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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

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    $\begingroup$ This question is 100% math and 0% physics. It belongs on Math SE. $\endgroup$
    – Ghoster
    Jun 4, 2023 at 4:23

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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 $\det (A\otimes B)=\det A\times \det B.$

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