# Proving $\int_{|x| \leq a} \mathrm d^n x \,\, x_\alpha^4 =\frac{3}{n(n+2)} \int_{|x| \leq a} \mathrm{d}^n x \, |x|^4$

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 $$x_\alpha$$.

• Can you state the name of the paper? Commented Jul 2, 2021 at 11:23
• @CalvinKhor it's a physics paper so not very relevant, but just in case: "Two-loop renormalization-group analysis of the Burgers-Kardar-Parisi-Zhang equation" Commented Jul 2, 2021 at 15:23

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

• This is similar to what I was expecting, but I don't quite understand the reasoning for $J=I/3$. are $x$ and $y$ any two components? is there a more systematic way to derive this and maybe also extend it to higher powers ($|x|^6$ etc)? thank you Commented Jul 5, 2021 at 13:27
• @SaMaSo sorry perhaps I was a bit terse. Yes x and y are two components. Just consider the n dimensional integral with x_1 and x_2 as the two most inner integrals, then these two inner integrals are over a ball (a 2d disc, where the radius is in terms of a, x_3, x_4,...), consider the integral of x_1^4 over this ball and x_1^2x_2^2 over this ball, one is 1/3 the other. I don't know about more systematic, I feel that Saad has posted the general answer. But yes you could for example do |x|^6 with the same approach, you will just need to know a trig identity or two. Commented Jul 5, 2021 at 16:58

$$\def\d{\mathrm{d}}\def\N{\mathbb{N}}\def\R{\mathbb{R}}\def\vec#1{\boldsymbol{#1}}\def\x{\vec{x}}\def\wx{\widetilde{\x}}\def\paren#1{\left(#1\right)}\def\dotint{\mathop{\intop\cdots\intop}\limits}$$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*}

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$$\frac{I}{I_0} = \frac{(n - 2)!! \prod\limits_{k = 1}^n (m_k - 1)!!}{\paren{\sum\limits_{k = 1}^n m_k + n - 2 }!!}.$$

• This is nicer than going to spherical coordinates but also only gets to the identity with $\int dx |x|^4$ after the full calculation! still +1... Commented Jul 3, 2021 at 11:18
• Nice approach to compute the integral! but I'm also looking for a more direct way of extracting the prefactors without computing the integral (and possibly extending it to integrals over a combination of components) Commented Jul 3, 2021 at 14:38
• @Saad combinations of components such as $x_{\alpha_1} x_{\alpha_2} \ldots$ as it is already mentioned in the question. Commented Jul 4, 2021 at 8:22
• Examples of nonvanishing ones would be integrals of $x_{a_1}^2 x_{a_2}^2$ and $x_{a_1}^2 x_{a_2}^2 x_{a_3} ^2$ etc, where the indices may or may not coincide Commented Jul 4, 2021 at 8:26
• @SaMaSo I still didn't find a tricky method for general cases but the naive method turned out less difficult than I thought. Commented Jul 4, 2021 at 15:37