# 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.

-
thank you, but integral on the left I can compute too) – Nimza Nov 9 '11 at 12:40
@Nimza I am not understanding your comment. If this post not-helpful, or have you figured out a different way to compute the integral ? – Sasha Nov 9 '11 at 13:10
Rather it is not helpful because I wanted to show that integrals on the left and on the right are equal. I don't need to count integral on the left – Nimza Nov 9 '11 at 17:35
@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


Note that $\ds{\pars{\verts{k} + \ic 0^{+}}^{2} = k^{2} + \ic 0^{+}}$ such that \begin{align} &\color{#c00000}{% \int_{{\mathbb R}^{3}}{\expo{i\vec{\xi}\cdot\vec{x}} \over \xi^{2} - k^{2} - \ic 0^{+}}\,d^{3}\vec{\xi}} =\int_{0}^{\infty}\dd\xi\, {4\pi\xi^{2} \over \xi^{2} - \pars{\verts{k} + \ic 0^{+}}^{2}}\ \overbrace{\int\expo{i\vec{\xi}\cdot\vec{x}}\,{\dd\Omega_{\vec{\xi}} \over 4\pi}} ^{\ds{\sin\pars{\xi x} \over \xi x}} \\[3mm]&=2\pi\int_{-\infty}^{\infty} {\bracks{\xi^{2} - \pars{\verts{k} + \ic 0^{+}}^{2}} + \pars{\verts{k} + \ic 0^{+}}^{2}\over \xi^{2} - \pars{\verts{k} + \ic 0^{+}}^{2}} \,{\sin\pars{x\xi} \over x\xi}\,\dd\xi \\[3mm]&={2\pi^{2} \over x} + 2\pi k^{2}\color{#00f}{\int_{-\infty}^{\infty} {1 \over \xi^{2} - \pars{\verts{k} + \ic 0^{+}}^{2}} \,{\sin\pars{x\xi} \over x\xi}\,\dd\xi}\tag{1} \end{align}

\begin{align}&\color{#00f}{% \int_{-\infty}^{\infty}{1 \over \xi^{2} - \pars{\verts{k} + \ic 0^{+}}^{2}} \,{\sin\pars{x\xi} \over x\xi}\,\dd\xi} =\Im\,\pp\int_{-\infty}^{\infty}{1 \over \xi^{2} - \pars{\verts{k} + \ic 0^{+}}^{2}} \,{\expo{\ic x\xi} \over x\xi}\,\dd\xi \\[3mm]&=\Im\bracks{% 2\pi\ic\,{1 \over 2\pars{\verts{k} + \ic 0^{+}}}\, {\exp\pars{\ic x\pars{\verts{k} + \ic 0^{+}}} \over x\pars{\verts{k} + \ic 0^{+}}} -\int_{\pi}^{0}{1 \over -\pars{\verts{k} + \ic 0^{+}}^{2}}\,{\ic \over x}\,\dd\theta} \\[3mm]&={\pi \over x}\, \Re\pars{\expo{\ic\verts{k}x} \over \bracks{\verts{k} + \ic 0^{+}}^{2}} -{\pi \over x}\,\Re\pars{1 \over \bracks{\verts{k} + \ic 0^{+}}^{2}} =\color{#00f}{{\pi \over xk^{2}}\bracks{\cos\pars{\verts{k}x} - 1}} \end{align}

Now, we'll replace this result in expression $\pars{1}$: $$\color{#00f}{\large% \int_{{\mathbb R}^{3}}{\expo{i\vec{\xi}\cdot\vec{x}} \over \xi^{2} - k^{2} - \ic 0^{+}}\,d^{3}\vec{\xi} =2\pi^{2}\,{\cos\pars{\verts{k}x} \over x}}\,,\qquad\quad x = \verts{\vec{x}}$$

-