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!
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.