Integrability of $\frac{1}{|x|^d}$

Question

For $x \in \mathfrak R^d$,why is $\int_\limits{\{x; |x|\geq 1\}} \frac{1}{|x|^d} dx = \infty$ in Lebesgue integral?

It's hinted to apply Tonelli Theorem (Fubini Theorem) and use the fact that $\frac 1 x$ is not integrable over $[1,\infty)$, but I don't know how.

Answer 1

Hint: Work in polar coordinates, and combine the facts that

• the integrand is a radial function

• the surface area of the sphere of radius $r$ in $\mathbb{R}^d$ scales like $r^{d - 1}$.

Answer 2

Make the change of variables $x = 2y.$ Then $dx= 2^ddy,$ and the new domain of integration is $\{|y|>1/2\}.$ We get

$$\int_{|x|\ge 1} \frac{1}{|x|^d}\, dx = \int_{|y|\ge 1/2} \frac{1}{2^d|y|^d}\, 2^d\,dy = \int_{|y|\ge 1/2} \frac{1}{|y|^d}\, dy.$$

That's a contradiction unless these integrals $= \infty.$