Lebesgue integral over a ball I want to ask about a result, which I do not know how to prove
Let $x_0 \in \mathbb{R}^3$. Is it true that
$$\int_{B(x_0,1)} \dfrac{1}{|x|} dx \le \int_{B(0,1)} \dfrac{1}{|x|} dx?$$
Thanks for your helps.
 A: Yes. This is obvious, although it took me some time to see why it's obvious...
Let $A=B(0,1)\setminus B(x_0,1)$, $B=B(0,1)\cap B(x_0,1)$, $C=B(x_0,1)\setminus B(0,1)$. It's enough to show $$\int_A\frac1{|x|}\ge\int_C\frac1{|x|}.$$ And that's obvious because $\frac1{|x|}\ge1$ for $x\in A$ while $\frac1{|x|}\le 1$ for $x\in C$ (and of course $A$ and $C$ have the same measure).
A: Not a very smart solution. If $B:=B(0,1)$, $$ I(x_0) = \int_{B(x_0,r)}\frac{dx}{|x|} = \int_{B}\frac{dx}{|x-x_0|}$$
note that the gradient of a radial function is easy to compute by rewriting $\nabla$ in spherical coordinates: $\nabla = \frac{x}{r} \partial_r $. Hence, 
$$\nabla \frac{1}{|x|} = \left.\left(\partial_r \frac{1}{r}\right)\frac{x}{r} \right|_{r=|x|}=\frac{-x}{|x|^3}$$
Differentiation under the integral sign is permitted, and we obtain
$$\nabla I=\int_B\nabla_{x_0} \frac{1}{|x-x_0|}dx=\int_B \frac{x-x_0}{|x-x_0|^3} dx$$
For an arbitrary choice of the last 2 components of $x_0=(x^1_0,x^2_0,x^3_0)$, we can see that we only get a full cancellation in the first component if $x_0^1 = 0$. Repeating the analysis for each component, we see that $\nabla I = 0$ iff $x_0$ is the zero vector. Finally, note that for $|x_0|\gg 1$, $I(x_0)<I(0)$. Hence, $I$ is maximized at $0$.
