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Azimuthal and polar integration of a 3D Gaussian

  1. Sep 28, 2012 #1

    Wox

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    Numerical evaluation of the following integral of a 3D gaussian [itex]G[/itex] seems to result in a 1D Gaussian [itex]g[/itex] (or at least very close to one)
    $$\int_{0}^{2\pi}\!\!\int_{0}^{\pi}G(R,\phi,\theta) \sin\theta\ \text{d}\theta \ \text{d}\phi= g(R)$$
    where the 3D Gaussian in spherical coordinates with mean [itex]\mu[/itex] and covariance matrix [itex]\Sigma[/itex]
    $$G(R,\phi,\theta)=H \exp\left[-\frac{1}{2}(X-\mu)^{T}\cdot \Sigma^{-1}\cdot (X-\mu)\right]$$
    $$X=\begin{bmatrix}R\cos\phi\sin\theta\\\ R\sin\phi\sin\theta\\\ R\cos\theta\end{bmatrix}$$
    and the 1D Gaussian with mean [itex]R_\mu[/itex] and variance [itex]\sigma^2[/itex]
    $$g(R)=H'\exp\left[-\frac{(R-R_\mu)}{2\sigma^2}\right]$$
    The question is: is this true and how do the parameters of [itex]g[/itex] relate to the parameters of [itex]G[/itex]?

    My attempt to solve this problem involved transforming [itex]G[/itex] so that [itex]\mu=(0\ 0\ R_\mu)[/itex] (adapting [itex]\Sigma[/itex] in the process) and substitute [itex]z=\cos\theta[/itex]. After some basic math I get the following
    $$\int_{0}^{2\pi}\!\!\int_{0}^{\pi}G(R,\phi,\theta) \sin\theta\ \text{d}\theta \ \text{d}\phi=\int_{0}^{2\pi}\!\!\int_{-1}^{1}\exp(Az^2 + (Bz+C)\sqrt{1-z^2}+Dz+E) \text{d}z \ \text{d}\phi$$
    where [itex]A,\ldots,E[/itex] depend on azimuth [itex]\phi[/itex]. Alternatively this can be written as
    $$\int_{0}^{2\pi}\!\!\int_{0}^{\pi}G(R,\phi,\theta) \sin\theta\ \text{d}\theta \ \text{d}\phi=\int_{-1}^{1}\!\!\int_{0}^{2\pi}\exp(A_{1}\cos(2\phi+ \delta_{1})+A_{2}\cos(2\phi+\delta_{2})+A_{3}) \text{d}\phi \ \text{d}z$$
    where [itex]A_1,\ldots,A_3[/itex] depend on [itex]z[/itex]. However I was unable to solve either of these integrals over [itex]z[/itex] or [itex]\phi[/itex], let alone solve the double integral.
     
  2. jcsd
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