Polar coordinates transformation

Question

Suppose I have a radial function in two variables $f(x_1,x_2)=g(\sqrt{x_{1}^{2}+x_{2}^{2}})$ with $(x_1,x_2)\in \mathbb{R}^2$. Is there a way to write the integral $\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}f(x_1,x_2)dx_1\Big)^2dx_2$ in polar coordinates? Thank you.

Answer 1

As it is a triple integral, expanding out and substituting yields: $$\int\limits_{-\infty}^\infty\left(\int\limits_{-\infty}^\infty f(x_1,x_2)dx_1\right)^2dx_2 \\=\int\limits_{-\infty}^\infty\left(\int\limits_{-\infty}^\infty f(x_1,x_2)dx_1\right)\left(\int\limits_{-\infty}^\infty f(x_1,x_2)dx_1\right)dx_2 \\=\int\limits_{-\infty}^\infty\left(\int\limits_{-\infty}^\infty f(x_0,x_2)dx_0\right)\left(\int\limits_{-\infty}^\infty f(x_1,x_2)dx_1\right)dx_2 \\=\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty f(x_0,x_2)f(x_1,x_2)dx_0dx_1dx_2 \\=\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty g\left(\sqrt{x_0^2+x_2^2}\right)g\left(\sqrt{x_1^2+x_2^2}\right)dx_0dx_1dx_2 \\=\int\limits_0^\infty\int\limits_0^\pi\int\limits_0^{2\pi}g(r\sin\theta)g\left(r\sqrt{\sin^2\theta\sin^2\varphi+\cos^2\theta}\right)r^2\sin\theta drd\theta d\varphi$$ Which only seems to make the problem worse

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What do you know about the function $g$?

Answer 2

Since this is a triple integral, it is hard to assign meaning to the statement "use polar coordinates". To get an idea of what we could mean by this, in two dimensions, integrating a radial function over all of $\mathbb{R}^2$ can be effectively reduced into a one-dimensional integral over the radial coordinate. We can do something similar here, and we will reduce the triple integral to a double.

We expand the triple integral out first in the form

$$I=\int_{-\infty}^\infty dx \int_{-\infty}^\infty dA~ g(\sqrt{x^2+A^2})\int_{-\infty}^\infty dB~ g(\sqrt{x^2+B^2})$$

Performing the changes of variables $r_1=\sqrt{x^2+A^2}, r_2=\sqrt{x^2+B^2}$ sequentially, we arrive at the form

$$I=8\int_{0}^\infty dx \int_{x}^\infty \frac{r_1~ dr_1}{\sqrt{r_1^2-x^2}} g(r_1)\int_{x}^\infty \frac{r_2~ dr_2}{\sqrt{r_2^2-x^2}} g(r_2)$$

We notice here that the $x$ dependence of the integrand is the same for any choice of function $g$ and we would like to exchange the order of integration to perform the $x$ integral. Exchanging the order of integration yields

$$I=8\int_0^\infty r_1~dr_1 g(r_1)\int_0^\infty r_2~dr_2 g(r_2)\int_0^{\min(r_1,r_2)}\frac{dx}{\sqrt{r_1^2-x^2}\sqrt{r_2^2-x^2}}$$

The $x-$integral can be done exactly and it's an elliptic integral. It can be shown that

$$\int_0^{\min(r_1,r_2)}\frac{dx}{\sqrt{r_1^2-x^2}\sqrt{r_2^2-x^2}}=\frac{1}{\max(r_1,r_2)}K\left(\frac{\max(r_1,r_2)}{\min(r_1,r_2)}\right)$$

and we succeeded in reducing the dimensionality of the integral by sacrificing the simplicity of the integrand. We also write down a form of the integral that takes advantage of the $r_1\leftrightarrow r_2$ symmetry and may be useful:

$$I=16\int_0^\infty~dr_1 g(r_1)\int_{0}^{r_1} r_2~dr_2 g(r_2)K(r_1/r_2)$$