Fourier transform of generalized function $\frac{1}{|x|^2-1+i0}$ on $ \mathbb{R}^3$ I am looking for an explicit computation of (or a reference to) the Fourier transform of the generalized function on $\mathbb{R}^3$
$$\frac{1}{|x|^2-1+i0}.$$
Sorry if this question is not appropriate for this site.
Remark. I believe that it should be computable and well known.
 A: Ok, since there are somewhat differing comments/answers, let's go through it explicitly - using the convention
$$
F(\vec{k} ) = \int d^3 x\ e^{-i\vec{k} \vec{x} } \frac{1}{|x|^2 -1 +i0}
$$
and choosing spherical coordinates for $\vec{x} $ such that $\vec{k} $ defines polar angle $\theta =0$, one has
$$
F(k) = \int_{0}^{2\pi} d\phi \int_{0}^{\pi} d\theta \sin \theta \int_{0}^{\infty } dx\ x^2 e^{-ikx\cos \theta } \frac{1}{x^2 -1 +i0}
$$
The angular integrals are elementary,
$$
\int_{0}^{2\pi} d\phi \int_{0}^{\pi} d\theta \sin \theta \ e^{-ikx\cos \theta } = \frac{2\pi }{ikx} (e^{ikx} - e^{-ikx} ) 
$$
and noting that the integrand is even in $x$, one can extend the integration over $x$ along the whole real axis, dividing by 2,
$$
F(k)=\frac{2\pi}{ik} \frac{1}{2} \int_{-\infty }^{\infty } dx \ x (e^{ikx} - e^{-ikx} ) \frac{1}{(x+1-i0)(x-1+i0)}
$$
To evaluate the term with the factor $e^{ikx}$, we close the contour over the upper half plane, picking up only the residue at $x=-1+i0$,
whereas for the term with the factor $e^{-ikx}$, we close the contour over the lower half plane, picking up only the residue at $x=1-i0$. Note the additional minus sign in the latter case since the integration is clockwise. The two terms give identical contributions,
$$
F(k) = \frac{2\pi}{ik} \frac{1}{2} 2\pi i \left( \frac{1}{2} e^{-ik} + \frac{1}{2} e^{-ik} \right)
$$
so, in terms of the original $|\vec{k} | =k$,
$$
F(\vec{k} ) = \frac{2\pi^{2} }{|\vec{k} |} e^{-i|\vec{k} |}
$$
as already noted by Carlo Beenakker.
A: Apart from a multiplicative constant, you should get $|x|^{-1}e^{-i|x|}$, the kernel of the resolvent at 1 of the Laplace operator, that is to say $(-\Delta-1)^{-1}$.
