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I'm trying to calculate the Fourier transform of $$ F(\vec{r}) = \frac{1}{|\vec{r}|^2 - 1}, $$

where $\vec{r}\in\mathbb{R}^3$, in the sense of tempered distributions. I want to get it to find the inhomogeneous wave equation fundamental solution with Fourier analysis. I think it is related to the principal value distribution, but I have not managed to solve it until now...

Thanks in advance.

Miguel Ibáñez
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  • Depends on how you define the distribution $F$. Let $r, \rho \in \mathbb R^3$ and $\mathcal F\phi = \int_{\mathbb R^3} \phi(r) e^{i \rho \cdot r} dr$ for a test function $\phi$. If $$(F, \phi) = \operatorname {v. ! p.} \int_0^\infty \frac {\phi_S(x)} {x^2 - 1} dx,$$ where $\phi_S(x) = \int_{|r| = x} \phi(r) dS$, then, similar to this, $$\mathcal FF = \frac {2 \pi^2 \cos |\rho|} {|\rho|}.$$ – Maxim Aug 13 '21 at 22:11
  • @Maxim I don't know how to define it either... I got F applying Fourier transform respect to position and time to the inhomogeneous wave equation with a Dirac delta source. I was trying to invert firstly the space coordinates, and that's how I got the previous expression. Maybe as you pointed out in the link it could be enough. – Miguel Ibáñez Aug 13 '21 at 22:22
  • @Maxim And also, the expression you propose is coherent with the one given in some Electrodynamics books when solving similar problems using different approaches – Miguel Ibáñez Aug 13 '21 at 22:25
  • Solving $(\Delta+1)G=\delta,$ where $G$ is spherically symmetric, using $\Delta=\frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r}$ gives $$G(r)=Ay_0(r) + Bj_0(r),$$ where $j_0,y_0$ are spherical Bessel functions and $A,B$ are constants. So the Fourier transform you are looking for should be of that form. – md2perpe Aug 14 '21 at 12:36
  • Use spherical co-ordinates for the Fourier integrals. The angular integrals are simple, and you are left with $\int\limits_0^\infty dr \ r \frac{e^{ikr}-e^{-ikr}}{r^2-1}$, which can be written $\int\limits_{-\infty}^\infty dr \ r \frac{e^{ikr}}{r^2-1}$. Then use contour integration, avoiding the poles on the real axis with little arcs. – Sal Aug 15 '21 at 22:15
  • See also the comment here. – Maxim Aug 31 '21 at 19:06

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