# Rederiving (known) Gaussian integral from (double-) spherical coordinates

I want to rederive the following result (see [Wikipedia Gaussian integral][1]) $$\int d^3r\,d^3R\, e^{-ar^2 + 2b \vec{r}\vec{R} - cR^2} = \left(\frac{\pi^2}{det A}\right)^{3/2}, \qquad A = \begin{pmatrix}a & b \\b & c \\\end{pmatrix}$$ using (double-) spherical coordinates. My aim is to add later additional terms in the integrand that only depend on the radial coordinates, hence I would like to understand this already known formula from the viewpoint of spherical coordinates.

My idea was to start with writing (and analogously for $$\vec{R}$$) $$\vec{r} = r \begin{pmatrix} \cos(\phi_r) \sin(\theta_r) \\ \sin(\phi_r) \sin(\theta_r) \\ \cos(\theta_r) \end{pmatrix}$$ As I integrate over two spheres simultaneously, I can choose $$\vec{r}$$ and $$\vec{R}$$ to be in the same $$x-z$$ plane, i.e. $$\phi_r = \phi_R$$. Then $$\vec{r}\vec{R} = r R \cos(\theta_r - \theta_R).$$ As there is no more dependence on the polar angles $$\phi_r$$ and $$\phi_R$$, integration over them simply yields a factor $$4\pi^2$$. So we are left with the integral $$4\pi^2 \int d r d R d\theta_r d \theta_R\, r^2 R^2 \sin(\theta_r) \sin(\theta_R) e^{-ar^2 + 2brR\cos(\theta_r - \theta_R) - cR^2}.$$

But here I got stuck. My memory tells me that there should be another argument with the angles $$\theta_r$$ and $$\theta_R$$, in order to reduce it to a simple integral over the radial variables, which I can then solve subsequently, however I fail to remember it exactly. I tried to substitute $$\Delta \theta = \theta_r - \theta_R$$, but this did not lead anywhere. I think somewhere I must overlook somehting basic... I'm happy for any hint :)

Edit: I found a way, which allows for easy generalizations with additional functions in the integrand depending only on $$r$$:

1. Complete the square in the exponent: $$-ar^2 + 2 b \vec{r}\vec{R} - c \vec{R}^2 = -a r^2 + \frac{b^2}{c} r^2 - c(\frac{b}{c}\vec{r} - \vec{R})^2$$

2. Substitute $$\vec{\rho} = \frac{b}{c}\vec{r} - \vec{R}$$. Then the integration over $$d^3 \rho$$ immediately yields $$\int d^3 \rho e^{-c \rho^2} = 4\pi \int d \rho \rho^2 e^{-c\rho^2} = \left(\frac{\pi}{c}\right)^{3/2}.$$

3. Angular integration over $$\phi_r, \theta_r$$:

We are left with $$\left(\frac{\pi}{c}\right)^{3/2} \int d^3 r e^{-(a - \frac{b^2}{c}) r^2}.$$ The angular integration is trivial and simply yields $$4\pi$$, hence we finally arrive at $$4\pi \left(\frac{\pi}{c}\right)^{3/2} \int d r r^2 e^{-(a - \frac{b^2}{c}) r^2}.$$

Now the integration can easily be extended to contain additional factors depending on the radial variabel $$r$$ alone. [1]: https://en.wikipedia.org/wiki/Gaussian_integral#n-dimensional_and_functional_generalization

• This question is 100% math and 0% physics. It belongs on Math SE. Jun 4, 2023 at 4:23

Let us replace 3 by $$k$$ and 2 by $$n.$$ Let $$A$$ be a positive definite matrix of order $$n$$, let $$x={x_1,\ldots, x_n}$$ be a sequence of $$n$$ vectors in $$R^k$$ and let $$Q(x)=\sum_{i,j}a_{ij}\langle x_i,x_j\rangle.$$ Then $$\int_{R^{nk}}e^{-Q(x)}dx= \left(\frac{\pi^n}{\det A}\right)^{k/2}.$$ This is a particular case of the Gaussian integral
$$\int_{R^N}e^{\frac{1}{2}y^T\Sigma^{-1}y}\frac{dy}{(\sqrt{2\pi})^N\sqrt{\det \Sigma}}dy=1\ \ (*)$$ applied to $$N=nk$$, and $$\Sigma ^{-1}=2\, A\otimes I_k$$ (here $$\otimes$$ mean the Kronecker product). For proving (*) just write $$\Sigma=U^TDU$$ with $$U$$ orthogonal and $$D$$ diagonal, and use the fact that $$y\mapsto Uy$$ preserves the Lebesgue measure. For the application use the fact that $$\det (A\otimes B)=\det A\times \det B.$$