Fourier transform of $\exp(-x^TAx)$

Question

Given $A$ symmetric, positive definite. Calculate $$\hat f(\omega)=\frac{1}{2\pi^{n/2}}\int_{\mathbb{R}^n} \exp(-x^TAx) \exp(-i\omega^Tx) \, dx$$

My approach: using $D=VAV^T$, $D$ diagonal matrix, $x=V^Ty$ one get $$ \hat f(\omega) = \frac{1}{2\pi^{n/2}}\int_{\mathbb{R}^n} \exp(-y^TDy) \exp(-i\omega^TV^Ty) det(V) \, dy$$

I am actually a bit stuck here, I could write $$\exp(-y^TDy) \exp(-i\omega^TV^Ty) = \exp(-y^TDy-i\omega^Ty) $$ but since the first term is bilinear, I am not sure how to proceed "optimally". I am very thankful for hints.

edit: $det(V)=1$ Greetings.

Answer 1

With $D = \operatorname{diag}(d_1, \ldots, d_n)$ we have $$ \exp(-y^TDy) = \exp(-\sum_{k=1}^{n} d_k y_k^2) = \prod_{k=1}^{n} \exp(-d_k y_k^2). $$

Setting $\tilde\omega = V\omega$ we also have $$ \exp(-i\omega^T V^T y) = \exp(-i\tilde\omega^T y) = \exp\left(-i \sum_{k=1}^{n} \tilde\omega_k y_k \right) = \prod_{k=1}^{n} \exp\left(-i \tilde\omega_k y_k \right) $$

Therefore, $$ \int_{\mathbb R^n} \exp(-y^TDy) \exp(-i\omega^T V^T y) \, dy \\ = \int_{\mathbb R^n} \prod_{k=1}^{n} \exp(-d_k y_k^2) \prod_{k=1}^{n} \exp\left(-i \tilde\omega_k y_k \right) \, dy \\ = \int_{\mathbb R^n} \prod_{k=1}^{n} \left( \exp(-d_k y_k^2) \exp\left(-i \tilde\omega_k y_k \right) \right) \, dy \\ = \prod_{k=1}^{n} \int_{\mathbb R} \left( \exp(-d_k y_k^2) \exp\left(-i \tilde\omega_k y_k \right) \right) \, dy \\ = \prod_{k=1}^{n} \exp(- \tilde\omega^2/(4 d_k)) \\ = \exp(- \sum_{k=1}^{n} \tilde\omega^2/(4 d_k)) \\ = \exp(- \frac14 \tilde\omega^T D^{-1} \tilde\omega) \\ = \exp(- \frac14 \omega^T V^T D^{-1} V \omega) \\ = \exp(- \frac14 \omega^T A^{-1} \omega) $$

Answer 2

I presume that you have taken $V$ to be orthogonal, so that $V^T = V^{-1}$. So far, we have (noting that $\det(V) = 1$) $$ \hat f(\omega) = \frac{1}{2\pi^{n/2}}\int_{\mathbb{R}^n} \exp(-y^TDy) \exp(-i\omega^TV^Ty) \, dy $$ Now, let $g(x) = \exp(-x^TDx)$. With the substitution $k = V \omega$, we can write $$ \hat f(\omega) = \frac{1}{2\pi^{n/2}}\int_{\mathbb{R}^n} \exp(-y^TDy) \exp(-i\omega^TV^Ty) \, dy \\ = \frac{1}{2\pi^{n/2}}\int_{\mathbb{R}^n} \exp(-y^TDy) \exp(-ik^Ty) \, dy \\ = \hat g(k) = \hat g(V\omega) $$ That is: once you've calculated $\hat g(\omega)$, you can calculate $\hat f(\omega) = \hat g(V\omega)$.

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Another hint: suppose that $$ D = \pmatrix{d_1\\&\ddots \\&&d_n} $$ with the careful application of some Fubini magic, we find that $$ \hat g(\omega) = \frac{1}{2\pi^{n/2}}\int_{\mathbb{R}^n} \exp(-x^TDx) \exp(-i\omega^Tx) \, dx \\ = \frac{1}{2\pi^{n/2}}\int_{\mathbb{R}^n} \left(\prod_{j=1}^{n}\exp^{-d_jx_j^2 }\exp^{- i\omega_j x_j} \right) dx \\ = \frac{1}{2\pi^{n/2}} \prod_{j=1}^{n}\left(\int_{\mathbb{R}}\exp^{-d_jx_j^2 - i\omega_j x_j}dx_j \right) $$ So, it suffices to answer the question in the 1-D case.