A detail about multivariable calculus concerning $\int_{\mathbb{S}^{n-1}}\theta^{\alpha}d\theta$ whether equal to $0$

Question

In the book of "Classical Fourier Analysis" by Loukas Grafakos, I am confused by the following detail, the author says that

$$\int_{\mathbb{S}^{n-1}}\theta^{\alpha}d\theta$$ is $0$ when at least one $\alpha_{j}$ is odd, why?

My attemption: suppose that $\alpha_{1}$ is odd then we can split the integral $\int_{\mathbb{S}^{n-1}}\theta^{\alpha}d\theta$ into two parts according the sign of $\theta_{1}$, we denote $\mathbb{S}^{n-1}_{+}$ the positive $\theta_{1}$ part and $\mathbb{S}^{n-1}_{-}$ the negtive $\theta_{1}$ part, and so we write $$\int_{\mathbb{S}^{n-1}}\theta^{\alpha}d\theta=\int_{\mathbb{S}_{+}^{n-1}}\theta^{\alpha}d\theta+\int_{\mathbb{S}_{-}^{n-1}}\theta^{\alpha}d\theta$$ We make the change of variable $\theta_{1}$ to $-\theta_{1}$ in the integral $$\int_{\mathbb{S}_{-}^{n-1}}\theta^{\alpha}d\theta$$ can we get $$\int_{\mathbb{S}_{-}^{n-1}}\theta^{\alpha}d\theta=-\int_{\mathbb{S}_{+}^{n-1}}\theta^{\alpha}d\theta$$?

Is sphere measure $d\theta$ invariant under the reflection $\theta_{1}\to-\theta_{1}$ transformation?

Notation: $\mathbb{S}^{n-1}$ is the sphere in $n$ dimension Euclidean space and $d\theta$ is surface measure on sphere, $\alpha=(\alpha_{1},\dots,\alpha_{n})\in\mathbb{Z}^{n}_{+}$,$\theta=(\theta_{1},\dots,\theta_{n})\in\mathbb{S}^{n-1}$, $\theta^{\alpha}=\theta_{1}^{\alpha_{1}}\theta_{2}^{\alpha_{2}}\cdots\theta_{n}^{\alpha_{n}}$

Answer 1

The sphere measure is invariant under the map $Rx = (-x_1, x_2, \dots, x_n)$, meaning that for any bounded measurable $f : S^{n - 1} \to \mathbb{R}$, $$\int_{S^{n - 1}}f(\theta)\,d\theta = \int_{S^{n - 1}}f(R\theta)\,d\theta.$$ A more general statement is true: the above equality holds if $R^TR = I$, i.e. $R$ is an orthogonal matrix. To prove it, use the local definition of $d\theta$.

Edit: I will write down the proof using the definition of $d\theta$. Suppose $O \subset \mathbb{R}^{n - 1}$ is open, and $U \subset S^{n - 1}$ is open, and $\phi : O \to U$ is a coordinate chart, meaning $\phi$ is a homeomorphism and $D\phi(x)$ is injective for all $x \in O$. Suppose $f : S^{n - 1} \to \mathbb{R}$ is measurable and supported in $U$. Then we set $$\int_{S^{n - 1}}f(\theta)\,d\theta = \int_{O}f(\phi(x))\sqrt{\det D\phi(x)^TD\phi(x)}\,dx.$$ Informally, $d\theta = \sqrt{\det D\phi(x)^TD\phi(x)}\,dx$. Now the change of variables theorem from calculus shows that if we have another chart $\psi : \Omega \to U$ parameterizing the same piece $U$, then $$\int_{O}f(\phi(x))\sqrt{\det D\phi(x)^TD\phi(x)}\,dx = \int_{\Omega}f(\psi(y))\sqrt{\det D\psi(y)^TD\psi(y)}\,dy,$$ which shows that $d\theta$ is well defined, at least locally. Using technical machinery, e.g. partition of unity or Riesz representation theorem, we can patch these local definitions of $d\theta$ to obtain a Borel measure $d\theta$ on $S^{n - 1}$. This is called the surface measure on $S^{n - 1}$. This exact same construction works to construct surface measure on any smooth surface in $\mathbb{R}^n$.

Now for proving the identity $\int_{S^{n - 1}}f(\theta)\,d\theta = \int_{S^{n - 1}}f(R\theta)\,d\theta$ for all bounded measurable $f$, we can reduce to the case where $f \in C_c^{\infty}(U)$, where $\phi : O \to U$ is a coordinate chart as above. One way to accomplish this reduction is to use the fact that $C^{\infty}(S^{n - 1})$ is dense in $L^1(S^{n - 1})$, and then use a partition of unity. So assume $f \in C_c^{\infty}(U)$. Then $f \circ R \in C_c^{\infty}(R^{-1}U)$. We have a natural chart $R^{-1} \circ \phi : O \to R^{-1}U \subset S^{n - 1}$. Thus $$\int_{S^{n - 1}}f(R\theta)\,d\theta = \int_{R^{-1}U}f(R\theta)\,d\theta = \int_{O}f(RR^{-1}\phi(x))\sqrt{\det D(R^{-1} \circ \phi)(x)^T D(R^{-1} \circ \phi)(x)}\,dx.$$ Now since $R^T R = I$, we have $R^{-1} = R^{T}$, so using the chain rule $D(R^{-1} \circ \phi)(x) = R^{-1}D\phi(x)$ we get $$D(R^{-1} \circ \phi)(x)^T D(R^{-1} \circ \phi)(x) = D\phi(x)^T RR^T D\phi(x) = D\phi(x)^T D\phi(x).$$ Plugging this back in shows that $$\int_{S^{n - 1}}f(R\theta)\,d\theta = \int_{O}f(\phi(x))\sqrt{\det D\phi(x)^T D\phi(x)}\,dx = \int_{U}f(\theta)d\theta = \int_{S^{n - 1}}f(\theta)\,d\theta.$$

Answer 2

For any $n\in\mathbb{N}$, let $B_n(\mathbf{0}_n;1)=\{\mathbf{x}\in\mathbb{R}^n:|\mathbf{x}|_2\leq1\}$ and $S_{n-1}=\{\mathbf{u}\in\mathbb{R}^n:|\mathbf{u}|_2=1\}$.
where $|\mathbf{x}|^2_2=\sum^n_{k=1}x^2_k$ for all $\mathbf{x}=[x_1,\ldots,x_n]^\intercal\in\mathbb{R}^n$. Notice that $$B_n(\mathbf{0}_n;1)=\{(\mathbf{y},z)\in\mathbf{R}^{n-1}\times\mathbb{R}: |\mathbf{y}|_2\leq1,\, |z|\leq\sqrt{1-|\mathbf{y}|_2}\}$$

For $\alpha\in\mathbb{Z}^n_+$, consider the integral $$I_\alpha:=\int_{B_n(\mathbf{0}_n;1)}\mathbf{x}^\alpha\,dx_1\ldots dx_n$$ where $\mathbf{x}^\alpha=x^{\alpha_1}_1\cdot\ldots\cdot x^{\alpha_n}_n$ Using spherical coordinates $$I_{\alpha}=\int^1_0r^{n+|\alpha|_1-1}\int_{\mathbb{S}_{n-1}}\mathbf{u}^\alpha\,\sigma_{n-1}(d\mathbf{u})=\frac{1}{n+|\alpha|_1}\int_{\mathbb{S}_{n-1}}\mathbf{u}^\alpha\sigma_{n-1}(d\mathbf{u})$$ where $|\alpha|_1=\sum^n_{k=1}\alpha_k$, and $\sigma_{n-1}$ is the Lebesgue (surface) measure on the unit sphere $\mathbb{S}_{n-1}$. Notice that for any unitary transformation $U\in L(\mathbb{R}^n)$ where $U^\intercal U=UU^\intercal =I_n$), $U(B(\mathbf{0}_n;1)=B_n(\mathbf{0}_n;1)$ and $|\operatorname{det}(U)|=1$; hence $$I_{\alpha}=\int_{B_n(\mathbf{0}_n;1)}\mathbf{u}^\alpha\,d\mathbf{u}=\int_{B_n(\mathbf{0}_n;1)}(U\mathbf{u})^\alpha\,d\mathbf{u}$$ In particular, for any permutation $\tau(\alpha_1,\ldots,\alpha_n)=(\alpha_{\tau(1)},\ldots,\alpha_{\tau(n)})$ we have that $$I_{\alpha}=I_{\tau(\alpha)}$$ Consequently, if $\alpha_k\equiv1\mod 2$, we may assume without loss of generality that $k=n$. By Fubini's theorem $$I_{\alpha}=\int_{B_{n-1}(\mathbf{0}_{n-1};1)} y^{\alpha_1}_1\cdot\ldots\cdot y^{\alpha_{n-1}}_{n-1}\Big(\int^{\sqrt{1-|\mathbf{y}|_2}}_{-\sqrt{1-|\mathbf{y}|_2}} z^{\alpha_n}\,dz\Big)\,dy_1\ldots dy_{n-1}=0 $$ since $\int^a_{-a} z^{\alpha_n}\,dz=0$ for all $a>0$. This implies that $$\int_{\mathbf{S}_{n-1}}\mathbf{u}^\alpha\sigma_{n-1}(d\mathbf{u})=0$$ whenever $\sigma\in\mathbb{Z}^n_+$ has an odd component.