Convergence of $ \int^\infty \int^\infty \cdots \int^\infty \frac{dx_1 dx_2 \cdots dx_n}{(x_1^2+x_2^2+\cdots+x_n^2)^a}$, where $a > \frac{1}{2} n$. I want to show that $ \int^\infty \int^\infty \cdots \int^\infty \frac{dx_1 dx_2 \cdots dx_n}{(x_1^2+x_2^2+\cdots+x_n^2)^a}$ converges, where $a > \frac{1}{2} n$. This problem is from Cambridge Math Tripos 1904. I am not sure why there is no lower bound for this integral. Anyone knows what is the default value for the empty lower bound? I can see if $n = 3$, we can use the spherical triple integrals to solve the problem. But how do we show that the general case for nth repeated integrals?
 A: Using the AG-GM  :
If $x_1,x_2,\cdots,x_n$ are positive real numbers, then
$$n\sqrt[n]{x_1^2x_2^2\cdots x_n^2}\leq x_1^2+x_2^2+ \cdots x_n^2$$
You can write:
$$
  \int_1^\infty \cdots \int_1^\infty
  \frac{dx_1 \cdots dx_n}{(x_1^{2}+\cdots + x_n^{2})^a}   
  \leq \frac{1}{n} \int_{1}^\infty \cdots \int_{1}^\infty
  \frac{dx_1 \cdots dx_n}{x_1^{2a/n}\cdots x_n^{2a/n}}
$$
But:
$$
\frac{1}{n} \int_{1}^\infty \cdots \int_{1}^\infty
\frac{dx_1 \cdots dx_n}{x_1^{2a/n}\cdots x_n^{2a/n}} =
\frac{1}{n} \int_{1}^\infty \frac{dx_1}{x_1^{2a/n}} \cdots \int_{1}^\infty \frac{dx_n}{x_n^{2a/n}}
$$
And, since $\frac{2a}{n}> 1$,for each $i \in [1,\infty)$:
$$
\int_{1}^\infty \frac{dx_i}{x_i^{2a/n}}  = \left[ \frac{x_i^{1-2a/n}}{1-2a/n} \right]_1^\infty = \frac{1}{1-2a/n}
$$
A: If the lower bound was the origin, there would be a problem.  However assuming a small sphere around (radius $=x$) the origin is excluded, then we can change the integral to n-dimensional spherical coordinates, integrate the finite surface term and get $K\int_x^\infty \frac{r^{n-1}}{r^{2a}}dr$.  Since $2a\gt n$, the integral =$K\frac{1}{2a-n}x^{n-2a}$ is finite.
A: I think one nice way to visualise this would be to use hyperspherical coordinates. We would have to first set up the definitions:
$$r^2=\sum_{i=1}^nx_i^2$$
$$1\le r<\infty,\,0\le\varphi_i\le \pi$$
and:
$$\mathrm{d}x_1\,\mathrm{d}x_2\cdots\,\mathrm{d}x_n=r^{n-1}\,\mathrm{d}r\prod_{i=1}^{n-1}\sin^{i-1}{\varphi_{i}\,\mathrm{d}\varphi_{i}}$$
and so with all of this combined we get:
$${\int_{1}^{+\infty}\cdots\int_{1}^{+\infty}\int_{1}^{+\infty}\frac{\mathrm{d}x_{1}\,\mathrm{d}x_{2}\cdots\,\mathrm{d}x_{n}}{\left(\sum\limits_{i=1}^{n}{x_i^2}\right)^a}}=\left(\int_1^\infty{\frac{r^{n-1}}{r^{2a}}\,\mathrm{d}r}\right)\left(\prod_{i=1}^{n-1}{\int_{0}^{\pi}{\sin^{i-1}\varphi_{i}\,\mathrm{d}\varphi_i}}\right)$$
Now if you look at the product of the integrals of sines you will notice that the maximum value an integral can take is $1$, so $n-1$ of these integrals as a product cannot be greater than one, and so:
$$I\le \int_1^\infty\frac{\mathrm{d}r}{r^{1+2a-n}}$$
and now finally, for this integral to converge we know the exponent of the integral must be greater than $1$ and so:
$$2a-n>1$$
$$a>\frac{n+1}{2}$$
I must have made a mistake somewhere because we have this extra factor of $1/2$ but I find this method quite nice
A: I'm going to assume the lower bounds of these integrals are all equal to $1$.
Obviously $\int_1^{\infty}\frac{dx}{x^{2\alpha}}$ converges when $\alpha>1/2$ by your standard $p-$test, so let's assume this statement is true when $n=k>1$. Choose $\alpha>\frac{k+1}{2}$ arbitrarily. Using the trigonometric substitution $$x_{k+1}=\tan(\theta)\sqrt{x_1^2+\dots +x_k^2}$$ it follows that for any fixed $(x_1, \ldots , x_k)\in (1,\infty)^k$ that $$\int_{1}^{\infty}\frac{dx_{k+1}}{\big(x_1^2+\dots+x_k^2+x_{k+1}^2\big)^{\alpha}}=\frac{1}{\big(x_1^2+\dots+x_k^2\big)^{\alpha-\frac{1}{2}}}\int_{\arctan\Big(1/\sqrt{x_1^2 + \dots + x_k^2}\Big)}^{\pi/2}\cos^{2\alpha-2}(\theta)d\theta \leq \frac{1}{\big(x_1^2+\dots+x_k^2\big)^{\alpha-\frac{1}{2}}}$$ since $\alpha>\frac{k+1}{2}>1$. Hence $$\int_{(1,\infty)^{k+1}}\frac{dx_1 \dots dx_k dx_{k+1}}{\big(x_1^2+\dots+x_k^2+x_{k+1}^2\big)^{\alpha}}\leq \int_{(1,\infty)^k}\frac{dx_1 \dots dx_k}{\big(x_1^2+\dots+x_k^2\big)^{\alpha-\frac{1}{2}}}$$ Using the induction hypothesis the RHS converges since $\alpha-\frac{1}{2}>\frac{k}{2}$.
