Suppose that $a_1,\ldots,a_n>0$ and consider the function $f:\mathbb R^n\rightarrow \mathbb R$ $$f(x_1,\ldots,x_n)=\frac{1}{x_1 ^{a_1}+\cdots+x_n ^{a_n}}$$ I'm trying to show that $$\int_{[0,\infty)^n \setminus[0,1]^n} f(\bar x)\lambda<\infty$$iff $$\sum_{i=1} ^n \frac{1}{a_i}<1$$I'm pretty sure that the direction in which the integral is finite has to do something with the Fubini - Tonelli theorem, yet I'm not sure how to use it. the other direction left me puzzled as well.
edit: I managed to prove the direction in which you assume the sum is less then $1$ by applying the inequality of arithmetic and geometric means on $f$.
Define $t_i$ by the relation $t_i^2=x_i^{a_i}$; with this substitution, the convergence of the integral $I:=\int_{[0,+\infty)^n}f(x_1,\dots,x_n)\mathrm dx_1\dots\mathrm dx_n$ is equivalent to that of $$\int_{[1,+\infty)^n}\frac{\prod_{i=1}^nt_i^{2/a_i-1}}{t_1^2+\dots+t_n^2}\mathrm dt_1\dots\mathrm dt_n.$$ Now using polar coordinates, since the part with $\cos$ and $\sin$ is not problematic, we can see that $I$ converges if and only if so does $$\int_1^{+\infty}r^{n-1}r^{\sum_{i=1}^n(2/a_i-1)}r^{-2}\mathrm dr.$$ Since $r^{n-1}r^{\sum_{i=1}^n(2/a_i-1)}r^{-2}=r^{-3+\sum_{i=1}^n2/a_i}$, we get the wanted result.