# Fourier transform in $\mathbb R^3$

I try to show that $$\int\limits_{R^3} \frac{e^{i\xi x} d\xi}{\xi^2 - k^2 - i0} = e^{ikx} \int\limits_{R^3} \frac{e^{i\xi x}d\xi}{\xi^2 + 2(k + i0\frac{k}{|k|})\xi}, \;\;\; k,x \in \mathbb R^3$$ I tried to make the change $\xi \mapsto \xi + k$ in second integral. I obtained $$\lim\limits_{\epsilon \to 0} \; \int\limits_{\mathbb R^d} \frac{e^{i\xi x} d\xi}{(\xi + i\epsilon \frac{k}{|k|})^2 - k^2 + \epsilon^2 - 2i|k|\epsilon}$$ So I don't know what to do...

-
Why write out $i0$? Isn't that just zero? –  Thomas Andrews Nov 8 '11 at 19:59
$\xi^2-k^2 - i 0$ is a way of saying that you consider $\lim_{\epsilon \downarrow 0} \frac{e^{i\xi x} }{\xi^2 - k^2 - i \epsilon} \mathrm{d} \xi$. –  Sasha Nov 8 '11 at 20:20
foil the square in the denominator ;) –  N. S. Nov 8 '11 at 20:21

Stuff does work out in spherical coordinates. Let $\kappa >0$ be such that $\kappa^2 = k^2$. Align north pole of spherical $\xi$-coordinates along $x$ vector. Let $\vert x \vert = \rho$. Then $\xi \cdot x = r \rho \cos(\theta)$, and $\mathrm{d} \xi = r^2 \sin \theta \mathrm{d} r \mathrm{d} \theta \mathrm{d} \phi$: $$\begin{eqnarray} \lim_{\epsilon \downarrow 0}\int_{\mathbb{R}^3} \frac{\mathrm{e}^{i\xi x} }{\xi^2 - k^2 - i \epsilon} \mathrm{d} \xi &=& \lim_{\epsilon \downarrow 0} \int_0^{2 \pi} \mathrm{d} \phi \int_0^\infty r^2 \mathrm{d} r \int_0^\pi \mathrm{d} \theta \, \sin(\theta) \cdot \frac{\mathrm{e}^{i r \rho \cos \theta} }{r^2 - \kappa^2 - i \epsilon} \\ &=& \lim_{\epsilon \downarrow 0} \left( 2 \pi \int_0^\infty r^2 \mathrm{d} r \frac{1}{r^2 - \kappa^2 - i \epsilon} \cdot \frac{2 \sin(r \rho)}{r \rho} \right) \\ &=& \lim_{\epsilon \downarrow 0} \left( 2 \pi \int_0^\infty r^2 \mathrm{d} r \frac{1}{r^2 - \kappa^2 - i \epsilon} \cdot \frac{2 \sin(r \rho)}{r \rho} \right) \\ &=& \lim_{\epsilon \downarrow 0} \left( \frac{2 \pi^2}{\rho} \mathrm{e}^{i \rho \sqrt{\kappa^2 + i \epsilon}} \right) = \frac{2 \pi^2}{\rho} \mathrm{e}^{i \kappa \rho} \end{eqnarray}$$
The integral with respect to $r$ was carried out my method of residues.
@Nimza Sorry, I missed your question, then. Notice that with formal substitution $\xi \to \xi + k + n \sqrt{\epsilon} \mathrm{e}^{i \pi/4}$, where $\vert n \vert =1$, the denominator on the l.h.s. becomes $\xi^2 + 2 \xi \cdot \left( k + n \sqrt{\epsilon} \mathrm{e}^{i \pi/4} \right) + 2 k \cdot n \sqrt{\epsilon} \mathrm{e}^{i \pi/4}$. Now you should figure out why $n = \frac{k}{\vert k \vert}$ is a good choice. –  Sasha Nov 9 '11 at 18:28
But we leave $R^3$ making this substitution –  Nimza Nov 9 '11 at 19:52