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The free resolvent in $\mathbb{R}^3$ has this rapresentation $$(R_0(z)f)(x)=\int_{\mathbb{R}^3}\frac{e^{i\sqrt{z}|x-y|}}{4\pi|x-y|}f(y)dy$$with $\Im \sqrt{z}>0$. Then its integral kernel is $$K(x,y)=\frac{e^{i\sqrt{z}|x-y|}}{4\pi|x-y|}$$ I need the Fourier transform of this integral kernel with respect to $x$; according to my calculation it should be $$\frac{1}{(2\pi)^{\frac{3}{2}}}\frac{e^{-i\xi\cdot y}}{|\xi|^2-z}$$ Is it right?

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  • $\begingroup$ Up to the normalization, it looks right to me. $\endgroup$ – Ron Gordon Mar 9 '13 at 11:21

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