integration over $\mathbf{R}^n$ of a radial function with translated argument

Let $f(x,t)$, $u(x,t)$, and $S(x,t)$ be radial in $x$, i.e. $f(x, t)=f(|x|, t)$ for all $x \in \mathbf{R}^n$, and $u(x,t) = \int_{0}^{t}\int_{\mathbf{R}^n} S(x-y, t-s)f(y,s)dyds$. I want to write this expression radially. I know that $\int_{\mathbf{R}^n}f(|x|)dx = \omega_{n-1} \int_{0}^{\infty}f(r)r^{n-1}dr$ where $\omega_{n-1}$ is the surface area of the $n-1$ unit ball. I was thinking about applying this to the integrand $S(x-y, t-s)f(y,s)$, but how does this translation work given the shifted argument $x-y$?

Solution 1:

For concreteness let us consider the example $$ S(x)=e^{-\lvert x\rvert^2}, $$ since the $t$ variable plays no role in these considerations. The integral to study is $$ \int_{\mathbb R^n} e^{-\lvert x-y\rvert^2}f(y)\, dy.$$ We introduce polar coordinates $$ x=rq, \qquad y=\rho \omega, $$ where $r, \rho\ge 0$ and $q, \omega\in\mathbb S^{n-1}$. Then the integral becomes $$ \int_0^\infty f(\rho)\rho^{n-1}\underbrace{\left(\int_{\mathbb S^{n-1}} e^{-\lvert rq-\rho \omega\rvert^2} d\omega\right)}_{K(r, \rho):=}\, d\rho. $$ Note that $K$ is independent on $q$. Indeed, if you replace $q$ with $Rq$ in the integral, where $R\in SO(n)$ is a rotation matrix, the change of variable $\omega\to R\omega$ shows that you get the same result.

So this is the conclusion; our integral equals $$ \int_0^\infty K(r, \rho)\rho^{n-1} f(\rho)\, d\rho.$$ All of this actually holds for an arbitrary $S=S(|x|)$. From now on, to say something more precise, we need more information on $S$. In our case, $$\tag{1} K(r, \rho)=\int_{\mathbb S^{n-1}} e^{-(r^2+\rho^2 - 2r\rho q\cdot \omega)}\,d\omega=e^{-(r^2+\rho^2)}F(r\rho), $$ where $$\tag{*} F(t)=\lvert \mathbb S^{n-2}\rvert \int_{-1}^1 e^{2t y}(1-y^2)^\frac{n-3}{2}\, dy.$$ To compute this, I have used the following reasoning. First, as we said, the integral in (1) is actually independent of $q\in\mathbb S^{n-1}\subset \mathbb R^n$ , so we can choose $q$ however we like; for example, $q=(0, \ldots, 0,1)$, so that $\omega\cdot q=\omega_n$. Now, the surface element $d\omega^{n-1}$ on $\mathbb S^{n-1}$ satisfies the following recursive formula; $$\tag{R} d\omega^{n-1}=(1-\omega_n^2)^\frac{n-3}{2}d\omega_n d\omega^{n-2}.$$ So we conclude that $$ F(r\rho)=\int_{\mathbb S^{n-1}} e^{2r\rho q\cdot \omega}\, d\omega=\left(\int_{\mathbb S^{n-2}} d\omega^{n-2}\right) \int_{-1}^1 e^{2r\rho\omega_n}(1-\omega_n^2)^{\frac{n-3}{2}}\, d\omega_n, $$ which is exactly (*).

Some ideas on (R). It is easier to visualize that formula by letting $\omega_n=\cos \theta$, for $\theta\in [0, \pi]$. Then it reads $$ d\omega^{n-1}=(\sin \theta)^{d-2}d\theta d\omega^{n-2}.$$ And this is basically slicing $\mathbb S^{n-1}$ into a superposition of $n-2$-dimensional spheres, each one of them having radius $\sin \theta$. For more details, you can consult, for example, the introduction of the book "Analysis of spherical symmetry in Euclidean spaces" by Claus Müller.