TITLE: Proving $\int_{|x| \leq a} \mathrm d^n x \,\, x_\alpha^4 =\frac{3}{n(n+2)} \int_{|x| \leq a} \mathrm{d}^n x \, |x|^4$
QUESTION [5 upvotes]: I am trying to compute the following $n$ dimensional integral
$$I_4(n)\equiv\int_{|x| \leq a} \mathrm d^n x \,\, x_\alpha^4$$
where $x=(x_1,x_2,\ldots,x_n)\in\mathbb{R}^{n}$ and $x_\alpha$ is one of its components. I have not been able to proceed, but have managed to compute an easier integral:
$$I_2 (n) \equiv \int_{|x| \leq a} \mathrm d^n x \,\, x_\alpha^2 = \frac{1}{n} \int_{|x| \leq a} \mathrm d^n {x} \,\, |x|^2 = \frac{a^{n+2} S_{n-1}}{n(n+2)}.$$
The first equality follows from the symmetry of exchanging the component $\alpha$ with any other of the components, and in the last result I have used the surface area of a unit sphere in $n$ dimensions.
I tried computing $I_4$ using spherical coordinates but it seems unnecessarily involved. From the paper I am reading, I know I should be able to show that
$$I_4 = \frac{3}{n(n+2)} \int \mathrm d^n x \, |x|^4,$$
from which the final result can be computed easily, but I haven't been able to do so. Are there any nice tricks to show this equality?
I was also wondering if, in general, we can say
$$\int \mathrm d^n x \, x_{\alpha_1} x_{\alpha_2} \cdots x_{\alpha_{2k}} 
= A_{\alpha_1 \alpha_2\cdots\alpha_{2k}} \int \mathrm d^n x \, |x|^{2k}$$
and if there is a way to compute the proportionality $A$ in this case (which is likely to involve some combinations of $\delta_{\alpha_i \alpha_j}$ due to symmetry - I have a very naive guess that this has some connection to constructing traceless symmetric tensors (see this question)…)
Edit
Here is an attempt using spherical coordinates:
From the symmetry, we know that $I_4$ will not depend on the specific index $\alpha$, so we can choose this as the direction with the first axis. In the $n$-dimensional spherical coordinates, we have
$$\mathrm d^n x = r^{n-1} \sin^{n-2}(\phi_1) \sin^{n-3}(\phi_2)\cdots\sin(\phi_{n-2}) \,\mathrm dr \mathrm d\phi_1 \cdots \mathrm d\phi_{n-1} = r^{n-1} \,\mathrm dS_{n-1} \mathrm dr.$$
Writing $x_\alpha = r \cos(\phi_1)$ gives
$$I_4(n) = \int_{|x|\leq a} r^{n+3} \,\mathrm dr \, \cos^4(\phi_1) \sin^{n-2}(\phi_1) \,\mathrm d\phi_1 \mathrm dS_{n-2}  = \frac{a^{n+4}}{n+4} \, \frac{3\sqrt{\pi}\, \Gamma(\frac{n-1}{2})}{n(n+2)\Gamma(\frac{n}{2})} S_{n-2},$$
where I used the integration result (thanks to Mathematica)
$$\int_0^\pi \cos^4(\phi_1) \sin^{n-2}(\phi_1) \,\mathrm d\phi_1 = \frac{3\sqrt{\pi}\, \Gamma(\frac{n-1}{2})}{n(n+2)\Gamma(\frac{n}{2})}.$$
The final expression for $I_4$ can be simplified since $S_{n-2} \sqrt{\pi} \,\Gamma((n-1)/2)=\Gamma(n/2) S_{n-1}$. This leaves us with
$$I_4 = \frac{a^{n+4}}{n+4} \frac{3 S_{n-1}}{n(n+2)} = \frac{3}{n(n+2)} 
\int \mathrm d^n x \, |x|^4.$$
However, this seems overkill to me, and I also don't know how to extend it to integrals over higher (even) powers of $x_\alpha$.

REPLY [4 votes]: By symmetry, the integral of $|x|^4$ over a ball is just $nI + n(n-1)J$, where $I$ is the integral of $x_1^4$ over a ball, and $J$ is the integral of $x_1^2x_2^2$ over a ball.
But we have $J = \frac{I}{3}$, implying $I(n + n(n-1)/3) = In(n+2)/3$ which is what you wanted.
Why is $J = \frac{I}{3}$?
Well there is probably a lot of ways, just consider the integral of $x^4 + y^4$ over a ball vs the integral of $6x^2y^2$, $x^4+y^4 - 6x^2y^2$
Use polar coordinates, implies $r^4(cos^4(\theta) + sin^4(\theta)-6cos^2(\theta)sin^2(\theta)) = r^4cos(4\theta)$
from which the integral $x^4+y^4 - 6x^2y^2$ over the ball is zero, implying the $I, J$ result