Difference between revisions of "Marks-Gravagne-Davis Fourier transform as a delta integral with classical exponential kernel"
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Revision as of 16:08, 15 January 2023
Theorem
If $0 \in \mathbb{T}$, then the Marks-Gravagne-Davis Fourier transform obeys $$\mathscr{F}\{f\}(z;0) = \displaystyle\int_{-\infty}^{\infty} f(t)e^{-2i\pi zt} \Delta t.$$
