What is the smallest $d$ such that $\int\cdots\int \frac{1}{(1+x_1^2 + \cdots + x_n^2)^d} \, dx_1\cdots dx_d$ converges? 
What power $d>0$ is the smallest integer such that in $\mathbb{R}^n$, 
  $$I(n) =\int_{-\infty}^\infty\int_{-\infty}^\infty \cdots\int_{-\infty}^\infty \frac{1}{(1+x_1^2 + \cdots + x_n^2)^d} \,dx_1\, dx_2 \cdots dx_n < +\infty$$
Hint: Think of the integral in "polar coordinates" where $r$ goes from $0$ to $\infty$, integrate over the sphere of radius $r$.

I have no idea how to start. More specifically, how do you do polar coordianates in $\mathbb{R}^n$? Some calculations show that if $n=1$ then $d=1$ would be good enough. But for $n=2$ and $n=3$, $d=2$ and the integrals would equal to $\pi^2$.
 A: Since
$$ \int_{x_1^2+\cdots+x_n^2 = R^2}1\,d\mu = \frac{2\pi^{n/2}}{\Gamma(n/2)} R^{n-1}$$

we have:
  $$ I(n) = \frac{2\pi^{n/2}}{\Gamma(n/2)}\int_0^{+\infty}\frac{R^{n-1}\,dR}{(1+R^2)^d}=\color{red}{\frac{\pi^{n/2}\,\Gamma\left(d-\frac{n}{2}\right)}{\Gamma(d)}} $$
  as soon as $\color{red}{d>\frac{n}{2}}$.

A: Think of $dx_1 \cdots dx_n$ as a volume. Given the the integrand has rotational symmetry, partition the space into spherical shells. The integrand is constant on the surface of this shell.
$$
I(n) = \int_0^\infty \frac{1}{(1+ r^2)^d} dV_r
$$
where $dV_r$ denotes the volume of the thin shell  spanning radius range $(r, r+dr)$. The volume of this thin shell scales as $r^{d-1} dr$ in radius. 
You should be able to take it from here.
A: In $\mathbb R^2$ the circle of radius $r$ has circumference $(\text{constant}\cdot r)$.
In $\mathbb R^3$ the sphere of radius $r$ has surface area $(\text{constant}\cdot r^2)$.
In $\mathbb R^n$ the sphere of radius $r$ has $(n-1)$-dimensional measure $(\text{constant}\cdot r^{n-1})$.
The integral
$$
\int_{-\infty}^\infty\int_{-\infty}^\infty \cdots\int_{-\infty}^\infty \frac{1}{(1+x_1^2 + \cdots + x_n^2)^d} \,dx_1\, dx_2 \cdots dx_n
$$
can be written as
$$
\int_0^\infty \left( \int \frac 1 {(1+r^2)^d} \, dV \right) \, dr
$$
where $dV$ is the element of $(n-1)$-dimensional volume on the surface of the sphere of radius $r$.
But the value of $r$ is constant on the sphere of radius $r$, so the inside integral is
$$
\frac 1 {(1+r^2)^d} \cdot\left( (n-1)\text{-dimensional volume} \right) = \frac 1 {(1+r^2)^d} \cdot (\text{constant}\cdot r^{n-1})
$$
Thus the question becomes this:
For which powers $d>0$ is the following integral finite?
$$
\int_0^\infty \frac{r^{n-1}}{(1+r^2)^d} \,dr
$$
