Consider the following Fourier transform: $$\hat{f}(\ln \xi) = \int_{-\infty}^{+\infty} f(x) e^{-2\pi i x \ln{\xi}} \, \, dx $$ Assuming I can calculate $\hat{f}(\ln \xi)$, how can I go from $\hat{f}(\ln \xi)$ to $\hat{f}(\xi)$?


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    $\begingroup$ Maybe substitute $\xi = e^u$? I mean, what exactly are you looking for here? $\endgroup$ – Shalop Aug 8 '18 at 21:52
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    $\begingroup$ To avoid confusion, give it a different name! Let's call $\hat{g}(\zeta) = \int f(x) e^{-2\pi i x \ln \zeta}{\rm d}x$ and let's call the normal Fourier transform $\hat{f}(\zeta) = \int f(x) e^{-2\pi i x \zeta}{\rm d}x$. Then if you know $\hat{g}(\zeta)$ we have $\hat{f}(\zeta) = \hat{g}(e^{\zeta})$. $\endgroup$ – Winther Aug 8 '18 at 22:11

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