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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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Does $f$ only depend on $r$? Then $\int_{\partial B_r} f(r) \, dS(x) = f(r) \cdot nr^{n-1}\pi^{n/2}/\Gamma(1+\frac n2)$. –  martini Oct 9 '12 at 10:13
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@martini: I guess there is a factor $r^{n-1}$ missing. –  Fabian Oct 9 '12 at 10:15
    
@Fabian Thx, edited. –  martini Oct 9 '12 at 10:16
    
Sorry, it should depend on other space variable $x$, of course. –  Nima Oct 9 '12 at 10:20

1 Answer 1

up vote 2 down vote accepted

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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Thank you for the reply. This will help me understand the subject (I am still reading your post). I am not sure if the formula I stated (not the one I asked) $\frac{\partial}{\partial r}\int_{B(0,r)}f(r)dx =\int_{\partial B(0,r)}f(r)dx+\int_{B(0,r)}\frac{\partial}{\partial r}f(r)dx$ holds or not. –  Nima Oct 9 '12 at 10:55
    
I am now clear about your argument. Thanks so much! One thing I am not yet sure is that how you interpreted the two terms in the answer. How can one think of $\frac{n-2}{r^2}$ as change of area form of the sphere? –  Nima Oct 9 '12 at 11:12
    
Well, I glossed over some details in my last paragraph, notably the fact that the integrand also contains an $r$. If the integrand contained no $r$ (just $f(x)$), the second term would be $(n-1)/r$ instead of $(n-2)/r^2$ which would be more of a correct reflection. In any case, you should think of this in terms of the Leibniz/product rule in differentiation. There is one term where $f$ is differentiated, and that measures the "change in $f$" when all else are held equal. The other term $f$ is not differentiated, that measures the "change of everything else" when $f$ is held equal. –  Willie Wong Oct 9 '12 at 11:16
    
I think $(n-2)/r^2$ is still correct. –  Nima Oct 9 '12 at 11:30
    
So that term contains both the change due to the $1/r$ factor in the integrand, and also the change due to the domain of integration moving. Now, about the formula you stated, I think what you mean to write is with the integrand being $f(x)$ instead of $f(r)$? As written, see martini's comment, and the answer is correct. If you compute it using the above method you need that the area of the $n-1$ sphere is $n/r$ times the volume of the $n$ ball (both radius $r$). –  Willie Wong Oct 9 '12 at 11:30

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