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

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 $x_\alpha$.

Answer 1

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

Answer 2

$\def\d{\mathrm{d}}\def\N{\mathbb{N}}\def\R{\mathbb{R}}\def\vec#1{\boldsymbol{#1}}\def\x{\vec{x}}\def\wx{\widetilde{\x}}\def\paren#1{\left(#1\right)}\def\dotint{\mathop{\intop\cdots\intop}\limits}$For any $n \in \N_+$ and $\x = (x_1, x_2, \cdots, x_n) \in \R^n$, define $\wx = (x_2, \cdots, x_n)$ as the cutoff of $\x$. For $m, n \in \N_+$ and $a > 0$, define$$ I_{m, n}(a) = \dotint_{|\x| \leqslant a} |x_1|^m \,\d x_1 \cdots \d x_n = \int\limits_{|\x| \leqslant a} |x_1|^m \,\d\x, $$ and $V_n = \dfrac{π^{\frac{n}{2}}}{Γ\paren{ \dfrac{n}{2} + 1 }}$ as the content of an $n$-dimensional unit ball (see here). Now,\begin{align*} I_{m, n}(a) &= \int_{-a}^a |x_1|^m \,\d x_1 \int\limits_{|\wx| \leqslant \sqrt{\smash[b]{ a^2 - x_1^2 }}} \d\wx = \int_{-a}^a |x_1|^m · (\sqrt{\smash[b]{ a^2 - x_1^2 }})^{n - 1} V_{n - 1} \,\d x_1\\ &=2V_{n - 1} \int_0^a x_1^m (a^2 - x_1^2)^{\frac{n}{2} - \frac{1}{2}} \,\d x_1 = 2a^{m + n} V_{n - 1} \int_0^1 y^m (1 - y^2)^{\frac{n}{2} - \frac{1}{2}} \,\d y\\ &= a^{m + n} V_{n - 1} \int_0^1 t^{\frac{m}{2} - \frac{1}{2}} (1 - t)^{\frac{n}{2} - \frac{1}{2}} \,\d t \quad (t = y^2)\\ &= a^{m + n} V_{n - 1} Β\paren{ \frac{m}{2} + \frac{1}{2}, \frac{n}{2} + \frac{1}{2} } = a^{m + n} π^{\frac{n}{2} - \frac{1}{2}} \frac{ Γ\paren{\dfrac{m}{2} + \dfrac{1}{2} }}{Γ\paren{ \dfrac{m}{2} + \dfrac{n}{2} + 1 }}. \end{align*}

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The following identity from here is needed for general cases.

Lemma: For any $u, v > -1$,$$ \int_0^{\frac{π}{2}} \sin^u θ \cos^v θ \, \d θ = \frac{1}{2} Β\paren{ \frac{u}{2} + \frac{1}{2}, \frac{v}{2} + \frac{1}{2} }. $$

Now for any $n \in \N_+$ and $m_1, \cdots, m_n > -1$, define$$ I = \dotint_{|\x| \leqslant 1} \prod_{k = 1}^n |x_k|^{m_k} \,\d x_1 \cdots \d x_n, \quad I_0 = \dotint_{|\x| \leqslant 1} |\x|^{\sum\limits_{k = 1}^n m_k} \,\d\x, $$ and the goal is to find the ratio $\dfrac{I}{I_0}$ since the radius $a$ is irrevalent here due to homogeneity. Using spherical coordinates,\begin{align*} I &= 2^n \dotint_{\substack{ |\x| \leqslant 1 \\ x_1, \cdots, x_n \geqslant 0 }} \prod_{k = 1}^n x_k^{m_k} \,\d x_1 \cdots \d x_n\\ &= 2^n \dotint_{\substack{ 0 \leqslant r \leqslant 1 \\ 0 \leqslant θ_1, \cdots, θ_{n - 1} \leqslant \frac{π}{2} }} \prod_{k = 1}^{n - 1} \paren{ r \prod_{j = 1}^{k - 1} \sin θ_j · \cos θ_k }^{m_k} · \paren{ r \prod_{j = 1}^{n - 1} \sin θ_j }^{m_n}\\ &\mathrel{\phantom=}{} · r^{n - 1} \prod_{k = 1}^{n - 2} (\sin θ_k)^{n - k - 1} \,\d r \d θ_1 \cdots \d θ_{n - 1}\\ &= 2^n \dotint_{\substack{ 0 \leqslant r \leqslant 1 \\ 0 \leqslant θ_1, \cdots, θ_{n - 1} \leqslant \frac{π}{2} }} r^{\sum\limits_{k = 1}^n m_k + n - 1} \prod_{k = 1}^{n - 1} (\sin θ_k)^{\sum\limits_{j = k + 1}^n m_j + n - k - 1} (\cos θ_k)^{m_k} \,\d r \d θ_1 \cdots \d θ_{n - 1}\\ &= 2^n \paren{ \int_0^1 r^{\sum\limits_{k = 1}^n m_k + n - 1} \,\d r } \prod_{k = 1}^{n - 1} \int_0^{\frac{π}{2}} (\sin θ_k)^{\sum\limits_{j = k + 1}^n m_j + n - k - 1} (\cos θ_k)^{m_k} \,\d θ_k\\ &= 2^n · \frac{1}{\sum\limits_{k = 1}^n m_k + n} · \prod_{k = 1}^{n - 1} \frac{1}{2} Β\paren{ \frac{1}{2} \paren{ \sum_{j = k + 1}^n m_j + n - k }, \frac{m_k}{2} + \frac{1}{2} }\\ &= \frac{2}{\sum\limits_{k = 1}^n m_k + n} \prod_{k = 1}^{n - 1} \frac{\color{blue}{Γ\paren{ \dfrac{1}{2} \paren{ \sum\limits_{j = k + 1}^n m_j + n - k } } } Γ\paren{ \dfrac{m_k}{2} + \dfrac{1}{2} } }{\color{blue}{Γ\paren{ \dfrac{1}{2} \paren{\sum\limits_{j = k}^n m_j + n - k + 1 } } }}\\ &= \frac{2}{\sum\limits_{k = 1}^n m_k + n} · \frac{\prod\limits_{k = 1}^n Γ\paren{ \dfrac{m_k}{2} + \dfrac{1}{2} }}{Γ\paren{ \dfrac{1}{2} \paren{ \sum\limits_{k = 1}^n m_k + n } }}, \quad (\text{telescoping blue terms}) \end{align*} and\begin{align*} I_0 &= \dotint_{\substack{ 0 \leqslant r \leqslant 1 \\ 0 \leqslant θ_1, \cdots, θ_{n - 1} \leqslant \frac{π}{2} }} r^{\sum\limits_{k = 1}^n m_k} · r^{n - 1} \prod_{k = 1}^{n - 2} (\sin θ_k)^{n - k - 1} \,\d r \d θ_1 \cdots \d θ_{n - 1}\\ &= \paren{ \int_0^1 r^{\sum\limits_{k = 1}^n m_k + n - 1} \,\d r } \Biggl( \dotint_{0 \leqslant θ_1, \cdots, θ_{n - 1} \leqslant \frac{π}{2}} \prod_{k = 1}^{n - 2} (\sin θ_k)^{n - k - 1} \,\d θ_1 \cdots \d θ_{n - 1} \Biggr)\\ &= \frac{1}{\sum\limits_{k = 1}^n m_k + n} · \frac{1}{\displaystyle\int_0^1 r^{n - 1} \,\d r } · \dotint_{\substack{ 0 \leqslant r \leqslant 1 \\ 0 \leqslant θ_1, \cdots, θ_{n - 1} \leqslant \frac{π}{2} }} r^{n - 1} \prod_{k = 1}^{n - 2} (\sin θ_k)^{n - k - 1} \,\d r \d θ_1 \cdots \d θ_{n - 1}\\ &= \frac{n V_n}{\sum\limits_{k = 1}^n m_k + n} = \frac{n}{\sum\limits_{k = 1}^n m_k + n} · \frac{π^{\frac{n}{2}}}{Γ\paren{ \dfrac{n}{2} + 1 }} = \frac{2}{\sum\limits_{k = 1}^n m_k + n} · \frac{π^{\frac{n}{2}}}{Γ\paren{ \dfrac{n}{2} }}. \end{align*} Therefore,$$ \frac{I}{I_0} = \frac{Γ\paren{ \dfrac{n}{2} } \prod\limits_{k = 1}^n Γ\paren{ \dfrac{m_k}{2} + \dfrac{1}{2} }}{π^{\frac{n}{2}} Γ\paren{ \dfrac{1}{2} \paren{ \sum\limits_{k = 1}^n m_k + n } }}. $$ When $m_1, \cdots, m_n$ are all even integers, the above expression reduces to be$$ \frac{I}{I_0} = \frac{(n - 2)!! \prod\limits_{k = 1}^n (m_k - 1)!!}{\paren{\sum\limits_{k = 1}^n m_k + n - 2 }!!}. $$