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 PF Gold P: 162 I think writing things out in sums for this problem obscures what's really going on. I think keeping things as matrices and vectors makes them a lot clearer. $$p({\bf x}) = (2\pi)^{N/2} |{\bf C}|^{-1} \exp\left(-\frac{1}{2}({\bf x}-{\bf \mu})^T {\bf C}^{-1}({\bf x}-{\bf \mu})\right)$$ $$p({\bf k}) = (2\pi)^{N/2} |{\bf C}|^{-1} \int \exp\left(-\frac{1}{2}({\bf x}-{\bf \mu})^T {\bf C}^{-1}({\bf x}-{\bf \mu})+ i {\bf k}^T {\bf x}\right) d {\bf x}$$ Now, you correctly notices that $${\bf C} = {\bf D}{\bf \sigma}{\bf D}^{-1}$$ Where ${\bf \sigma}$ is a diagonal matrix and ${\bf D}$ is orthogonal. However did you know that ${\bf D}$ being orthogonal means that $${\bf D}^{-1} = {\bf D}^{T}$$ Okay so define $${\bf y} = {\bf D}({\bf x}-{\bf \mu})$$ Hence $${\bf x} = {\bf D}^T{\bf y}+{\bf \mu}$$ And so after a change of variables in the integral we get $$p({\bf k}) = (2\pi)^{N/2} |{\bf C}|^{-1} \int \exp\left(-\frac{1}{2}{\bf y}^T {\bf \sigma}^{-1}{\bf y}+ i {\bf k}^T ({\bf D}^T{\bf y}+{\bf \mu}))\right) d {\bf y}$$ Pull out the mean $$p({\bf k}) = (2\pi)^{N/2} |{\bf C}|^{-1} \exp( i {\bf k}^T{\bf \mu})\int \exp\left(-\frac{1}{2}{\bf y}^T {\bf \sigma}^{-1}{\bf y}+ i {\bf k}^T{\bf D}^T{\bf y})\right) d {\bf y}$$ Okay now hint: the ${\bf D}$ in the integral is a rotation. Think about it rotating in k-space as opposed to real space. So define $${\bf \kappa} = {\bf D} {\bf k}$$