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the following is drawn from a rather rough set of lecture notes and I am not sure I understand it.

the goal is to determine for which values of $p$ we have $$ \int_{|x|\leq 1} \frac{1}{|x|^p} \,dx < \infty \qquad $$

where $d\mathbf{x} = dx_1 \cdots dx_n$ denotes Lebesgue measure.

according to the notes, the first step is to the change variables to $\mathbf{x} = r\mathbf{w}$ where $r > 0$ and $\mathbf{w}$ lies on the unit sphere $\mathbb{S}^{n-1}$. we then have that

$$d\mathbf{x} = r^{n-1}drd\mathbf{w}$$

and this is where I get stuck - how do I compute this? I tried to look up spherical coordinates, but the formulas seem different. and if I try out an example, say in $\mathbb{R}^2$ then I have \begin{align*} d\mathbf{x} &= dx_1dx_2 \\ &= (\frac{\partial x_1}{\partial r} dr + \frac{\partial x_1}{\partial w_1} dw_1 + \frac{\partial x_1}{\partial w_2} dw_2) (\frac{\partial x_2}{\partial r} dr + \frac{\partial x_2}{\partial w_1} dw_1 + \frac{\partial x_2}{\partial w_2} dw_2) \\ &= (w_1 dr + r dw_1 )(w_2 dr + r dw_2) \\ &= w_1rdrdw_2 + rw_2dw_1dr + r^2dw_1dw_2 \\ &= (w_2 dw_1 - w_1dw_2)rdr + r^2d\mathbf{w} \end{align*} which does look different. Many thanks for help and hints on how to understand the above formula!

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I suspect a missing Jacobian... –  copper.hat Nov 16 '12 at 1:40

3 Answers 3

up vote 1 down vote accepted

Indeed $d\mathrm{w}$ vanishes since $dw_2$ is a linear multiple of $dw_1$ via the relation $w_1^2 + w_2^2 = 1$. To be specific, we have

$$ 2w_1 dw_1 + 2w_2 dw_2 = d(1) = 0 \quad \Longrightarrow \quad dw_2 = -\frac{w_1}{w_2} \, dw_1. $$

Thus we have

$$ dx_1 dx_2 = \frac{r}{w_2} dw_1 dr = -\frac{r}{w_2} dr dw_1. $$

With our friendly arc-length parametrization $\theta$ of $S^1$, we have $w_1 = \cos \theta$ and thus

$$ dw_1 = -\sin\theta \,d\theta = -w_2 d\theta \quad \Longrightarrow \quad dx_1 dx_2 = r drd\theta. $$

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We use the following result: when f is a radial integrable function, that is, we can write $f(x)=g(∥x∥)$ where $∥⋅∥$ is the Euclidian norm, we have $$\int_{\Bbb R^n}f(x)dx=nV_n\int_0^{+\infty}r^{n-1}g(r)\,dr,$$ Back to the problem: let $f(x):=|x|^{-p},|x|≥1.$ It's a radial function. so $$\int_{|x|\leq 1} \frac{1}{|x|^p} \,dx =\int_{0}^{1}r^{n-1}r^{-p}dr=\frac{r^{n-p}}{n-p}|_{0}^{1}=\frac{1}{n-p}$$

therefore,$p$ can not be $n$.

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We may refer to Rudin, Real and Complex Analysis, Chapter 8, Problem 6.

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It might be helpful to expound on this a bit... –  Arkamis Nov 16 '12 at 5:14

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