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I am looking for a higher-dimensional analogue of the convergence of the improper integral $\int_{x \geq 1} x^{-\alpha}dx$. To be more precise, I want to understand under what conditions on $\alpha, \beta > 0$, the integral

$$\int_{\mathbf R^{n+m} \setminus B_1} \frac{dxdy}{|x|^\alpha |y|^\beta}$$

converges, with $x \in \mathbf R^n$ and $y \in \mathbf R^m$. Here $B_1$ is the unit ball in $\mathbf R^{n+m}$. I expect that the condition $\alpha + \beta > n+m$ is sufficient. Clearly, it is necessary because we can estimate

$$\frac 1{|x|^\alpha |y|^\beta} \geq \frac 1{(|x|^2 + |y|^2)^\frac{\alpha+\beta}2} = \frac 1{|(x,y)|^{\alpha+\beta}}.$$

Please advise!

QA Ngô
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  • as in the answer below, $\mathbf R^{n+m}\setminus B_1$ might allow things like $\frac1{|x|^\alpha+|y|^\beta}$ to converge, but for $\frac1{|x|^\alpha|y|^\alpha}$ you will need something like ${ (x,y) \in\mathbf R^{n+m} : |x|>1, |y|>1}$. Your function is "singular along the axes", so just cutting out the origin isnt enough – Calvin Khor Jul 18 '20 at 03:31

1 Answers1

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Let $E = \{x \in \mathbf R^n : |x| > 1\}$. Use the fact that $\mathbf R^{n+m} \setminus B_1 \supset E \times \mathbf R^m$ and Fubini's theorem to find $$\int_{\mathbf R^{n+m} \setminus B_1} \frac{dxdy}{|x|^\alpha |y|^\beta} \ge \int_{E \times \mathbf R^m} \frac{dxdy}{|x|^\alpha |y|^\beta} = \int_E \frac{dx}{|x|^\alpha} \int_{\mathbf R^m} \frac{dy}{|y|^\beta}.$$

Regardless of the value of $\beta$ you have $\displaystyle \int_{\mathbf R^m} \frac{dy}{|y|^\beta} = \infty$ so the original integral is divergent.

Umberto P.
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