Is the function $f(x)=\frac{1}{|x|^2-1}$ locally integrable for $x \in \mathbb{R}^3$?

Question

I'm not sure because of the singularities.

Answer 1

If $B_R(0)$ denotes the ball with radius $0<R< 1$ centered at $0$ one sees using spherical coordinates that

$\int_{B_R(o)}\frac{1}{|x|^2-1}dx=\int_0^R\frac{4\pi r^2}{r^2-1}dr=4\pi[r+\frac{ln|r-1|-ln|r+1|}{2}]_0^R=4\pi(R+\frac{ln|R-1|-ln|R+1|}{2})$.

Thus

$\int_{B_1(o)}\frac{1}{|x|^2-1}dx=\lim_{R\rightarrow 1}4\pi(R+\frac{ln|R-1|-ln|R+1|}{2})=-\infty$.

So the answer to Your question is negative and You were right to be not sure about the singularities. $g(x)=\frac{1}{|x|^2}$ is locally integrable over $\mathbb{R}^3$ but it has just one singularity at $0$ whilst Your function $f$ blows up on the whole surface of the unit ball.