Integrate over $g(\vec{v} \cdot \vec{x}) \ h(|x|)$ using solid angle and polar coordinates

Question

Relating to a question about fourier transforms I want to solve the n-dimensional integral for $n > 2$ $$ \int_{x \in \mathbb{R}^n} f(x) = \int_{x \in \mathbb{R}^n} g(\vec{v} \cdot \vec{x}) \ h(|x|) $$ for some fixed $\vec{v}$, and $g,h: \mathbb{R} \rightarrow \mathbb{C}$ .
Now I searched for polar coordinates, but did not find a similar enough question, if there is one I would be happy if you could link it to this one.
As @fatalerrorfunc pointed out I should use n-dimensional polar coordinates but I am not that familiar with it. The idea is to choose $\phi$ such that $$ \vec{v} \cdot \vec{x} = \cos(\phi) \ |\vec{v}| |\vec{x}| $$ and the $n-2$ dimensional solid angle $\omega$ normal to $\vec{v}$ . Then it should be something like: $$ \int_{x \in \mathbb{R}^n} f(x) = \int_{x \in \mathbb{R}^n} g(\vec{v} \cdot \vec{x}) \ h(|x|) \\ = \int_{r = -\infty}^{\infty} \mathrm{d} r \ h(r) \ \int_{\phi = 0}^{\pi} \mathrm{d}\phi \ g(\cos(\phi) \ |\vec{v}| |\vec{x}|) \\ \int_{\omega \in \mathbb{S^{n-2}}} \mathrm{d}\omega \ \sin(\phi)^{n-2} r^{n-1} \\ = \int_{r = -\infty}^{\infty} r^{n-1} \ h(r) \ \int_{\phi = 0}^{\pi} g(\cos(\phi) \ |\vec{v}| |\vec{x}|) \ \sin(\phi)^{n-2} \ \omega_{n-2} $$ with $\omega_{n-2}$ being the area over the $n-2$ dimensional sphere.
I have a geometric understanding of why I can do that, but am lacking a strict argument as well as am not sure about the determinant of the Jacobian. Can someone explain the interactions between solid angles and n-dimensional polar coordinates?

Answer 1

Using a rotation, you can always assume that $v = e_n := (0,\ldots, 0, 1)$; let's assume that $|v| = 1$ for simplicity. The change to polar coordinates gives the identity $$\int_{\mathbb{R}^n} f(x)\,dx = \int_0^r r^{n-1}\int_{S^{n-1}}f(r\theta)\,dS(\theta)\,dr, \qquad(*)$$ where $dS$ is the standard measure over the sphere. You can justify this formula using the change of variables $(r,\theta)\in\mathbb{R}_+\times S^{n-1}\mapsto r\theta\in\mathbb{R}^n\setminus\{0\}$; notice that the integral doesn't change by removing the origin or any other set with null measure.

More formally, you can parametrize the sphere using $y\in B_1 \mapsto \big(y,\pm\sqrt{1-|y|^2}\big)$, where $B_1:=\{|y|<1\}\subset\mathbb{R}^{n-1}$. For example, for the upper half-plane $\{x_n>0\}$ we have the change of variables $$x(r,y) = \Big(ry,\, r\sqrt{1-|y|^2}\Big),$$ so the Jacobian is $$\frac{dx}{drdy} = \det\begin{pmatrix}y & rI_{n-1\times n-1} \\ \sqrt{1-|y|^2} & -ry\,/\,\sqrt{1-|y|^2} \end{pmatrix} = r^{n-1}\frac{1}{\sqrt{1-|y|^2}}.$$ Hence, we have that $$\int_{\{x_n>0\}} f(x)\,dx = \int_0^\infty r^{n-1}\int_{B_1}f(r,y)\frac{dy}{\sqrt{1-|y|^2}}\,dr,$$ and $dy/\sqrt{1-|y|^2}$ is the volume element of the sphere. Similarly we can compute the integral of the lower half-space, which surely convinces you of the formula (*) above.

Now in you case we get (for the upper half-space) $$\int_{\{x_n>0\}}f\,dx = \int_0^\infty r^{n-1}h(r)\int_{B_1}g\Big(r\sqrt{1-|y|^2}\Big)\frac{dy}{\sqrt{1-|y|^2}}dr.$$ To compute the integral in $B_1$ you can use again polar coordinates $$\int_{B_1}g\Big(r\sqrt{1-|y|^2}\Big)\frac{dy}{\sqrt{1-|y|^2}} = \omega_{n-2}\int_0^1\rho^{n-2}g\Big(r\sqrt{1-\rho^2}\Big)\frac{d\rho}{\sqrt{1-\rho^2}},$$ where $\omega_{n-2}$ is the volume of $S^{n-2}$. You may want to call $\rho = \sin\phi$ so that $$\int_{B_1}g\Big(r\sqrt{1-|y|^2}\Big)\frac{dy}{\sqrt{1-|y|^2}} = \omega_{n-2}\int_0^{\pi/2}\sin^{n-2}(\phi)g\big(r\cos\phi\big)\,d\phi.$$ The lower half-space is similar, so at the end we have $$\int_{\mathbb{R}^n}f(x)\,dx = \omega_{n-2}\int_0^\infty r^{n-1}h(r)\int_0^\pi \sin^{n-2}(\phi)g\big(r\cos\phi\big)\,d\phi dr. $$

Hopefully everything is right.