Given $n\in\mathbb{N}$, I'm stuck in the calculation of the Fourier transform in $\mathbb{R}^n$ of $$(x_1,\dots,x_n)\mapsto\exp(-\sqrt{x_1^2+\dots+x_n^2}).$$ I know how to compute the result for each $n$ odd (see e.g. [1]: Fourier transform of $e^{-|x|}$), but also in this case I don't know a closed formula and, for $n$ even, I don't know how to procede. Can anyone give a closed formula in $n\in\mathbb{N}$ and show how to calculate it? Thanks in advance.
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The idea is to use $e^{-a}=\frac2{\sqrt\pi}\int_0^\infty\exp\big(-t^2-\frac{a^2}{4t^2}\big)\,dt$ (see e.g. here) at $a=\|x\|$: \begin{align} \int_{\mathbb{R}^n}e^{-i\omega x-\|x\|}\,dx&=\frac2{\sqrt\pi}\int_0^\infty e^{-t^2}\int_{\mathbb{R}^n}\exp\left(-i\omega x-\frac{\|x\|^2}{4t^2}\right)dx\,dt \\\color{gray}{[x=2ty]}\qquad&=\frac2{\sqrt\pi}\int_0^\infty e^{-t^2}(2t)^n\int_{\mathbb{R}^n}e^{-2i\omega ty-\|y\|^2}\,dy\,dt \\\color{gray}{[\text{gaussian integral}]}&=\frac2{\sqrt\pi}\int_0^\infty e^{-t^2}(2t)^n\pi^{n/2}e^{-\|\omega\|^2 t^2}\,dt \\\color{gray}{[\text{gamma integral}]}&=\frac{2^n\pi^{(n-1)/2}\Gamma\left(\frac{n+1}2\right)}{\left(1+\|\omega\|^2\right)^{(n+1)/2}}. \end{align}
metamorphy
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$$ \int_{\mathbb{R}} e^{-i\omega x_1}\int_{\mathbb{R}^{n-1}}e^{-\sqrt{x_1^2+\ldots+x_n^2}},dx_2,dx_3\ldots dx_ndx_1$$ and the measure of $x_2^2+\ldots+x_n^2=\rho^2$ is given by $\frac{2\pi^{\frac{n-1}{2}}}{\Gamma\left(\frac{n-1}{2}\right)}\rho^{n-2}$, so the previous integral equals $$\int_{-\infty}^{+\infty}e^{-i\omega x_1}\int_{0}^{+\infty}\frac{2\pi^{\frac{n-1}{2}}}{\Gamma\left(\frac{n-1}{2}\right)}\rho^{n-2} e^{-\sqrt{x_1^2+\rho^2}},d\rho,dx_1 $$
– Jack D'Aurizio Dec 17 '17 at 01:42