ makes $ \int_{\partial B(0,r)} \frac{f(x)}{r} \mathrm{d} x = \int_{\partial B(0,1)} \frac{f(ry)}{r} \mathrm{d}(ry) = \int_{\partial B(0,1)} f(ry)r^{n-2} ~\mathrm{d}y$ where in the change of variable we treat $r$ as a given constant. (Note that $\partial B(0,r)$ is an $n-1$ dimensional surface, so the Jacobian is $r^{n-1}$ from the change of variables.
  2. Now we can take the derivative easily, since the domain of integration is fixed, we can (assuming various regularity properties about $f$) interchange integration and differentiation, and we end up with $ \partial_r \int_{\partial B(0,r)} \frac{f(x)}{r} \mathrm{d} x = \int_{\partial B(0,1)} \partial_r [r^{n-2}f(ry)] \mathrm{d}y = \int_{\partial B(0,1)} r^{n-2}y\cdot (\nabla f)(ry) + (n-2)r^{n-3} f(ry) ~\mathrm{d}y $ using chain rule.
  3. If we want, we can change variables back $ \int_{\partial B(0,1)} r^{n-2} y\cdot (\nabla f)(ry) + (n-2)r^{n-3} f(ry) ~\mathrm{d}y = \int_{\partial B(0,r)} \frac{x}{r^2} \cdot (\nabla f)(x) + \frac{n-2}{r^2} f(x)~\mathrm{d}x $ and get the final answer.

Observe that in the final answer there are two terms: the first term corresponds to how the value of $f$ changes when we move from $\partial B(0,r)$ to the slightly larger $\partial B(0,r+ \Delta r)$; the second term reflects the fact that when we move from $\partial B(0,r) \to \partial B(0,r+\Delta r)$, the sphere becomes bigger and the area element increases.

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    I see. Many thanks again!2012-10-09