Determine under for what values $a_1, a_2, \cdots, a_n$ are this integral is finite. 
Determine for which real numbers $a_1,a_2,\dots,a_n$ the following integral is finite:
  $$\int_{\mathbb{R}^n} \frac{1}{1+|x_1|^{a_1}+|x_2|^{a_2}+\dots+|x_n|^{a_n}}dx. $$ 

I was stuck on this homework problem and looking for some help. 
I tried to use the change of variable by defining $G(x_1,x_2,\dots,x_n) = y_i $ for $ |x_i|^{a_i}$. then the integral $$\int_{\mathbb{R}_+^n} = \int_{G^{-1}(\mathbb{R}_+^n)}f(G_n(x))|det\,
 D_xG|dx \\= \frac{1}{a_1a_2\dots a_n}\int_{\mathbb{R}^n_+}\frac{1}{1+x_1+x_2+\dots+x_n} x_1^{\frac{1}{a_1}-1}x_2^{\frac{1}{a_2}-1}\dots 
x_n^{\frac{1}{a_n}-1}dx$$
and then I don't know how to proceed. I feel like this way is an dead end.
 A: Assume that $a_1,a_2,\dots,a_n$ are positive real numbers and for $r\geq 0$ consider the set
$$B(r):=\{(x_1,\dots,x_n)\in\mathbb{R}^n: |x_1|^{a_1}+|x_2|^{a_2}+\dots+|x_n|^{a_n}\leq r \}.$$
$B(r)$ is bounded and its finite volume satisfies the property
\begin{align}|B(r)|&=\left|\left\{(x_1,\dots,x_n)\in\mathbb{R}^n: \left|\frac{x_1}{r^{1/a_1}}\right|^{a_1}+\left|\frac{x_2}{r^{1/a_2}}\right|^{a_2}+\dots+\left|\frac{x_n}{r^{1/a_n}}\right|^{a_n}\leq1 \right\}\right|\\
&=|B(1)|r^{\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}}.
\end{align}
Therefore have that
\begin{align}\int_{\mathbb{R}^n} &\frac{dx_1dx_2\cdots dx_n}{1+|x_1|^{a_1}+|x_2|^{a_2}+\dots+|x_n|^{a_n}}=
\int_{0}^{\infty} \frac{D_r(|B(r)|)dr}{1+r}\\
&=|B(1)|\left(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}\right)\int_{0}^{\infty}
\frac{r^{\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}-1}}{1+r}dr.
\end{align}
Now note that 
$$\frac{r^{\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}-1}}{1+r}\sim
r^{\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}-2}$$
and therefore the integral is convergent if and only if
$$\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}<1.$$
A: This is related to problem 3 of the 2016 NCUMC Competition. It asked to show that the integral, when convergent, is $>1$. In fact, you can calculate it explicitly, and the convergence criterion follows from the calculation. (What follows is not one of the official solutions.)
We have (this is the trick!) $$\frac1{1+x}=\int_0^1t^xdt\qquad(x\geq0)$$
so using Fubini (all functions are positive)
$$\int_{\mathbb{R}^n} \frac{1}{1+\sum|x_i|^{a_i}}dx = \int_0^1\int_{\mathbb{R}^n} t^{\sum|x_i|^{a_i}}dxdt=2^n\int_0^1\prod_i\int_0^\infty t^{x^{a_i}}dxdt$$
Now use $$\int_0^\infty t^{x^p}dx = (-\log t)^{-1/p}\cdot\Gamma(1+1/p)\qquad(0<t<1, p>0)$$
Moreover, this is divergent for $p\leq 0$ because by substituting $x=u\cdot(-\log t)^{-1/p}$ (which is how you can compute it) it transforms into $\int_0^\infty\exp(-u^p)du$, so you need $u^p\to\infty$. So it is necessary that all $a_i>0$.
Continuiing, we find that the integral equals
$$2^n\int_0^1(-\log t)^{-\sum1/a_i}dt\cdot\prod_i\Gamma(1+1/a_i)$$
where the integral equals $\Gamma(1-\sum1/a_i)$ (substitute $t=e^{-u}$), convergent iff $\sum1/a_i<1$; in summary:
The integral converges iff all $a_i>0$ and $\sum_i\frac1{a_i}<1$, in which case:
$$\bbox[5px,border:2px solid red]{\int_{\mathbb{R}^n} \frac{1}{1+\sum|x_i|^{a_i}}dx = 2^n\cdot\Gamma\left(1-\sum_i\frac1{a_i}\right)\cdot\prod_i\Gamma\left(1+\frac1{a_i}\right)}$$
the fact that it is $>2^n$ (the question from the competition) now follows from the log-convexity of $\Gamma(1+x)$.
