Question:
Prove the Fourier Inversion Formula for the specific function $\phi_{\Sigma, \mu}(x)$:
$$\phi_{\Sigma, \mu}(x) = (2\pi)^{-k} \int_{R^k}\hat{\phi}_{\Sigma,\mu}(\xi)e^{-i\xi\cdot x}d\xi$$
where $\Sigma$ is a $k \times k$ positive-definite symmetric matrix, $\mu, \xi \in R^k$, and $$\phi_{\Sigma, \mu}(x) = (2\pi)^{-\frac{k}{2}}(\text{det}\Sigma)^{-\frac{1}{2}}\exp \left\{-\frac{1}{2}(x-\mu)\cdot\Sigma^{-1}(x-\mu)\right\}$$ $$\hat{\phi}_{\Sigma,\mu}(\xi) = \exp\left\{i\xi\cdot\mu - \frac{1}{2}\xi\cdot\Sigma\xi\right\}$$
My attempt at a solution:
I've been trying to prove this by looking at the RHS of the equation, and manually calculating the integral. So, substituting in for $\hat{\phi}_{\Sigma, \mu}(x)$, we have:
$$(2\pi)^{-k} \int_{R^k}\hat{\phi}_{\Sigma,\mu}(\xi)e^{-i\xi\cdot x}d\xi = (2\pi)^{-k} \int_{R^k}\exp\left\{i\xi\cdot\mu - \frac{1}{2}\xi\cdot\Sigma\xi\right\}e^{-i\xi\cdot x}d\xi$$
$$ = (2\pi)^{-k}\int_{R^k}e^{i\xi\cdot\mu-\frac{1}{2}\xi\cdot\Sigma\xi-i\xi \cdot x}d\xi$$
$$ =(2\pi)^{-k}\int_{R^k}e^{-\frac{1}{2}[(2i\xi\cdot\mu)-\xi^T\Sigma\xi]-i\xi\cdot x}d\xi$$
I am unsure of where to go from here - I simplified it down to this point because I was hoping that I would end up with a term inside the integral that was a product of terms involving a single component of $\xi$, so that I could then use the Fubini-Tonelli theorem and integrate over each component separately. But the $\xi^T\Sigma\xi$ term throws it off.
Thanks for the help!