For part a, how does the change of variables affect the domain of integration? Since x ranges from 1 to infinity does letting y = ux mean y now ranges from u to infinity?
However this integral gives something in terms of u which doesn't help towards the proof since u isn't specified.
For the second part of a), is that just reversing the order of integration?
Any help would be greatly appreciated!
Thank you!
We have $0 \leqslant y < \infty$ and $1 \leqslant x < \infty$. Hence, $0 < 1/x \leqslant 1$ and $0 \leqslant u =y/x < \infty$. With this change of variables, the lower limit for integration with respect to $u$ will be $0$ , and
\begin{align}&\int_1^\infty \left(\int_0^\infty (x^2 + y^2)^{-\alpha} \, dy \right)\,dx \\&= \int_1^\infty \left(\int_0^\infty x^{-2 \alpha}(1 + u^2)^{-\alpha} x\, du \right)\,dx \\&= \int_1^\infty x^{-2 \alpha+1}\,dx \int_0^\infty (1 + u^2)^{-\alpha}\, du.\end{align}
Therefore, $f$ is integrable on $[1,\infty) \times [0, \infty)$ since the single integrals above are convergent for $\alpha > 1$.
We can switch the order of integration since $f$ is nonnegative and integrable:
$$\int_1^\infty \left(\int_0^\infty (x^2 + y^2)^{-\alpha} \, dy \right)\,dx \\ = \int_0^\infty \left(\int_1^\infty (x^2 + y^2)^{-\alpha} \, dx \right)\,dy. $$
Since the integrand is symmetric with respect to $x$ and $y$ we see that $f$ is integrable on $[0,\infty) \times [1,\infty)$ and $[0,1] \times [1,\infty) = \left( [0,\infty) \times [1,\infty) \right) \setminus \left( (1,\infty) \times [1,\infty) \right).$
Note that $g$ is continuous and, hence, integrable on $[-1,1]^2$. Since $g(x,y) = (1 + x^2 + y^2)^{-\alpha} \leqslant (x^2 + y^2)^{-\alpha} = f(x,y)$ then the integrability of $f$ implies that $g$ is integrable on $\mathbb{R}^2 \setminus [-1,1]^2$. Thus $g$ is integrable on $\mathbb{R}^2.$
