5.2.A. Multidimensional Fourier transform, Fourier integral

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5.2.A. Multidimensional Fourier transform, Fourier integral

Definition 1. Multidimensional Fourier transform is defined as \begin{align} & \hat{f}(\mathbf{k})= \left(\frac{\kappa}{2\pi}\right)^n\iiint_{\mathbb{R}^n} f(\mathbf{x}) e^{-i\mathbf{k} \cdot \mathbf{x}}\,d^nx \tag{FT}\\ & \check{F}(\mathbf{x})= \left(\frac{1}{\kappa}\right)^n \iiint_{\mathbf{R}^n} F(\mathbf{k}) e^{i\mathbf{k} \cdot \mathbf{x}} \,d^n \mathbf{k} \tag{IFT} \end{align} with $\kappa=1$ (but here we will be a bit more flexible).

All the main properties of $1$-dimensional Fourier transform are preserved (with obvious modifications) but some less obvious modifications are mentioned:

Remark 1. Theorem 5.2.3[5] is replaced by \begin{equation} g(\mathbf{x})=f(Q \mathbf{x})\implies \hat{g}(\mathbf{k})= |\det Q|^{-1}\hat{f}(Q^{*\,-1} \mathbf{k}) \label{eq-5.2A.1} \end{equation} where $Q$ is a non-degenerate linear transformation.

Remark 2. Example 5.2.2 is replaced by the following: Let $f(x)=e^{-\frac{1}{2}A\mathbf{x}\cdot \mathbf{x}}$ where $A$ is a symmetric (but not necessarily real matrix) $A^T=A$ with positive definite real part: \begin{equation} \Re (A\mathbf{x}\cdot \mathbf{x}) \ge \epsilon |\mathbf{x}|^2 \qquad\forall \mathbf{x} \label{eq-5.2A.2} \end{equation} with $\epsilon >0$. One can prove that inverse matrix $A^{-1}$ has the same property and \begin{equation} \hat{f}(\mathbf{k})= \left(\frac{\kappa}{\sqrt{2\pi}}\right)^n \det (A^{-\frac{1}{2}}) e^{-\frac{1}{2}A^{-1}\mathbf{k}\cdot\mathbf{k}}; \label{eq-5.2A.3} \end{equation} as long as a complex matrix $A$ does not have eigenvalues on $(-\infty,0]$, one can define properly $A^z$, $\ln (A)$ etc.
In the case of Hermitian matrix $A$ we have \begin{equation} \hat{f}(\mathbf{k})= \left(\frac{\kappa}{\sqrt{2\pi}}\right)^n |\det A|^{-\frac{1}{2}} e^{-\frac{1}{2}A^{-1}\mathbf{k}\cdot\mathbf{k}}. \label{eq-5.2A.4} \end{equation} This could be generalized to the matrices satisfying condition \begin{equation} \Re (A\mathbf{x}\cdot \mathbf{x}) \ge 0 \qquad\forall \mathbf{x},\qquad \det A\ne 0 \label{eq-5.2A.5} \end{equation} rather than (\ref{eq-5.2A.2}). In particular, for anti-Hermitian matrix $A=iB$ we have \begin{equation} \hat{f}(\mathbf{k})= \left(\frac{\kappa}{\sqrt{2\pi}}\right)^n |\det B|^{-\frac{1}{2}} e^{-i\sigma(B)\pi/4}e^{i\frac{1}{2}B^{-1}\mathbf{k},\cdot\mathbf{k}}. \label{eq-5.2A.6} \end{equation} where $\sigma(B)=\sigma_+(B)-\sigma_-(B)$, $\sigma_\pm (B)$ is the number of positive (negative) eigenvalues of $B$.

Remark 3. Poisson summation formula (Theorem 5.2.5) is replaced by \begin{equation} \sum_{\mathbf{m}\in \Gamma} f(\mathbf{m}) = \sum_{\mathbf{k}\in \Gamma^*} (2\pi )^n |\Omega|^{-1} \hat{f}(\mathbf{k}) . \label{eq-5.2A.7} \end{equation} (in notations of Section 4.B).

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