Calculation of the Fourier transform of a function I have calculated the Fourier transform of this function
$$f(x)=\frac{e^{i\sqrt{z}|x-y|}}{4\pi|x-y|}$$ with $x\in\mathbb{R}^3$, $\Im \sqrt{z}>0$ e $y$ fixed point in $\mathbb{R}^3$.
I have obtained the following result
$$\tilde{f}(\xi)=\frac{1}{(2\pi)^{\frac{3}{2}}}\frac{e^{-i\xi\cdot y}}{|\xi|^2-z}$$
Is it right?

My calculations are:
$$\int dx\frac{e^{i\sqrt{z}|x-y|}}{4\pi|x-y|}e^{-ix\cdot\xi}=\int dx\frac{e^{i\sqrt{z}|t|}}{4\pi|t|}e^{-i(y+t)\cdot\xi}=\frac{e^{-iy\cdot\xi}}{4\pi}\int dx\frac{e^{i\sqrt{z}|t|}}{4\pi|t|}e^{-it\cdot\xi}$$
Using spherical coordinates and choosing axes in which $t\cdot\xi=|t||\xi|\cos\theta$ I have
$$\frac{e^{-iy\cdot\xi}}{2}\int_{0}^{\infty}dr e^{i\sqrt{z}r}\int_{0}^{\pi}d\theta e^{-ir|\xi|\cos\theta}\sin\theta$$
But
$$\int_{0}^{\pi}d\theta e^{-ir|\xi|\cos\theta}\sin\theta=\frac{i}{|\xi|r}(e^{-ir|\xi|}-e^{ir|\xi|})$$
So I have to calculate
$$\frac{ie^{-i\xi\cdot y}}{2|\xi|}\bigg[\int_{0}^{\infty}dre^{i(\sqrt{z}-|\xi|)r}-\int_{0}^{\infty}dre^{i(\sqrt{z}+|\xi|)r}\bigg]$$
A simple calculation give the thesis.
